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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 12 Atoms. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 12 Important Extra Questions Atoms

Atoms Important Extra Questions Very Short Answer Type

Question 1.
Name the spectral series which lies in the visible region.
Answer:
Balmer series.

Question 2.
What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the third excited state? (CBSE AI 2013C)
Answer:
Six.

Question 3.
When is H_α line of the Balmer series in the emission spectrum of hydrogen atom obtained? (CBSE Delhi 2013C)
Answer:
It is obtained when an electron jumps from n =3 to n = 2 level.

Question 4.
A mass of lead is embedded in a block of wood. Radiations from a radioactive source incident on the side of the block produce a shadow on a fluorescent screen placed beyond the block. The shadow of the wood is faint but the shadow of lead is dark. Give a reason for this difference.
Answer:
The shadow of the wood is faint because only the a-radiations are stopped by the wood (since a-radiations are least penetrating). The shadow of lead is dark because p and y-radiations are also stopped by lead.

Question 5.
What was the source of alpha particles in Rutherford’s alpha scattering experiment?
Answer:
The source was ²¹⁴₈₃Bi.

Question 6.
If the radius of the ground level of a hydrogen atom is 5.3 nm, what is the radius of the first excited state?
Answer:
It is 4 × 5.3 = 21.2 nm ( ∵ r = n²r_o)

Question 7.
Calculate the ratio of energies of photons produced due to the transition of electron of a hydrogen atom from its
(a) Second permitted energy level to the first level, and
Answer:
energy of photon E₁ = – 3.4 – (-13.6) = 10.2 eV

(b) Highest permitted energy level to the second permitted level.
Answer:
energy of photon E₂ = 0 – (-3.4) = 3.4 eV
Ratio \(\frac{E_{1}}{E_{2}}=\frac{10.2}{3.4}\) = 3

Question 8.
The mass of an H-atom is less than the sum of the masses of a proton and electron. Why is this? (NCERT Exemplar) Answer:
Einstein’s mass-energy equivalence gives E = mc². Thus the mass of an H-atom is m_p + m_e – B/c² where B ≈ 13.6 eV

Atoms Important Extra Questions Short Answer Type

Question 1.
Define electron-volt and atomic mass unit. Calculate the energy in joule equivalent to the mass of one proton.
Answer:
Electron volt: It is defined as the energy gained by an electron when accelerated through a potential difference of 1 volt. Atomic mass unit: It is defined as one-twelfth of the mass of one atom of carbon 12.

The mass of a proton is 1.67 × 10^-27 kg. Therefore, energy equivalent of this mass is E = mc² = 1.67 × 10^-27 × (3 × 10⁸)² = 1.5 × 10^-10 J

Question 2.
State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. (CBSE Delhi 2019)
Or
Calculate the orbital period of the electron in the first excited state of the hydrogen atom.
Answer:
Bohr’s Quantisation condition: Only those orbits are permitted in which the angular momentum of the electron is an integral multiple of h/2π.

For Brackett Series,
The shortest wavelength is for the transition of electrons from n_i = ∞ to n_f = 4

Using the equation

Question 3.
Write two important limitations of the Rutherford nuclear model of the atom. (CBSE AI2018, Delhi 2018)
Answer:

1. Rutherford’s model fails to explain the line spectra of the atom.
2. Rutherford’s model cannot explain the stability of the nucleus.

Question 4.
Find out the wavelength of the electron orbiting in the ground state of the hydrogen atom. (CBSEAI 2018, Delhi 2018)
Answer:
The wavelength of an electron in the ground state of hydrogen atom is given by
E = \(\frac{hc}{λ}\)
or
λ = \(\frac{hc}{E}\)

For ground state
E = – 13.6 eV = 13.6 × 1.6 × 10^-19 J

Hence wavelength of electron in the first orbit
λ = \(\frac{h c}{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.6 \times 1.6 \times 10^{-19}}\) = 0.9 × 10-7 J

Question 5.
(a) State Bohr’s postulate to define stable orbits in a hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits?
Answer:
Bohr’s postulate for stable orbits states the electron in an atom revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of h/2π (h = Planck’s constant), (n = 1, 2, 3 …)

As per de Broglie’s hypothesis λ = h/p = h/mv
For a stable orbit, we must have a circumference of the orbit = nλ (n = 1, 2, 3,…)
∴ 2πr = nλ
or
mvr = nh/2π

Thus de-Broglie showed that the formation . of stationary patterns for integral “n” gives rise to the stability of the atom.
This is nothing but Bohr’s postulate.

(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. (CBSE AI 2018, Delhi 2018)
Answer:
Energy in the n = 4 level n₁ = 1 and n₂ = 4


Question 6.
An alpha particle moving with initial kinetic energy K towards a nucleus of atomic number Z approaches a distance ‘d’ at which it reverses its direction. Obtain an expression for the distance of closest approach ‘d’ in terms of the kinetic energy of the alpha particle, K. (CBSEAI2016C)
Answer:
At the distance of the closest approach, the kinetic energy of the alpha particle is converted into the electrostatic potential energy of the alpha particle-nucleus system. Therefore, at the distance of the closest approach

we have
Kinetic energy = Potential energy
Therefore,

where K is the kinetic energy.

Question 7.
The figure shows the energy level diagram of the hydrogen atom.

(a) Find out the transition which results in the emission of a photon of wavelength 496 nm.
Answer:
The wavelength of photon emitted is given by \(\frac{1}{λ}\) = R_H\(\left(\frac{1}{n_{\mathrm{f}}^{2}}-\frac{1}{n_{\mathrm{i}}^{2}}\right)\)

None of these transitions correspond to a wavelength of 496 nm. The closest is 4 to 2 of 489 nm

(b) Which transition corresponds to the emission of radiation of maximum wavelength? Justify your answer. (CBSE AI 2015 C)
Answer:
Transition 4 to 3 as the frequency of this radiation is maximum.

Question 8.
A nucleus makes a transition from one permitted energy level to another level of lower energy. Name the region of the electromagnetic spectrum to which the emitted photon belongs. What is the order of its energy in electron-volts? Write four characteristics of nuclear forces.
Answer:
(a) Emitted photon belongs to gamma-rays part of the electromagnetic spectrum.
(b) the energy is of the order of MeV.
(c) Four characteristics of nuclear forces are:

1. Nuclear forces are independent of charges.
2. Nuclear forces are short-range forces.
3. Nuclear forces are the strongest forces in nature, in their own small range of few fermis.
4. Nuclear forces are saturated forces.

Atoms Important Extra Questions Long Answer Type

Question 1.
Explain Rutherford’s experiment on the scattering of alpha particles and state the significance of the results.
Answer:
The schematic arrangement in the Geiger Marsden experiment is shown in the figure.

Alpha-particles emitted by a Bismuth (²¹⁴₈₃Bi) radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10^-7 m. The scattered alpha-particles were observed through a rotatable detector consisting of a zinc sulfide screen and a microscope. The scattered alpha-particles on striking the screen produced bright light flashes or scintillations. These scintillations could be viewed through the microscope and counted at different angles from the direction of the incident beam.

Significance: The experiment established the existence of a nucleus that contained the entire positive charge and about 99.95% of the mass.

Question 2.
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to the Balmer series occur due to the transition between energy levels. (CBSE Delhi 2013)
Answer:
The electron revolving around the nucleus has two types of energy:

Kinetic energy due to its motion.
Potential energy due to it lying in the electric field of the nucleus.

Thus the total energy of the electron is given by
E = K. E. + P. E. …(1)

An electron of mass m moving around the nucleus with an orbital velocity v has kinetic energy given by
K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2} \frac{k e^{2}}{r}\) …(2)

Now the potential energy of the electron at a distance r from the nucleus is given by
PE = potential due to the nucleus at a distance r × charge on the electron = V × – e …(3)

Now the potential at a distance r from the nucleus having a charge e is given by
V = k \(\frac{e}{r}\) …..(4)

Substituting in equation (3) we have
P.E. = V × – e = -k\(\frac{e^{2}}{r}\) …(5)

Substituting equations (2) and (3) in equation 1 we have

But the radius of the nth orbit is given by
r_n = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)

Substituting in equation (6) we have
E = – \(\frac{2 \pi^{2} m e^{4} \mathrm{k}^{2}}{n^{2} \mathrm{~h}^{2}}\) …(7)

This gives the expression for the energy possessed by the electron in the nth orbit of the hydrogen atom.

Question 3.
Hydrogen atoms are excited with an electron beam of energy of 12.5 eV. Find
(a) The highest energy level up to which the hydrogen atoms will be excited.
Answer:
The maximum energy that the excited hydrogen atom can have is

For n=4, E4 = \(\frac{-13.6}{16}\) = – 0.85eV
(> – 1.1 eV)

∴ The eLectron can only be excited up to n = 3 states.

(b) The longest wavelengths in the (i) Lyman series, (ii) Balmer series of the spectrum of these hydrogen atoms. (CBSE 2019C)
Answer:
From energy tevet of hydrogen atom,
we have
\(\frac{1}{λ}\) = R\(\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)

Longest wavelength of Lyman senes

Question 4.
Using Bohr’s postulates of the atomic model derive the expression for the radius of the 11^th electron orbit. Hence obtain the expression for Bohr’s radius. (CBSE AI 2014)
Answer:
Let us consider a mechanical model of the hydrogen atom as shown in the figure that incorporates this quantization assumption.

This atom consists of a single electron with mass m and charge – e revolving around a single proton of charge + e. The proton is nearly 2000 times as massive as the electron, so we can assume that the proton does not move. As the electron revolves around the nucleus the electrostatic force of attraction between the electron and the proton provides the necessary centripetal force. Therefore, we have

This gives the radius of the nth orbit of the hydrogen atom.
If n = 1 we have r = a_o which is called Bohr’s radius.
a_o = \(\frac{h^{2}}{4 \pi^{2} m e^{2} k}\)

Question 5.
State Bohr’s postulate of the hydrogen atom successfully explains the emission lines in the spectrum of the hydrogen atoms.
Use the Rydberg formula to determine the wavelength of Ha line. [Given Rydberg constant R = 1.03 × 10⁷ m^-1] (CBSE AI 2015)
Answer:
It states that an electron might make a transition from one of its specified non¬radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by

hv = E_i – E_f where E_i and E_f are the energies of the initial and final states
Using the formula \(\frac{1}{λ}\) = R_H\(\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\) we have for Ha Line n_i = 3 and n_f = 2

Question 6.
Using Bohr’s postulates derive the expression for the frequency of radiation emitted when an electron In a hydrogen atom undergoes a transition from a higher energy state (quantum number n-) to the towering state (n,). When an electron in a hydrogen atom jumps from the energy state n_i =4 to n = 3, 2, 1, identify the spectral series to which the emission lines belong. (CBSE Delhi 201 1C)
Answer:
According to Bohr’s frequency condition, if an electron jumps from an energy Level E to E₁, then the frequency of the emitted radiation is given by
hv = E – E₁ …(1)

Let ni and nf be the corresponding orbits then

This gives the frequency of the emitted radiation.
When n_i =4 and n_f = 3, Paschen series
When n_i = 4 and n_f = 2, Balmer series
When n_i = 4 and n_f = 1, Lyman senes

Question 7.
Calculate the ratio of the frequencies of the radiation emitted due to the transition of the electron In a hydrogen atom from Its (i) second permitted energy level to the first level and (ii) highest permitted energy level to the second permitted level. (CBSE Delhi 2018C)
Answer:
We have

Question 8.
Monochromatic radiation of wavelength 975 A excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible In the resulting spectrum? Which transition corresponds to the longest wavelength amongst them? (CBSE Sample Paper 201819)
Answer:
The energy corresponding to the given wavelength:

The longest wavelength Will correspond to the transition n = 4 to n = 3

Question 9.
(a) Using postulates of Bohr’s theory of hydrogen atom, show that
(i) the radii of orbits increases as n², and
Answer:
Let us consider a mechanical. model of the hydrogen atom as shown in the figure.

This atom consists of a single electron with mass m and charge – e revolving around a single proton of charge + e. As the electron revolves around the nucleus the electrostatic force of attraction between the electron and the proton provides the necessary centripetal force. Therefore we have,

Substituting equation 3 in equation 2 we have

This gives the radius of the n^th orbit of the hydrogen atom which shows that E ∝ \(\frac{1}{n^{2}}\)

(ii) the total energy of the electron increases as 1/n², where n is the principal quantum number of the atom.
Answer:
the total energy possessed by an electron in the nth orbit of the hydrogen atom is given by
E = T + U …(1)
i.e. the sum of its kinetic and electrostatic potential energies.

An electron of mass m moving around the nucleus with an orbital velocity v has kinetic energy given by
K.E. = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\frac{k e^{2}}{r}\) ….(2)

Now the potential energy of the electron at a distance r from the nucleus is given by
PE = potential due to the nucleus at a distance r × charge on the electron
= V × – e …(3)

Now the potential at a distance r from the nucleus having a charge e is given by
V = k\(\frac{e}{r}\) ….(4)

Substituting in equation 2 we have
P.E. = V × – e = – k \(\frac{e^{2}}{r}\) …(5)

Substituting equations 2 and 5 in equation 1 we have

But the radius of the nth orbit is given by r_n = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)

Substituting in equation 6 we have
E = – \(\frac{2 \pi^{2} m e^{4} k^{2}}{n^{2} h^{2}}\) …(7)

This gives the expression for the energy possessed by the eLectron in the nth orbit of the hydrogen atom which shows that E ∝ \(\frac{1}{n^{2}}\)

(b) Calculate the wavelength of H2 line In Balmer series of hydrogen atom, given Rydberg constant R = 1.097 × 10⁷ m^-1. (CBSE AI 2011C)
Answer:
For H₂ Line in Balmer series n₁ = 2 and n₂ = 3

Question 10.
State Bohr’s quantization condition for defining stationary orbits. How does de Brogue hypothesis explain the stationary orbits?
Find the relation between the three wavelengths λ1 λ2 and λ3 from the energy level diagram shown below. (CBSE Delhi 2016)

Answer:
It states that only those orbits are permitted in which the angular momentum of the electron about the nucleus is an integral multiple of \(\frac{h}{2π}\), where his Planck’s constant.

According to de Broglie, an electron of mass m moving with speed v would have a wavelength λ given by
λ = h/mv.

Now according to Bohr’s postulate,
mvr_n = \(\frac{n h}{2 \pi}\)
or
2πr_n = \(\frac{nh}{mv}\)

But h / mv = A is the de BrogUe wavelength of the electron, therefore, the above equation becomes 2πr_n = nλ where 2πr_n is the circumference of the permitted orbit. If the wavelength of a wave does not close upon itself, destructive interference takes place as the wave travels around the loop and quickly dies out. Thus only waves that persist are those for which the circumference of the circular orbit contains a whole number of wavelengths.

Numerical Problem :
Formulae for solving numerical problems

• Distance of closest approach r_o = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 z E^{2}}{E_{k}}\)
• Radius of the nth orbit of hydrogen atom rn = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\)
• Velocity of eLectron in the ntt orbit v = \(v=\frac{c}{137 n}\)
• Wavelength of radiation emitted when electron jumps form ni to nf \(\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_{\mathrm{f}}^{2}}-\frac{1}{n_{\mathrm{i}}^{2}}\right)\)
• Energy of electron in the nth orbit of hydrogen atom
  E = – \(\frac{2 \pi^{2} m e^{4} k^{2}}{n^{2} h^{2}}\)
  or
  E = – \(\frac{13.6}{n^{2}}\) eV

Question 1.
A hydrogen atom in its excited state emits radiations of wavelengths 1218 Å and 974.3 Å when it finally comes to the ground state. Identify the energy levels from where transitions occur. Given Rydberg constant R = 1.1 × 10⁷ m^-1. Also, specify the spectral series to which these lines belong. (CBSE AI 2019)
Answer:
We know that


On solving n = 4
Lyman series.

Question 2.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, i.e. an atom where the electron is replaced by a negatively charged muon (μ^–) of mass about 207 me that orbits around a proton. (Given for hydrogen atom, the radius of first orbit and ground state energy are 0.53 × 10^-10m and – 13.6 eV respectively) (CBSE AI 2019)
Answer:
In Bohr’s model of hydrogen atom the radius of n^th orbit is given by
r_n = \(\frac{n^{2} h^{2}}{4 \pi^{2} e^{2} m_{e} k}\)

As n = 1
Therefore, we have 1

Question 3.
The electron in a given Bohr orbit has a total energy of – 1.5 eV. Calculate Its
(a) kinetic energy.
(b) potential energy.
(C) the wavelength of radiation emitted, when this electron makes a transition to the ground state.
(Given Energy in the ground state = – 13.6 eV and Rydberg’s constant = 1.09 × 1o⁷ m^-1) (CBSE Delhi 2011C)
Answer:
Total energy of the electron In a Bohr’s orbit is – 1.5 eV

We know that kinetic energy of the electron in any orbt is half of the potential energy in magnitude and potential energy is negative
(a) Total energy = kinetic energy + potential energy
– 1.5 = E_k – 2E_k
1.5 = E_k

(b) E_p = – 2 × 1.5 = – 3 eV

(c) Energy released when the transition of this electron takes place from this orbit to the ground state
= – 1.5 – (- 13.6)
= 12.1 eV
= 12.1 × 1.6 × 10¹⁹ = 1.936 × 10^-18 J

Let be the wavelength of the Light emitted then,

Question 4.
In a Geiger—Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an a-particle of 8 MeV energy Impinges on it before It comes momentarily to rest and reverses its direction. How will the distance of the closest approach be affected when the kinetic energy of the a-particle is doubled? (CBSEAI 2012)
Answer:
Given Z = 80, E = 8 MeV = 8 × 10⁶ × 1.6 × 10¹⁹ J, r_o =?
Using the expression

When the kinetic energy of a-particle has doubled the distance of the closest approach becomes half its previous value, i.e. 1.44 × 10¹⁴ m

Question 5.
The ground state energy of the hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to – 3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? (CBSE AI 2012)
Answer:
Energy released = – 0.85 – (- 3.4) = 2.55 eV = 2.55 × 1.6 × 10^-19 J
Using E = hc/λ we have
λ = \(\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.55 \times 1.6 \times 10^{-19}}\) = 4.87 × 10^-7 m

It belongs to the Balmer series.

Question 6.
A hydrogen atom in the third excited state de-excites to the first excited state. Obtain the expressions for the frequency of radiation emitted in this process.
Also, determine the ratio of the wavelengths of the emitted radiations when the atom de-excites from the third excited state to the second excited state and from the third excited state to the first excited state. (CBSEAI2019)
Answer:
We know that


Question 7.
Obtain the expression for the ratio of the de Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom. (CBSE Delhi 2019)
Answer:
We know that
2πr = nλ ….(i)
For the second excited state (n = 3)
r = 0.529(n)² A = 0.529(3)²

Putting in (i) we get 2π(0.529)(3)² = 3 λ₂

For third excited state n = 4
r = 0.529 (4)²
Putting in (i) we get 2π (0.529)(4)² = 4 λ₃

Question 8.
(a) The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-10 m. Calculate its radius in n = 3 orbit.
(b) The total energy of an electron in the first excited state of the hydrogen atom is 3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state. (CBSE Delhi 2014C)
Answer:
(a) The radius is given by
r_n = \(\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2} k}\) = n²ao

where n is the number of orbit, hence r3 = 5.3 × 10^-10 × 32 = 4.77 × 10^-9 m

(b) Total kinetic energy = + 3.4 eV
Total potential energy = – 6.8 eV

Question 9.
Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Answer:
The de-Broglie wavelength is given by 2πrn = nλ.
In ground state, n = 1 and r_o = 0.53 Å, therefore, λ_o = 2 × 3.14 × 0.53 = 3.33 Å

In first excited state, n = 2 and r₁ = 4 × 0.53 Å = 2.12 Å, therefore,
r₁ = (2 × 3.14 × 2.12)/2 = 6.66Å

Therefore, λ₁ – λ_o = 6.66 – 3.33 = 3.33 Å.
In other words, the de-Broglie wavelength becomes double.

Question 10.
When is the H_α line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition. (CBSE AI 2016)
Answer:
Hα is obtained when n_i = 3 and n_f = 2

Question 11.
A 12.5 eV beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelength and the corresponding series of lines emitted. (CBSE AI 2017)
Answer:
Given the energy of electron beam E = 12.5 eV
This will make the energy of the electrons 12.5 – 13.6 = – 1.1 eV.
Thus, the electrons of hydrogen will be raised to the third orbit.
Lines emitted will be n_i = 3, n_f = 2 Balmer series.

The wavelength is given by

For first member n_i = 2, therefore, the wavelength of the first member of the Lyman series.

Question 12.
Calculate the longest wavelength of the photons emitted in the Balmer series of the hydrogen spectrum. Which part of the e.m. spectrum does it belong to? [Given Rydberg constant, R = 1.1 × 10⁷ m^-1 J] (CBSE Delhi 2017C)
Answer:
The longest wavelength is obtained when the electron jumps from n_i = 3 to n_f = 2 in the Balmer series. Therefore we have,

It lies in the visible region.

Question 13.
10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. ‘ Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite. (NCERT Exemplar)
Answer:
From Bohr’s postulate, we have mv_n r_n = nh/2π
Here m = 10 kg and r_n = 8 × 10⁶ m

We have the time period T of the circling satellite as 2 h.
That is T = 7200 s
Thus the velocity v_n = 2πr_n/T

The quantum number of the orbit of the satellite
N = (2πr)² × m/(T × h).

Substituting the values,
n = (2π × 8 × 106 )² × 10/(7200 × 6.64 × 10^-34)
= 5.3 × 1045

Note that the quantum number for the satellite motion is extremely large. In fact, for such large quantum numbers, the results of quantization conditions tend to those of classical physics.

Question 14.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
Answer:
No, because according to the Bohr model, E = – 13.6 /n² eV and electrons having different energies belong to different levels having different values of n.

So, their angular momenta will be different, as nh/2π = mvr

Question 15.
Using the Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state. (NCEFU Exemplar)
Answer:
Let v be the velocity of the electron in the ground state. Let a0 be the Bohr radius.
Now time is taken to complete one revolution

Question 16.
What is the minimum energy that must be given to an H atom in the ground state so that it can emit a H_y line in the Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such H_y photon? (NCERJ Exemplar)
Answer:
The H_y in Balmer series corresponds to transition n = 5 to n = 2. So, the electron in ground state n = 1 must first be put in state n = 5.

Hence energy required for the same.
Energy required = E₁ – E₅ = 13.6 – 0.54 = 13.06 eV.

If angular momentum is conserved, angular momentum of photon = change in angular momentum of electron = L₅ – L₂ = 5h – 2h = 3h = 3 × 1.06 × 10^-34 = 3.18 × 10^-34 kg m² s^-1

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 16 Chemistry in Everyday Life. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 16 Important Extra Questions Chemistry in Everyday Life

Chemistry in Everyday Life Important Extra Questions Very Short Answer Type

Question 1.
Name a substance that can be used as an antiseptic as well as a disinfectant. (CBSE 2019C)
Answer:
Phenol / C₆H₆₅OH

Question 2.
Differentiate between disinfectants and antiseptics. (CBSE 2012)
Answer:
Both antiseptics and disinfectants kill or prevent the growth of microorganisms. Antiseptics are applied to living tissues (cuts, wounds). But disinfectants are applied to inanimate objects only (floor, instruments).

The same substance can act as an antiseptic as well as a disinfectant by varying the concentration of the solution used. For example, 0.2% solution of phenol acts as an antiseptic and its 1% solution is a disinfectant.

Question 3.
What is the cause of a feeling of depression in human beings? Name a drug that can be useful in treating this depression. (CBSE 2012)
Answer:
The cause of depression in human beings is a low level of noradrenaline. Because of the low level of noradrenaline, the signal-sending activity of the hormones becomes low and the person suffers from depression. Phenelzine is useful in treating this depression.

Question 4.
Name one substance that can act as both:
(i) Analgesic and antipyretic. (CBSE Sample Paper 2012)
Answer:
Aspirin

(ii) Antiseptic and disinfectant. (CBSE Sample Paper 2011)
Answer:
Phenol

Question 5.
What is the tincture of iodine? What is its use? (CBSE 2019C)
Answer:
2 – 3% iodine solution of alcohol-water is called tincture of iodine. It is a powerful antiseptic and is applied on wounds.

Question 6.
(a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate. (CBSE Delhi 2013)
Answer:
Sodium benzoate

(b) If the water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes? (CBSE AI 2013)
Answer:
Synthetic detergent.

Question 7.
Among the following which one acts as a food preservative?
Aspartame, Aspirin, Sodium benzoate, Paracetamol
Answer:
Sodium benzoate.

Chemistry in Everyday Life Important Extra Questions Short Answer Type

Question 1.
What are food preservatives? Name two such substances. (CBSE 2012)
Answer:
Food preservatives are chemical substances that are added to food materials to prevent spoilage and to retain their nutritive value for long periods. For example, sodium benzoate, potassium metabisulphite.

Question 2.
(i) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Which class of drugs is used in sleeping pills? (CBSE AI 2018)
Answer:
Tranquilizers.

Question 3.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) Which of the following is an antiseptic?
0. 2% phenol, 1% phenol(CBSE AI 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 4.
Give one example for each of the following:
(i) An artificial sweetener whose use is limited to cold drinks.
Answer:
Aspartame.

(ii) A non-ionic detergent. (CBSE Sample Paper 2011)
Answer:
Ester of stearic acid and polyethylene glycol
CH₃(CH₂)₁₆COO(CH₂CH₂O)_nCH₂CH₂OH

Question 5.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why? (CBSE Sample Paper 2011)
Answer:
Most of the drugs taken in doses higher than recommended may cause harmful effects and act as poison leading to death. Therefore, a doctor must always be consulted before taking any medicine, who will advise the patient for proper and safe doses of the drug.

Question 6.
Why do soaps not work in hard water? (CBSE Delhi 2011)
Answer:
Hard water contains calcium and magnesium salts. Therefore, in hard water soap gets precipitated as insoluble calcium and magnesium soaps which being insoluble stick to the cloth as gummy mass and blocks the ability of soap to remove oil or grease from the cloth.

Question 7.
Explain the following term with one suitable example:
Antifertility drugs (CBSE 2010)
Answer:
Antifertility drugs. These are the chemical substances used to control the pregnancy. These are also called oral contraceptives birth control pills. The common drugs used as antifertility drugs are norethindrone, ethynylestradiol (nostril), mifepristone, ormeloxifene, etc.

Chemistry in Everyday Life Important Extra Questions Long Answer Type

Question 1.
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents is used in toothpaste?
(iii) Why is the use of alitame as an artificial sweetener not recommended?
OR
Define the following terms with a suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents (CBSE Delhi 2019)
Answer:
(i) Tranquilizers
(ii) Anionic detergents
(iii) With alitame, it is difficult to control the sweetness of food to which they are added.

OR
(i) Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful microorganisms and are used for curing a variety of diseases, for example, chloramphenicol.

(ii) Disinfectants: The chemical substances which are used to kill microorganisms but cannot be applied on living tissues are called disinfectants, for example, phenol (1% solution).

(iii) Cationic detergents: The substances which have a long hydrocarbon chain with a positive charge on the nitrogen atom (cationic part) which is involved in cleaning action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions, for example, cetyltrimethylammonium bromide.

Question 2.
What is the following substance? Give one example of it:
Antacids (CBSE2011, CBSE Delhi 2012)
Answer:
Antacids: These are the chemical substances that neutralize excess acid in the gastric juices and give relief from acid indigestion, acidity, heartburns, and gastric ulcers. Until 1970, the antacids such as sodium bicarbonate or a mixture of aluminum and magnesium hydroxide have been commonly used for the treatment of acidity.

However, excessive bicarbonate can make the stomach alkaline and trigger the production of even more acid. Nowadays, acidity is cured by drugs such as cimetidine (Tagamet), ranitidine (Zantac), omeprazole, lansoprazole, etc.

Question 3.
(i) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
Answer:
Sodium benzoate

(ii) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(iii) Which class of drugs is used in sleeping pills? (CBSE Delhi 2013)
Answer:
Tranquilizers.

Question 4.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) If the water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes?
Answer:
Synthetic detergent

(iii) Which of the following is an antiseptic:
0. 2% phenol, 1% phenol? (CBSE 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 5.
(i) Give two examples of macromolecules that are chosen as drug targets.
Answer:
Carbohydrates, proteins.

(ii) What are antiseptics? Give an example.
Answer:
The chemical substances which are used to either kill or prevent the growth of micro-organisms are called antiseptics. These are not harmful to live tissues and can be safely applied on wounds, cuts, ulcers, diseased skin surfaces, etc.
For example, Dettol.

(iii) Why is the use of aspartame limited to cold foods and soft drinks? (CBSE Delhi 2014)
Answer:
Aspartame is unstable at cooking temperatures and therefore, it is used as sugar substitute for cold foods and
soft drinks.

Question 6.
(i) Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Saccharin

(ii) What are antibiotics? Give an example.
Answer:
Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, moulds) and can inhibit the growth or even destroy other micro-organisms. For example, Penicillin.

(iii) Give two examples of macromolecules that are chosen as drug targets. (CBSE Delhi 2014)
Answer:
Proteins, lipids.

Question 7.
(a) Differentiate between antiseptic and disinfectant. Give one example of each.
(b) Why do we require artificial sweetening agents?
OR
Define the following terms:
(a) Tranquilizers
(b) Antacids
(c) Analgesics (CBSE2019C)
Answer:
(a) Antiseptics: Antiseptics are the chemical substances that are used to either kill or prevent the growth of micro-organisms. These are not harmful to live tissues and can be safely applied on wounds, cuts, diseased skin surfaces. For example, Dettol, Savlon, furnace, tobramycin, etc.

Disinfectants: Disinfectants are chemical substances that kill micro-organisms but cannot be applied to living tissues. In other words, they also kill micro-organisms like antiseptics but are not safe for living tissues. These are commonly applied to inanimate objects such as the floor, drainage systems, instruments, etc. Some common examples of disinfectants are phenol (1% solution), chlorine (0.2 to 0.4 ppm), etc.

(b) Artificial sweetening agents are used to reduce calorie intake. These also protect teeth from decay.
OR
(a) The chemical substances used for the treatment of stress, fatigue, mild and severe mental diseases are called tranquilizers.

Tranquilizers are neurologically active drugs that affect the message transfer mechanism from nerve to receptor. These are used to relieve or reduce mental tension, irritability, excitement, and anxiety leading to calmness. These form an essential component of sleeping pills.

(b) Acidic stomach is necessary for good health, but excessive acidity in the stomach can cause discomforts such as acid indigestion, heartburn, irritation, or pain of gastric ulcers. The chemical substances which neutralize excess acid in the gastric juices and give relief from add indigestion, acidity, heartburns, and gastric ulcers are called antacids.

(c) The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

1. Non-narcotic (non-addictive) drugs
2. Narcotic drugs

Question 8.
Explain the following terms with one suitable example for each:
(i) A sweetening agent for diabetic patients
Answer:
A sweetening agent for diabetic patients: The chemical substances which give a sweetening effect to food but do not add any calorie to our body are called artificial sweetening agents. The sweetening agent for diabetic patients is saccharin.

(ii) Enzymes
Answer:
Enzymes: Enzymes are biological catalysts produced by living cells that catalyze the biochemical reaction in living organisms. Chemically enzymes are naturally occurring simple or conjugated proteins and some enzymes are non-proteins also. These increase the rates of biochemical reactions by providing alternative paths of lower energy. For example, maltase, which catalyzes the hydrolysis of maltose to glucose.

(iii) Analgesics (CBSE AI 2011)
Answer:
Analgesics: These are medicines used to relieve pains. Aspirin and some other antipyretics are also used as analgesics. Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine, codeine, heroin, marijuana. These are known to be habit-forming.

Question 9.
Define and write an example for the following:
(a) Broad-spectrum antibiotics.
Answer:
Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful micro-organisms. Therefore, these are used for curing a variety of diseases. The common examples are tetracycline, Chloromycetin, and chloramphenicol which are effective against a variety of diseases.

Other important broad-spectrum antibiotics used are vancomycin and ofloxacin. The antibiotic dysidazirine is found to be toxic to certain strains of cancer cells.

(b) Analgesics (CBSE Sample Paper 2019)
Answer:
Analgesics: The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

1. Non-narcotic (non-addictive) drugs
2. Narcotic drugs

1. Non-narcotic drugs: The common non-addictive analgesics are aspirin and paracetamol. Aspirin (2-acetoxy benzoic acid) is the most familiar example. It inhibits the synthesis of compounds known as prostaglandins which stimulate inflammation in the tissues and cause pain. These drugs are effective in relieving skeletal pain such as that due to arthritis.

Aspirin has also been very popular because it has antipyretic (temperature lowering) properties. Now, aspirin also finds use in the prevention of heart attack because it has anti-blood-clotting action. In addition, many other potential applications of aspirin, presently under investigation, include pregnancy-related complications, viral inflammation in AIDS patients, Alzheimer’s disease, dementia, cancer, etc.

Because of the shortcomings of aspirin, other analgesics like naproxen, ibuprofen, and diclofenac sodium or potassium find use as alternatives.

2. Narcotic drugs: Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine and its derivatives codeine, heroin, and marijuana are used in severe pain as analgesics. These are known to be habit-forming. When used in medicinal doses, these relieve pain and produce sleep. However, in excessive (poisonous) doses these produce stupor coma, convulsions and ultimately leading to death.

These analgesics are mainly used for relief in postoperative pains, cardiac pain, and pains related to childbirth and terminal cancer.

Question 10.
(a) Why are metal hydroxides better alternatives than sodium hydrogen carbonate in antacids?
(b) Why is aspirin used in the prevention of heart attacks?
(c) Why antihistamines do not affect the secretion of acid in the stomach?
OR
Define the following terms with a suitable example of each:
(a) Tranquilizers
(b) Antibiotics
(c) Non-ionic detergents (CBSE AI 2019)
Answer:
(a) Metal hydroxides are better alternatives than sodium hydrogen carbonates because of being insoluble. These do not increase the pH above neutrality. However, excessive hydrogen carbonates can make the stomach alkaline and may trigger the production of even more acid.

(b) Aspirin is a blood thinner and has anti-blood-clotting action and therefore, used in the prevention of heart attacks.

(c) Antihistamines do not affect the secretion of acid in the stomach because anti-allergic and antacid drugs work on different receptors. The receptors present in the stomach do not react with antihistamines.
OR
(a) Tranquilizers: These are substances used to relieve mental diseases. They reduce tension and anxiety. They act on higher centers of the nervous system. These are the constituents of sleeping pills. The common examples are Derivatives of barbituric acid, Equanil, luminal, diazepam, methedrine, etc.

(b) Antibiotics: Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, molds) and can inhibit the growth or even destroy other micro-organisms. For example, penicillin.

(c) Non-ionic detergents: These detergents do not contain any iron in their constitution and therefore, are non-ionic like the esters of high molecular mass. However, these contain polar groups which can form hydrogen bonds with water. For example, polyethylene glycol stearate.

Question 11.
(i) Why bithional is added in soap?
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(iii) Why soaps are biodegradable whereas detergents are non-biodegradable?
OR
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics (CBSE Delhi 2019)
Answer:
(i) Bithional is added to soap because it acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Magnesium hydroxide is insoluble and does not increase the pH above neutrality. But excess sodium bicarbonate can make the stomach alkaline and may trigger the production of even more acid.

(iii) The synthetic detergents have hydrocarbon chains that are highly branched. Bacteria cannot degrade them easily. Therefore, detergents are non-biodegradable. On the other hand, soaps have unbranched chains which can be biodegraded more easily. Hence, soaps are biodegradable.
OR
(i) Antibiotics: Antibiotics are chemical substances that are produced by microorganisms (bacteria, fungi, and molds) and can inhibit the growth or even destroy other micro-organisms, for example, penicillin, ampicillin, Amoxycillin, etc.

(ii) Artificial sweeteners: These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor, for example, saccharin, aspartame, alitame, etc.

(iii) Analgesics: These are neurologically active pain-killing drugs that reduce or abolish pain without causing impairment of consciousness, incoordination, mental confusion, paralysis, or some other disturbances of the nervous system, for example, aspirin, paracetamol, naproxen, ibuprofen, etc.

Question 12.
Give reasons for the following:
(i) Use of aspartame as an artificial sweetener is limited to cold foods.
Answer:
It is unstable at cooking temperature.

(ii) Metal hydroxides are better alternatives than sodium hydrogen carbonate for the treatment of acidity.
Answer:
Excessive hydrogen carbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides being insoluble do not increase the pH above neutrality,

(iii) Aspirin is used in the prevention of heart attacks. (CBSE Sample Paper 2018, 2019)
Answer:
Aspirin has anti-blood-clotting action.

Question 13.
(i) Why is bithional added to soap?
Answer:
To impart antiseptic properties

(ii) What is the tincture of iodine? Write its one use.
Answer:
2-3% solution of iodine in the alcohol-water mixture is the tincture of iodine. Iodine dissolved in alcohol, used as an antiseptic, and applied on wounds.

(iii) Among the following, which one acts as a food preservative?
Sodium benzoate, Aspartame (CBSE 2016)
Answer:
Sodium benzoate

Question 14.
(a) Which one of the following is a disinfectant?
0-2% solution of phenol or 1% solution of phenol
(b) What is the difference between agonists and antagonists?
(c) Write one example of each of
(i) Artificial sweetener
(ii) Antacids
OR
Define the following terms with a suitable example of each:
(a) Antiseptics
(b) Bactericidal antibiotics
(c) Cationic detergents (CBSE Al 2019)
Answer:
(a) 1% solution of phenol

(b) Agonists are substances that bind the receptor and produce a biological response.
Antagonists are the substances that bind to the receptor but inhibit its natural biological response.

(c) (i) Saccharin
(ii) Ranitidine
OR
(a) Antiseptics: The chemicals which either kill or prevent the growth of microorganisms when applied on living tissues are called antiseptics, for example, tobramycin.

(b) Bactericidal antibiotics: The chemicals which have a killing effect on microbes are called bactericidal antibiotics, for example, penicillin.

(c) Cationic detergents: The molecules which have long hydrocarbon chains with a positive charge on the nitrogen atom (cationic part) which is involved in cleansing action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions. The cationic part of the molecule is involved in the cleansing action, for example, cetyltrimethylammonium bromide.

Question 15.
(i) How are synthetic detergents better than soaps?
Answer:
The detergents are better than soaps because of the following reasons:
(a) Detergents can be used for washing even in hard water. On the other hand, soaps cannot be used in hard water.
(b) Detergents can be used in acidic solutions because they are not readily decomposed in an acidic medium. On the other hand, soaps cannot be used in an acidic medium because they are decomposed into carboxylic acids in an acidic medium.
(c) Detergents have a stronger cleansing action than soap.

(ii) Can you use soaps and synthetic detergents to check the hardness of water? (CBSE 2015)
Answer:
Soaps give an insoluble precipitate of calcium and magnesium in hard water whereas detergents do not give a precipitate. Therefore, soaps but not detergents can be used to check the hardness of the water.

Question 16.
Explain the role of the allosteric site in enzyme inhibition?
Answer:
Some drugs do not bind to the enzyme’s active site. These bind to a different site of enzyme which is called the allosteric site. This bonding of inhibitor at the allosteric site changes the shape of the active site in such a way that the substrate cannot recognize it. As a result, the affinity of the substrate for the active site is reduced.

It may be noted that if the bond formed between enzyme and inhibitor is a strong covalent bond and therefore cannot be broken easily, then the enzyme gets blocked permanently. The body then degrades the enzyme-inhibitor complex and synthesizes a new enzyme.

Question 17.
What are food additives? Describe the following with suitable examples:
(i) Preservatives
(ii) Artificial sweetening agents
Answer:
Food additives are the chemicals that are added to food for their preservation and enhancing their appeal. Some common food additives are:
(a) Flavours and sweeteners
(b) Food colors (dyes)
(c) Stabilising agents
(d) Antioxidants
(e) Preservatives
(f) Nutritional supplements such as vitamins, minerals, and amino acids.

(i) Preservatives: These are the chemical substances that are added to the food materials to prevent their spoilage and to retain their nutritive value for long periods. These preservatives prevent the rancidity of food and inhibit the growth or kill the micro-organisms.

The common salt is generally added to resist the activity of micro-organisms in food. The preservation of food by adding a sufficient amount of salt to it is called salting. It is used for the preservation of raw mango, amla, beans, tamarind, fish, meat, etc. The salt prevents the water from being available for microbial growth.

Sugar syrup is also used for preserving many fruits such as apples, mango, strawberry, carrot, etc. Besides these vinegar, oils, spices, citric acid is also used as food preservatives, which are used for pickles, ketchup, jams, squashes, etc.

The growth of microbes in food material can also be prevented by adding certain chemical substances. The most common preservative used is sodium benzoate (C₆H₅COONa), salts of propanoic acid, sorbic acid, and potassium metabisulphite (source of sulfur dioxide). Certain food preservatives such as butylated hydroxyanisole (BHA) and butylated hydroxytoluene (BHT) for edible oils also act as antioxidants.

(ii) Artificial sweetening agents. These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor. Ortho- sulphobenzimide known as saccharin is the most popular sweetening agent and has been used for many articles of food. It has a very sweet taste and is about 550 times sweeter than sucrose.

Other artificial sweeteners commercially used in food articles are aspartame (methyl ester), alitame, dulcin (urea sweetener), dihydrochalcones (DHC), sucralose, etc.

We have given detailed NCERT Solutions for Class 10 Sanskrit Shemushi Chapter 3 व्यायामः सर्वदा पथ्यः Questions and Answers come in handy for quickly completing your homework.

Shemushi Sanskrit Class 10 Solutions Chapter 3 व्यायामः सर्वदा पथ्यः

Class 10 Sanskrit Shemushi Chapter 3 व्यायामः सर्वदा पथ्यः Textbook Questions and Answers

अभ्यासः

प्रश्न 1.
एकपदेन उत्तरं लिखत-

(क) परमम् आरोग्यं कस्मात् उपजायते?
उत्तराणि:
व्यायामात्

(ख) कस्य मांस स्थिरीभवति?
उत्तराणि:
व्यायामाभिरतस्य

(ग) सदा कः पथ्यः?
उत्तराणि:
व्यायामः

(घ) कै: पुभिः सर्वेषु ऋतुषु व्यायामः कर्तव्यः?
उत्तराणि:
आत्महितैषिभि

(ङ) व्यायामस्विन्नगात्रस्य समीपं के न उपसर्पन्ति?
उत्तराणि:
व्याधयो

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत-

(क) कीदृशं कर्म व्यायामसंज्ञितम् कथ्यते?
उत्तराणि:
शरीरायासजननम् कर्म व्यायामसज्ञितम् कथ्यते।

(ख) व्यायामात् कि किमुपजायते?
उत्तराणि:
व्यायामात् श्रमक्लमपिपासा ऊष्म-शीतादीनां सहिष्णुता-परमं च आरोग्यम् उपजायते।

(ग) जरा कस्य सकाशं सहसा न समधिरोहति?
उत्तराणि:
जरा व्यायामिनस्य जनस्य सकाशं सहसा न समधिगच्छति।

(घ) कस्य विरुद्धमपि भोजनं परिपच्यते?
उत्तराणि:
व्यायाम कुर्वतो नित्य विरुद्धमपि भोजनं परिपच्यते।

(ङ) कियता बलेन व्यायामः कर्तव्यः?
उत्तराणि:
अर्धन बलेन व्यायामः कर्तव्यः।

(च) अर्धबलस्य लक्षणम् किम्?
उत्तराणि:
व्यायाम कुर्वतः जन्तोः यदा हृदिस्थानास्थितः वायुः वस्त्र प्रपद्यते तद् अर्धबलस्य लक्षणम् अस्ति।

प्रश्न 3.
उदाहरणमनुसृत्य कोष्ठकगतेषु पदेषु तृतीयाविभक्तिं प्रयुज्य रिक्तस्थानानि पुरयत-

यथा- व्यायामः _________ होनमपि सुदर्शनं करोति (गुण)
व्यायामः गुणैः हीनमपि सुदर्शनं करोति।

(क) ________ व्यायामः कर्त्तव्यः। (बलस्याधं)
उत्तराणि:
बलस्यार्धन

(ख) ________ सदृशं किञ्चित् स्थौल्यापकर्षणं नास्ति। (व्यायाम)
उत्तराणि:
व्यायामेन

(ग) ________ विना जीवनं नास्ति। (विद्या)
उत्तराणि:
विद्यया

(घ) सः ________ खञ्जः अस्ति। (चरण)
उत्तराणि:
चरणेन

(ङ) सूपकारः ________ भोजनं जिघ्रति। (नासिका)
उत्तराणि:
नासिकया

प्रश्न 4(अ).
स्थूलपदमाधृत्य प्रश्ननिर्माणं कुरुत-

(क) शरीरस्य आयासजननं कर्म व्यायामः इति कथ्यते।
उत्तराणि:
कस्य आयासजननं कर्म व्यायामः इति कथ्यते?

(ख) अरयः व्यायामिनं न अर्दयन्ति।
उत्तराणि:
के व्यायामिनं न अर्दयन्ति?

(ग) आत्महितैषिभिः सर्वदा व्यायामः कर्तव्यः।
उत्तराणि:
कैः सर्वदा व्यायामः कर्तव्यः?

(घ) व्यायाम कुर्वतः विरुद्धं भोजनम् अपि परिपच्यते।
उत्तराणि:
व्यायाम कुर्वतः कीदृशम् भोजनम् अपि परिपच्यते?

(ङ) गात्राणां सुविभक्तता व्यायामेन संभवति।
उत्तराणि:
केषाम् सुविभक्तता व्यायामेन संभवति?

प्रश्न 4(आ).
षष्ठ श्लोकस्य भावमाश्रित्य रिक्तस्थानानि पूरयत-

यथा- ________ समीपे उरगाः न ________ एवमेव व्यायामिनः जनस्य समीपं ________ न गच्छन्ति। व्यायामः वयोरूपेगुणहीनम् अपि जनम् ________ करोति।
उत्तराणि:
वैनतेयस्य, गच्छन्ति, व्याधयः (रोगाः), सुदर्शन (दर्शननीयम्)।

प्रश्न 5(अ).
‘व्यायामस्य लाभाः’ इति विषयमधिकृत्य पञ्चवाक्येषु ‘संस्कृतभाषया’ एकम् अनुच्छेद लिखत।
उत्तराणि:
(क) व्यायामः जनेभ्यः स्वस्थं यच्छति।
(ख) नित्यं व्यायाम कुर्वन्तः जनाः कदापि रोगिणः न भवन्ति।
(ग) व्यायामेन जनैः खादितं भोजनं पूर्णरुपेण पचति।
(घ) व्यायामः जनाम् सुदर्शनाम् करोति।
(ङ) जनैः यथाशक्ति एव व्यायामः करणीयः।

प्रश्न 5(आ).
यथानिर्देशमुत्तरत-

(क) ‘तत्कृत्वा तु सुखं देहम्’ अत्र विशेषणपदं किम्?
उत्तराणि:
सुखं

(ख) ‘व्याधयो नोपसर्पन्ति वैनतेयमिवोरगाः’ अस्मिन् वाक्ये क्रियापदं किम्?
उत्तराणि:
उपसर्यन्ति

(ग) ‘पुम्भिरात्महितैषिभिः’ अत्र ‘पुरुषैः’ इत्यर्थे कि पदं प्रयुक्तम्?
उत्तराणि:
पुम्भि

(घ) ‘दीप्ताग्नित्वमनालस्य स्थिरत्वं लाघवं मजा’ इति वाक्यात ‘गौरवम्’ इति पदस्य विपरीतार्थकं पदं चित्वा लिखत।
उत्तराणि:
लाघवं

(ङ) न चास्ति सदृशं तेन किञ्चित् स्थौल्यापकर्षणम्’ अस्मिन् वाक्ये ‘तेन’ इति सर्वनामपदं कस्मै प्रयुक्तम्?
उत्तराणि:
व्यायामम्

प्रश्न 6(अ).
निम्नलिखितानाम् अव्ययानाम् रिक्तस्थानेषु प्रयोगं कुरुत-

सहसा, अपि, सदृशं, सर्वदा, यदा, सदा, अन्यथा

(क) _______ व्यायामः कर्त्तव्यः।
उत्तराणि:
सर्वदा

(ख) _______ मनुष्यः सम्यकूरूपेण व्यायाम करोति तदा सः _______ स्वस्थः तिष्ठति।
उत्तराणि:
यदा, सदा

(ग) व्यायामेन असुन्दराः _______ सुन्दराः भवति।
उत्तराणि:
अपि

(घ) व्यायामिनः जनस्य सकाशं वार्धक्यं _______ नायाति।
उत्तराणि:
सहसा

(ङ) व्यायामेन _______ किञ्चित् स्थौल्यापकर्षणं नास्ति।
उत्तराणि:
सदृश

(च) व्यायाम समीक्ष्य एवं कर्तव्यम् _______ व्याधयः आयान्ति।
उत्तराणि:
अन्यथा

प्रश्न 6(आ).
उदाहरणमनुसृत्य वाच्यपरिवर्तनं कुरुत-

उत्तराणि:
(क) बलवान् विरुद्धमपि भोजन पचति।
(ख) जनाः व्यायामेन कान्ति लभन्ते।
(ग) मोहनः पाठं पठति।
(घ) लता गीत गायति।

प्रश्न 7.
अधोलिखितेषु तद्धितपदेषु प्रकृति/प्रत्ययं च पृथक् कृत्वा लिखत-

उत्तराणि:
(क) पथ्य + तमप्
(ख) सहिष्णु + तल्
(ग) अग्नि – त्व
(घ) स्थिर + त्व
(ङ) लाघु + ष्यञ्

योग्यताविस्तारः
यह पाठ आयुर्वेद के प्रसिद्ध ग्रन्थ ‘सुश्रुतसंहिता’ के चिकित्सा स्थान में वर्णित 24वें अध्याय से संकलित है। इसमें आचार्य सुश्रुत ने व्यायाम की परिभाषा बताते हुए उससे होने वाले लाभों की चर्चा की है। शरीर में सुगठन, कान्ति, स्फूर्ति, सहिष्णुता, नीरोगता आदि व्यायाम के प्रमुख लाभ हैं।

(क) सुश्रुतः आयुर्वेदस्य, ‘सुश्रुतसंहिता’ इत्याख्यस्य ग्रन्थस्य रचयिता। अस्मिन् ग्रन्थे शल्यचिकित्सायाः प्राधान्यमस्ति। सुश्रुतः शल्यशास्त्रज्ञस्य दिवोदासस्य शिष्यः आसीत्। दिवोदासः सुश्रुत वाराणस्याम् आयुर्वेदम् अपाठयत्। सुश्रुतः दिवोदासस्य उपदेशान् स्वग्रन्थेऽलिखत्।

(ख) उपलब्धासु आयुर्वेदीय-संहितासु ‘सुश्रुतसंहिता’ सर्वश्रेष्ठः शल्यचिकित्साप्रधानो ग्रन्थः। अस्मिन् ग्रन्थे 120 अध्यायेषु क्रमेण सूत्रस्थाने मौलिकसिद्धान्तानां शल्यकर्मापयोगि-यन्त्रदीना, निदानस्थाने प्रमुखाणां रोगाणां, शरीरस्थाने शरीरशास्त्रस्य चिकित्सास्थाने, शल्यचिकित्सायाः कल्पस्थाने च विषाणां प्रकरणानि वर्णितानि। अस्य उत्तरतन्त्रे 66 अध्यायाः सन्ति।

(ग) वैनतेयमिवोरगा: – कश्यप ऋषि की दो पत्नियाँ थीं-कट्ठ और विनता। विनता का पुत्र गरुड़ था और कद्रु के पुत्र सर्प थे। विनता का पुत्र होने के कारण गरुड़ को वैनतेय कहा जाता है। (विनताया: अयम् वैनतेयः, ढक् (एय) प्रत्यये कृते)। गरुड़ सर्प से अधिक ताकतवर होता है, भयवश साँप गरुड़ के पास जाने का साहस नहीं करता। यहाँ व्यायाम करने वाले मनुष्य की तुलना गरुड़ से तथा व्याधियों की तुलना साँप से की गई है। जिस प्रकार गरुड़ के समक्ष साँप नहीं जाते। उसी प्रकार व्यायाम करने वाले व्यक्ति के पास रोग नहीं फटकते।

भाषिकविस्तारः
गुणवाचक शब्दों से भाव अर्थ में ष्यञ् अर्थात् य प्रत्यय लगाकर भाववाची पदों का निर्माण किया जाता है। शब्द के प्रथम स्वर में वृद्धि होती है और अन्तिम अ का लोप होता है।
(क) शूरस्य भावः शौर्यम् – शूर + ष्यञ्
(ख) सुन्दरस्य भावः सौन्दर्यम् – सुन्दर + ष्यञ्
(ग) सुखस्य भावः सौख्यम् – सुख + ष्यञ्
(घ) विदुषः भावः वैदुष्यम् – विद्वस् + ष्यञ्
(ङ) मधुरस्य भावः माधुर्यम् – मधुर + ष्यञ्
(च) स्थूलस्य भावः स्थौल्यम् – स्थूल + ष्यञ्
(छ) अरोगस्य भावः आरोग्यम् – अरोग + ष्यञ्
(ज) सहितस्य भावः साहित्यम् – सहित + ष्यञ्

थाल्-प्रत्ययः – ‘प्रकार’ अर्थ में थाल् प्रत्यय का प्रयोग होता है। जैसे-
तेन प्रकारेण – तथा
येन प्रकारेण – यथा
अन्येन प्रकारेण – अन्यथा
सर्व प्रकारेण – सर्वथा
उभय प्रकारेण – उभयथा

भावविस्तारः
(क) शरीरमाद्यं खलु धर्मसाधनम्।

(ख) लाघवं कर्मसामर्थ्य स्थैर्य क्लेशसहिष्णुता।
दोषक्षयोऽग्निवृद्धिश्च व्यायामादुपजायते।।

(ग) यथा शरीरस्य रक्षायै उचितं भोजनम्, उचितश्च व्यवहारः आवश्यकोऽस्ति तथैव शरीरस्य स्वास्थ्याय व्यायामः अपि आवश्यकः।

(घ) युक्ताहारविहारस्य युक्तचेष्टस्य कर्मसु।
युक्तस्वप्नावबोधस्य योगो भवति दुःखहा।।

(ङ) पक्षिणः आकाशे उड्डीयन्ते तेषाम् उड्डयनमेव तेषां व्यायामः। पशवोऽपि इतस्ततः पलायन्ते, पलायनमेव तेषां व्यायामः। शैशवे शिशुः स्वहस्तपादौ चालयति, अयमेव तस्य व्यायामः।

वि + आ + यम् धातो: घञ् प्रत्ययात् निष्पन्नः व्यायाम शब्द: विस्तारस्य विकासस्य च वाचकः। यतो हि व्यायामेन अङ्गानां विकासः भवति। अत: सुखपूर्वकं जीवन यापयितुं मनुष्य: नित्यं व्यायामः करणीयः।

Class 10 Sanskrit Shemushi Chapter 3 व्यायामः सर्वदा पथ्यः Additional Important Questions and Answers

अतिरिक्त प्रश्नाः

1. निम्न श्लोकान् पठित्वा तदाधारितानां प्रश्नानाम् उत्तराणि लिखत-

(क) शरीरायासजननं कर्म व्यायामसंज्ञितम्।
तत्कृत्वा तु सुखं देहं विमृद्नीयात् समन्ततः॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कस्य आयासजननं कर्म व्यायामः इति कथ्यते?
2. किम् कृत्वा सुखं प्रप्नोति?
3. व्यायामः कीदृशं कर्म अस्ति?

उत्तराणि:

1. शरीरस्य
2. व्यायामम्
3. शरीरायासजननम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. व्यायामात् पुरः किम् कर्त्तव्यम्?
2. व्यायामः कः कथितः?

उत्तराणि:

1. व्यायामात् पुरः विमृद्नीयात्।
2. शरीरायसजननम् कर्म व्यायामः कथितः।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘सर्वतः’ इत्यर्थे किम् पदं अत्र प्रयुक्तम्?
2. ‘दु:खम्’ इत्यस्य पदस्य विलोमपदं कि लिखितम्।
3. अत्र श्लोके ‘परिश्रमः’ इत्यस्य पदस्य कः पर्यायः लिखितः?
4. श्लोके ‘सुखं देहम्’ अत्र विशेषणपदं किम्?

उत्तराणि:

1. समन्ततः
2. सुखम्
3. आयासः
4. सुखम्

(ख) शरीरोपचयः कान्तिर्गात्राणां सुविभक्तता।
दीप्ताग्नित्वमनालस्यं स्थिरत्वं लाघवं मजा॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. कान्तिः गात्राणां कथम् भवति?
2. शरीरस्य मृजायै कः कर्त्तव्यम्?
3. कस्य उपचयः व्यायामेन भवति?

उत्तराणि:

1. व्यायामेन
2. व्यायामः
3. शरीरस्य

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)

1. व्यायामेन किम्-किम् भवति?
2. व्यायामेन कि दीप्तं भवति?

उत्तराणि:

1. व्यायामेन शरीरोपचयः, कान्तिर्गात्राणां, सुविभक्तता दीप्ताग्नित्वमनालस्य स्थिरत्वं लाघवं मजा।
2. व्यायामेन अग्नित्वं दीप्तं भवति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)

1. ‘अस्थिरत्वम्’ इति पदस्य विपर्ययपदं किम्?
2. ‘सुविभक्ता’ अस्मिन् पदे उपसर्ग पृथक् कृत्वा लिखत।
3. श्लोक ‘स्वच्छता’ अस्य पदस्य पर्यायपदं किम् अस्ति?
4. ‘सुन्दरता’ इति पदस्य अर्थे श्लोके किं पदं प्रयुक्तम्?

उत्तराणि:

1. स्थिरत्वम्
2. सु + विभक्तता
3. मृजा
4. कान्ति:

(ग) श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता।
आरोग्यं चापि परमं व्यायामादुपजायते॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. उष्णशीतादीनाम् सहिष्णुता कथम् जायते?
2. श्रमेण किम् भवति?
3. व्यायामेन केषां सहिष्णता भवति?

उत्तराणि:

1. व्यायामेन
2. क्लमम्
3. श्रमक्लमपिपासाण्ण-शीतादीनाम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. शरीर व्यायामात् किं किमुपजायते?
2. परमम् आरोग्य केन भवति?

उत्तराणि:

1. श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता आरोग्यं चापि परमं व्यायामादुपजायते।
2. परमम् आरोग्यं व्यायामेन भवति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘निरोग्यम्’ इति पदस्य किम् पर्यायपदम् अत्र प्रयुक्तम्?
2. ‘व्यायामादुपजायते’ अत्र क्रियापदं किम् अस्ति?
3. ‘आरोग्यं चापि परमं’ अत्र विशेष्यपदं किम् अस्ति?
4. श्लोके ‘उष्ण’ पदस्य विपर्ययः को वर्तते?

उत्तराणि:

1. आरोग्यम्
2. उपजायते
3. आरोग्यम्
4. शीत

(घ) न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम्।
न च व्यायामिनं मर्त्यमर्दयन्त्यरयो बलात्॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. स्थौल्यापकर्षणम् कथम् भवति?
2. के व्यायामिनं न अर्दयन्ति?
3. व्यायामेन सदृशं किम् नास्ति?

उत्तराणि:

1. व्यायामेन
2. अरयः
3. स्थौल्पायकर्षणम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. किम् शत्रवः बलपूर्वकं व्यायामिनम् अर्दयन्ति?
2. व्यायामः कस्य आकर्षणं करोति?

उत्तराणि:

1. न शत्रवः बलपूर्वक व्यायमिनम् न अर्दयन्ति।
2. व्यायामः स्थौल्यस्य आकर्षणं करोति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘बलपूर्वकम्’ इत्यर्थे किं पदम् अत्र प्रयुक्तम्?
2. ‘मर्त्यमर्दयन्त्यरयो’ अस्मिन् पदे क्रियापदं किम् अस्ति?
3. ‘मित्राणि’ इति पदस्य कि विलोमपदं अत्र प्रयुक्तम्?
4. श्लोके ‘शत्रवः’ इति पदस्य कः पर्यायः आगतः?

उत्तराणि:

1. बलात्
2. अर्दयन्ति
3. अरयः
4. अरयः

(ङ) न चैनं सहसाक्रम्य जरा समधिरोहति।
स्थिरीभवति मांसं च व्यायामाभिरतस्य च॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. मनुष्यस्य जीवने का सहसा आक्रम्यति?
2. जरा कीदृशस्य जनस्य समीपम् सहसा न समधिगच्छति?
3. जरा कथं मनुष्यजीवनम् आक्रम्यति?

उत्तराणि:

1. जरा
2. व्यायामाभिरतस्य
3. सहसा

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. जनस्य मांस कथम् स्थिरी भवति?
2. केन कारणंन मनुष्यस्य मांस स्थिर जायत?

उत्तराणि:

1. व्यायामेन मांस स्थिरीभवति।
2. व्यायामाभिरतेन मनुष्यस्य मांस स्थिर जायते।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. अत्र ‘चैनं’ एनम् पदं कस्मै प्रयुक्तम्?
2. ‘अवरोहति’ इति पदस्य विपर्ययपदं श्लोके किम्?
3. संलग्नस्य’ इति पदस्य पर्यायपदं श्लोके किम्?
4. श्लोके ‘जरा’ इति कर्तृपदस्य क्रियापदं किमस्ति?

उत्तराणि:

1. व्यायामशीलाय
2. आरोहति
3. अभिरतस्य
4. समधिरोहति

(च) व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य च।
व्याधयो नोपसर्पन्ति वैनतेयमिवोरगा:
वयोरूपगुणीनमपि कुर्यात्सुदर्शनम्॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. वैनतेयस्य समीपे कः न आगच्छति?
2. गुणहीनः जनः केन सुदर्शनः क्रियत?
3. पद्भ्याम् उद्वर्तितस्य जनस्य समीप के न आगच्छन्ति?

उत्तराणि:

1. सर्पः
2. व्यायामेन
3. व्याधयः

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. जना रोगैः आक्रान्ताः कदा न भवन्ति?
2. व्यायामः कान् जनान् सुदर्शनान् करोति?

उत्तराणि:

1. यदा जनाः नियमित रूपेण व्यायाम कुवन्ति तदा ते रोगः आक्रान्ताः न भवन्ति।
2. व्यायामः वयोरूपगुणैः हीनान् जनान् सुदर्शनान् करोति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘कुर्यात्सुदर्शनम्’ अत्र क्रियापदं किम्?
2. ‘गुणैयुक्तः’ इति पदस्य विपर्ययपदं गद्यांशे किम्?
3. ‘सर्पाः इति पदस्य पर्यायपदं गद्यांशे किम्?
4. ‘व्याधयो नोपसर्पन्ति’। अत्र कर्तृपदं किम्?

उत्तराणि:

1. कुर्यात्
2. गुणहीनः
3. उरगाः
4. व्याधयः

(छ) व्यायाम कुर्वतो नित्यं विरुद्धमपि भोजनम्।
विदग्धमविदग्धं वा निर्दोष परिपच्यते॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. प्रतिदिनं कम् कर्त्तव्यम्?
2. सुचारुरूपेण भोजनम् पाचयितुम् कः कर्त्तव्यः?
3. व्यायामेन विदग्धमविदग्धं वा भोजनं कथं परिपच्यते?

उत्तराणि:

1. व्यायामम्
2. व्यायामः
3. निर्दोषम्

प्रश्न 2.
पूर्णवाक्येनउत्तरत (केवलं प्रश्नमेकमेव)-

1. अस्माकं शरीरे व्यायामेन कीदृशं भोजनं परिपच्यते?
2. जनः नित्यं कम् कुर्यात्?

उत्तराणि:

1. अस्माकं शरीरे व्यायामेन विदग्धम् अविदग्धम् वा भोजनम् परिपच्यते।
2. जनाः नित्यं व्यायाम कुर्यात्।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘विदग्धम्’ इति पदस्य विपर्ययपदं गद्यांशे किम्?
2. ‘पच्यते’ अत्र क्रियापदस्य कर्तृपदं किम्?
3. ‘सुपक्वम्’ इति पदस्य पर्यायपदं किम्?
4. विरुद्धमपि भोजनम्’ अत्र विशेष्यपदं किम्?

उत्तराणि:

1. अविदग्धम्
2. भोजनम्
3. विदग्धम्
4. भोजनम्

(ज) व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम्।
स च शीते वसन्ते च तेषां पथ्यतमः स्मृतः॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. बलवते पथ्यः कः अस्ति?
2. स्निग्ध भोजिने’ क: औषिधि-सदृशं अस्ति?
3. केषां सदा व्यायामः पथ्यः वर्तते?

उत्तराणि:

1. व्यायामः
2. व्यायामः
3. बलिनाम्

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. व्यायामः कदा पथ्यतमः भवति?
2. कः सर्वेभ्यः ओषधिः वर्तते?

उत्तराणि:

1. शीते वसन्ते च व्यायामः सदा पथ्यतमः भवति।
2. व्यायामः सर्वेभ्यः औषधिः वर्तते।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘लाभदायकः’ इति पदस्य पर्यायपदं किम्?
2. ‘अपथ्यः’ इति पदस्य विपर्ययपदं किम् प्रयुक्तम्?
3. ‘तेषां पथ्यतमः स्मृतः’ अत्र क्रियापदं किम्?
4. ‘व्यायामों’ हि सदा पथ्यः’। अत्र विशेषणपदं किमस्ति?

उत्तराणि:

1. पथ्यतमः
2. पथ्यः
3. स्मृतः
4. पध्यः

(झ) सर्वेष्वृतुष्वहरहः पुम्भिरात्महितैषिभिः।
बलस्यार्धेन कर्त्तव्यो व्यायामो हन्त्यतोऽन्यथा।।

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)

1. व्यायामः केषु ऋतुषु कर्त्तव्यः?
2. मनुष्यः कस्य हितैषी अस्ति?
3. कस्य हितैषिभिः व्यायामः कर्तव्य?

उत्तराणि:

1. सर्वेषु
2. आत्मनः
3. आत्मनः

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. व्यायामः कदा हानिकारकः न भवति?
2. व्यायामः कदा जनान् हन्ति?

उत्तराणि:

1. सर्वेषु ऋतुषु बलस्यार्धन व्यायामः कर्त्तव्यः तदा व्यायामः हानिकारकः न भवति।
2. यदा व्यायामः पूर्णबले क्रियते तदा सः जनान् हन्ति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘पुरुषैः’ इति पदस्य पर्यायपदं लिखत।
2. सर्वेष्वृतुषु’ अत्र विशेष्यपदं किम् अस्ति?
3. ‘पूर्णतया’ इति पदस्य विलोमपदं किम्।
4. ‘बलस्यार्धन व्यायामः कर्तव्यः।’ अत्र क्रियापदं किम्?

उत्तराणि:

1. पुम्भिः
2. ऋतुषु
3. अर्धन
4. कर्तव्यः

(ञ) विस्थानस्थितो वायुर्यदा वक्त्रं प्रपद्यते।
व्यायाम कुर्वतो जन्तोस्तबलार्धस्य लक्षणम्॥

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. हृदिस्थाने कः भवति?
2. हदिस्थाने स्थितो वायुः कुत्र प्रपद्यते?
3. अत्र कस्य लक्षणम् कथितम्?

उत्तराणि:

1. वायुः
2. वक्त्रम्
3. बलार्धस्य

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवलं प्रश्नमेकमेव)-

1. जन्तोः बलार्धस्य लक्षणम् किम् भवति?
2. वायुः कुत्र स्थितो भवति?

उत्तराणि:

1. हदिस्थान स्थितो वायुयंदा वक्त्रं प्रपद्यते व्यायाम कुर्वता जन्तोः तद् बलार्धस्य लक्षणम्।
2. वायुः हृदिस्थाने स्थिता भवति।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. अत्र श्लोके ‘प्रपद्यते’ इति क्रियापदस्य कर्तृपदं किम्?
2. श्लोके ‘पूर्णबलस्य’ इति पदस्य विलोमपदं किम्?
3. अत्र ‘जीवस्य’ इति पदस्य पर्यायपदं किम्?
4. ‘कुर्वतः जन्तोः’ अनयोः विशेषणपदं किम्?

उत्तराणि:

1. वायुः
2. बलार्धस्य
3. जन्तोः
4. कुवंत:

(ट) वयोबलशरीराणि देशकालाशनानि च।
समीक्ष्य कुर्याद् व्यायाममन्यथा रोगमाप्नुयात्।।

प्रश्न 1.
एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव)-

1. व्यायामशीलः कीदृशं भोजनं कुर्यात्?
2. सूर्यासनम् कदा करणीयम्?
3. कानि दृष्ट्वा व्यानाम कुर्यात्?

उत्तराणि:

1. अनुकूलम्
2. प्रात:काले
3. वयोबलशरीराणि

प्रश्न 2.
पूर्णवाक्येन उत्तरत (केवल प्रश्नमेकमेव)-

1. व्यायामशीलः कदा रुग्णः न भवति?
2. यदि नियमत: व्यायामः न क्रियते तदा कि प्राप्यते?

उत्तराणि:

1. वयोबलशरीराणि देशकालाशनानि च समीक्ष्य व्यायाम कुर्यात् तदा व्यायामशीलः रुग्णः न भवति।
2. यदि नियमतः व्यायामः न क्रियते तदा रोगं प्राप्यते।

प्रश्न 3.
भाषिककार्यम् (केवलं प्रश्नत्रयमेव)-

1. ‘कुर्याद्’ इति क्रियापदस्य कर्तृपदं किम्?
2. ‘आयुः’ इति पदस्य पर्यायपदं किम्?
3. ‘रोगमाप्नुयात्’ अत्र क्रियापदं किम्?
4. श्लोके ‘स्वास्थ्यम्’ इति पदस्य कः विपर्ययः आगतः?

उत्तराणि:

1. व्यायामम्
2. वयः
3. आप्नुयात्
4. रोगम्

2. निम्नवाक्येषु रेखांकितपदानां स्थानेषु प्रश्नवाचकपदं लिखत-

(क) व्यायामं कृत्वा सुखं प्राप्नोति।
(i) किम्
(ii) कम्
(iii) काम्
(iv) के
उत्तराणि:
(ii) कम्

(ख) व्यायामात् आरोग्यम् उपजायते।
(i) कम्
(ii) किम्
(iii) कथम्
(iv) केषाम्
उत्तराणि:
(ii) किम्

(ग) शरीरस्य मृजायै व्यायामः कर्त्तव्यः।
(i) कः
(ii) कम्
(iii) किम्
(iv) केन्
उत्तराणि:
(i) कः

(घ) गात्राणां सुविभक्तता व्यायामेन संभवति।
(i) कम्
(ii) कान्
(iii) केन
(iv) कः
उत्तराणि:
(iii) केन

(ङ) जनैः व्यायामेन कान्तिः लभ्यते।
(i) कः
(ii) को
(iii) के
(iv) केन
उत्तराणि:
(iv) केन

(च) सर्वदा व्यायाम् कर्त्तव्यः।
(i) का
(ii) कः
(iii) कदा
(iv) कस्मिन्
उत्तराणि:
(iii) कदा

(छ) व्यायामेन सुन्दराः भवन्ति।
(ii) का:
(iii) किम
(iv) क:
उत्तराणि:
(ii) काः

(ज) अरयः व्यायामिनं न अर्दयन्ति।
(i) कयाः
(ii) के
(iii) कः
(iv) काः
उत्तराणि:
(ii) के

(झ) शरीरस्य मृजायै व्यायामः कर्त्तव्यः।
(i) कः
(ii) कम्
(iii) कने
(iv) कस्य
उत्तराणि:
(iv) कस्य

(ज) व्यायामेन सुन्दराः किञ्चित् स्थौल्यापकर्षणं नास्ति।
(i) कम्
(ii) कान्
(iii) केन
(iv) कः
उत्तराणि:
(iii) केन

(ट) व्यायामः गुणैः हीनमपि सुदर्शनं करोति।
(i) कः
(ii) कौ
(iii) कैः
(iv) के
उत्तराणि:
(iii) कैः

(ठ) बलस्यार्धन व्यायामः कर्त्तव्यः।
(i) क:
(ii) केन
(iii) कान्
(iv) कम्
उत्तराणि:
(ii) केन

(ड) मनुष्यस्य जीवने जरा सहसा आक्रम्यति?
(i) कः
(ii) का
(iii) किम्
(iv) काम्
उत्तराणि:
(ii) का

(ढ) मनुष्यस्य मासं व्यायामेन परिपक्वं भवति।
(i) कीदृशं
(ii) किम्
(iii) कीदृशः
(iv) कीदृशी
उत्तराणि:
(i) कीदृशं

(ण) व्यायामिनः विरुद्धम् भोजनम् अपि परिपच्यते।
(i) कः
(ii) काः
(iii) के
(iv) कस्य
उत्तराणि:
(iii) के

(त) व्यायामः वसन्तऋतौ अतीव लाभदायकः भवति।
(i) को
(ii) कदा
(iii) काम्
(iv) कैः
उत्तराणि:
(ii) कदा

(थ) वैनतेयस्य समीपे सर्पः न आगच्छति।
(i) कस्य
(ii) कः
(iii) कदा
(iv) किम्
उत्तराणि:
(i) कस्य

(द) व्यायामः बलस्यार्धन कर्त्तव्यः।
(i) कः
(ii) कम्
(iii) केन
(iv) कान्
उत्तराणि:
(ii) केन

(ध) हृदिस्थाने वायुः भवन्ति।
(i) कुत्र
(ii) का
(iv) किम्
(iv) के
उत्तराणि:
(i) कुत्र

(न) व्यायामशीलः पौष्टिकं भोजनं कुर्यात।
(i) कीदृशः
(ii) कीदृश
(iv) कीदृशी
(iv) कम्
उत्तराणि:
(ii) कीदृश

3. अधोलिखितश्लोकानाम् अन्वयं मञ्जूषातः उचितं पद चित्वा पूरयत-

(क) शरीरायासजननं कर्म व्यायामसंज्ञितम्।
तत्कृत्वा तु सुखं देहं विमृद्नीयात् समन्ततः॥

अन्वयः
शरीर आयासजननम् (i) ________ व्यायामसंज्ञितम् (ii) ________ तु (iii) ________ सुख (iv) ________ विमृनीयात्।
मञ्जूषा- देहम्, समन्ततः कर्म, तत्कृत्वा
उत्तराणि:
(i) कर्म
(ii) तत्कृत्वा
(iii) देहम्
(iv) समन्ततः

(ख) शरीरोपचयः कान्तिर्गात्राणां सुविभक्तता।
दीप्ताग्नित्वमनालस्यं स्थिरत्वं लाघवं मजा।।

अन्वयः
(i) ________ गात्राणाम् (ii) ________ सुविभक्तता दीप्ताग्नित्वम् (iii) ________ स्थिरत्वं (iv) ________ मजा।
मञ्जूषा- लाघवं, कान्तिः, अनालस्यं, शरीरोपचयः
उत्तराणि:
(i) शरीरोपचयः
(ii) कान्तिः
(iii) अनालस्यं
(iv) लाघवं

(ग) श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता।
आरोग्यं चापि परमं व्यायामादुपजायते॥

अन्वयः
श्रमक्लमपिपासोष्ण (i) ________ सहिष्णता (ii) ________ आरोग्य (iii) ________ व्यायामाद् (iv) ________।
मञ्जूषा- परमं, उपजायते, शीतादीनां, चापि
उत्तराणि:
(i) शीतादीनां
(ii) परमं
(iii) चापि
(iv) उपजायते

(घ) न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम्।
न च व्यायामिनं मर्त्यमर्दयन्त्यरयो बलात्॥

अन्वयः
(i) ________ तेन (ii) ________ किञ्चित्स्थौल्यापकर्षणम् न (iii) ________ च व्यायामिनं मर्त्यम् अरयः (iv) ________ न अर्दयन्ति।
मञ्जूषा- बलात्, सदृशं, अस्ति, च
उत्तराणि:
(i) च
(ii) सदृशं
(ii) अस्ति
(iv) बलात्

(ङ) न चैनं सहसाक्रम्य जरा समधिरोहति।
स्थिरीभवति मांसं च व्यायामाभिरतस्य च॥

अन्वयः
चैन (i) ________ सहसाक्रम्य (ii) ________ समधिरोहति च व्यायामाभिरतस्य (iii) ________ च (iv) ________
मञ्जूषा- मांस, जरा, न, स्थिरीभवति
उत्तराणि:
(i) जरा
(ii) न
(iii) मांस
(iv) स्थिरीभवति

(च) व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य च।
व्याधयो नोपसर्पन्ति वैनतेयमिवोरगाः
वयोरूपगुणहीनमपि कुर्यात्सुदर्शनम्॥

अन्वयः
व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य (i) ________ (ii) ________ वैनतेयमिवोरगाः (iii) ________ वयोरूपगुणैीनमपि (iv) ________ कुरयात्।
मञ्जूषा- नोपसर्पन्ति, सुदर्शनम्, व्याधयो, च
उत्तराणि:
(i) च
(ii) व्याधयो
(iii) नोपसर्पन्ति
(iv) सुदर्शनम्

(छ) व्यायाम कुर्वतो नित्यं विरुद्धमपि भोजनम्।
विदग्धमविदग्धं वा निर्दोष परिपच्यते॥

अन्वयः
नित्यं (i) ________ कुर्वतो विरुद्धमपि (ii) ________ विदग्धमविदग्धं (iii) ________ निर्दोष (iv) ________|
मञ्जूषा- वा, व्यायाम, परिपच्यते, भोजनम्
उत्तराणि:
(i) व्यायाम
(ii) भोजनम्
(iii) वा
(iv) परिपच्यते

(ज) व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम्।
स च शीते वसन्ते च तेषां पथ्यतमः समृतः॥

अन्वयः
हि व्यायामो (i) ________ बलिनां स्निग्धभोजिनाम् (ii) ________ शीते च (iii) ________ च तेषां (iv) ________ स्मृतः।
मञ्जूषा- सदा, पथ्यो, वसन्ते, पथ्यतमः
उत्तराणि:
(i) सदा
(ii) पथ्यो
(iii) वसन्ते
(iv) पथ्यतमः

(झ) सर्वेष्वृतुष्वहरहः पुम्भिरात्महितैषिभिः।
बलस्यार्धेन कर्त्तव्यो व्यायामो हन्त्यतोऽन्यथा॥

अन्वयः
आत्महितैषिभिः (i) ________ सर्वेषु ऋतुष अरहरः (ii) ________ अर्धन (iii) ________ कर्त्तव्यो (iv) ________ हन्त्यतः।
मञ्जूषा- अन्यथा, पुम्भिः, बलस्य, व्यायामो
उत्तराणि:
(i) पुम्भिः
(ii) बलस्य
(iii) व्यायामो
(iv) अन्यथा

(ञ) हृदिस्थानस्थितो वायुर्यदा वक्त्रं प्रपद्यते।
व्यायाम कुर्वतो जन्तोस्तबलार्धस्य लक्षणम्॥

अन्वयः
यदा हदिस्थानस्थितो (i) ________ वक्त्र (ii) ________ व्यायाम कुर्वतो (ii) ________ तबलार्धस्य (iv) ________।
मञ्जूषा- वायुः, प्रपद्यते, लक्षणम्, जन्तोः
उत्तराणि:
(i) वायुः
(ii) प्रपद्यते
(iii) जन्तोः
(iv) लक्षणम्

(ट) वयोबलशरीराणि देशकालाशनानि च।
समीक्ष्य कुर्याद् व्यायाममन्यथा रोगमाप्नुयात्॥

अन्वयः
वयोबलशरीराणि (i) ________ च व्यायाम (ii) ________ कुर्याद् (ii) ________ रोगम् (iv) ________
मञ्जूषा- देशकालाशनानि, आप्नुयात्, समीक्ष्य, अन्यथा
उत्तराणि:
(i) देशकालाशनानि
(ii) समीक्ष्य
(iii) अन्यथा
(iv) आप्नुयात्

4. अधोलिखित श्लोकानाम् भावार्थम् मञ्जूषायाः सहायत्या उचित-क्रमेण पूरयत्-

(क) शरीरायासजननं कर्म व्यायामसंज्ञितम्।
तत्कृत्वा तु सुखं देह विमृद्नीयात् समन्ततः॥

भावार्थ:
अस्यभावेऽस्ति यत् शरीरस्य (i) _______ कार्यं (ii) _______ कथ्यते। तं कृत्वा जनाः (iii) _______ सुखं प्राप्नुवन्ति अतः समन्ततः शरीरस्य (iv) _______ अवश्यमेव नित्यं कर्तव्यम्।
मञ्जूषा- मर्दनम्, परिश्रमस्य, व्यायामः, दैहिकं
उत्तराणि:
(i) परिश्रमस्य
(ii) व्यायामः
(iii) दैहिक
(iv) मर्दनम्

(ख) शरीरोपचयः कान्तिर्गात्राणां सुविभक्तता।
दीप्ताग्नित्वमनालस्य स्थिरत्वं लाघवं मजा॥

भावार्थ:
व्यायात् जनानां (i) _______ शारीरिकसौन्दर्य जठाराग्नेश्च (ii) _______वति। सहैव निरालस्यता (iii) _______ सूक्ष्मता (iv) _______ स्वच्छता चापिव्यायामात् भवति।
मञ्जूषा- स्थिरता, शरीरवृद्धि:, शरीरस्य, प्रकाशः (तेजः)
उत्तराणि:
(i) शरीरवृद्धि:
(ii) प्रकाशः (तेज:)
(iii) स्थिरता
(iv) शरीरस्य

(ग) श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता।
आरोग्यं चापि परमं व्यायामादुपजायते॥

भावार्थ:
आचार्यः सुश्रुतः कथयति यत् व्यायामात् (i) _______ क्लमम् (ii) _______ ऊष्म तापशीतादीनां (iii) _______ एवम् (iv) _______ स्वास्थ्यं भति।
मञ्जूषा- पिपासा, शारीरिकपरिश्रम, उत्तम, सहनम्
उत्तराणि:
(i) शारीरिकपरिश्रमं
(ii) पिपासा
(iii) सहनम्
(iv) उत्तम

(घ) न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम्।
न च व्यायामिनं मर्त्यमर्दयन्त्यरयो बलात्।।

भावार्थ:
व्यायामात् अतिरिक्तं (i) _______ दूरीकरणस्य अन्यः कोऽपि (ii) _______ नास्ति। व्यायाम च कुर्वन्तं (iii) _______ शत्रवः बलात् (iv) _______ समर्थाः न भवन्ति।
मञ्जूषा- मर्दयितुम्, उपायः, जनं, स्थूलताम्।
उत्तराणि:
(i) स्थूलताम्
(ii) उपाय:
(iii) जन
(iv) मर्दयितुम्

(ङ) न चैनं सहसाक्रम्य जरा समधिरोहति।
स्थिरीभवति मांसं च व्यायामाभिरतस्य च॥

भावार्थ:
व्यायामे नित्यं (i) _______ जगस्य समीपे (ii) _______ सहसैव कदापि आक्रमणं न करोति, एवमेव (iii) _______ जनस्य मांसम् अपि (iv) _______ निरन्तरम् भवन्ति।
मञ्जूषा- रतस्य, व्यायामिनः, जरावस्था (वृद्धावस्था), परिपक्वम्
उत्तराणि:
(i) रतस्य
(ii) जरावस्था (वृद्धावस्था)
(iii) व्यायामिनः
(iv) परिपक्वम्

(च) व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य च।
व्याधयो नोपसर्पन्ति वैनतेयमिवोरगाः
वयोरुपगुणहीनमपि कुर्यात्सुदर्शनम्॥

भावार्थ:
अस्यभावोऽस्ति यत् येजनाः व्यायाम कुर्वन्तः स्वशरीराणि (i) _______ कुर्वन्ति एवं पादाभ्याम् (ii) _______ भवन्ति। तेषां समीपे रोगाः तथैव नागच्छन्ति यथा (iii) _______  समीपे सर्पाः न आयान्ति। व्यायामन जनाः आयुषा, रुपेण (iv) _______ च हीनाः भूत्वा अपि दर्शनीयाः (सुन्दराः) भवन्ति।
मञ्जूषा- गृध्रस्य, स्वेदयुक्तानि, गुणैः, उत्थिताः
उत्तराणि:
(i) स्वेदयुक्तानि
(ii) उत्थिताः
(iii) गृध्रस्य
(iv) गुणः

(छ) व्यायाम कुर्वतो नित्यं विरुद्धमपि भोजनम्
विदग्धमविदग्धं वा निर्दोषं परिपच्यते॥

भावार्थ:
ये जनाः (i) _______ व्यायाम कुर्वन्ति तेषां (ii) _______ पूर्णतया पक्वं (iii) _______ वा अन्नं (iv) _______ विना एव पचति। मञ्जूषा- कष्टम्, अपक्वं, नित्यं, अनुपयोगि
उत्तराणि:
(i) नित्यं
(ii) अनुपयोगि
(iii) अपक्वं
(iv) कष्टम्

(ज) व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम्।
स च शीते वसन्ते च तेषां पथ्यतमः समृतः॥

भावार्थ:
नूनम् व्यायामः सदैव (i) _______ स्निग्धभोजिनाम् च (ii) _______ वर्तते। सः च (iii) _______ वसन्तौ च तेभ्यः अतीव (iv) ______ कथितः।
मञ्जूषा- लाभदायकः, औषधिः, शीतकाले, बलशालिनां
उत्तराणि:
(i) बलशालिनां
(ii) ओषधिः
(iii) शीतकाले
(iv) लाभदायक:

(झ) सर्वेष्वतुष्वहरहः पुम्भिरात्महितैषिभिः।
बलस्यार्धेन कर्त्तव्यो व्यायामो हन्त्यतोऽन्यथा।।

भावार्थ:
स्वहितं इच्छन् (i) _______ सदैव सर्वेषु (ii) _______ अर्घन बलेन एव (iii) _______ करणीयः। अन्यथा अत्यधिकस्य बलस्य प्रयोगण व्यायामः (iv) _______ भवति।
मञ्जूषा- हानिकरः, नरः, ऋतुषु, व्यायामः
उत्तराणि:
(i) नरः
(ii) ऋतुषु
(iii) व्यायामः
(iv) हानिकरः

(ज) हृदिस्थानस्थितो वायुर्यदा वक्त्रं प्रपद्यते।
व्यायाम कुर्वतो जन्तोस्तबलार्धस्य लक्षणम्॥

भावार्थ:
यदा व्यायाम-काले (i) _______ स्थितः वायु मुखं (ii) _______ अधिगच्छति। तदा सः व्यायामिनः (iii) _______ अर्धस्य बलस्य (iv) _______ भवति।
मञ्जूषा- यावत्:, हृदयस्थाने, लक्षणं, जनस्य
उत्तराणि:
(i) हृदयस्थाने
(ii) यावत्
(iii) जनस्य
(iv) लक्षण

(ट) वयोबलशरीराणि देशकालाशनानि च।
समीक्ष्य कुर्याद् व्यायाममन्यथा रोगमाप्नुयात्॥

भावार्थ:
अस्यभवोऽस्ति यत् जनस्य कर्तव्यम् अस्तियत् सः स्व आयुः (i) _______ शरीरं देश (ii) _______ भोजनञ्च दृष्ट्वा एवं (iii) _______ कुर्यात्। अन्यथा तु सः शीघ्रमेव (iv) _______ प्राप्स्यर्यत।
मञ्जूषा- रोगान्, बलम्, व्यायाम, समयम्
उत्तराणि:
(i) बलम्
(ii) समयम्
(iii) व्यायामम्
(iv) रोगान्

5. निम्न ‘क’ वर्गीयपदानाम् ‘ख’ वर्गीयपर्यायपवैः सह मेलनं कुर्यात-


उत्तराणि:

6. ‘क’ स्तम्भे विशेषणपदं लिखितम् ‘ख’ स्तम्भे पुनः विशेष्यपदम्। तयोः मेलनं कुरुत-

उत्तराणि:
(क) (v), (ख) (ii), (ग) (ii), (घ) (iii), (ङ) (iv)

7. अधोलिखितपदानां तेषां विपर्ययपदैः सह मेलनं कुरुतः-

उत्तराणि:

8. अधोलिखितेभ्यः पदेभ्यः उपसर्गान् पृथक् कृत्वा लिखत-

यथा- सुदर्शनम् – सु + दर्शनम्
(क) उपजायते – ______ + ______
(ख) अपकर्षणम् – ______ + ______
(ग) अधिरोहति – ______ + ______
(घ) प्रपद्यते – ______ + ______
(ङ) सुविभक्तता – ______ + ______
उत्तराणि:
(क) उप + जायते
(ख) अप + कर्षणम्
(ग) अधि + रोहति
(घ) प्र + पद्यते
(ङ) सु + वि + भक्तता

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 6 General Principles and Processes of Isolation of Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 6 Important Extra Questions General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Extra Questions Very Short Answer Type

Question 1.
What is the composition of copper matte? (CBSE Delhi 2013)
Answer:
Copper matte contains cuprous sulphide (Cu₂S) and iron sulphide (FeS).

Question 2.
Name the method used for refining of
(i) Nickel
(ii) Zirconium (CBSE Sample Paper 2007, 2014)
Answer:
(i) Mond’s process
(ii) Van Arkel method

Question 3.
Write the overall reaction taking place in the process used for the electrolysis of alumina by Hall-Heroult process. (CBSE Sample Paper 2011)
Answer:
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 4.
Write a non-exothermic reaction taking place in the blast furnace during extraction of iron. (CBSE Sample Paper 2011)
Answer:

Question 5.
Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
Answer:
Zone refining.

Question 6.
In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of the process? (CBSE Delhi 2016)
Answer:
Al₂O₃ + 2NaOH + 3H₂O → 2Na [Al(OH)₄]
This process is called leaching.

Question 7.
Name the method of refining which is based on the principle of adsorption.
(CBSE AI 2016)
Answer:
Chromatography.

Question 8.
Out of PbS and PbCO₃ (ores of lead), which one is concentrated by froth floatation process preferably? (CBSE Delhi 2017)
Answer:
PbS

Question 9.
What is the role of depressants in the froth floatation process? (CBSE Al 2017)
Answer:
The depressants in the froth floatation process selectively prevent one of the sulphide ores from forming the froth with air bubbles.

Question 10.
What is the role of collector and froth stabilizer in froth floatation process?
(CBSE AI 2012)
Answer:

• Collector enhances non-wettability of the mineral particles.
• Froth stabilisers stabilise the froth.

Question 11.
Name the method used for refining of copper. (CBSE AI 2013)
Answer:
Electrolytic refining

Question 12.
What is the role of graphite in the electrometallurgy of aluminium? (CBSE Delhi 2012)
Answer:
The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium.
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 13.
Why is it that only sulphide ores are concentrated by ‘froth floatation process’? (CBSE Delhi 2011)
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and the gangue particles by water.

Question 14.
What is the role of collectors in Froth- Floatation process? (CBSE AI 2012, 2017)
Answer:
The collectors such as pine oil, eucalyptus oil and fatty acids enhance the non – wettability of the mineral particles in froth – floatation process.

Question 15.
Name the substance used as depressant in the separation of two sulphide ores in froth floatation method. (CBSE Sample Paper 2019)
Answer:
Sodium cyanide.

Question 16.
Which reducing agent is employed to get copper from the leached low grade copper ore? (CBSE Delhi 2014)
Answer:
Iron scraps.

Question 17.
What is the role of zinc metal in the extraction of silver? (CBSE 2014)
Answer:
Zinc acts as a reducing agent.

General Principles and Processes of Isolation of Elements Important Extra Questions Short Answer Type

Question 1.
Out of C and CO which is a better reducing agent for FeO?
(i) In the lower part of blast furnace (Higher temperature)
(ii) In the upper part of blast furnace (Lower temperature) (CBSE Sample Paper 2011)
Answer:
(i) C is better reducing agent at higher temperature (lower part of blast furnace).
(ii) CO is better reducing agent at lower temperature (higher part of blast furnace).

Question 2.
What is the role of limestone in the extraction of iron from its oxides? (CBSE Al 2016)
Answer:
Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with silica (present as impurity in oxides of iron) to produce slag.

Question 3.
How is copper extracted from a low grade ore of it? (CBSE Al 2012)
Answer:
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions (Cu²⁺) is treated with scrap iron or H₂ as:
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)
In this way, copper is obtained.

Question 4.
(i) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver?
(ii) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore? (CBSE Delhi 2013)
Answer:
(i) Dilute solution (0.5%) of NaCN or KCN in the presence of air.
(ii) CO

Question 5.
(i) Which of the following ores can be concentrated by froth floatation method and why?
Fe₂O₃, ZnS, Al₂O₃
(ii) What is the role of silica in the metallurgy of copper? (CBSE Delhi 2013)
Answer:
(i) ZnS, because in sulphide ores, the sulphide ore particles are preferentially wetted by oil and gangue particles by water.
(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

Question 6.
(i) Name the method used for removing gangue from sulphide ores.
(ii) How is wrought iron different from steel? (CBSE Al 2013)
Answer:
(i) Froth floatation method.
(ii) Wrought iron is the pure form of iron and contains carbon and other impurities not more than 0.5%.
Steel contains 0.5 to 1.5% carbon along with small amounts of other elements such as Mn, Cr, Ni, etc. and other impurities.

Question 7.
Outline the principles behind the refining of metals by the following methods: (CBSE Delhi 2014)
(i) Zone refining method
(ii) Chromatographic method
Answer:
(i) Zone refining method. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Chromatographic method. This is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 8.
Write the principle behind the froth floatation process. What is the role of collectors in this process? (CBSE 2014)
Answer:
The froth floatation process is based on the principle of difference in the wetting properties of the ore and gangue particles with water and oil. The collectors such as pine oil, eucalyptus oil, fatty acids, etc. are used to enhance the non-wettability of the mineral particles.

Question 9.
What is meant by vapour phase refining? Write any one example of the process which illustrates this technique, giving the chemical equations involved.
OR
Write and explain the reactions involved in the extraction of gold. (CBSE Sample Paper 2019)
Answer:
Vapour phase refining: This method is based on the fact that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. Thus, the two requirements are:

• The metal should form a suitable compound with a suitable reagent.
• The volatile compound should be easily decomposable so that the metal can be easily recovered.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a stream of carbon monoxide to form volatile nickel tetracarbonyl, Ni(CO)₄, complex.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

OR
The extraction of gold involves leaching of metal present in the ore with CN^– ions. This is an oxidation-reaction because during the leaching process, Au is oxidised to Au⁺ which then combines with CN^– ions to form their respective soluble complexes.

The metal is then recovered from these complexes by reduction or displacement method using a more electropositive zinc metal.
2 [Au (CN)₂]^– (aq) + Zn(s) → 2 Au (s) + [Zn (CN)₄]^2- (aq)
Zinc acts as a reducing agent.
This process is called hydro metallurgy.

General Principles and Processes of Isolation of Elements Important Extra Questions Long Answer Type

Question 1.
(i) Give one example of each the following:
(a) Acidic flux
(b) Basic flux
(ii) What happens when
(a) Cu₂O undergoes self reduction in a silica line converter
(b) Haematite oxidises carbon to carbon monoxide? (CBSE Sample Paper 2012)
Answer:
(i) (a) Acidic flux: SiO₂
(b) Basic flux: CaO

(ii) (a) Cu₂O undergoes self reduction to form blister copper as:

(b) Haematite oxidises carbon to carbon monoxide forming iron.
Fe₂O₃ + 3C → 3CO + 2Fe

Question 2.
What is a flux? What is the role of flux in the metallurgy of iron and copper? (CBSE Sample Paper 2011)
Answer:
Flux is a substance which combines with gangue which may still be present in the calcined or roasted ore to form an easily fusible material called the slag.
Flux + Gangue → Slag (fusible)
In the metallurgy of copper, most of the ferrous sulphide present as impurity gets oxidised to ferrous oxide which combines with silica (flux) to form fusible slag.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer which is removed from the slag hole from time to time.
In the blast furnace for metallurgy of iron, lime acts as a flux and combines with silica present as impurity to form slag.

Question 3.
Describe the principle involved in each of the following processes.
(i) Mond process for refining of Nickel.
(ii) Column chromatography for purification of rare elements. (CBSE 2012)
Answer:
(i) Mond’s process for refining of nickel. Impure nickel is heated in a stream of carbon monoxide forming volatile complex, tetra carbonylnickel (0).

The volatile complex is decomposed at high temperature to give pure metal

(ii) Column chromatography for purification of rare elements. This technique is based on the differences in the adsorbing capacities of the metal and its impurities. The metal and the impurities are differently adsorbed. Impure metal is put in a liquid or gaseous medium (called moving phase) which is moved through an adsorbent (stationary phase).
Metal and impurities are adsorbed at different levels in column.

Question 4.
Which methods are usually employed for purifying the following metals?
(i) Nickel
(ii) Germanium
Mention the principle behind each one of them. (CBSE 2012)
Answer:
(i) Nickel. By vapour phase refining. This method is based on the principle that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a steam of carbon monoxide to form volatile nickel carbonyl Ni(CO)₄.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

(ii) Germanium. By Zone refining method. This method is based on the principle that the impurities are more soluble in melt than in the solid state of the metal. Therefore, an impure metal on solidification will deposit pure metal and the impurities will remain behind in the molten part of the metal.

Question 5.
Explain the principle of the method of electrolytic refining of metals. Give one example. (CBSE 2014)
Answer:
In electrolytic refining method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metal. On passing the electrical current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal, while equivalent amount of metal dissolves from the anode into the elctrolyte in the form of metal ions. The impurities fall down below the anode as anode mud. The reactions occurring at the electrodes are:
At cathode: Mⁿ⁺ + ne^– → M
At anode: M → Mⁿ⁺ + ne^–
Copper is refined using electrolytic refining method.

Question 6.
(i) Name the method used for the refining of zirconium. (CBSE 2015)
(ii) What is the role of CO in the extraction of iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
Answer:
(i) Van Arkel method
(ii) CO acts as reducing agent. It reduces oxides of iron to iron.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

(iii) This is because if the metal is in liquid state, its entropy is higher than when it is in solid state. Therefore, the value of entropy change (ΔS) of the reduction process is more on +ve side when the metal formed is in the liquid state and the metal oxide being reduced is in solid state. As a result, the value of ΔG° becomes more on negative side and hence the reduction becomes easier.

Question 7.
(i) Name the method of refining which is based on the principle of adsorption. (CBSE 2008, 2016)
(ii) What is the role of depressant in froth floatation process?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer:
(i) Chromatography

(ii) The depressants are used to prevent certain types of particles from forming the froth with bubbles in froth floatation process. This helps to separate two sulphide ores. For example, in case of an ore containing zinc sulphide (ZnS) and lead sulphide (PbS), sodium cyanide (NaCN) is used as a depressant. It forms a layer of zinc complex Na₂[Zn(CN)₄] with ZnS on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth and allows it to come with the froth.

(iii) Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with the silica (present as impurity in oxides of iron) to produce slag.

Question 8.
(i) Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
(ii) In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(ii) What is the role of coke in the extraction of iron from its oxides?
Answer:
(i) Zone refining
(ii) Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]
This process is called leaching.
(iii) The coke serves as a fuel as well as a reducing agent. It burns to produce CO₂ and heat which raises the temperature to about 2200 K (in combustion zone).
C + O₂ → CO₂, ΔH = -393.4 kJ
It also reduces Fe₂O₃ to iron.
Fe₂O₃ + 3C → 2Fe + 3CO + Heat

Question 9.
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron? (CBSE Delhi 2015)
Answer:
(i) Zinc is refined by electrolytic refining:
In this method, the impure zinc is made anode, while a plate of pure zinc is made the cathode. These are placed in a suitable electrolytic bath containing suitable salt of the same metal (e.g. zinc sulphate) containing a small amount of dilute sulphuric acid. On passing current, zinc from the solution is deposited at the cathode, while an equivalent amount of zinc from anode goes into the electrolytic solution. Therefore, pure zinc is obtained at cathode.

(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

(iii) Wrought iron is the purest form of commercial iron.

Question 10.
(a) What is the difference between calamine and malachite?
(b) Why is zinc used instead of Cu for recovery of Ag from [Ag(CN)₂]?
(c) What is the role of cryolite in metallurgy of Al? (CBSE 2019C)
OR
(a) Give two points of differences between pig iron and cast iron.
(b) Outline the principle of zone refining.
Answer:
(a) Calamine is an ore of Zn while malachite is an ore of copper. Calamine is ZnCO₃ while malachite is CuCO₃.Cu(OH)₂

(b) Zinc is more electropositive than silver and therefore, zinc displaces silver from its solution.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag
On the other hand, copper is less electropositive than silver and therefore, cannot displace silver from its solution. Zn is more reactive than Cu, so reduction will be faster in case of Zn.

(c) Fused alumina is a bad conductor of electricity. Purified alumina is mixed with molten cryolite (Na₃AlF₆) and is electrolysed in an iron tank lined inside with carbon. The molten cryolite decreases the melting point and increases the electrical conductivity. It acts as a solvent.

OR

(a)

Cast Iron	Pig Iron
1. It is moulded pig iron.	1. It is obtained directly from the blast furnace.
2. It is made by melting pig iron with scrap iron and coke using hot air blast.	2. It dissolves some sulphur, silicon, phosphorus and manganese as impurities.
3. It contains about 3% carbon.	3. It contains about 4% carbon.
4. It is extremely hard and brittle.	4. It is more brittle.

(b) Zone refining – It is based on the principle that impurities are more soluble in the melt than the solid state of the metal. Therefore, an impure metal on solidification will deposit crystals of pure metal and the impurities will remain behind in the molten part of the metal.

Question 11.
How will you convert the following:
(i) Impure Nickel to pure Nickel
(ii) Zinc blende to Zinc metal
(iii) [Ag(CN)₂]^– to Ag (CBSE Delhi 2019)
Answer:
(i) Impure nickel is heated in a stream of carbon monoxide forming volatile complex tetracarbonylnickel (0). The volatile complex is decomposed at high tempreature to give pure nickel.

This is called Mond’s process.

(ii) The zinc blende ore is concentrated by froth floatation process and then roasted in the presence of excess air at about 1200 K.
2ZnS + 3O₂ → 2ZnO + 2SO₂
Zinc oxide formed is reduced to zinc by heating with crushed coke at 1673 K in vertical fire clay retorts.

(iii) [Ag(CN)₂]^– complex is heated with zinc to get silver.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

Question 12.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu₂S. (CBSE Delhi 2019)
Answer:
(a) (i) Zone refining
(ii) Distillation

(b) Cu₂S is first roasted and converted to oxide
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide is then easily reduced to metallic copper with coke.
Cu₂O + C → 2Cu + CO

Question 13.
Describe how the following steps can be carried out.
(i) Recovery of Gold from leached gold metal complex.
(ii) Conversion of Zirconium iodide to pure Zirconium.
(iii) Formation of slag in the extraction of copper.
(Write the chemical equations also for the reactions involved)
OR
Explain the use of the following:
(i) NaCN in Froth Floatation Method.
(ii) Carbon monoxide in Mond process.
(iii) Coke in the extraction of Zinc from Zinc Oxide.
Answer:
(i) Leached gold complex is treated with Zinc and gold is recovered by displacement method.
2Au[(CN)₂]r(aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2-(aq)

(ii) Zirconium iodide is decomposed on a tungsten filament, electrically heated to 1800 K. Pure Zr metal is deposited on the filament.
ZrI₄ → Zr + I₂

(iii) Silica is added to the ore and heated. It helps to slag off iron oxide as iron silicate.
FeO + SiO₂ → FeSiO₃ (slag)

OR

(i) NaCN is used as depressants to separate two sulphide ores (ZnS and PbS) in Froth Floatation Method.
(ii) Carbon monoxide forms a volatile complex of nickel, nickel tetracarbonyl.
(iii) Coke is used as a reducing agent to reduce zinc oxide to zinc.

Question 14.
Write the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of Nickel. (CBSE 2018)
Answer:
(i) Gold is leached out in the form of a complex with dil. solution of NaCN in the presence of air. Here NaCN acts as leaching agent.
(ii) It lowers the melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with nickel which is further decomposed to give pure Ni metal.

Question 15.
(i) Write the role of ‘CO’ in the purification of nickel.
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method? (CBSE Delhi 2019)
Answer:
(i) Carbon monoxide forms a volatile complex, Ni(CO)₄, on heating nickel in a stream of CO while the impurities remain unaffected. The compound on further heating decomposes to give pure nickel.

Thus, CO helps to purify nickel.

(ii) During roasting, the copper pyrites are converted into a mixture of Cu₂O and FeO.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with FeO to form fusible slag of iron silicate.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer and is removed. Thus, silica helps to remove FeO.

(iii) Very reactive metals such as sodium, potassium, calcium, aluminium, etc. are extracted by electrolytic methods. Certain less reactive metals such as copper are also purified by using electro-refining method.

Question 16.
Write down the reactions taking place in blast furnace related to the metallurgy of iron in the temperature range 500 K to 800 K. What is the role of limestone in the metallurgy of iron?
OR
What happens when
(a) Silver is leached with NaCN in the presence of air?
(b) Copper matte is charged into silica-lined converter and hot air blast is blown?
(c) NaCN is added in an ore containing PbS and ZnS during concentration by froth floatation method? (CBSE Delhi Al 2019)
Answer:
Reactions taking place in blast furnace at 500K to 800K: The oxides of iron are reduced by CO.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂
Role of Limestone: It acts as a flux which decomposes to calcium oxide.
CaCO₃ → CaO + CO₂
Limestone
CaO combines with impurity (e.g. Si02) to form slag which is then removed.
CaO + SiO₂ → CaSiO₃

OR

(a) Silver is leached with dilute solution (0.5%) of NaCN in the presence of atmospheric oxygen. The metal dissolves forming the complex.
4Ag + 8CN^– + 2H₂O + O₂ → 4[Ag(CN)₂]^– + 4OH^–
The metal is extracted from water-soluble complex by zinc metal.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

(b) The copper matte containing Cu₂S and FeS is put in silica-lined convertor and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₃
Cu2S or CuO gets converted to copper
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(c) NaCN acts as a depressant in preventing ZnS from forming the froth. If forms a layer of zinc complex Na₂[Zn(CN)₄] on the surface of ZnS and thereby prevents it from forming the froth.
4NaCN + ZnS → Na₂[Zn(CN)₄] + Na₂S

Long Answer Type Questions (LA-II)

Question 1.
Explain the following:
(i) Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?
(ii) Why is zinc and not copper used for the recovery of silver from the complex [Ag(CN)₂]^–?
(iii) The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equations. (CBSE Sample Paper 2011)
(iv) Lime-stone is used in the manufacture of pig iron from haematite. Why?
Answer:
(i) Inspection of Ellingham diagram shows that ∆G vs T curves for Al₂O₃ and MgO intersect at a point corresponding to very high temperature of the order of 2000 K. This means above this temperature, ∆G for the reaction:
Al₂O₃ + 3Mg → 2Al + 3Mgo
would become negative and hence reduction will be feasible. However, this temperature is very high so that the process is uneconomical and technologically difficult.

(ii) Zinc is a stronger reducing agent (E° = – 0.76 V) and more electropositive than copper (E° = + 0.34 V). Therefore, zinc is used for recovery of Ag from [Ag(CN₂)]^– complex.

(iii) During leaching process, gold (Au) is first oxidised by O₂ of the air to Au⁺ which then combines with CN^– ions to form the soluble complex, dicyanoaurate (I).

Gold is then extracted from this complex by displacement method by using a more electropositive zinc metal. In this method, zinc acts as a reducing agent and it reduces Au⁺ to Au. Zinc itself gets oxidised to Zn²⁺ ions which combine with CN^– ions to form soluble complex, sodium tetracyanozincate (II).

Thus, extraction of gold by leaching with NaCN involves both oxidation and reduction.

(iv) Haematite is an ore of iron and contains silica (SiO₂) as the main impurity. The purpose of limestone is to remove SiO₂ as calcium silicate (CaSiO₃) slag.

Question 2.
Name the principal ore of aluminium. Explain the significance of leaching in the extraction of aluminium. (CBSE AI 2013)
Answer:
The principal ore of aluminium is bauxite, Al₂O₃.2H₂O. Leaching process is used to concentrate the ore of aluminium, bauxite, which is contaminated with impurities of silica (SiO₂), iron oxide (Fe₂O₃), titanium oxide (TiO₂), etc. Leaching is done by treating the powdered ore with hot cone. (45%) solution of NaOH at about 473-523 K and 35-36 bar pressure.
Al₂O₃.2H₂O(s) + 2NaOH(aq) + H₂O(l) → 2Na [Al(OH)₄](aq)
The impurities of ferric oxide and silica are insoluble and are removed by filtration.
The solution containing sodium aluminate is neutralised by passing C02 gas and hydrated alumina is precipitated.
2Na [Al(OH)₄] (aq) + CO₂ (g) → Al₂O₃. x H₂O (s) + 2Na HCO₃(s)
Hydrated alumina is heated to 1473 K to get pure alumina.

Question 3.
Describe the principle controlling each of the following processes:
(i) Vapour phase refining of titanium metal.
(ii) Froth floatation method of concentration of sulphide ore.
(iii) Recovery of silver after silver ore was leached with NaCN. (CBSE AI 2011, CBSE Delhi 2011)
Answer:
(i) The principle of vapour phase refining is that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. During refining of titanium metal, the titanium metal is heated with l₂ to form a volatile compound Til₄, which on further heating at higher temperature decomposes to give pure titanium metal.

(ii) This is based upon the different wetting properties of the ore and the gangue particles with oil and water. The mineral particles become wet by oil and the gangue particles by water. When the ore is mixed with water containing small quantities of pine oil and then by agitating the water by blowing air violently, the froth is formed. The froth carries the lighter ore particles to the surface and the heavier impurities sink to the bottom.

(iii) This is a chemical method and is useful for the ores which are soluble in certain suitable solvents but the impurities are not soluble. The impurities left undissolved are removed by filtration. The ore of silver, Ag₂S (argentite), reacts with NaCN as

Sodium sulphide thus formed is oxidised to sodium sulphate by blowing air into the solution. This helps the reaction to occur in the forward direction.
4Na₂S + 5O₂ + 2H₂O → 2Na₂SO₄ + 4NaOH + 2S

The above solution after filtration and removing insoluble impurities is heated with zinc to get silver.

Question 4.
Describe the role of the following:
(i) NaCN in the extraction of silver from a silver ore
(ii) Iodine in the refining of titanium
(iii) Cryolite in the metallurgy of aluminium (CBSE 2010)
Answer:
(i) The role of NaCN in the extraction of silver is to do the leaching of silver ore in the presence of atmospheric air from which silver is obtained later by replacement with zinc as:

(ii) Iodine is used to form a volatile unstable compound of the metal, which is decomposed to get the pure metal. For example, titanium is heated with iodine to 523 K forming unstable titanium iodide which is decomposed to Ti as

(iii) Cryolite is added to bauxite ore before electrolysis because of the following reasons:
(a) It acts as a solvent.
(b) It lowers the melting point of alumina to about 1173 K.
(c) Addition of cryolite to alumina increases the electrical conductivity.

Question 5.
Describe the role of the following:
(i) SiO₂ in the extraction of copper from copper matte
(ii) NaCN in froth floatation process
Answer:
(i) The copper matte containing Cu₂S and FeS is put in silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + O₂ → 2FeO + 2SO₂
FeO + Si0₂ → FeSiO₃
Cu₂S or CuO gets converted to copper.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(ii) NaCN in froth floatation process is used to separate one sulphide ore from another and is known as depressant.
For example, sodium cyanide can be used as a depressant in the separation of zinc sulphide ore (ZnS) and lead sulphide ore (PbS). Sodium cyanide forms a layer of zinc complex, Na₂[Zn(CN)₄], on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth. Thus, it selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. Thus, the two ores can be separated by the use of a depressant.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 5 Surface. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 5 Important Extra Questions Surface Chemistry

Surface Chemistry Important Extra Questions Very Short Answer Type

Question 1.
Give an example of a shape selective catalyst. (CBSE Delhi 2010, 2011)
Answer:
Zeolite catalyst known as ZSM-5.

Question 2.
Why are medicines more effective in colloidal state? (CBSE Delhi 2019)
Answer:
Medicines are most effective in colloidal state because colloids have large surface area and hence the medicines can be easily adsorbed and assimilated.

Question 3.
What is difference between an emulsion and a gel? (CBSE Delhi 2019)
Answer:
Emulsions are colloidal solutions in which both the dispersed phase and the dispersion medium are liquids. Gel is a colloidal system in which a liquid is dispersed in a solid.

Question 4.
Define ‘peptization’. (CBSE2012)
Answer:
The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization.

Question 5.
What is meant by ‘shape-selective catalysis? (CBSE 2012)
Answer:
The catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis.

Question 6.
What happens when a freshly precipitated
Fe(OH)₃ is shaken with little amount of dilute solution of FeCl₃? (CBSE 2013)
Answer:
A reddish brown colloidal solution of Fe(OH)₃ is obtained. This process is called peptization.

Question 7.
What is especially observed when a beam of light is passed through a colloidal solution? (CBSE 2013)
Answer:
Tyndall effect

Question 8.
Give one example each of sol and gel. (CBSE Delhi 2014)
Answer:
Sol: As₂S₃ sol.
Gel: Cheese.

Question 9.
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion. (CBSE 2014)
Answer:
Oil in water emulsion:
Milk Water in oil emulsion: Butter

Question 10.
Write a method by which lyophobic colloids can be coagulated. (CBSE 2015)
Answer:
By electrophoresis lyophobic colloids can be coagulated. Other methods are by mixing two oppositely charged sols or by boiling or by addition of electrolyte.

Question 11.
Write the main reason for the stability of colloidal sols. (CBSE Delhi 2016)
Answer:
The stability of colloidal sols is because of the presence of electrical charge on the particles which prevents the particles to come close together.

Question 12.
What are lyophobic colloids? Give one example for them. (CBSE Delhi 2011)
Answer:
The colloidal solutions in which the particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. For example, metal sulphides like As₂S₃.

Question 13.
Write one similarity between physisorption and chemisorption. (CBSE Delhi 2017, CBSE AI 2017)
Answer:
Both are surface phenomena and increase with increase in surface area.

Question 14.
Why are lyophilic colloidal sols more stable than lyophobic colloidal sols? (CBSE Delhi 2015)
Answer:
The lyophilic colloidal sols are more stable because they are highly hydrated in solution.

Question 15.
What is the difference between a sol and a gel? (CBSE Delhi 2017)
Answer:
In a sol, dispersion medium is liquid and dispersed phase is solid. On the other hand, in a gel, dispersion medium is solid and dispersed phase is liquid.

Question 16.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. (CBSE AI 2017)
Answer:
Gel
For example: Cheese

Question 17.
Which of the following is most effective electrolyte in the coagulation of Fe₂O₃.x H₂O/Fe³⁺ sol?
KCl, AlCl₃, MgCl₂, K₄[Fe(CN)₆] (CBSE Sample Paper 2011)
Answer:
Since Fe(OH)₃ sol is positively charged, the anion having highest charge will be most effective, i.e. [Fe(CN)₆]⁴.

Question 18.
Why is colloidal gold used for intramuscular injection? (CBSE Sample Paper 2011)
Answer:
Colloidal gold is more effective because of larger surface area and therefore, is easily assimilated with blood which is colloidal.

Question 19.
What is colloidion? (CBSE Sample Paper 2011)
Answer:
4% solution of nitrocellulose in a mixture of alcohol and ether.

Question 20.
Leather gets hardened after tanning. Why? (CBSE Delhi 2015)
Answer:
Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

Question 21.
It is necessary to remove CO when ammonia is prepared by Haber’s process. Explain. (CBSE Delhi 2015)
Answer:
Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst. Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 22.
Addition of alum purifies water. Why? (CBSE AI 2015)
Answer:
Alum coagulates the impurities present in water by neutralising the charge.

Question 23.
Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 24.
Write the chemical method by which Fe(OH)₃ sol is prepared from FeCl₃. (CBSE AI 2017)
Answer:

Question 25.
Out of MgCl₂ and AlCl₃ which one is more effective in causing coagulation of negatively charged sol and why? (CBSE Delhi 2016)
Answer:
According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus, AlCl₃ (Al³⁺ ions) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

Question 26.
Give one example each of lyophobic sol and lyophilic sol. (CBSE Delhi 2014)
Answer:
Lyophobic : As₂S₃, Lyophilic : Gelatin

Question 27.
Out of BaCl₂ and KCl which one is more effective in causing coagulation of negatively charged colloidal sol. Give reason. (CBSE Delhi 2015)
Answer:
BaCl₂ because greater the valency of the coagulating ion (positive ion), greater is its tendency to coagulate.

Question 28.
What are the dispersed phase and . dispersion medium in milk? (CBSE AI 2014)
Answer:
Liquid (dispersed phase), Liquid (dispersion medium)

Question 29.
What type of colloid is formed when a solid is dispersed in liquid? Give an example. (CBSE AI 2017)
Answer:
Sol, paints

Question 30.
CO (g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? (CBSE AI 2018)
Answer:
Selectivity of a catalyst. It is the ability of a catalyst to selectively form a particular product.

Surface Chemistry Important Extra Questions Short Answer Type

Question 1.
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol. (C.B.S.E. Delhi 2012)
Answer:
1. Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

2. When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of light by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

Question 2.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved. (CBSE Delhi 2012)
Answer:
Catalysis can be divided into two groups
(i) Homogeneous catalysis
(ii) Heterogeneous catalysis

(i) Homogenous catalysis: When the catalyst is present in the same phase as the reactants and products, it is called homogeneous catalysis. For example,

SO₂(g) is oxidised to SO₃(g) in the presence of gaseous nitric oxide as catalyst.

(ii) Heterogeneous catalysis: When the catalyst is in different phase than the
reactants, it is called heterogeneous catalysis. For example,

Question 3.
Write the dispersed phase and dispersion medium of the following colloidal systems: (CBSE Delhi 2013)
(i) Smoke
(ii) Milk
Answer:
Colloidal system: Dispersed phase/Dispersion medium
(i) Smoke: Solid/Gas
(ii) Milk: Liquid/Liquid

Question 4.
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, solutions of . metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.

Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 5.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. (CBSE Delhi 2013)
Answer:
1. Oil-in-water emulsion (o/w). In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

2. Water-in-oil emulsion (w/o). In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 6.
Peptizing agent is added to convert precipitate into colloidal solution. Explain. (CBSE Sample Paper 2011)
Answer:
Peptization is a process of converting a freshly prepared precipitate into colloidal form by the addition of an electrolyte called peptizing agent. The suitable ions from the peptizing agent (electrolyte) are adsorbed by the particles of the precipitate giving it positive or negative charge. The charged particles repel one another and break up the precipitate into smaller particles of the size of the colloid. Therefore, it results into the formation of colloid.

For example, on treating a precipitate of iron (III) oxide with a small amount of FeCl₃ solution gives a reddish brown coloured colloidal solution.

Question 7.
Cottrell’s smoke precipitator is fitted at the mouth of chimney used in factories. Give reasons. (CBSE Sample Paper 2011)
Answer:
Smoke coming out of chimney of a factory is a colloidal solution of solid carbon particles which are charged in nature.

The mouth of the chimneys used in factories is fitted with Cottrell smoke precipitator. In this method, the smoke is allowed to pass through a chamber having a series of plates charged to very high potential (20,000 to 70,000 V).

Charged particles of smoke get attracted by charged plates, get precipitated and the gases coming out of chimney become free of charged carbon and dust particles.

Question 8.
Differentiate between peptization and coagulation.
(CBSE Sample Paper 2011, CBSE AI 2017)
Answer:
Peptization is the process of converting a freshly prepared precipitate into collodial form by the the addition of a suitable electrolyte. The electrolytes used for the purpose are called peptizing agents. On the other hand, coagulation is the phenomenon of precipitation of a collodial solution by the addition of excess of an electrolyte.

Question 9.
Explain:
(i) Sky appears blue in colour.
(ii) A freshly formed precipitate of ferric hydroxide can be converted to a colloidal sol by shaking it with a small quantity of ferric chloride. (CBSE Sample Paper 2011)
Answer:
1. Dust particles along with water suspended in air scatter blue light which reaches our eyes and therefore, sky looks blue to us.

2. When we add FeCl₃ to a freshly formed precipitate of Fe(OH)₃, peptization occurs. The Fe³⁺ ions are adsorbed on the surface of the precipitate which ‘ ultimately breaks down into smaller particles of colloidal size.

Question 10.
What happens when
(i) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(ii) persistent dialysis of a colloidal solution is carried out?
(iii) an emulsion is centrifuged? (CBSE 2018)
Answer:
(i) Peptization occurs / Colloidal solution of Fe(OH)₃ is formed.
(ii) Coagulation occurs
(iii) Demulsification or breaks into constituent liquids

Surface Chemistry Important Extra Questions Long Answer Type

Question 1.
Classify colloids where the dispersion medium is water. State their characteristics and write one example of each of these classes.
OR
Explain what is observed when
(i) an electric current is passed through a sol.
(ii) a beam of light is passed through a sol.
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol. (CBSE 2011)
Answer:
These colloids are of two types:
(i) Hydrophilic
(ii) Hydrophobic

Characteristics: In lyophilic colloids, there is affinity between dispersed phase and the dispersion medium while in the lyophobic colloids, there is no affinity rather hatred between the phases.

The main points of distinction between lyophilic and lyophobic sols are summarized below:

Lyophilic colloids (Hydrophilic)	Lyophobic collodis (Hydrophobic)
1. These are easily formed by direct mixing.	1. These are formed only by special methods.
2. These are reversible in nature.	2. These are irreversible in nature.
3. The particles of colloids are true molecules and are big in size.	3. The particles are aggregates of many molecules.
4. The particles are not easily visible even under ultramicroscope.	4. The particles are easily detected under ultramicroscope.
5. These are very stable.	5. These are unstable and require traces of stabilizers.
6. The addition of small amount of electrolytes causes precipitation (called coagulation) of colloidal solution.	6. The addition of small amount of electrolytes has less effect. Larger quantities of electrolytes are required to cause coagulation.
7. The particles do not carry any charge. The particles may migrate in any direction or even not under the influence of an electric field.	7. The particles move in a specific direction i.e., either towards anode or cathode depending upon their charge.
8. The particles of colloids are heavily hydrated due to the attraction for the solvent.	8. The particles of colloids are not appreciably hydrated due to the hatred for the solvent.
9. The viscosity and surface tension of the sols are much higher than that of the dispersion medium.	9. The viscosity and surface tension are nearly the same as that of the dispersion medium.
10. They do not show Tyndall effect.	10. They show Tyndall effect.

Example: Hydrophilic: Starch,
Hydrophobic : Metal sulphide

OR

(i) On passing electric current through colloidal solution, the colloidal particles move towards the oppositely charged electrodes, where they lose their charge and get coagulated. This is electrophoresis process.

(ii) When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of tight by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

(iii) Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

Question 2.
Explain how the phenomenon of adsorption finds application in each of the following processes:
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth floatation process (CBSE Delhi 2011)
Answer:
(i) Production of vacuum: The adsorption of air in liquid air helps to create a high vacuum in a vessel. This process is used in high vacuum instruments such as Dewar flask for storage of liquid air or liquid hydrogen. The remaining traces of air can be adsorbed by charcoal from the vessel evacuated by a vacuum pump to give a very high vacuum.

(ii) Heterogeneous catalysis: The phenomenon of adsorption is useful in the heterogeneous catalysis. Adsorption of reactants on the solid surface of catalysts increases the rate of reaction. The metals such as Fe, Ni, Pt, Pd, etc. are used in the manufacturing processes. Manufacture of ammonia using iron as catalyst (Haber process), manufacture of sulphuric acid by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis. Its use is based upon the phenomenon of adsorption.

(iii) In froth floatation process: A low grade sulphide ore is concentrated by separating it from silica and other earthly matter by adsorption using pine oil and frothing agent.

Question 3.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out. (CBSE 2012)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol. Coagulation of lyophobic sols can be carried out by the following methods:

1. By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

2. By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

3. By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 4.
Explain the following terms giving a suitable example for each:
(i) Aerosol
(ii) Emulsion
(iii) Micelle (CBSE 2012)
Answer:
(i) Aerosol: It is a colloidal dispersion of a liquid in a gas. For example, fog, mist, cloud.

(ii) Emulsion. Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Types of emulsion. These are of two types:

(a) Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(b) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

(iii) Micelle: The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations.

Question 5.
Write three distinct features of chemisorption which are not found in physisorption. (CBSE 2012)
Answer:
Characteristics of Chemisorption:

1. The forces between the adsorbate molecules and the adsorbent are strong chemical forces similar to chemical bonds. But no chemical bonds exist in physisorption.
2. Chemisorption is irreversible in nature but physisorption is reversible.
3. Chemisorption forms mono – molecular layers, but in physisorption multimolecular layers are present.

Question 6.
What are the characteristics of the following colloids? Give one example of each.
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions (CBSE 2013)
Answer:
(i) The colloidal solutions in which atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions. The particles thus formed are called multimolecular colloids. For example, sols of gold atoms and sulphur (S₈) molecules.

(ii) The colloidal solutions in which there in no affinity (or love rather they have hatred) between the particles of the dispersed phase and the dispersion medium are called lyophobic sols.
For example, metal sulphides like As₂S₃.

(iii) The colloidal solutions in which both the dispersed phase and the dispersion medium are liquids are called emulsions. For example, milk.

Question 7.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:

Colloidol System	Dispersed phase	Dispersed medium
(i) Smoke	Solid	Gas
(ii) Milk	Liquid	Liquid

OR
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, ‘ solutions of metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.
Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 8.
What are emulsions? What are their different types? Give one example of each type. (CBSE 2014)
Answer:
Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium.

Types of emulsions. These are of two types:
1. Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(ii) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 9.
(i) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(ii) Write an important characteristic of lyophilic sols.
(iff) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid. (CBSE Delhi 2014)
Answer:
(i) \(\frac{x}{m}\) = kp^1/n
where \(\frac{x}{m}\) is extent of adsorption, x is the mass of adsorbate, m is the mass of adsorbent, p is pressure at a particular temperature, k is a constant and n takes any whole number value.

(ii) in lyophilic sols, the particles of dispersed phase have a great affinity for the dispersion medium and are reversible in nature.

(iii) Associated colloids: Soap (e.g. sodium stearate).
Multimolecular colloids: Sol of gold (atoms).

Question 10.
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. (CBSE Delhi 2015)
Answer:
(i) Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

(ii) Lyophilic sol is more stable than lyophobic sol because it is highly hydrated or solvated in solution.

(iii) Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst.
Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 11.
Give reasons for the following observations:
(i) Physisorption decreases with increase in temperature.
(ii) Addition of alum purifies the water.
(iii) Brownian movement provides stability to the colloidal solution. (CBSE 2015)
Answer:
(i) In physisorption, the attractive forces between adsorbent and adsorbate molecules are weak vander Waals forces. When temperature is increased, the kinetic energy of the molecules of the gas increases and they can easily leave the surface of adsorbent.
Therefore, physisorption decreases with increase in temperature.

(ii) Alum coagulates the impurities present in water by neutralising the charge.

(iii) The Brownian movement is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. This results in zigzag motion. Because of brownian movement, the particles do not settle down and hence is responsible for the stability of the sols.

Question 12.
Define the following terms:
(i) O/W Emulsion
(ii) Zeta potential
(iii) Multimolecular colloids (CBSE 2016)
Answer:
(i) O/W Emulsion: It is the emulsion in which oil is the dispersed phase and water is the dispersion medium. For example, milk is an emulsion of soluble fats in water.

(ii) Zeta potential: The potential difference between the fixed layer and the diffused layer of opposite charges is called Zeta potential.

(iii) Multimolecular colloids: When a dissolution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimension, the particles thus formed are called multimolecular colloids.

Question 13.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl₂ and AlCl₂, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
(i)

Absorption	Adsorption
1. It is the phenomenon in which the particles of gas or liquid get uniformly distributed throughout the body of the solid.	1. It is the phenomenon of higher concentration of particles of gas or liquid on the surface than in the bulk of the solid.
2. The concentration is the same throughout the material. Therefore, it is a bulk phenomenon.	2. The concentration on the surface of the adsorbent is different from that in the bulk.  Therefore, it is a surface phenomenon.
3. Absorption occurs at uniform rate.	3. Adsorption is rapid in the beginning and its rate slowly decreases.

(ii) According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus AlCl₃ (Al³⁺ ion) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

(iii) Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 14.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
(CBSE Delhi 2013)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol.
Coagulation of lyophobic sols can be carried out by the following methods:
(i) By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

(ii) By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

(iii) By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 15.
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature? (CBSE Delhi 2019) Answer:
(i) Activated charcoalin gas mask adsorbs the poisonous or toxic gases from air. Therefore, these masks help to purify the air for breathing.

(ii) Negatively charged sol
Representation: Fe₂O₃. xH₂O/OH^–

(iii) In chemisorption, adsorption initially increases with rise in temperature and then decreases.

Question 16.
Define the following terms with an example in each case:
(i) Macromolecular sol
(ii) Peptization
(iii) Emulsion (CBSE Al 2013)
Answer:
(i) Macromolecular sol: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. In this case, a large number of small molecules are joined together through their primary valencies to form giant molecules. These molecules are called macromolecules and each macromolecule may consist of hundreds or thousands of simple molecules. The solutions of such molecules are called macromolecular solutions. For example, colloidal solution of starch, cellulose, etc.

(ii) Peptization: The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization. The electrolytes used for the purpose are called peptizing agents.

(iii) Emulsions: They are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Normally, these are obtained by mixing an oil with water. Since the two do not mix well, the emulsion is generally unstable and is stabilised by adding a suitable reagent called emulsifier or emulsifying agent. The substances that are commonly employed for the purpose are gum, soap, glass powder, etc.

Question 17.
Define the following with a suitable example of each:
(a) Coagulation
(b) Multimolecular colloid
(c) Gel
Answer:
(a) Coagulation. The phenomenon of precipitation of a colloidal solution by the addition of excess of an . electrolyte is called coagulation. For
example, Fe(OH)₃ colloidal solution is coagulated by adding HCl.

(b) Multimolecular colloid. When on dilution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions, then the particles thus formed are called multimolecular colloids, for example, sols of gold atoms.

(c) Gel. A gel is a colloidal system in which a liquid is dispersed in a solid. For example, jellies, cheese.

OR

(a) Out of starch and ferric hydroxide sol, which one can easily be coagulated and why?
(b) What is observed when an emulsion is centrifuged?
(c) What is the role of promoters and poisons in catalysis? (CBSE Al 2019)
Answer:
(a) Ferric hydroxide sol is easier to coagulate because it is a lyophobic sol.
(b) De-emulsification occurs.
(c) Promoters increase the efficiency of catalyst whereas poisons inhibit the efficiency of catalyst.

Question 18.
Give reason for the following observations:
(i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Lyophilic colloids are also called reversible sols.
Answer:
(i) The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution.
(ii) Due to large surface area
(iii) If the dispersion medium is separated from the dispersed phase, the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.

Question 19.
Answer the following questions:
(a) Which of the following electrolytes is most effective for the coagulation of Agl/Ag⁺ sol?
MgCl₂, K₂SO₄, K₄[Fe(CN)₆]
(b) What happens when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃?
(c) Out of sulphur sol and proteins, which one forms macromolecular colloids? (CBSE Sample Paper 2019)
Answer:
(a) K₄[Fe(CN)₆]
Hardy Schulze rule: The coagulation tendency of different electrolytes is different. It depends upon the valency of the active ion called flocculatins ion, which is the ion-carrying charge opposite to the charge on the colloidal particles. According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. For example, to coagulate negative sol of AS₂S₃, the coagulating power of different cations has been found to decrease in the order as:
Al³⁺ > Mg²⁺ > Na⁺
Similarly, to coagulate a positive sol such as Fe(OH)₃, the coagulating power of different anions has been found to decrease in the order:
[Fe(CN)₆]4^– > PO₄^3- > SO₄^2- > Cl^–

(b) Due to the phenomenon of peptization, a reddish brown positively charged sol of ferric hydroxide is obtained when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃. It involves the adsorption of suitable ions from the electrolyte by . the particles of the precipitate. The charged particles repel one another and form a colloidal sol. Ferric ions from ferric chloride are adsorbed by Fe(OH)₃, precipitate and get converted into colloidal sol.

(c) Proteins are macromolecules which cannot form multi-molecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multi-molecular colloids.

Question 20.
Answer the following:
(a) Why is ester hydrolysis slow in the beginning and then becomes faster after some time?
(b) In a solution of methylene blue, animal charcoal is added, the solution is then well shaken. What will be observed and why?
(c) Give an example of oil in water emulsion.
Answer:
(a) This is because of the process of autocatalysis. Ester on hydrolysis gives an acid which starts acting as a catalyst after sometime and therefore, the reaction becomes fast.

(b) Adsorption of a dye by charcoal. When animal charcoal is shaken with a solution of an organic dye such as methylene blue it is observed that the solution turns colourless. The discharge of the colour is due to the fact that the coloured component (generally an organic dye) gets adsorbed on the surface of animal charcoal. Therefore, animal charcoal is used for decolourising a number of organic substances in the form of their solutions.
(c) Milk, vanishing cream.

OR

Define the following:
(a) Associated colloids
(b) Electrophoresis
(c) Zeta potential (CBSE 2019C)
Answer:
(a) Associated colloids: These are the substances which when dissolved
in a medium behave as normal electrolytes at low concentration but behave as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregate particles thus formed are called micelles. For example, in aqueous solution, soap (sodium stearate) ionises as:

In concentrated solution, these ions get associated to form an aggregate of colloidal size.

(b) The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative.

(c) The combination of two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed diffused layer. The charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Question 21.
Write the differences between physisorption and chemisorption With respect to the following:
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility and
(iv) Enthalpy change (CBSE Delhi 2013)
Answer:

Property	Physisorption	Chemisorption
(i) Specificity:	not specific	highly specific
(ii) Temperature dependence:	Occurs at low temperature and decreases with increasing temperature	Occurs at high temperature and increases with increasing temperature
(iii) Reversibility:	Reversible in nature	Irreversible in nature
(iv) Enthalpy change:	Low enthalpy of adsorption (20 – 40 kJ mol-1)	High enthalpy of adsorption (80 – 240 kJ mol-1)

Question 22.
Describe some features of catalysis by zeolites.
Answer:
Zeolites are microporous aluminosilicates of the general formula M_x/n [(AlO₂)_x (SiO₂)y] mH₂O. These are most important oxide catalysts. These are used in petrochemical industries for cracking of hydrocarbons and isomerisation. The reactions in zeolites depend upon the size of the cavities (cages) or pores (apertures) present in them.

The most remarkable feature of zeolite catalysis is the shape selectivity. Therefore, the selectivity of catalyst depends on the pores structure. It has been observed that the pore size in zeolites generally varies between 260 pm and 740 pm. Depending upon the size of the reactants and products compared to the size of the cages or pores of zeolite, reactions proceed in specific manner.

A zeolite catalyst called ZSM-5 converts alcohols to gasoline, by first dehydrating the alcohol by loss of water.

Question 23.
What do you understand by adsorption isotherm? Discuss briefly Freundlich adsorption isotherm.
Answer:
Adsorption isotherm. A graph between amount of adsorption and gas pressure keeping temperature constant is called an adsorption isotherm.

Freundlich adsorption isotherm. The behaviour of adsorption with pressure can be expressed by adsorption isotherm as shown in figure.

The inspection of the graph reveals the following facts:
(i) At low pressure, the graph is almost straight line which indicates that x/m is directly proportional to the pressure. This may be expressed as:
\(\frac{x}{m}\) ∝ P or \(\frac{x}{m}\) = kP ……. (i)
where k is constant.

(ii) At high pressure. The graph becomes almost constant which means that x/m becomes independent of pressure. This may be expressed as:

(iii) Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between 0 and 1, i.e. fractional power of pressure. This may be expressed as:

where n can take any value between 1 and a larger integer depending upon the pressure. The above relationship is also called Freundlich’s adsorption isotherm.

Question 24.
Explain the following in connection with colloids:
(i) Hardy-Schulze rule
(ii) Dialysis.
Answer:
(i) Hardy-Schulze rule: This rule states that greater the valency of the active ion or flocculating ion, greater will be its coagulating power. According to this rule:

(a) The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.

(b) Coagulation power of an electrolyte is directly proportional to the valency of the active ions.
For example, for negative As₂S₃ sol, the coagulating power of cations decreases as Al³⁺ > Mg²⁺ > K⁺
Similarly, for positively charged sol such as Fe (OH)₃, the coagulating power decreases as
[Fe(CN)₃]^4- > PO₄^3- > SO₄^2- > Cr^–

(ii) Dialysis. The method used to separate the impurities from the colloidal solution is called dialysis. Its principle is based upon the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment. The bag is suspended in fresh water. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution. Dialysis can be used for removing HCl from the ferric hydroxide sol. The method is shown in figure.

The ordinary process of dialysis is slow. To increase the process of purification, the dialysis is carried out by applying electric field. This process is called electrodialysis.

Question 25.
Explain the following terms:
(i) Electrodialysis
(ii) Tyndall effect.
Answer:
(i) Electrodialysis. It is the process of separating the impurities from the colloidal solution by applying electric field. It is based on the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment and is suspended in a fresh water and electric current is applied. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution.

(ii) Tyndall effect. When a strong beam of light is passed through a true solution placed in a beaker, in a dark room, the path of the light does not become visible. However, if the light is passed through a sol, placed in the same room, the path of the light becomes visible when viewed from a direction at right angle to that of the incident beam.

This phenomenon was studied for the first time by Tyndall and therefore it is called Tyndall effect. The cause to Tyndall effect is the scattering of light by colloidal particles, i.e. these particles first absorb the incident light and then a part of it gets scattered by them. Since the intensity of the scattered light is at right angle to the plane of the incident light, the path becomes visible only when seen in that direction. The particles in true solution are too small in size to cause any scattering, i.e. the Tyndall effect is not noticed in true solutions.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 5 Magnetism and Matter. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 5 Important Extra Questions Magnetism and Matter

Magnetism and Matter Important Extra Questions Very Short Answer Type

Question 1.
A small magnetic needle pivoted at the center is free to rotate In a magnetic meridian. At what place will the needle
be vertical?
Answer:
At the potes

Question 2.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
450

Question 3.
How does the intensity of a paramagnetic sample vary with temperature?
Answer:
it decreases with the increase in temperature.

Question 4.
What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?
Answer:
It should be anti-parallel to the applied magnetic field.

Question 5.
What is the value of angle of dip at a place on the surface of the earth where the ratio of the vertical component to the horizontal component of the earth’s magnetic field is 1/\(\sqrt{3}\)?
Answer:
Using the expression tan δ = \(\frac{B_{V}}{B_{H}}=\frac{1}{\sqrt{3}}\)
Therefore, δ = 30°

Question 6.
Where on the surface of the earth is the angle of dip 90°? (CBSE Al 2011)
Answer:
Poles.

Question 7.
Where on the surface of the earth is the angle dip zero? (CBSE Al 2011)
Answer:
Magnetic equator

Question 8.
What are permanent magnets? Give one example. (CBSE Delhi 2013)
Answer:
It is an arrangement that has a permanent dipole moment, e.g. bar magnet.

Question 9.
At a place, the horizontal component of the earth’s magnetic field is B, and the angle of dip is 60°. What is the value of the horizontal component of the earth’s magnetic field at the equator? (CBSE Delhi 2017)
Answer:
Zero.

Question 10.
Is the steady electric current the only source of the magnetic field? Justify your answer. (CBSE Delhi 2013C)
Answer:
No, the magnetic field is also produced by alternating current.

Question 11.
Write one important property of a paramagnetic material. (CBSEAI2019)
OR
Do the diamagnetic substances have a resultant magnetic moment in an atom in the absence of an external magnetic field?
Answer:
It moves from the weaker to the stronger regions of the magnetic field.
OR
No

Question 12.
Steel is preferred for making permanent magnets, whereas soft iron is preferred for making electromagnets. Give one reason.
Answer:

1. The Retentivity of steel is more than that of soft iron.
2. Soft iron has a smaller value of coercivity than steel.

Question 13.
Define angle of inclination at a given place due to the earth’s magnetic field. (CBSE 2019C)
Answer:
The angle of inclination is the angle at which the earth’s magnetic field at a given place makes with the horizontal line in the magnetic meridian.

Question 14.
Relative permeability of a material p = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (CBSE Delhi 2014C)
Answer:
Paramagnetic.
Susceptibility is small and greater than one.

Magnetism and Matter Important Extra Questions Short Answer Type

Question 1.
(a) Define the term magnetic susceptibility and write its relation in terms of relative magnetic permeability.
Answer:
It refers to the ease with which a substance can be magnetized. It is defined as the ratio of the intensity of magnetization to the magnetizing field. The required relation is µ_r = 1 + χ_m

(b) Two magnetic materials A and B have relative magnetic permeabilities of 0. 96 and 500. Identify the magnetic materials A and B. (CBSE Al, Delhi 2018C)
Answer:
A: Paramagnetic,
B: Ferromagnetic

Question 2.
A magnetic needle free to rotate in a vertical position orients itself with its axis vertical at a certain place on the earth. What are the values of
(a) the angle of dip and
(b) the horizontal component of the earth’s magnetic field at this place? Where will this place be on the earth?
Answer:
The angle of dip is 90° and the horizontal component of the earth’s magnetic field is zero. This place is the magnetic pole of the earth.

Question 3.
Out of the two magnetic materials ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative? (CBSE Delhi 2014)
Answer:

• ‘A’ is paramagnetic and ‘B’ is diamagnetic.
• ‘A’ will have positive susceptibility while
• ‘B’ will have negative susceptibility.

Question 4.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its northern tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (CBSE Delhi 2011)
Answer:
Given δ = 30°, B_H = 0.4 G, B = ?
Using the expression
B_H = Bcos δ we have
B = \(\frac{B_{H}}{\cos \delta}=\frac{0.4}{\cos 30^{\circ}}=\frac{0.4}{\sqrt{3} / 2}=\frac{0.8}{\sqrt{3}}\) G

Question 5.
The susceptibility of a magnetic material is -0.085. Identify the type of magnetic material. A specimen of this material is kept in a non-uniform magnetic field. Draw the modified field pattern.
Answer:
The material is a diamagnetic material as diamagnetic materials have negative susceptibility. The modified field pattern is as shown below.

Question 6.
A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it.

(a) Identify the two specimens X and Y.
Answer:
X is a diamagnetic substance and Y is a paramagnetic substance.

(b) State the reason for the behavior of the field lines in X and Y.
Answer:
This is because the permeability of a diamagnetic substance is less than one and that of a paramagnetic substance is greater than one.

Question 7.
Three identical specimens of magnetic materials nickel, antimony, and aluminum are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.
Answer:
Nickel is ferromagnetic, antimony is diamagnetic and aluminium is paramagnetic. Therefore, they will show the behavior as shown in the following figures.

Question 8.
Define neutral point. Draw lines of force when two identical magnets are placed at a finite distance apart with their N-poles facing each other. Locate the neutral points.
Answer:
It is a point near a magnet where the magnetic field of the earth is completely balanced by the magnetic field of the magnet. The figure is as shown below.

The cross indicates the neutral point.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties. (CBSE Delhi 2012)
Answer:
Diamagnetic:

1. Very small and negative susceptibility
2. Permeability is less than one.

Magnetism and Matter Important Extra Questions Long Answer Type

Question 1.
Write the expression for the magnetic dipole moment for a closed current loop. Give its SI unit. Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field.
Answer:
The required expression is m = nIA.
It is measured in A m².

Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle 6 with the field as shown in the figure below. If ‘m’ is the strength of each pole, the two poles experience two equal and opposite force ‘B’ each. These forces constitute a couple that tends to rotate the dipole. Suppose the couple exerts a torque of magnitude τ.

Then

τ = either force × arm of the couple
= mB × AN = mB × 2 L sin θ
or
Since m × 2L is the magnetic dipole moment of the magnet.
Therefore τ = MB sin θ in vector form

we have \(\vec{τ}\) = \(\vec{M}\) × \(\vec{B}\)

Question 2.
(a) State Gauss’s law for magnetism. Explain Its significance.
Answer:
(a) Gauss’s Law for magnetism states that “The total flux of the magnetic field, through any closed surface, is aLways
zero, i.e. ∮\(\vec{B}\).\(\vec{dL}\) = 0

This law implies that magnetic monopoles do not exist” or magnetic field lines form closed loops.

(b) Write the four Important properties of the magnetic field lines due to a bar magnet. (CBSE Delhi 2019)
Answer:
Four properties of magnetic field lines are as follows:

1. Magnetic field lines always form continuous closed loops.
2. The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at that point.
3. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
4. Magnetic field lines do not intersect.

Question 3.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet?
Answer:
(a) From equation τ = mB sin θ,
For θ = 30°, sin θ = 0.5

Thus, 0.016 = m × 800 × 10^-4 × 0.5 or m = 160 × 2/800 = 0.40 Am²

(b) What is the work done in moving it from its most stable to a most unstable position?
Answer:
From equation U= -mB cos θ, the most stable position is for θ = 0° and the most unstable position is for θ = 180°.

Work done is given by
W = U_m (θ = 180°) – Um for (θ = 0°)
= 2 mB = 2 × 0.40 × 800 × 10^-4
= 0.064 J

(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns but of the same magnetic moment. Determine the current flowing through the solenoid. (NCERT Delhi 2005C)
Answer:
From part (i), m = 0.40 Am² Using equation m = NIA.
0.40 = 1000 × l × 2 × 10^-4
l = 2 A

Question 4.
(a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length and (ii) along its length?
Answer:
(a) In either case one gets two magnets each with a north pole and a south pole.

(b) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to torque. Why?
Answer:
The needle experiences no net force because the field is uniform.

The iron nail experiences a non¬uniform field due to the bar magnet. There is an induced magnetic moment in the nail; therefore, it experiences both force and torque.

The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of the magnet than the induced north pole.

Question 5.
Name the elements of the earth’s magnetic field at a place. Explain their meaning.
Answer:
The elements required to completely specify the earth’s magnetic field are called magnetic elements. These elements are:
(a) Declination (θ)
(b) Dip or inclination (δ)
(c) Horizontal component of the earth’s magnetic field (B_H).

(a) Declination (θ): Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.
In the figure above ∠BAB’ = θ is called the angle of declination. There are several periodic variations in the value of declination for a given location.

(b) Dip (δ): Dip at a place is defined as the angle between the direction of the total intensity of the earth’s magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by 8.

Suppose a freely suspended magnet aligns itself along the direction AB as shown in the figure, then this gives the direction of the earth’s total magnetic field at that place. Then the angle ∠BAO = δ is called the angle of dip or simply dip at that point.

(c) Horizontal component of the earth’s magnetic field (BH): It is the component of the total intensity of the earth’s magnetic field along a horizontal line in the magnetic meridian. It is denoted by BH or H.

Question 6.
Name the three types of magnetic materials which behave differently when placed in a non-uniform magnetic field. Give two properties for each of them.
Answer:
The three types of magnetic materials are
(a) Diamagnetic
(b) Paramagnetic and
(c) Ferromagnetic

Diamagnetic	Paramagnetic	Ferromagnetic
These are the substances that are feebly repelled by a magnet, e.g., Bi, Zn, Cu, Ag, Au, diamond, C, NaCI, H20, Hg, N2, H2, etc. Exhibited by solids, liquids, and gases.	These are the substances that are feebly attracted by a magnet, e.g., AL, Na, Pt, Mn, CuCL2, O2, etc. Exhibited by solids, liquids, and gases.	These are the substances that are strongly attracted by a magnet, e.g., Fe, Ni, Co, Fe304, etc. Exhibited by solids only, that too crystalline.
When a diamagnetic substance is placed in a magnetizing field, the Lines of force prefer not to pass through the substance.	When a paramagnetic substance is placed in a magnetizing field, the Lines of force prefer to pass through the substance.	When a ferromagnetic substance is placed in a magnetizing field, the Lines of force prefer to pass through the substance.

Question 7.
The susceptibility of a magnetic material is 0-9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. (CBSEAI, Delhi 2018)
Answer:
We know that the susceptibility of a paramagnetic substance is positive but small. Hence, the given magnetic material is paramagnetic in nature.

When a specimen of a paramagnetic substance is placed in a magnetizing field, the magnetic field lines prefer to pass through the specimen rather than through air. Thus, magnetic induction 6 inside the sample is more than the magnetic induction B0 outside the sample.

Question 8.
(a) Draw the magnetic field lines due to a circular loop of area \(\vec{A}\) carrying current l. Show that it acts as a bar magnet of magnetic moment \(\vec{m}\) = I \(\vec{A}\).
Answer:
(a) The magnetic field lines are as shown.

Magnetic field due to circular loop on its axis at far off points B = \(\frac{\mu_{0}}{4 \pi} \frac{2 l A}{x^{3}}\)

Magnetic field due to a bar magnet at its axial point is B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)
Comparing the above two we have m = IA

(b) Derive the expression for the magnetic field due to a solenoid of length ‘21’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current T at a point on the axial line, distant ‘r’ from the center of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of the magnetic moment ‘m’? (CBSEAI 2015)
Answer:
(b) By the-Biot-Savarts law, the magnetic field at point P on the axis of the solenoid due to a small element of length dx at a distance ‘x’ from the center of the solenoid is

dB = \(\frac{\mu_{0} n d x / a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

Hence total magnetic field is given by
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2} \int_{-L}^{d x} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

For r >> a (R >> L)
We have
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2 r^{3}} \int_{-L}^{+L} d x=\frac{\mu_{0} \ln }{2} \frac{2 L a^{2}}{r^{3}}\)

Magnetic moment of a solenoid
m = (n × 2L) I (πa²)

Therefore,
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\) as that of a bar magnet.

Numerical Problems :

• Torque on a magnetic dipole τ = MB sin θ
• Potential energy of a magnetic dipole U = – MB cos θ
• \(B_{H}^{2}+B_{v}^{2}=B^{2}\)
• tan θ = \(\frac{B_{V}}{B_{H}}\)
• H = B cos δ
• V = B sin δ

Question 1.
A bar magnet of the magnetic moment 6 J T¹ is aligned at 60° with a uniform external magnetic field of 0 – 44 T. Calculate
(a) the work is done in turning the magnet to align its magnetic moment (i) normal to the magnetic field and (ii) opposite to the magnetic field and
Answer:

(b) the torque on the magnet in the final orientation in case (ii). (CBSEAI, Delhi 2018)
Answer:
τ = M × B sin θ₂
= 6 × 0.44 × sin 180° = 0

Question 2.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10^-2 J. What is the magnitude of the magnetic moment of the magnet?
Answer:
Given θ = 30°, B = 0.25 T, τ = 4.5 × 10^-2 J,M = ?
Using the expression τ = MB sin θ we have
M = \(\frac{\tau}{B \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\) = 0.36 J T^-1