Category: CBSE Sample Papers

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 7 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Study the change of concentration of reactant with respect to time as depicted in the graphical representation below: (2)
(A) Predict the order of the reaction
Answer:
The variation in the concentration (R) vs. time (t) plot shown here represents a zero order reaction, for which the rate of the reaction is proportional to zero power of the concentration of the reactants.

(B) What does the slope represent?

Answer:
For a zero order reaction, rate constant is given as [R] = [R]₀ – kt
So, the slope of the curve for the variation in the concentration (R) vs. time (t) plot is equal to the negative of the rate constant for the reaction.

Question 2.
Arrange the following compounds in increasing order of their property as indicated. to-carbon bond of alkenes? (Any two)
(A) CH₃COCH₃, C₆H₅—CO—C₆H₅, CH₃CHO (Reactivity towards nucleophilic addition reactions)
Answer:
C₆H₅—CO—C₆H₅, < CH₃COCH₃ < CH₃CHO

Explanation: Ketones are less reactive towards nucleophilic addition reactions due to more steric hinderance.

(B)

Answer:

Explanation: More the number of electron withdrawing groups (-Cl), more the acidic character.

(C) C₂H₅OH, CH₃CHO, CH₃COOH (Boiling points) (2)
Answer:
CH₃—CHO < C₂H₅OH, < CH₃COOH

Explanation: Carboxylic acids have higher boiling points then alcohols due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bonds are not broken completely even in the vapour phase.

Question 3.
Answer the following questions.
(A) What feature of their structure makes aldehydes easier to oxidize than Ketones?
Answer:
The H on the carbonyl carbon atom of aldehyde makes it easier to oxidize.

(B) How does the carbon-to-oxygen bond of aldehydes and ketones differ from the carbon-to-carbon bond of alkenes? (2)
Answer:
The carbon-to-oxygen double bond is polar due resonance; the carbon-to-carbon double bond is non-polar.

Related Theory:
There is a significant contribution from the resonance structure which puts a formal negative charge on the oxygen atom and a formal positive charge on the carbon atom and there by increases the polarity of the bond.

SECTION – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
Draw diagram to show splitting of d – orbital in octahedral crystal field. Explain the two patterns of filling d⁴ in octahedral crystal FIeld. (3)
Answer:

The splitting of the d orbitals in an octahedral field takes place in such a way that dx² – y², dz² experience a rise in energy and form the eg level, while d_xy, d_yz and d_zx experience a fall in energy and form the t_2g level.
For electric configuration of d⁴

(i) When Δ₀ > P Electronic configuration is t⁴2g eg⁰

When Δ₀ < P
Electronic configuration is t³_2g eg¹

Question 5.
A colloidal solution of Agl is prepared by two methods. (3)

(A) What is the charge on Agl colloidal method.
Answer:
Test tube (A) has negative charge whereas test tube (B) has positive on the colloidal particles.

(B) Give the reason for origin of change.
Answer:
According to preferential adsorption theory, test tube (A) I^– is absorbed on precipitate Agl [or Agl / I^– is formed] and in test tube (B). Ag⁺ is absorbed on precipitate Agl [or Agl/Ag⁺ is formed].

Related Theory:
The electric double layer theory describes the interaction between surface of colloidal particles and ions that are present in the fluid in which the colloidal particles are dispersed.

(C) What ¡s zeta potential?
Answer:
Zeta potential is the potential difference between fixed layer (primary layer) and diffused layer (secondary layer) of colloidal particle. It is also called as electrokinetic potential.

Question 6.
A first order reaction takes 10 minutes for 25% decomposition. Calculate t_1/2 for the reaction. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021).
OR
Consider the reaction
2A + B → C + D
Following results were obtained in experiments designed to study the rate of reaction:

(A) Write the rate law for the reaction.
(B) Calculate the value of rate constant for the reaction. (3)
Answer:

OR

Question 7.
(A) Describe the mechanism of the addition of Grignard reagent to the carbonyl group of compound to form an adduct which on hydrolysis yields an alcohol.
(B) Write the IUPAC name of the compound.

OR
(A) Illustrate the following name reactions:
(i) Hell-Vothard-Zelinsky reaction.
(ii) Wolff-Kishner reduction reaction
(B) Write the structural formula of 1-phenylpentan-l-one. (3)
Answer:
(A) a. Mechanism:
(i) Nucleophilic addition of Grignard reagent to carbonyl group to form an adduct.

(ii) Hydrolysis of the adduct to alcohol.

Caution:
Students should draw the curved arrows in proper directions and proper charges on the atoms in the mechanism.

(B) 4-Hydroxy-4-methyl pentan-2-one.
OR
(A) (i) Hell-Volhard-Zelinsky reaction – Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give αx-halocarboxylic acids. The reaction is known as Hell-Volhard- Zelinsky reaction.

(ii) Wolff-Kishner reduction reaction – The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with hydrazine followed by heating with potassium hydroxide in a high boiling solvent such as ethylene glycol.

Caution:
Students should write the chemical reaction also otherwise marks would be deducted during evaluation.

(B)

Question 8.
State Reasons:
(A) Aniline is a weaker base than cyclohexylamine. (3)
Answer:
Aniline is a weaker base than cyclohexylamine because the lone pair of electrons on the N-atom is delocalised over the benzene ring in aniline which results in the decrease in electron density on Nitrogen. In cyclohexylamine, the lone pair of the electrons on N-atom is readily available due to absence of π-electrons.

Caution:
Basicity of the amino group means it is unsuitable for reactions with acids (e.g. H₂SO₄ or AlCl₃) such as nitration, sulfonation and Friedel-Crafts alkylation or acylation.

(B) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Answer:
It is difficult to prepare pure amines by ammonolysis of alkyl halides as the primary amine formed by ammonolysis acts as a nucleophile. It further produces 2° and 3° alkyl amine.

(C) Etectrophilic substitution in aromatic amines takes place more readily than benzene.
Answer:
Arylamines are potentially very reactive towards electrophilic aromatic substitution. This is because -NH₂, -NHR₂ and -NR₂ are very strong activators and are ortho, para-directing.

Question 9.
Observe the graph and answer the questions that follow:

(A) Which transition metaL of 3d series has positive E°(M²⁺/M) value and why?
Answer:
The E°_(M²⁺/M) for copper is positive. This is because high energy is required to transform Cu to Cu²⁺ which is not balanced by its hydration enthalpy.

(B) E° value for the Mn⁺³/Mn⁺² couple is positive (+1.5 V) whereas that of Cr⁺³/Cr⁺² is negative (-0.4 V). Why?
Answer:
The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/ Cr²⁺ couple. This is because Mn²⁺ ion is particularly stable due to extra stability of its half filled valence electronic configuration (d⁵). Thus Mn³⁺ ion has a very high tendency to gain an electron and form the much more stable Mn²⁺ ion.

(C) E° values are not regular for first row transition elements? (3)
Answer:
The (M²⁺/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (ΔiH₁ + ΔiH₂) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question 10.
What is the lanthanoid contraction? What are its causes and consequences? (3)
OR
Explain the following:
(A) Copper (I) ion is not stable in an aqueous solution.
(B) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(C) Transition metals in general act as good catalysts.
Answer:
Lanthanoid contractions – The cumulative effect of the regular decrease in size or radii of Lanthanoid with increase in atomic number is called Lanthanoid contraction.

Causes: With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell.
The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell.
With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction.

Consequences: Due to Lanthanoid contraction.
1. Radii of the members of the third transition series is similar to those of second transition series.
2. It becomes difficult to separate Lanthanoids.
OR
(A) In an aqueous medium, Cu²⁺ is more stable than Cu⁺. This is because although energy is required to remove one electron from Cu⁺ to Cu²⁺, high hydration energy of Cu²⁺ compensates for it. Therefore, C^u+ ion in an aqueous solution is unstable. It disproportionates to give Cu²⁺ and Cu.

(B) The density of elements from titanium to copper increase in the first series of transition elements. This is due to decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density.
Explanation: As the atomic radii decreases moving across from titanium to Cu, So its volume will decrease and density is expected to increase.

(C) They have variable valencies and show multiple oxidation states and transition metals sometime form unstable intermediate compounds and provide a new path with lower activation energy for the reaction. In some cases transition elements provide a suitable surface for the reaction to take place.

Question 11.
An aromatic compound ‘A’ of moLucutar formula C₇H₅O₂ undergoes a series of reactions an shown below. Write the structures of A, B, C, D and E in the following reactions.

OR
How will you convert the following –
(A) Aniline to chtorobenzene
(B) Ethanoic acid to methanamine
(C) Methyl chloride to ethanamine (3)
Answer:

OR

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the questions that follow:
The potential of each electrode is known as electrode potential. Standard electrode potential is the potential when concentration of each species taking part in electrocde reaction is unity and the reaction is taking place at 298 K. By convention, the standard electrode potential of hydrogen (SHE) is 0.0 V. The electrode potential value for each electrode process is a measure of relative tendency of the active species in the process to remain in the oxidised/reduced form. The negative electrode potential means that the redox couple is stronger reducing agent than H⁺/H₂ couple. A positive electrode potential means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple. Metals which have higher positive value of standard reduction potential form the oxides of greater thermal stability.
(A) What is meant by reference electrode?
(B) Platinum is used in the standard hydrogen electrode. Give reason.
(C) Explain the term Standard electrode potential.
(D) Calculate the emf of the following cell at 298 K: Fe(s) | Fe²⁺ (0.001 M) || H⁺ (1M) | H₂(g)_{(1 bar)}, Pt_(s) (Given E°cell = +0.44V)
OR
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. (5)
Answer:
(A) Standard Hydrogen electrode: It is a reference electrode against which the electrode potentials of all electrodes are measured.
Explanation: The Standard electrode potential of SHE is zero volt. This SHE can act as anode or cathode, depending upon the half cell that is attached to it.

(B) Platinum is a less reactive metal. So, it doesn’t easily react with other metals. As a result, it provides the surface for the redox reaction.

(C) The potential difference developed between metal electrode and solution of ions of unit molarity (1M) at 1 atm pressure and 25°C (298 K) is called standard electrode potential. It is denoted by E°.

(D) The cell reaction:

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 5 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks.
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the chokes in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

SECTION – A
(Section A contains 3 questions of 2 marks each.)

Question 1.
Explain the formation of secondary rainbow. (2)
Answer:
It is the result of fur-step process. Refraction with dispersion followed by internal reflection, and finally reflection.

It is found that red light emerges at an angle of 50° related to the incoming sunlight and violet light emerges at an angle of 53°. Thus, the observer sees secondary rainbow with violet colour on top and red colour on the bottom. The intensity of light is reduced at second internal reflection and hence it is fainter than primary rainbow.

Question 2.
What is the shape of fringe pattern on the screen? (2)
Answer:
The locus of point P in the plane of the screen, such that S2P-SiP is a constant, is a hyperbola. Thus, fringe pattern are hyperbola.

Question 3.
A parallel beam of light of 450 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.5 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit.
OR
Draw the diagram of a solar cell and draw its V-l characteristics. (2)
Answer:
The distance of the n^th minimum from the centre of the screen,
X_n = \(\frac{n \mathrm{D} \lambda}{a}\)
Where, D → distance of slit from
screen
λ → wavelength of the light
a → width of the slit
For first minimum,
n = 1,
X_n = 3 × 10^-3 m
D = 1.5 m
λ = 450 × 10^-9m
3 × 10^-3 = \(\frac{1 \times(1.5) \times\left(450 \times 10^{-9}\right)}{a} \frac{n D \lambda}{a}\)
a = \(\frac{1 \times(1.5) \times\left(450 \times 10^{-9}\right)}{3 \times 10^{-3}}\)
a = 0.255 mm
OR

Caution
V-I characteristics of solar cell is drawn in fourth quadrant of the coordinate axes, This is because a solar cell does not draw current but supplies the same to the load.

SECTION – B
(Section B consists of 8 questions of 3 marks each.)

Question 4.
The trajectories, traced by different a-particles, in Geiger Marsden experiment were observed as shown in the figure.

(A) What names are given to the symbols ‘b’ and ‘θ’ shown here.
(B) What can we say about the values of b for
(i) θ = 0°
(ii) θ = π radians. (3)
Answer:
(A) The smbo1b’ represents impact parameter and ‘θ’ represents the scattering angle.
(B) (i) When θ = 0_o. the impact parameter will be maximum and represent the atomic size.

(ii) When θ = π radians, the impact parameter ‘b’ will be minimum and represent the nuclear size.

Question 5.
Draw reflected or refracted wavefront when a plane wavefront is incident on prism, convex lens and concave mirror.
OR
What difference between an intrinsic semiconductor and p-type extrinsic semiconductor? Why p-type semiconductor is electrically neutral even if number of holes are larger than the number of electrons? (3)
Answer:
(A) (i) When a plane wave is incident on prism

(ii) Convex lens

(iii) Concave mirror

OR

Intrinsic Semiconductors	p-Type Extrinsic Semiconductor
These pure semiconducting materials and known variety is added to them. They have equal number of holes and electrons.	These are prepared by doping a small quantity of impurity of trivalent material to the pure semiconductor material which results in higher number of holes compared to electrons.
The electrical conductivity is generally small.	The electrical conductivity is comparatively very high.

A p-type semiconductor crystal is electrically neutral because the charge of additional charge carriers is just equal and opposite to that of the ionized cores in the crystal lattice.

Question 6.
(A) Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. Estimate the number of photons emitted per second on an average by the source.

(B) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface. (3)
Answer:
(A) The energy of a photon of frequency v is
E = hv
= (6.63 × 10^-34 Js) × (6 × 10¹⁴ s^-1)
= 3.98 × 10^-19 J = 4 × 10^-19J

If n be the number of photons emitted by the source per second, then the power P transmitted in the beam is given by,
P = nE
∴ n = \(\frac{P}{E}\)
⇒ n = \(\frac{2 \times 10^{-3}}{4 \times 10^{-19}}\)
5 × 10¹⁵ photons/sec

Related Theory
Photoelectric current is directly proportional to the intensity of incident radiation.

Question 7.
The oscillating magnetic field in a plane electromagnetic wave is given by:
B_y = (8 × 10^-6) sin[2 × 10¹¹t + 300π X]T
(A) Calculate the wavelength of the electromagnetic wave.
(B) Write down the expression for the oscillating electric field. (3)
Answer:
(A) Comparing above equation by the standard equation of magnetic field is:
B_y = B₀ sin(ωt + kx)T
We get B₀ = 8 × 10^-6 T,
ω = 2 × 10¹¹ rad/s,
k = \(\frac{2 \pi}{\lambda}\) = 300π
Wavelength
λ = \(\frac{2 \pi}{300 \pi}\)
= \(\frac{1}{150}\) m = 6.67 mm

(B) E₀ = B₀C
= 8 × 10^-6 × 3 × 10⁸
= 2.4 × 10³V/m.
According to right hand system of vector E, vector B, vector K, the electric field oscillates along negative Z-axis, so equation is
E_z = -2.4 × 10³sin(2 × 10¹¹t + 300π X)\(\frac{\mathrm{V}}{\mathrm{m}}\)

Question 8.
(A) Write the relationship between angle of incidence ‘F, angle of prism ‘A’ and angle of minimum deviations δ_m for a triangular prism.

(B) A ray of light suffers minimum deviation, while passing through a prism of refractive index 1.5 and refracting angle 60°. Calculate the angle of deviation and angle of incidence. (3)
Answer:
(A) The relation between the angle of incidence i, angle of prism A, and the angle of minimum deviation δ_m, for a triangular prism is given by
i = \(\frac{\left(A+\delta_{m}\right)}{2}\)

(B) Here,
μ = 1.5,
A = 60°,
δ_m = ?
In the position of minimum deviation
I = \(\frac{A+\delta_{m}}{2}\),
r = \(\frac{\mathrm{A}}{2}\)
As
μ = \(\frac{\sin i}{\sin r}\)
∴ 1.5 = \(\frac{\sin i}{\sin 30^{\circ}}\)
⇒ sin i = 1.5 × 0.5
= 0.75
⇒ i = sin^-1 (0.75)
= 48.60°
δ_m = 2i – A
= 2 × 48.6 – 60 = 37.2°

Question 9.
(A) In a nuclear reaction
\({ }_{2}^{3} \mathrm{He}\) + \({ }_{2}^{3} \mathrm{He}\) → \({ }_{2}^{4} \mathrm{He}\) + \({ }_{1}^{1} \mathrm{He}\) + \({ }_{1}^{1} \mathrm{He}\) + 12.86 MeV though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How ? Explain. (B) Draw a plot of potential energy between a pair of nucleoni as a function of their separation. Mark the regions where
potential energy is
(i) positive and
(ii) negative. (3)
Answer:
(A) in a nuclear reaction, the sum of the masses of the target nucleus \({ }_{2}^{3} \mathrm{He}\) may be greater or less then sum of the masses of the product nucleus \({ }_{2}^{4} \mathrm{He}\) and the \({ }_{1}^{1} \mathrm{He} .\) So from the law of conservation of mass energy, some energy, (12.86 MeV) is evolved in nuclear reaction. This energy is called Q-value of the nuclear reaction. The binding energy of the nucleus on the left side is not equal to the right side. The difference in the binding energies on two sides appears as energy released or absorbed in the nuclear reaction.

(B) The potential energy is minimum at to- For distance larger than r₀ the negative potential energy goes on decreasing and for the distances Less than r₀ the negative potential energy decrease to zero and then becomes positive and increases abruptly. Thus, A to B is the positive potential energy region and B to C is the negative potential energy region.

Question 10.
(A) Give the significance of the sentence “a light ray, while undergoing refraction at the interface of two media, bends towards the normal in the second media”.

(B) Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 × 10⁸ m/s. Find the focal length of the lens? (3)
Answer:
(A) The second medium is optically denser with respect to first medium and speed of Light in second medium is less.
(B)

n = \(\frac{3}{2}\) [∵ n = \(\frac{C}{V}\)]
3² + (R – 3 mm)² = R²
⇒ 3² + R² – 2R (3 mm) + (3 mm)² = R²
⇒ R ≈ 15 cm

⇒ f = 30 cm

Caution
Students are often confused to find the value ofR. In this question they have to use Pythagoras theorem.

Question 11.
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -0.85 eV to -1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
OR
Give reasons for the following:
(A) Why semiconductor diodes are preferred over vacuum diodes?
(B) A photodiode, when used as a detector of optical signals is operated under reverse bias.
(C) The band gap of the semiconductor used for fabrication of visible LEDs must be at least 1.8 eV. (3)
Answer:
As we know, E_n = \(\frac{-13.6}{n^{2}}\)eV …………. (i)
For n = 1, E₁ = – 13.6 eV
When electron undergoes transition from EA = – 0.85 eV Then, from equation (i),
– 0.85 = – \(\frac{13.6}{n_{A}^{2}}\) ⇒ n_A = 4
Similarly, -1.51 = \(\frac{-13.6}{n_{B}^{2}}\) ⇒ n_B

Hence, electron transits from n = 4 to n = 3 which corresponds to Paschen series of hydrogen atom.
As, \(\frac{1}{\lambda}\) = R(\(\frac{1}{n_{B}^{2}}-\frac{1}{n_{A}^{2}}\))
Here n_A = 4, n_B = 3, R = 1.0974 × 10⁷
Then \(\frac{1}{\lambda}\) = 1.097 × 10⁷(\(\frac{1}{3^{2}}-\frac{1}{4^{2}}\))^m-1
⇒ λ = 1875 nm
OR
(A) In semiconductor diodes, no cathode heating is required in junction diode for production of charge carriers. Voltage drop across a junction diode is much less compared to that across a vacuum diode. Semiconductor diodes can be used for much higher frequencies.

(B) When operated under reverse bias, the photodiode can detect changes in current with changes in light intensity more easily because the fractional change in minority charge carriers is more than that in majority charge carriers.

(C) The Photon energy, visible light photons vary from about 1.8 eV to 3 eV. Hence for visible LEDs, the semiconductor must have a band gap of 1.8 eV.

SECTION – C
(Section C consists one case study-based question of 5 marks.)

Question 12.
Wave theory of light was followed while explanation of photoelectric effect. There were many failures of wave theory of light. According to wave theory when light incident on a surface, energy is distributed continuously over the surface. So that electron must take a time interval to accumulate suffiicient energy to come out. But in experiment there is no time lag. When intensity is increased, more energetic electrons should be emitted. So that stopping potential should be intensity dependent. But it is not observed. According to wave theory, if intensity is suffiicient then, at each frequency, electron emission is possible. It means there should not be existence of threshold frequency.

(A) The stopping potential necessary to reduce the photoelectric current of zero:
(i) is directly proportional to wavelength of incident light.
(ii) uniformly increases with the wavelength of incident light.
(iii) directly proportional to frequency of incident light.
(iv) uniformly increases with the frequency of incident light.

(B) The photoelectric threshold frequency of a material is u. When light of frequency 4 u is incident on the metal, the maximum kinetic energy of the emitted photoelectron is:
(i) \(\frac{5}{2}\)hu
(ii) 3 hu
(iii) 4 hu
(iv) 5 hu

(C) When photons of energy hv fall on an aluminium plate (of work function E_o), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:
(i) K + hv
(ii) K + E₀
(iii) 2K
(iv) K

(D) To understand photoelectric effect a graph is plotted between two factors, which is a straight line. The factors are:
(i) Intensity of radiation and photoelectric current
(ii) Potential of anode and photoelectric current
(iii) Threshold frequency and velocity of photoelectrons
(iv) Intensity of radiation and stopping potential.

(E) Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function 1.9 eV. The maximum kinetic energy of the photoelectrons emitted will be:
(i) 0.58 eV
(ii) 2.48 eV
(iii) 1.24 eV
(iv) 1.16 eV (5)
Answer:
(A) (d) uniformly increases with the frequency of incident light
Explanation: The negative potential at which photoelectric current becomes zero is called stopping potential.
By _eV_s = hv – Φ
Stopping potential required to reduce the photoelectric current to zero “decreases uniformly with the frequency of the incident radiation”.

(B) (b) 3hu
Explanation: The maximum kinetic energy of the emitted electrons is given by K_max = hu-Φ₀
= h(4u) – h(u) = 3hu

Related Theory
Maximum kinetic energy of the emitted electron = hv – Φ₀
(C) (a) K + hv
Explanation: Let K and K’ be the maximum kinetic energy of photoelectrons for incident light of frequency v and 2v respectively. According to Einstein’s photoelectric equation,
K = hv – E₀ ………….. (i)
and K’ = h(2v) – E₀0 ………….. (ii)
= 2hv – E₀ = hv + hv – E₀
K’ = hv + K [using (i)]

Related Theory
The kinetic energy of the emitted electrons is given by K= hv – Φ₀

(D) (a) Intensity of radiation and photoelectric current
Explanation: Intensity of radiation linearly changes the photoelectriccurrent. Photoelectric current ∝ intensity of radiation.

Caution
Students are often confused about the relation between photoelectric current and intensity of radiation. They are directly proportional to each other.
(E) (a) 0.58 eV
Explanation: hv = \(\frac{h c}{\lambda}\)
= \(\frac{1240}{500}\)
= 2.48 eV
\(\frac{1}{2}\) mv_max² = hv – Φ0 = 2.48 – 1.90
= 0.58 eV

Related Theory
Work function is as the minimum amount of energy which is required to remove an electron to infinity from its source. It was defined in the concept of photoelectric effect in which a light is incident upon the metal and the electrons are ejected from the metal.

Students can access the CBSE Sample Papers for Class 12 Economics with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Economics Term 2 Set 4 with Solutions

Time allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This is a Subjective Question Paper containing 13 questions.
• This paper contains 5 questions of 2 marks each, 5 questions of 3 marks each and 3 questions of 5 marks each.
• 2 marks questions are Short Answer Type Questions and are to be answered in 30-50 words.
• 3 marks questions are Short Answer Type Questions and are to be answered in 50-80 words.
• 5 marks questions are Long Answer Type Questions and are to be answered in 80-120 words.
• This question paper contains Case/Source Based Questions.

Question 1.
“Final goods include only those goods which are consumed by the households”. Defend or refute the given statement with valid reason. (2)
Answer:
The given statement is refuted as final goods include those goods which are either consumed by the households or purchased by a producer for investment purposes.

Question 2.
When are actual stocks greater than desired stocks?
OR
Answer the following questions based on the figure given below:

At which level of income, Average Propensity to Consume will be equal to one and why? (2)
Answer:
When aggregate demand falls short of the expectations of the producers, actual stocks are greater than desired stocks. Some Output remains unsold.
OR
Average Propensity to Consume will be equal to one at point B, as here Average Propensity to Save is zero.

Question 3.
If a consumption function of a hypothetical economy is given as C = 100 + 0.6 Y, then
(A) What will be the values of marginal propensity to consume and marginal propensity to save for the economy?
(B) Write the corresponding saving functions.
OR
From the following data calculate investment expenditure:
(i) Marginal propensity to save = 0.2
(ii) Equilibrium level of income = ₹22,500
(iii) Autonomous consumption = ₹500 (2)
Answer:
(A) C = 100 + 0.6 Y (given)
So, MPC = 0.6
MPS = 1 – MPC = 1
– 0.6 = 0.4

(B) S = – C + (l- b)Y
S = – 100 + 0.4 Y
OR
Y = \(\bar{C}\) + (1 – MPS)Y + I
₹22,500 = ₹500 + (1 – 0.2)
₹22.500 + I
I = ₹22.500 – ₹500 – ₹18.000
= ₹4.000

Question 4.
‘Water has become an economic commodity.’ Justify the statement.
OR
You might have seen and heard on TV news or read in newspapers about foreigners flocking to India for surgeries, liver transplants, dental and even cosmetic care. Why? (2)
Answer:
Water has become an economic commodity because drinking water is not readily available in the world. We pay monthly water bills for the consumption of water. We buy water for drinking in the form of bottled water. Since we pay for water on the basis of supply and demand, it would not be technically wrong to view water as an economic commodity. Although water is not yet a formal commodity for sale on a formal basis, it will likely be within the next two decades.
OR
Because our health services combine latest medical technologies with qualified professionals and is cheaper for foreigners as compared to costs of similar health care services in their own countries. In the year 2004-05, as many as 1,50,000 foreigners visited India for medical treatment. And this figure is likely to increase by 15 percent each year.

Question 5.
Raj is going to school. When he is not in school, you will find him working on his farm. Can you consider him as a worker? Why? (2)
Answer:
Yes, he will be considered as a worker because all those who are engaged in economic activities, in whatever capacity – high or low, are workers. Raj is also engaged in an economic activity as he is contributing to the gross national product of the economy.

Question 6.
Compare and analyse the given data of India and China with valid arguments.
Annual Growth of Gross Domestic Product (%), 1980 – 2017

Country	1980 – 90	2015 – 2017
India	5.7	7.3
China	10.3	6.8

Source: Key Indicators for Asia and Pacific 2016, Asian Development Bank, Philippines: World Development Indicators 2018
Answer:
The given data shows that China has gained economic strength over the years. When many developed countries were finding it difficult to maintain a growth rate of even 5%, China was able to maintain near double-digit growth during the decade of 1980s. The growth rate o’ China has decelerated to an average of 6.8%, over the period 2015-17.

In the recent past India has posted a decent rise in the growth rate. While India had maintained a reasonable growth rate of 5.7% in the decade of 1980’s it has shown great caliber and character in the period 2015-17 by registering an average of 7.3%, over the period 2015-17. Nevertheless, Indian elephant has to travel a long distance before it could present itself as a real threat to the growth story of the Chinese dragon.

Question 7.
“The one-child policy produced consequences beyond the goal of reducing population growth.” Explain.
OR
Group the following features pertaining to the economies of India, China and Pakistan under three heads:

• One-child norm
• Low fertility rate
• High degree of urbanisation
• Mixed economy
• Very high fertility rate
• Large population
• High density of population
• Growth due to manufacturing sector
• Growth due to service sector (3)

Read the following case carefully and answer question number 8 and 9 given below :
Ramesh is a tailor and he also owns a shop. He is famous quality and comfortable women’s and men’s clothing for a casual lifestyle. Ramesh stitches only for households. In a single day he earns ₹1,000 from stitching the clothes. Over this day, his equipment such as swing machines, scissors, measuring tape, etc. depreciates in the value by ₹100. Out of his remaining income, i.e., ₹900, Ramesh pays sales tax worth ₹60. He takes ₹400 to his home and keeps ₹440 for improvement and buying of new equipment. He further pays ₹40 as income tax from his income.
Answer:
One-child norm introduced in China in the late 1970s is the major reason for low population growth. It is stated that this measure led to a decline in the sex ratio, that is, the proportion a females per 1000 males. One child norm and the resultant arrest in the growth of pollution also have other implications. For instance after a few decades, in China, there will be more elderly people in proportion to young people, and also reduced the fertility rate considerably.
OR

India	China	Pakistan
Mixed economy	One-child norm	High degree of urbanisation Mixed economy
Large population	Low fertility rate	Very high fertility rate
High density of population	High degree of urbanisation
	Large population
	Growth due manufacturing sector

Question 8.
Calculate the value of net value added at factor cost by Ramesh. (3)
Answer:
Gross value added at market price (GVA_MP) = Ramesh’s contribution to GDP = ₹1.000 NVA_MP = GVA_MP – Depreciation = ₹1.000 – ₹100 = ₹900
NVA_FC = NVA_MP – Net Indirect Taxes = ₹900 – ₹60 = ₹840

Question 9.
What is the difference between ‘market price’ or ‘factor cost’ in reference to value-added? When will be market price and factor cost equal? (3)
Answer:
Factor cost is the total cost of all the factors of production consumed or used in producing a good or service. Whereas, market price is the price at which a product is sold in the market. Market price and Factor cost will be equal when there is no indirect taxes and subsidy.

Factor cost = Market price – Indirect tax + Subsidies
Factor cost = Market price – 0 + 0
Factor cost = Market price

Question 10.
What will happen if there is no additional employment generated in the economy even though we are able to produce goods and services in the economy? How could jobless growth happen? (3)
Answer:
The country is going through a structural change and shifting to the modern techniques using the capital-intensive method of production, which is cost-effective and gives more output by employing less labor. Thus, in India employment growth started declining and reached the level of growth that India had in the early stages of planning. This means that in the Indian economy, without generating employment, we have been able to produce more goods and services. This phenomenon is known as jobless growth.

Question 11.
“Ruling out any impact of stimulus on the price situation, Chief Economic Advisor K.V. Subramanian on Thursday said the COVID-19 pandemic has severely dented the demand for non-essential or discretionary goods.”

Identify the situation stated in the above lines and explain any two measures to correct them. (5)
Answer:
The situation stated in the given lines is a Deflationary condition. It is that situation when Aggregate Demand is lesser than Aggregate Supply corresponding to full employment level.

Two fiscal measures to control it are:

1. Decrease in Taxes: To curb the deflationary gap the government may decrease the taxes. This may increase the purchasing power in the hands of the people which in turn may increase the Aggregate Demand in the economy to bring it equal to the Aggregate Supply.
2. Increase in Government Expenditure: To curb the deflationary gap the government may increase its expenditure. This may increase the purchasing power in the hands of the people which in turn may increase the Aggregate Demand in the economy to bring it equal to the Aggregate Supply.

Question 12.
(A) Find gross national product at market price. (₹ in crores)
(i) Private final consumption expenditure 800
(ii) Net current transfers to abroad 20
(iii) Net factor income to abroad (-)10
(iv) Government final consumption expenditure 300
(v) Net indirect tax 150
(vi) Net domestic capital formation 200
(vii) Depreciation 100
(viii) Net imports 30
(ix) Income accruing to government 90
(B) State any two precautions to be taken while using the expenditure method of measuring national income?
OR
(A) Calculate Net Domestic Product at Factor Cost: (₹ crore)
(i) Gross National Disposable Income 600
00 Net current transfers to abroad (-) 20
(iii) Consumption of fixed capital 60
(iv) Indirect tax 100
(v) Subsidies 10
(vi) Net factor income to abroad (-) 10
(vii) National debt interest 40

(B) When will be the domestic factor income greater than national income? (5)
Answer:
(A) GNP_MP = Private final consumption expenditure + Government final consumption expenditure + Net domestic capital formation + Depreciation – Net imports – Net factor income to abroad = 800 + 300 + 200 + 100 – 30 – (-)10 = ₹1380 crore

(B) The following precautions to be taken while using the expenditure method of measuring national income:

• Expenditure on second-hand goods is not to be included.
• Expenditure on shares and bonds is not to be included.

OR

(A) NDP_FC = Gross National Disposable Income – Consumption of fixed capital + Net current transfers to abroad-indirect tax + Subsidies + Net factor income to abroad
= 600 – 60 + (-20) – 100 + 10 + (-10)
= ₹420 Crore

(B) National Income = Domestic Factor Income + Net Factor Income from Abroad
Thus, when net factor income from abroad is negative, domestic factor income is greater than national income.

Question 13.
(A) “Ujjwala Yojana has been a game-changer for rural India.” State any three conventional fuels being targeted under the LPG cylinder distribution scheme (Ujjwala Yojana).

(B) “The Indian Health System needs a stronger does of public expenditure to cure itself.” Justify the given statements with valid arguments. (5)
Answer:
(A) The conventional sources of energy cause environmental pollution therefore the government have introduced the ‘Ujjwala Yojana as a game-changer for rural India by providing free LPG gas cylinders (cleaner fuel) to rural households.

The three conventional fuels targeted under Ujjawala Yojana are:

• Agricultural waste and dried dung
• Firewood
• Coal

(B) The statement is defended as the improvement in the health system in India has been unreasonably slow since independence. Indian health system has been a victim of a relatively low public expenditure.

The health expenditure as a percentage of GDP is abysmally low as compared to some of the major developing countries. It stood at around 4.7% of the total GDP in the year 2014-15. Thus, actually Indian health system needs the increased dose of public expenditure to cure itself.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 8 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
What is meant by the following terms?
(i) Cyanohydrin
Answer:
-OH group and cyano group are present on same carbon atom. Addition of HCN to carbonyl group in weakly acidic medium forms cyanohydrin.

(ii) Acetal (2)
Answer:
Acetal : Terminal C atom has two alkoxy groups. Two equivalents of monohydric alcohol add to 1 equivalent of aldehyde in presence of dry HCl gas.

Caution:
Students should draw the related structures to get full marks.

Question 2.
Arrange the following as instructed: (Any two)
(A) In an increasing order of basic strength CH₆H₅NH₂, C₆H₅N(CH₃)₂ (C₂H₆)₂ NH and CH₃ NH₂.
Answer:
C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₆₂NH.

Explanation:
In C₆H₅N(CH₃)₂ presence of two CH₃ groups (increases the electron density of N-atom) makes it . more basic than C₆H₅NH₃ but less basic than (C₂H₅)₂NH₂ and CH₃NH₂ due to the presence of aromatic ring which is responsible for the delocalisation of lone pair of electrons of N-atom over the benzene ring (decreases the electron density of N-atom).

(B) In a decreasing order of basic strength Aniline, p-nitroaniline and p-toluidine.
Answer:
p-nitro aniline < Aniline < p-toluidine

Explanation:
In p-toluidine, the presence of electron-donating-CH₃ group increases the electron density on the N-atom.
Thus, p-toluidine is more basic than aniline. On the other hand, the presence of electron – withdrawing – NO₂ group decreases the electron density over the N-atom in p-nitroaniline. Thus, p-nitroaniline is less basic than aniline.

(C) In an increasing order of pKb values
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂ NH and C₆H₅NH₂ (2)
Answer:
(C₂H₅)₂ NH < C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂.

Explanation:
In C₂H₅NH₂, only one -C₂H₅ group is present while in (C₂H₅)2₂NH, two – C₂H₅ groups are present. Thus, the +1 effect is more in (C₂H₅)₂NH than in C₂H₅NH₂. Therefore, the electron density over the N-atom is more in (C₂H₅)₂NH than in C₂H₅NH₂. Hence, (C₂H₅)₂NH is more basic than C₂H₅NH₂.

Also, both C₆H₅NHCH₃ and C₆H₅NH₂ are less basic than (C₂H₅)₂NH and C₂H₅NH₂ due to the delocalization of the lone pair in the former two. Further, among C₆H₅NHCH₃ and C₆H₅NH₂, the former will be more basic due to the +1 effect of -CH₃ group.

Question 3.
What is meant by rate of a reaction? Differentiate between average rate and instantaneous rate of a reaction. (2)
Answer:
Rate of a chemical reaction is the change in the concentration of any one of the reactants or products per unit time. It is expressed in mol L^-1 s^-1 or Ms^-1 or atm time^-1 units.

The average rate of a reaction is defined as the rate of change of concentration of a reactant (or of a product) over a specified measurable period of time.
Instantaneous rate of reaction gives the tendency of the reaction at a particular point of time during its course (or) The time derivative of the concentration of a reactant (or product) converted to a positive number is called the instantaneous rate of reaction.

Rate of disappearance of R

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
Write the IUPAC names of the following coordination compounds:
(i) [Co(NH₃)₆]Cl₃
(ii) [Cr(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
OR
Explain different types of ligands with one example of each. (3)
Answer:
(i) Hexaamminecobalt (III) chloride.
(ii) Pentaamminechloridochromium (III) chloride.
(iii) Potassium hexacyanidoferrate (III).
OR
The ions or molecules bound to central atom or ion in the coordination entity are ligands e.g. [Fe(CN)₆]^4- has six CN- ligands.

Types:

1. On the basis of charges on them ligands can be negative, positive (e.g. H₃O⁺. NH₄⁺ etc) or neutral (eg. CO, NH₃, H₂O).
2. On the basis of their donor atoms ligands can be monodentate or unidentate (one donor atom) e.g- NH₃, H₂O, Cl- etc or didentgte (two donor atoms) H₂NCH₂CH₂NH₂ or C₂O₄^2- etc or polydentate (several donor atoms) eg. [EDTA]^4- is a hexadentate ligand.
3. Ligands which can ligate through two different atoms are called ambidentate ligands eg. NO₂^– and SCN^– ions whereas when a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is called chelate ligand.

Question 5.
The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.905 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (a).
Given: λ°(H⁺) = 349.6 S cm² mol^-1 and λ° (CH₃COO^–) = 40.9 S cm² mol^-1 (3)
Answer:
5. C = 0.001 mol L^-1, k = 3.905 x 10^-5 S cm^-1

Molar conductivity
K = 3.905 × 10^-5 S cm^-1 ∧^c_m = ?

Using formula,
∧^c_m = \(\frac{K \times 1000}{C}=\frac{3.905 \times 10^{-5} \times 1000}{0.001}\)
=39.05 S cm² mol^-1

Molar conductivity at infinite dilution:
CH₃COOH → CH3COC^– + H⁺
∧°_m = λ°H⁺+ λ°CH₃COO
= (349.6 + 40.9)
∧°_m = 390.5 S cm² mol^-1

Degree of dissociation:
α = \(\frac{\wedge^{c} \mathrm{~m}}{\wedge^{\circ} \mathrm{m}}=\frac{39.05}{390.5}\)
∴α = 0.1

Question 6.
A first order gas phase reaction : A₂B₂(g) → 2A(g) + 2B(g) at the temperature 400°C
has the rate constant k = 2.0 × 10^-4 sec^-1. What percentage of A₂B₂ is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197).
OR
The thermal decomposition of HC0₂H is a first order reaction with a rate constant of 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will it take for three-fourths of initial quantity of HCOOH to decompose, (log 0.25 = -0.6021) (3)
Answer:
Since the reaction is of the first order

OR
Given : K = 2.4 × 10^-4 sec^-1

Question 7.
(A) Write the electronic configuration of Ce³⁺ ion, and calculate the magnetic moment on the basis of the ‘spin-only’ formula. [Atomic No. of Ce = 58]
Answer:
Ce(58) : [Xe]4 f⁴5d¹6s²
Ce³⁺(58): [Xe]4 f¹

Spin only Formula
Magentic moments

= 1.732 B.M.
where ‘S’ is spin quantum number, Ce³⁺ has one unpaired electron which has spin quantum number equal to 1/2.

(B) Account for the following:
(i) The entholpies of atomisation of the transition metals are high.
Answer:
It is because of strong metallic bonds due to large number of unpaired electrons in d-orbitals.

Explanation:
The enthalpies of atomization of the transition metals are high because of large number of unpaired electrons in their atoms, they have stronger inter atomic interaction and hence strong metallic bonding is present between atoms.

(ii) The lowest oxide of o transition metal is basic, the highest is amphoteric/ acidic. (3)
Answer:
It is because transition metals in lowest oxidation state are more metallic and in higher oxidation state are least metallic, therefore, oxides in Lower oxidation state are basic, whereas in higher oxidation state are amphoteric/acidic

Explanation:
In Low oxidation state of the metaL, some of the valence electrons of the metal atom are not involved in bonding. Hence it can donate electrons and behave as a base. On the other hand, in higher oxidation state of the metal, valence electrons are involved in bonding and are not available. Instead effective nuclear charge is high and hence it can accept electrons and behave as an acid.

Question 8.
(A) Arrange the following complex ions in increasing order of crystal field splitting energy ():
[Cr(Cl)₆]^3-, [Cr(CN)₆]^3-, [Cr(NH₃)₆]³⁺.
Answer:
Crystal field splitting energy AO increases
in the order:
[Cr(Cl)₆]^3- < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3-

Explanation:
Among the ligands Cl, NH3 and CN, Cl is the weakest ligand and CN is the strongest one. Therefore, increasing order of crystal field splitting energy A0 is:
[Cr(Cl)₆]^3- < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3-

(B) Why do compounds having similar geometry have different magnetic moments?
Answer:
Strong field ligand can easily cause the pairing of electrons while the weak field Ligands are not able to form pairs. The magnetic moment of a compound depends on the number of unpaired or paired electrons. Hence it is different for compounds having similar geometry.
Explanation – If CFSE is high, the complex will show low value of magnetic moment and vice versa, e.g. [CoF₆]^3- and [Co(NH₃)₆]³⁺, the former is paramagnetic and the latter is diamagnetic.

(C) Name the type of Isomerism when ambidentate Ligands are attached to o central metal ion. Give one example this isomerism. (3)
Answer:
Linkage Isomerism is a type of isomerism when ambidentate ligands get attached to central metal ions.

Examples:
[Co(NH₃)₅ONO]Cl₂(Pentaammine nitrito cobalt (III) chloride) – O – attached. (Red in colour) and [CO(NH₃)₅NO₂]Cl₂ (Pentaammine nitro cobalt (III) chloride) – N – attached (Yellow-brown in colour)

Related Theory:
The Ligands which have two binding sites are known as Ambidendate ligands. Some of the examples are: Nitrite-N, Nitrito-0 and Isothidcyanato, Thiocyanato.

Question 9.
Predict the products in the followng reactions

OR
Two motes of compound (A) on treatment with a strong base gives two compounds (B) and (C). The compound (B) on dehydrogenation with Cu gives (A) white acidification of (C) gives carboxytic acid (D) having molecular formulo CH₂O₂. Identify (A) to (D). (3)
Answer:

Since (D) is a carboxylic acid with one carbon only, it is HCOOH. As it is obtained from (C) acidification, (C) COONa and (A) is HCHO which on treatment with strong base (NaOH) gives CH₃OH & HCOONa (Cannizaro’s reaction).

The reactions are:

Question 10.
How would you account for the following:
(A) Etectrophitic susbstitution ¡n case of aromatic amines takes place more readily than benzene.
(B) Ethanamide is a weaker base than ethonamine.
(C) It is difficult to prepare pure amines by ammonotysis of atkyl halides. (3)
OR
Give the structures of A, B and C in the following reactions:

Answer:
(A) Due to the strong activating effect (+1) of the – NH₂ group, aromatic amines undergo electrophilic substitution reactions readiLy than benzene.

(B) In case of acetamide due to resonance, the lone pair of electrons on the nitrogen atom is delocalized over keto group which decreases electron density hence less basic while in ethylamine due to +1 effect of ethyl group electron density increases on N-atom and hence basic character increases.

(C) By ammonolysis of alkyl halides, a mixture of primary, secondary and tertiary amines is formed. The separation of these amines is very diffcult.
Thus, it is very difficult to prepare pure amines by ammonolysis of alkyl halides.
OR

Question 11.
Answer the following questions:
(A) Benzaldehyde can be obtained from benzol chloride. Write reactions for obtaining benzol chloride and then benzaldehyde from it.
Answer:

Explanation:
It is the commercial method for preparing benzaldehyde. Benzal chloride can be obtained by photochlorination of toluene, i.e. chlorination of toluene in the presence of sunlight. Then, benzal chloride on heating with boiling water produces benzaldehyde.

(B) Name the electrophile produced in the reaction of benzene with benzoyt chloride in the presence of anhydrous AlCl₃. Name the reaction also.
Answer:
The electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃ is benzoylinium cation (C₆H₅CO⁺). The product formed in this reaction is benzophenone. This reaction is called Friedel Craft’s acylation or benzoylation reaction.

Explanation:

(C) Give the structure of the following compound – 4-Nitro Propiophenone (3)
Answer:
4-Nitro Propiophenone

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the following questions:
Hardly Schulze rule states that the precipitating effect of an ion on dispersed phase of opposite charge increases with the valency of the ion. The higher the valency of the flocculating ion, the greater is its precipitating power. Thus, for the precipitation of AS₂S₃ sol (-ve sol) the precipitating power of Al³⁺ > Ba²⁺ > Na and Na⁺ ion is of the orde Al³⁺ > Na²⁺ Na⁺. Similarly, for precipitating Fe(OH)₃ sol (+ ve sol) the precipitating power of [Fe(CN)₆]^3-, SO₄4^2- and Cl^– is of the order, [Fe(CN)₆]^3- > <SO₄2- > Cl^–. The minimum concentration of an electrolyte in million per litre required to cause precipitation of a sol in 2 hours is called flocculation value.

The smaller the flocculation value, the higher will be the coagulating power of the ion. The minimum mass of the protective colloid (lyophilic colloid) in milligrams that must be added to 10 ml. of a standard red gold sol so that no coagulation occurs when I ml. of 10% NaCl solution is rapidly added to it is called the gold number of the protective colloid.

(A) Arrange the following ions in the increasing order of their flocculation value:
Cl^–, P0₄^3-, S0₄^2-, [Fe(CN)₆]^4-
(B) What is gold number?
(C) State Hardy Schulze rule.
(D) (i) Out of AlCl₃ and KCl, which is more effective in causing coagulation of a negative sol and why?
(ii) What happens when an electrolyte is added to a hydrated ferric oxide sol in water and why?
OR
Explain the following:
(i) Deltas are formed when river and seawater meet.
(ii) Artificial rain is caused by spraying salt over clouds. (5)
Answer:
(A) [Fe(CN)₆]^4- < P0₄^3- < S0₄^2- < Cl^–

Explanation:
More the charge on flocculating ion, less the flocculation value.

(B) The gold number is defined as the minimum mass of the colloid in milligram that is added to the 10ml of red gold sol to protect it from the coagulation when 1 ml of 10% NaCl is added.

(C) Hardy Schulze rule states that the amount of electrolyte required for the coagulation of a definite amount of a colloidal solution is dependent on the valency of the coagulating ion. Coagulating ion is the ion which has the charge opposite to the charge of the colloidat particles).

(D) (i) AlCl₃, because Al³⁺ has higher charge than K⁺ ion. Higher the charge, more effective it will be for coagulation. Explanation: The greater the magnitude of the opposite charge, the higher the ability of a salt to coagulate the sol. Thus, trivalent salt AlCl₃ is more effective in causing the coagulation of a negatively charged sol than monovalent salt KCl.

(ii) The positively charged colloidal particles of Fe(OH)₃ get coagulated by the negatively charged ions provided by electrolyte. Explanation: When an electrolyte like KCl is added of Fe(OH)₃ sol, the positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl^– ions provided by KCl.
OR
(D) (i) River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition with the formation of deltas.

Explanation:
River water is the negatively charged colloidal solution of clay and sands, whereas sea water contains a number of electrolytes. At the meeting point of sea water and river water, the electrolytes present in sea water neutralize and coagulate the colloidal solution of clay and sand resulting in their deposition leading to the formation of delta.

(ii) Clouds are colloidal dispersion of water particles in air carrying some charge over them.
It is possible to cause artificial rain by throwing electrified sand or spraying a sol carrying charge opposite to the one on clouds from an aeroplane. The colloidal water particles present in the clouds will get neutralized and as result they will come closer and grow in size to form bigger water drops and ultimately cause artificial rain.

Students can access the CBSE Sample Papers for Class 12 Geography with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Geography Term 2 Set 5 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35
Roll No. ___________

General Instructions:

• Question paper is divided into 5 sections A, B, C, D & E
• In section A, question number 1 to 3 are Very Short Answer type questions. Attempt any 3 questions.
• In section B, question number 4 is Source based question.
• In section C, question number 5 & 6 are Short Answer type questions.
• In section D, question number 7 to 9 are Long Answer type questions.
• In section E, question number 10 is a Map based question.

Section – A
Very Short Answer Questions

Question 1.
Identify and explain any two factors affecting tourism. (2)
Or
Comment on the role of communication in today’s world.
Answer:
Two factors that affect tourism are

1. Demand Since the last century, the demand for holidays has increased rapidly. Improvements in the standard of living and increased leisure time, permit many more people to go on holidays for leisure.
2. Transport The opening-up of tourist areas has been aided by improvement in transport facilities. Travel is easier by car, with better road systems. More significant in recent years has been the expansion in air transport.

Or
The role of communication in today’s world is

1. Telecommunication has revolutionised communications because of the speed with which messages are sent. The time reduced is from weeks to minutes.
2. Radio and television through satellite communication also help to relay news, pictures, and telephone calls to vast audiences around the world and hence they are termed as mass media.
3. They are vital for advertising and entertainment.

Question 2.
State about state highways and district roads. (2)
Answer:
State Highways are constructed and maintained by state governments. They join the state . capitals with district headquarters and other important towns. These roads are connected to the National Highways. These constitute 4 per cent of total road length in the country. District Roads are the connecting link between District Headquarters and the other important nodes in the district. They account for 14 per cent of the total road length of the country.

Question 3.
Which are the most polluted stretches of the Ganga river and what are the sources contributing to the pollution?
Answer:
The most polluted stretches of the Ganga river are downstream of Kanpur, downstream of Varanasi and Farkka Barrage. The sources of pollution in the Ganga River are industrial pollution from towns like Kanpur, domestic wastes from urban centres and dumping of carcasses in the river. Cities of Kanpur, Allahabad, Varanasi, Patna and Kolkata release domestic waste into the river flowing through Uttar Pradesh, Bihar and West Bengal in India.

Section – B
Source Based Question

Question 4.
Read the source given below and answer the following questions by choosing the correct option.
Bauxite is the ore, which is used in manufacturing of aluminium. Bauxite is found mainly in tertiary deposits and is associated with laterite rocks occurring extensively either on the plateau or hill ranges of peninsular India and also in the coastal tracts of the country.

Odisha happens to be the largest producer of Bauxite. Kalahandi and Sambalpur are the leading producers. The other two areas which have been increasing their production are Bolangir and Koraput. The patlands of Lohardaga in Jharkhand have rich deposits. Gujarat, Chhattisgarh, Madhya Pradesh and Maharashtra are other major producers.

Bhavanagar, and Jamnagar in Gujarat have the major deposits. Chhattisgarh has bauxite deposits in Amarkantak Plateau while Katni Jabalpur area and Balaghat in M.P. have important deposits of bauxite. Kolaba, Thane, Ratnagiri, Satara, Pune and Kolhapur in Maharashtra are important producers. Tamil Nadu, Kamataka and Goa are minor producers of bauxite. Copper is an indispensable metal in the electrical industry for making wires, electric motors, transformers and generators.

It is alloyable, malleable and ductile. It is also mixed with gold to provide strength to jewellery. The Copper deposits mainly occur in Singhbhum district in Jharkhand, Balaghat district in Madhya Pradesh and Jhunjhunu and Alwar districts in Rajasthan. Minor producers of Copper are Agnigundala in Guntur District (Andhra Pradesh), Chitradurg and Hasan districts (Karnataka) and South Arcot district (Tamil Nadu).

i. State important features of copper. (1)
Answer:
Important features of copper are Copper is alloyable and melleable i.e. can be converted into thin sheets. . It is ductile i.e. can be drawn into thin wires.

ii. In which locations, bauxite is found? (1)
Answer:
Bauxite is found both on high elevations and near the shore. For example, Odisha, Jharkhand, Gujarat, Chhattisgarh and so on.

iii. The Balaghat district in Madhya Pradesh, and Jhunjhunu and Alwar districts in Rajasthan in India have deposits of which metal? (1)
Answer:
The Balaghat district in Madhya Pradesh, and Jhunjhunu and Alwar districts in Rahasthan in India have deposits of copper metal.

Section – C
Short Answer Questions

Question 5.
Outline the distribution of iron and steel industry across the world. Or Identify the three subsectors of the cotton textile industry. Outline the distribution of cotton textile industry in the world. (3)
Answer:
The iron and steel industry is one of the most complex and capital-intensive industries and is concentrated in the advanced countries of North America, Europe and Asia. The distrubution of iron and steel industry across the world is

In USA, most of the production comes from the north Appalachian region (Pittsburgh), Great Lake region (Chicago-Gary, Erie, Cleveland, Lorain, Buffalo and Duluth) and the Atlantic Coast (Sparrows Point and Morisville). The industry has also moved towards the southern state of Alabama. Pittsburg area is now losing ground. It has now become the “rust bowl” of U.S.A.

In Europe, UK, Germany, France, Belgium, Luxembourgh, the Netherlands and Russia are the leading producers. The important steel centres are Scun Thorpe, Port Talbot, Birmingham and Sheffield in the U.K; Duisburg, Dortmund, Dusseldorf and Essen in Germany; Le Creusot and St. Ettienne in France; and Moscow, St. Petersburgh, Lipetsk, Tula, in Russia and Krivoi Rog, and Donetsk in Ukraine.

In Asia, the important centres include Nagasaki and Tokyo-Yokohama in Japan; Shanghai, Tienstin and Wuhan in China; and Jamshedpur, Kulti-Burnpur, Durgapur, Rourkela, Bhilai, Bokaro, Salem, Visakhapatnam and Bhadravati in India.
Or
Cotton textile industry has three sub-sectors i.e., handloom, powerloom and mill sectors

1. Handloom sector is labour-intensive and provides employment to semi-skilled workers. It requires small capital investment. This sector involves spinning, weaving and finishing of the fabrics.
2. The powerloom sector introduces machines and becomes less labour intensive and the volume of production increases.
3. Cotton textile mill sector is highly capital intensive and produces fine clothes in bulk Cotton textile manufacturing requires good quality cotton as raw material.
4. India, China, USA, Pakistan, Uzbekistan, Egypt produces more than half of the world’s raw cotton.
5. The UK, NW European countries and Japan also produce cotton textile made from imported yarn Europe alone accounts for nearly half of the world’s cotton imports.

Question 6.
Briefly outline the geographical location of the Bharmaur region of Himachal Pradesh. (3)
Answer:
Bharmaur is a tehsil in Chamba district of Himachal Pradesh. It is a notified tribal area since 21st November, 1975. The geographical location and features of the Bharmaur region are

• This region lies between 32° 11′ N and 32°41′ N latitudes and 76° 22′ E and 76° 53’E longitudes.
• It is spread over an area of about 1,818 sq km, the region mostly lies between 1,500 m to 3,700 m above the mean sea level.
• This region popularly known as the homeland of Gaddis is surrounded by lofty mountains on all sides. It has Pir Panjal in the north and Dhaula Dhar in the south.
• In the east, the extension of Dhaula Dhar converges with Pir Panjal near Rohtang Pass.
• The river Ravi and its tributaries, the Budhil and the Tundahen, drain this territory, and carve out deep gorges. These rivers divide the region into four physiographic divisions called Holi, Khani, Kugti and Tundah areas.
• Bharmaur experiences freezing weather conditions and snowfall in winter. Its mean monthly temperature in January remains 4°C and in July 26°C.

Section – D
Long Answer Questions

Question 7.
Industries are classified based on their sizes depending on the investme and output. Explain the basis of size based classification in industries. (5)
Or
Raw material is critical for functioning of an industry. How are industries classified based on raw materials?
Answer:
The size of an industry is determined by the capital invested, number of workers employed and volume of production determine the size of industry. Accordingly, industries may be classified into household or cottage, small-scale and large-scale on the basis of size

Household or Cottage Industries It is the smallest manufacturing unit. The artisans use local raw materials and simple tools to produce everyday goods in their homes with the help of their family members or part-time labour. Finished products may be for consumption in the same household or, for sale in local (village) markets, or, for barter.

Some common everyday products produced in this sector of manufacturing include foodstuffs, fabrics, mats, containers, tools, furniture, shoes, and figurines from wood lot and forest, shoes, thongs and other articles from leather, pottery and bricks from clays and stones.

Small Scale Manufacturing It is distinguished from household industries by its production techniques and place of manufacture (a workshop outside the home/cottage of the producer). This type of manufacturing uses local raw material, simple power-driven machines and semi-skilled labour. It provides employment and raises local purchasing power. Therefore, countries like India, China, Indonesia and Brazil, etc. have developed labour-intensive small scale manufacturing in order to provide employment to their population.

Large Scale Manufacturing It involves a large market, various raw materials, enormous energy, specialised workers, advanced technology, assembly-line mass production and large capital. This kind of manufacturing developed in the last 200 years, in the United Kingdom, NorthEastern USA and Europe. Now it has diffused to almost all over the world. The industries can be classified on the basis of raw material as

Agro Based Industries Agro processing involves the processing of raw materials from the field and the farm into finished products for rural and urban markets. Major Agro-processing industries are food processing, sugar, pickles, fruits juices, beverages (tea, coffee and cocoa), spices and oils fats and textiles (cotton, jute, silk), rubber, etc.

Mineral Based Industries These industries use minerals as a raw material. Some industries use ferrous metallic minerals which contain ferrous (iron), such as iron and steel industries but some use non-ferrous metallic minerals, such as aluminium, copper and jewellery industries. Many industries use non-metallic minerals such as cement and pottery industries.

Chemical Based Industries Such industries use natural chemical minerals, e.g. mineral-oil (petroleum) is used in petrochemical industry. Salts, sulphur and potash industries also use natural minerals. Synthetic fibre, plastic, etc. are other examples of chemical based industries.

Forest Based Industries Forests provide many major and minor products which are used as raw material. Timber for furniture industry, wood, bamboo and grass for paper industry, lac for lac industries come from forests.

Animal Based Industries Leather for leather industry and wool for woollen textiles are obtained from animals. Besides, ivory is also obtained from elephant’s tusks.

Question 8.
Explain the concept of Trans-continental Railways with any two relevant and well known examples from world. (5)
Answer:
Trans-continental railways run across the continent and link its two ends. They were constructed for economic and political reasons to facilitate long runs in different directions. Since, trans-continental railways run across a continent, they cover large land mass and connects interior parts of the continent to major industrial and transportation hubs on the continent borders and ocean shores. These railways forms the backbone of many countries as they move a large amount of freight and passenger across large distances. Two of the well-known trans-continental railways in world are

(i) Trans-Siberian Railway This is a major rail route of Russia which runs from St. Petersburg in the west to Vladivostok on the Pacific Coast in the east passing through Moscow, Ufa, Novosibirsk, Irkutsk, Chita and Khabarovsk. It is the most important route in Asia and the longest (9,332 km) tracked and electrified trans-continental railway in the world.

It has helped in opening up its Asian region to West European markets. It runs across the Ural Mountains Ob and Yenisei rivers. There are connecting links to the south, namely, to Odessa (Ukraine), Baku on the Caspian Coast, Tashkent (Uzbekistan), Ulan Bator (Mongolia), and Shenyang (Mukden) and Beijing in China.

(ii) Trans-Canadian Railways This 7,050 km long rail-line in Canada runs from Halifax in the east to Vancouver on the Pacific Coast passing through Montreal, Ottawa, Winnipeg and Calgary. It was constructed in 1886, initially as part of an agreement to make British Columbia on the west coast join the Federation of States.

Later on, it gained economic significance because it connected the Quebec-Montreal Industrial Region with the wheat belt of the Prairie Region and the Coniferous Forest region in the north. Thus, each of these regions became complementary to the other. A loop line from Winnipeg to Thunder Bay (Lake Superior) connects this rail-line with one of the important waterways of the world. This line is the economic artery of Canada. Wheat and meat are the important exports on this route.

Question 9.
Waterways have been used for transport of people and cargo since medieval times. Elucidate the advantages and types of waterways transport with examples. (5)
Answer:
Water transport is an essential form of transport. It is associated with waterbodies such as lakes, rivers, seas and oceans. The advantages of waterways are

Transport through waterways doesn’t require construction of a new route as an already existing water body is used for the transport. Movement through water offers less friction compared to roads and hence is more energy efficient.

Compared to land and air, ocean transport is a cheaper means of haulage (carrying of load) of bulky material over long distances from one continent to another. Water transport is broadly divided into sea routes and inland waterways

Sea routes The oceans offer a smooth highway traversable in all directions with minimal maintenance costs. Some of the important sea routes are the Northern Atlantic sea route linking USA to Europe and the Mediterranean-Indian ocean route linking regions along the South China Sea, Strait of Malacca, Indian Ocean, Red Sea, Suez Canal and Mediterranean.

Inland Waterways Rivers, canals, lakes and coastal areas are important inland waterways. The development of inland waterways is dependent on the navigability width and depth of the channel, continuity in the water flow, and transport technology in use. Important waterways are the Rhine Waterways through Germany and the Netherlands. It is navigable for 700 km; and Volga Waterway in Russia with a navigable waterway of 11,200 km.

Section – E
Map Based Question

Question 10.
On the political map of the world, identify the places (Attempt any 5). (1 × 5 = 5)
A. A place of secondary activities
B. A major sea port
C. A major sea port
D. An international airport
E. A major airport
F. The Terminal Station of Trans-Continental Railway

Answer:

Students can access the CBSE Sample Papers for Class 12 Geography with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Geography Term 2 Set 4 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35
Roll No. ___________

General Instructions:

• Question paper is divided into 5 sections A, B, C, D & E
• In section A, question number 1 to 3 are Very Short Answer type questions. Attempt any 3 questions.
• In section B, question number 4 is Source based question.
• In section C, question number 5 & 6 are Short Answer type questions.
• In section D, question number 7 to 9 are Long Answer type questions.
• In section E, question number 10 is a Map based question.

Section – A
Very Short Answer Questions

Question 1.
Differentiate between the characteristics of large scale manufacturing and small scale manufacturing (2)
Or
Differentiate between private sector industries and joint sector industries.
Answer:
The large scale manufacturing is characterised by a number of supporting factors like a large market, various raw materials, enormous use of energy, specialised workers, advanced technology, assembly line mass production and large capital investments. On the other hand, the characteristics of small scale manufacturing are use of local raw material, simple power driven machines and semi-skilled labour.
Or
Private sector industries are owned by individual investors. These are managed by private organisations. On the other hand, Joint sector industries are managed by joint stock companies or sometimes the private and public sectors together establish and manage the industries.

Question 2.
How has the Border Road Organisation (BRO) contributed to road transport in India? (2)
Answer:
Border Road Organisation (BRO) has contributed to road transport in India in many ways, some of which are . It has constructed roads in high altitude mountainous terrain joining Chandigarh with Manali (Himachal Pradesh) and Leh (Ladakh). This road runs at an average altitude of 4,270 metres above the mean sea level. Apart from the construction and maintenance of roads in strategically sensitive areas, the BRO also undertakes snow clearance in high altitude areas.

Question 3.
State in brief the types of pollutants involved in water pollution. Also mention sources of pollution for this. (2)
Answer:
In water pollution, the types of pollutants involved are dissolved and suspended solids, ammonia and urea, nitrate and nitrites, chloride, fluoride, carbonates, oil and grease, insecticide and pesticide residue, tannin, coliform MPM (bacterial count) sulphates and sulphides, heavy metals like lead, arsenic, mercury and manganese, radioactive substances, etc. Some of the major sources of water pollution are sewage disposal, urban run-off, toxic effluents from industries, run-off over cultivated lands and nuclear power plants.

Section – B
Source Based Question

Question 4.
Read the source given below and answer the following questions by choosing the correct option.
Railways are a mode of land transport for bulky goods and passengers over long distances. The railway gauges vary in different countries and are roughly classified as broad (more than 1.5 m), standard (1.44 m), metre gauge (1 m) and smaller gauges.

The standard gauge is used in the UK Commuter trains are very popular in UK, USA, Japan and India. These carry millions of passengers daily to and from in the city. There are about 13 lakh km of railways open for traffic in the world. Europe has one of the most dense rail networks in the world. There are about 4,40,000 km of railways, most of which is double or multiple-tracked.

Belgium has the highest density of 1 km of railway for every 6.5 sq. kms area. The industrial regions exhibit some of the highest densities in the world. The important rail heads are London, Paris, Brussels, Milan, Berlin and Warsaw. Passenger transport is more important than freight in many of these countries. Underground railways are important in London and Paris. Channel Tunnel, operated by Euro Tunnel Group through England, connects London with Paris.

Trans-continental railway lines have now lost their importance to quicker and more flexible transport systems of airways and roadways. In Russia, railways account for about 90 per cent of the country’s total transport with a very dense network west of the Urals. Moscow is the most important rail head with major lines radiating to different parts of the country’s vast geographical area.

Underground railways and commuter trains are also important in Moscow. North America has one of the most extensive rail networks accounting for nearly 40 per cent of the world’s total. In contrast to many European countries, the railways are used more for long-distance bulky freight like ores, grains, timber and machinery than for passengers. The most dense rail network is found in the highly industrialised and urbanised region of East Central USA and adjoining Canada.

i. The West-East Australian railway line runs across the country in which regions? (1)
Answer:
The West-East Australian railway line runs across the country from Perth to Sydney.

ii. Which continent has the largest network of railways in the world? (1)
Answer:
Asia has the largest network of railways in the world.

iii. What is the common form of public transport in London and Moscow? (1)
Answer:
A common form of public transport in London and Moscow is underground railways.

Section – C
Short Answer Questions

Question 5.
Differentiate between rural marketing centres and urban marketing centres. (3)
Answer:
Differences between rural marketing centres and urban marketing centres are

Rural Marketing	Urban Marketing
Rural marketing centres cater to nearby settlements. These are quasi-urban centres They serve as trading centres of the most rudimentary type.	Urban marketing centres have more  widely specialised urban services. They provide ordinary goods and services as well as many of the specialised goods and services required by people.
In these centres, personal and professional services are not well-developed. These form local collecting and distributing centres. Most of these have mandis (wholesale markets) and also retailing areas.	Urban centres, therefore, offer manufactured goods as well as many specialised markets develop, e.g.,  markets for labour, housing, semi or finished products.
They are not actually urban centres but are significant centres for making available goods and services which are most frequently demanded by rural folk.	Services of educational institutions and professionals such as teachers, lawyers, consultants, physicians, dentists  and veterinary doctors are available.

Question 6.
Outline the distribution of sources of bauxite in India and mention any one use of this mineral. Classify minerals on the basis of chemical and physical properties. (3)
Answer:
Bauxite is found mainly in tertiary deposits and is associated with laterite rocks occurring extensively either on the plateau or hill ranges of peninsular India and also in the coastal tracts of the country. The distribution of bauxite in India is

→ Odisha happens to be the largest producer of Bauxite. Kalahandi and Sambalpur are the leading producers. The other two areas which have been increasing their production are Bolangir and Koraput.

→ The patlands of Lohardaga in Jharkhand have rich deposits of bauxite.

→ Gujarat, Chhattisgarh, Madhya Pradesh and Maharashtra are other major producers. Bhavanagar, and Jamnagar in Gujarat have the major deposits. Chhattisgarh has bauxite deposits in Amarkantak Plateau while Katni Jabalpur area and Balaghat in M.P. have important deposits of bauxite. Kolaba, Thane, Ratnagiri, Satara, Pune and Kolhapur in Maharashtra are important producers.

→ Tamil Nadu, Karnataka and Goa are minor producers of bauxite. Bauxite is the ore, which is used in manufacturing of aluminium.
Or
On the basis of chemical and physical properties, minerals may be grouped under two main categories of metallics and non-metallics. Metallic minerals are the sources of metals. Iron ore, copper, gold produce metal and are included in this category. Metallic minerals are further divided into ferrous and non-ferrous metallic minerals. Ferrous, as you know, refers to iron.

All those minerals which have iron content are ferrous such as iron ore itself and those which do not have iron content are non-ferrous such as copper, bauxite, etc. Non-metallic minerals are either organic in origin such as fossil fuels also known as mineral fuels which are derived from the buried animal and plant life such as coal and petroleum. Other type of non-metallic minerals are inorganic in origin such as mica, limestone and graphite, etc.

Section – D
Long Answer Questions

Question 7.
What are the characteristics of modern large scale manufacturing? (5)
Or
What are traditional large scale industrial regions? Discuss the example of Ruhr coal field located in Germany.
Answer:
Modern large scale manufacturing has the following characteristics

→ Specialisation of Skills/Methods of Production Under the ‘craft’ method factories produce only a few pieces which are made-to-order. So, the costs are high. On the other hand, mass production involves production of large quantities of standardised parts by each worker performing only one task repeatedly and thereby reducing the overall cost.

→ Mechanisation It refers to using gadgets which accomplish tasks. Automation (without aid of human thinking during the manufacturing process) is the advanced stage of mechanisation. Automatic factories with feedback and closed loop computer control systems where machines are developed to think’, have sprung up all over the world. It reduces error due to human negligence, increase output due to use of machines and can work at any time due to automated instructions and minimal supervision. Overall, mechanisation of manufacturing process reduces error and increase output in a given period of time.

→ Technological Innovation Innovations through research and development strategy are an important aspect of modern manufacturing for quality control, eliminating waste and inefficiency, and combating pollution. With technological advancements, the machines and automated processes are regularly being upgraded, to reduce human interventions.
Or
Traditional Large-Scale Industrial Regions are based on heavy industry, often located near coal-fields and engaged in metal smelting, heavy engineering, chemical manufacture or textile production. These industries are now known as smokestack industries.

Traditional industrial regions can be recognised by –

• High proportion of employment in manufacturing industry. High-density housing, often of inferior type, and poor services. Unattractive environment, for example, pollution, waste heaps, and so on.
• Problems of unemployment, emigration and derelict land areas caused by closure of factories because of a worldwide fall in demand.

The Ruhr coal field is located in the East of Germany and represents one of the largest coal reserves in the country. This coal field is estimated to have over 42 billion tonnes of coal reserves. This has been one of the major industrial regions of Europe for a long time. Coal, iron and steel formed on the basis of the economy, but as the demand for coal declined, the industry started shrinking.

Even after the iron ore was exhausted, the industry remained, using imported ore brought by waterways to the Ruhr. The Ruhr region is responsible for 80 per cent of Germany’s total steel production Changes in the industrial structure have led to the decay of some areas, and there are problems of industrial waste and pollution.

The future prosperity of the Ruhr is based less on the products of coal and steel, for which it was initially famous, and more on the new industries like the huge Opel car assembly plant, new chemical plants, universities.

Question 8.
Discuss the distribution of minerals in India. (5)
Answer:
Most of the metallic minerals in India occur in the peninsular plateau region in the old crystalline rocks. Over 97 per cent of coal reserves occur in the valleys of Damodar, Sone, Mahanadi and Godavari. Petroleum reserves are located in the sedimentary basis of Assam, Gujarat and Mumbai High ie., off-shore region in the Arabian Sea. New reserves have been located in the Krishna-Godavari and Kaveri basins. Most of the major mineral resources occur to the east of a line linking Mangaluru and Kanpur. Minerals are generally concentrated in three broad belts in India. These belts are

1. The North-Eastern Plateau Region This belt covers Chhotanagpur (harkhand), Odisha Plateau, West Bengal and parts of Chhattisgarh
2. The South-Western Plateau Region This belt extends over Kamataka, Goa and contiguous Tamil Nadu uplands and Kerala. This belt is rich in ferrous metals and bauxite. It also contains high grade iron ore, manganese and limestone.
3. The North-Western Region This belt extends along Aravali in Rajasthan and part of Gujarat. Rajasthan is rich in building stones i.e. sandstone, granite, marble. Gypsum and Fuller’s earth deposits are also extensive.

Question 9.
Petroleum is an essential fuel for several industries in an economy. State its uses in an economy and production in India. (5)
Answer:
Petroleum is a conventional fossil fuel. Its uses are .

• It is an essential source of energy for all internal combustion engines in automobiles, railways and aircraft.
• Its numerous by-products are processed in petrochemical industries, such as fertiliser, synthetic rubber, synthetic fibre, medicines, vaseline, lubricants, wax, soap and cosmetics.

The production of crude petroleum is

• Digboi in Assam was the only oil producing region till 1956 but the scenario changed after 1956.
• In Assam, Digboi, Naharkatiya and Moran are important oil producing areas.
• The major oilfields of Gujarat are Ankaleshwar, Kalol, Mehsana, Nawagam, Kosamba and Lunej.
• Mumbai High which lies 160 km off Mumbai was discovered in 1973 and production commenced in 1976.
• Oil and natural gas have been found in exploratory wells in Krishna Godavari and Kaveri basin on the East coast.

Section – E
Map Based Question

Question 10.
On the political map of the world, identify the following features (Attempt any 5) (1 × 5 = 5)
A. A region of secondary activities
B. A major seaport
C. A secondary activity region in South-West USA
D. A major seaport
E. An important airport
F. The Terminal Station of Trans-Continental Railway

Answer:

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 3 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B, and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks.
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the chokes in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

SECTION – A
(Section A contains 3 questions of 2 marks each.)

Question 1.
Under what conditions does the phenomenon of total internal reflection take place? Draw a ray diagram showing how a rag of light deviates by 90° after passing through a right-angled isosceles prism.
OR
What are the (A) momentum, (B) de Broglie wavelength of an electron with kinetic energy of 120 eV? (2)
Answer:
The phenomenon of total internal reflection occurs when,
(A) Angle of incidence is equaL or greater than critical angle that is i ≥ C
(B) Angle of incidence is equal or greater than critical angle that is i ≥ C In case of right-angle isosceles triangle, if light rays fall normally on face AB the light ray incident of face AC with angle of incidence greater than the critical angle.

Hence, total Internat reflection will occur with normal to the surface of BC. The phenomenon in which a ray of light travelling at an angle of incidence greater than the critical angle from denser to a rarer medium is totally reflected back into the denser medium is called total internal reflection.
OR
Here, kinetic energy of electron
E = 120 eV
= 120 x 1.6 x 10^-19 J
= 192 x 10^-19J

(A) Momentum of electron,
P = \(\sqrt{2 m_{e} \mathrm{E}} \)
= \(\sqrt{2 \times 9.1 \times 10^{-31} \mathrm{~kg} \times 192 \times 10^{-19} \mathrm{~J}}\)
= 5.9 x 10^-24 Kg ms^-1

(B) de – Brogue wavelength,
P = \(\frac{h}{\lambda}\)
λ = \(\frac{h}{p}\)
= \(\frac{6.63 \times 10^{34} \mathrm{Js}}{5.9 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}}\)
= 1.12 x 10^-10
λ = 1.12 Å.

Question 2.
From the relation R = R_o A^1/3 where R_o is a constant and A is the mass number of a nucleus, show that the nuclear matter density is independent of A? (2)
Answer:
If m is the overage mass of a nucleon and R is the nucLear radius, then mass of nucteus = mA,
where A is the mass number of the element Volume of the nucleus
V = \(\frac{4}{3} \pi R^{3}\)
Given R = R₀ A^1/3
⇒ V= \(\frac{4}{3} \pi\left(R_{0} A^{1 / 3}\right)^{3}\)
⇒ V = \(\frac{4}{3} \pi R_{0}^{3} A\)

Density of nuclear matter,
ρ = \(\frac{m \mathrm{~A}}{\mathrm{~V}}\)
⇒ ρ = \(\frac{m \mathrm{~A}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}}\)
ρ = \(\frac{3 m}{4 \pi R_{0}^{3}}\)
This shows that nuclear density is independent of A.

Question 3.
The given graph shows the V-I characteristics of a semiconductor diode.

(A) Identify the semiconductor diode used.
(B) Draw the circuit diagram to obtain the given characteristics of this device.(2)
Answer:
(A) Since the voltages quoted in the given diagram are much more than actual values and It is usually < 0.6 V. The semiconductor the diode used is Zener diode.
(B) The circuit diagram for the characteristics of Zener diode is shown as: For forward bias Zener diode

SECTION – B
(Section B consists of 8 questions of 3 marks each.)

Question 4.
Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photo-electrons gets doubled when the wavelength of light incident on the surface changes from λ₁ to λ₂. Derive the expressions for the threshold wavelength λ₀ and work function for the metal surface. (3)
Answer:
Einstein’s photoelectric equation is given by
hv =Φ₀ + K_max
where K_max = Maximum kinetic energy of the photoelectron
v = Frequency of the Light radiation
h = Planck’s constant
Φ₀= Work function

We have.
K_max = \(\frac{1}{2} m v^{2}_{\max }\)
where, v_max = Maximum velocity of the emitted photoelectron
m = Mass of the photoelectron
Therefore.
hv =Φ₀= \(\frac{1}{2} m v^{2}_{\max }\)
If v₀ is the thresho[d frequencj, then the work function can be written as
Φ₀ = hv

Hence,
hv = hv₀ + \(\frac{1}{2} m v^{2}_{\max }\)
hv – hv₀ = \(\frac{1}{2} m v^{2}_{\max }\)
\(\frac{1}{2} m v^{2}\) = h (v -v₀)

The above equations explains the following results:
(i) If v < v₀, then the maximum kinetic energy is negative, which is impossible.
Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if v > v₀.

(ii) The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectrons depends only on the frequency of incident light not on the intensity According to the photoelectric equation.
K₁ = \(\frac{1}{2}\) mv²
= hv – Φ₀
Let the maximum kinetic energy for the wavelength λ₂
K₁ = \(\frac{h c}{\lambda_{1}}-\phi_{0}\) ……………………… (i)
Let the maximum kinetic energy for the waveLength of the incident λ₂ be K₂
K₂ = \(\frac{h c}{\lambda_{2}}\) – Φ₀ …………………. (ii)
(Kmax)2 = (Kmax)1 (given)

Work function is the energy required to eject o photoeLectron from the metal.
Φ₀ = \(\frac{h c}{\lambda_{0}} [latex]
∴ Φ₀ = [latex]\frac{h c\left(2 \lambda_{2}-\lambda_{1}\right)}{\lambda_{1} \lambda_{2}}\).

Question 5.
Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity. (3)
Answer:
A compound microscope consists of two convex parallel Lenses separated by some distance. The Lens nearer to the object is called the objective lens. The lens through which the final image is viewed is called the eyepiece or eye lens.

The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.
M = M_e X M₀

Where M_e and M₀ are the magnifying powers of the eyepiece and objective respectively. If u₀ is the distance of the object from the objective and v₀ is the distance of the image from the objective, then the magnifying power of the objective is:
M₀ = \(\frac{h^{\prime}}{h}=\frac{L}{f_{0}}\)

Where, h, h’ are object and image heights respectively and f₀ is the focal length of the objective and L is the tube length Le., the distance between the second focal point of the objective and the first focal point of the eyepiece.
When the final image is at infinity,
M_e = \(\frac{D}{f_{e}}\)
Magnifying power of compound microscope.
M = M₀ x M_e
= \(\frac{L}{f_{0}} \times \frac{D}{f_{e}}\)

Question 6.
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double-slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
OR
In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when an a-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverse its direction.
How will the distance of closest approach be affected when the kinetic energy of the a particle is doubled? (3)
Answer:
Given:
λ₁ = 800 nm
=800x 10^-9m
= 600 nm
λ₂ =600 x 10^-9m
D=14m
d = 0.28 mm
= 0.28 x 10^-3m
Suppose n₁th, maximum corresponds to wavelength λ₁ and it coincides with n₂th maximum corresponding to wavelength λ₂.
∴ \(n_{1} \frac{\lambda_{1} D}{d}=n_{2} \frac{\lambda_{2} D}{d}\)
Thus, 3^rd maximum corresponding to wavelength 800 nm coincides with 4^th maximum corresponding to wavelength 600
nm.
And the minimum distance is given by,
X_min = n₁\(\frac{\lambda_{1} D}{d}\)
= \(\frac{3 \times 800 \times 10^{-9} \times 1.4}{0.28 \times 10^{-3}}\)
X_min = 12 mm
OR
Let r₀ be the centre to centre distance between the α-particle and nucleus when the α-particle is at its stopping point.
Now, E_k = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{r_{0}}\)
r₀ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{E_{K}}\)
Given, E_k = 5 x 10⁶ eV
= 5 x 10⁶ x 1.6 x 10^-19 V
Z = 75
r₀ = \(\frac{9 \times 10^{9} \times 75 \times 2 \times\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{-19}}\)
= \(\frac{3456 \times 10^{9} \times 10^{-38}}{8 \times 10^{-13}}\)
= 432 x 10^-16 m
= 43.2 x 10^-15 m
= 43.2 fm
Since, distance of cLosest approach (r₀) is given
r₀ =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{E_{K}}\)
⇒ r₀ ∝ \(\frac{1}{E_{K}}\)
So, when kinetic energy of α-particle is doubled the distance between closest approach r₀ is halved.

Related Theory:
kinetic energy of’ α-particle is inverseLy proportional the distance of closest approach.

Question 7.
(A) What type of wavefront will emerge from a point source and a distant light source?
(B) Make a labelled diagram showing the wavefronts in reflection from:
(i) Plane mirror
(ii) Curved mirror (3)
Answer:
(A) From o point source, the wavefront is diverging spherical wavefront and from a distant Light source, the wavefront is plane front.
(B) (i) Reflection from plane mirror:

(ii) Reflection from curved mirror

Question 8.
Distinguish on the basis of features between conductors, insulators and semiconductors. (3)
Answer:

Question 9.
Describe the role of processes involved in the formation of p-n junction with the help of a diagram. (3)
Answer:
Two process occur during the formation of p-n junction are diffusion and drift.
(i) Diffusion: In n-type semiconductors, the concentration of electrons is much greater as compared to concentration of holes arid in p-type semiconductors, the concentration of holes is much greater than concentration of electrons. When p-n junction is formed, concentration gradient is set up. The holes diffuse n-side and electrons dislikes the top-side. This motion of charge carriers gives rise to diffusion current across the junction.

(ii) Drift: The drift charge carriers occur due to the electric field. As potential barrier is built, on electric field is directed from n-side to p-side of the junction. This field causes motion of electrons on p-side of the junction to n-side of the junction and motion of holes on n-side to p-side of the junction. This is known as drift current which is opposite direction of diffusion current

Question 10.
(A) Can a charge moving with a constant velocity act as a source of electromagnetic waves?
(B) In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 2 x 10¹⁰Hz and amplitude 48 vm’1.
(i) What is the wavelength of the wave?
(ii) What is the amplitude of the oscillating magnetic field? (3)
Answer:
(A) A charge moving with a constant velocity has zero acceleration and according to the classical theory of electromagnetism only accelerated charge can produce electromagnetic wave. Therefore, a charge moving with a constant velocity cannot produce electromagnetic wave. Frequency of the electromagnetic wave,
v=2.0x 1010Hz
Electric field amplitude, E₀ = 48 Vm-1
Speed of light, c = 3 x 108 m/s
E₀ = 48 \(\frac{\mathrm{V}}{m}\)
and c=3 x 10⁸ms^-1
(i) λ = \(\frac{c}{v}\)
= \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{2 \times 10^{10} \mathrm{~Hz}}\)
= 1.5 x 10^-2 m
(ii) c = \(\frac{E_{0}}{B_{0}}\)
B₀ = \(\frac{E_{0}}{c}\)
= \(\frac{48 \frac{\mathrm{V}}{m}}{3 \times 10^{8} \mathrm{~ms}^{-1}}\)
= 1.6 x 10^-7 T

Question 11.
Give reason for following observations in YDSE.
(A) The resultant intensity at any point on the screen varies between 0 and 4 times the intensity, due to one slit.
(B) A few coloured fringes, around a central white region, are observed on the screen, find the source of monochromatic light is replaced by white light.
(C) Why no interference is seen when two coherent sources are infinitely close to each other or when they are far apart from each other?
OR
Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.(3)
Answer:
(A) The resuLtant intensity, at any point on the screen is given by.
I = 4I₀, cos²\(\frac{\phi_{0}}{2}\)
For constructive interference;
Φ₀= O, 2π. 4π and so on
I = O for minimum intensity
For destructive interference:
Φ₀ = π, 3π, 5π and so on
I = 4I₀ for maximum intensity.
Therefore, intensity varies from zero to four times the intensity due to single slit.

(B) The interference patterns due to different colours of white light overtop incoherently.
The central bright fringes for different colours are at the same position. Therefore the central fringe is white and the fringes closest, on either side of central white fringe, are red and farthest wilt appear blue. After a few fringes, no clear fringe pattern is seen.

(C) When two coherent sources are placed infinitely close to each other, the fringe width becomes very large. Even a single fringe width may occupy the entire screen. When the sources are far apart, the fringe width increases. At a very large separation. it becomes too small to be detected. Hence, the interference pattern cannot be detected when the sources are infinitely close to each other or kept far apart from each other.
OR
A plot of binding energy per nucleon moss number is shown in the figure given below

(A) When we move from the heavy nuclei region to the middle region of the plot, we find that there will be again in the overall binding energy and hence results in release of energy. This indicates that energy can be released when a heavy nucleus (A ~ 240) breaks into two roughly equal fragments.
This process is called nuclear fission.

(B) Similarly. when we move from lighter nuclei to heavier nuclei, we again find that there will be again in the overall binding energy and hence release of energy takes place. This indicates that energy can be released when two or more Lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

SECTION – C
(Section C consists one case study-based question of 5 marks.)

Question 12.
Dispersion is the splitting of white light into its constituent colours. The dispersion occurs in prisms but not in glass slabs because of its geometric design. In a slab, the opposite sides are parallel to each other whereas in case of the prism, the sides are not parallel to each other.

In prism, the dispersion occurring at one face and is enhanced at the other end but in glass slabs, dispersion occurring at one end is neutralized by refraction at the other end. When we put two prisms in series but in opposite direction, they will act as a glass slab together.

Based on the above facts, answer the following questions:
(A) Two parallel rays of red and violet colour pass through a glass stab as shown In the figure. Which of the following is correct?

(i) 3 and 4 are parallel
(ii) 4 and 5 are parallel
(iii) 6 and 3 are parallel
(iv) 2 and 5 are parallel.
(B) A convex lens is made up of two types of material Upper half is made by the refractive index n₁ and Lower half is made by the refractive index n₂. Which of the following is true for rays coming from infinity?
(ï) Two images are formed
(ii) One image is formed
(iii) Continuous ¡mage is formed between the focal points of upper and tower tens.
(iv) Image is not formed

(C) Maximum Lateral displacement of ray Light incident on a stab of thickness t is:
(i) \(\frac{t}{2}\)
(ii) \(\frac{t}{3}\)
(iii) \(\frac{t}{4}\) (iv) t
(D) The refractive index of water, glass, and diamond are 1.33, 1.50, and 2.40, respectively. What is the refractive index of the diamond relative to water and of glass relative to diamond, respectively are nearly?
(i) 1.80, 1.6
(ii) 0.5 54, 1.6
(iii) 1.80, 0.625
(iv) 0.554, 0.625

(E) An air bubble in a glass slab with refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed
from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is:
(i) 8
(ii) 10
(iii) 12
(iv) 16 (5)
Answer:
(A) (d) 2 and 5 are parallel Explanation: When refraction occurs through a parallel glass slab, the emergent ray is parallel to incident ray. From the given diagram
6||2,5||1
Because 1 || 2
⇒ 2 || 5
(B) (a) Two images are formed
Explanation: The given lens is made of two materials of different refractive indices, so it has two focal lengths. Hence, two images are formed.

‘Related Theory
Due to the two materials of different refractive indices lens will form two images (C) (d)t
Explanation: For I = 90⁰ and r=0⁰, Lateral shift is maximum.
Lateral shift for a sLob of t thickness
LS = t x sin\(\frac{(i-r)}{\cos r}\)
LS = t
Hence, the Lateral shift is equaL to thickness t.

Caution:
Students are often confused about the values of i and r when lateral shift is maximum. in this case i = 90° and r= 0°
(D) (c) 1.80, 0.625
Explanation: n_w = 1.33,
n_g = 1.5
and n_d= 2.4
Now, _wn_d = \(\frac{n_{d}}{n_{w}}\)
= \(\frac{2.4}{1.33}\) = 1.80
Further, _dn_g = \(\frac{n_{g}}{n_{d}}\)
= \(\frac{1.50}{2.4}\) = 0.625

(E) (c)12
Explanation: Refractive index
n= \(\frac{\text { Real thickness }}{\text { Apparent”thickness }}\)
ReaL thickness = n x Apparent thickness
= 1.5 x (5 + 3) = 12 cm

Students can access the CBSE Sample Papers for Class 12 Economics with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Economics Term 2 Set 3 with Solutions

Time allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This is a Subjective Question Paper containing 13 questions.
• This paper contains 5 questions of 2 marks each, 5 questions of 3 marks each and 3 questions of 5 marks each.
• 2 marks questions are Short Answer Type Questions and are to be answered in 30-50 words.
• 3 marks questions are Short Answer Type Questions and are to be answered in 50-80 words.
• 5 marks questions are Long Answer Type Questions and are to be answered in 80-120 words.
• This question paper contains Case/Source Based Questions.

Question 1.
Distinguish between intermediate consumption and final consumption.
OR
Distinguish between domestic product and national product. (2)
Answer:
Intermediate Consumption refers to the expenditure incurred by a production unit on purchasing those goods and services from other production units which are meant for resale or for using up completely during the same year whereas Final consumption refers to the expenditure of those goods which are used either for consumption or for investment.
OR
Sum of the value of final products that take place within the domestic territory of a country is called domestic product, whereas the sum of contribution of residents of a country both within domestic territory or abroad is called national product.

Question 2.
If in an economy:
Change in initial investment (Δl) = ₹1,000 crores
Marginal propensity to save (MPS) = 0.2.
Find the value of the following:
(A) Investment multiplier (K)
(B) Change in final income (ΔY)
OR
Calculate change in final income, if Marginal Propensity to Consume (MPC) is 0 – 8 and change in initial investment is ₹1,000 crores. (2)
Answer:
(A) Investment multiplier (K) = 1/MPS
K = 10.2
K = 5
(B) Investment multiplier (K) = Δ Y/Δl
Substituting
K = 5
5 = ΔY/1000
ΔY = ₹5000 crores
OR
Investment Multiplier (K)

Change in final income = ₹5,000 crores

Question 3.
“An economy facing an unintended accumulation of inventories would try to reduce aggregate demand.’’ Do you agree with the given statement? Support your answer with valid reasons. (2)
Answer:
The given statement is not correct.
The situation of unintended accumulation of inventories arises when ex-ante aggregate demand is lesser than the ex-ante aggregate supply. This would pile up the stock with the producers, thus to tackle this situation the economy must increase aggregate demand (AD).

Question 4.
“Sustainable development is defined as an approach to developing or growing by using resources in a way that allows for them to renew or continue to exist for others.” In the light of this statement explain any one strategy of sustainable development. (2)
Answer:
Use of non-conventional sources of energy in India should be promoted. If we continue to depend upon the conventional sources like coal or petroleum products, we may end up with higher levels of pollution and an unsustainable environment. We should promote the use of non-conventional sources of energy like solar and wind power for a safer planet, to be passed on to the coming generations.

Question 5.
Discuss the need for on-the-job training for an employee.
OR
Analyze the distribution of employment by gender based in the following information:

Answer:
Employers use on-the-job training methods for its employees to target an overall increase in the skills and efficiencies of the workers. On-the-job training leads to an increase in productivity of labor and production of goods. Such training keep the employees updated with the latest changes in their field of working.

OR
Based on the given information, self-employment is a major source of livelihood for both men and women as this category accounts for more than 50 percent of the workforce in both diagrams. Casual wage work is the second major source for both men and women, a little more so for the latter (31 percent). When it comes to regular salaried employment, men are found to be so engaged in greater proportion. They form 20 percent whereas women form only 13 percent.

Question 6.
Giving reason state how are the following treated in estimation of national income.
(A) Payment of interest by an individual to a bank on a loan to buy a car.
(B) Expenditure by government on providing free educational services.
(C) Expenditure on purchasing a machine installed in a production unit.
OR
If in an economy the value of Net Factor Income from Abroad is ₹200 crores and the value of Factor Income to Abroad is ₹40 crores. Calculate the value of Factor Income from Abroad. (3)
Answer:
(A) Payment of interest to a bank by an individual is not included because the individual is a consumer.
(B) Expenditure by the government on free education is included because it is a final expenditure.
(C) Expenditure on machine installed in a production unit is included because it is an investment expenditure.
OR
Net Factor Income from Abroad = Factor Income from Abroad – Factor Income to Abroad ₹200 = Factor Income from Abroad – ₹40 Factor Income from Abroad = ₹200 + ₹40 = ₹240 crores

Question 7.
Study the following information and compare the Economies of India and Singapore on the grounds of ‘Investment in Infrastructure as a percentage of GDP’. (3)

Some Infrastructure in India and Other Countries, 2018

Sources: World Development Indicators 2019, World Bank website: www.worldbank.org.:
BP Statistical Review of World Energy 2019, 69th Edition

Note : (*) refers to Grass Capital Formation.

Read the following text carefully and answer questions number 8 and 9 given below:

India vs Pakistan: A tale of two economies
NEW DELHI : India and Pakistan – two of the biggest South Asian nations – started their economic journey around the same time after gaining Independence. Not many know that there was a point in the 1960s when Pakistan’s per capita GDP used to be higher than that of India’s. However, over the years, India has not only surpassed Pakistan’s per capita GDP but taken a commanding lead on almost every economic front.

Pakistan today is facing global criticism for failing to rein in homegrown terror groups even as it grapples with an ailing economy and mounting global debt. India, on the other hand, has become a global frontrunner which recently surpassed the UK to become the fifth-largest economy in the world. The stark economic contrast between the two neighbors shows that Pakistan’s unstable government, conflicting power centres, covert support to terror groups, and involvement in global terror strikes such as the 26/11 Mumbai attacks, have bled its economy.

From a higher per capita GDP of $83.33 in 1960, Pakistan’s today lags much behind India which witnessed a multi-fold raise its per capita GDP over the years. From 2007, India’s per capita GDP has consistently stayed higher than that of Pakistan’s. India is today the fastest-growing trillion-dollar economy in the world and the fifth-largest overall, according to data compiled by IMF’s World Economic Outlook.

It jumped from the 9th spot in 2010 to the 5th spot in a span of just 9 years. The GDP of India is almost 10 times that of Pakistan, which is placed at the 45th position.

India’s rise has been even more dramatic across the past couple of years. Since 2008, India’s GDP has risen almost 140 per cent as against an increase of about 63 percent for Pakistan.

https://timesofindia.indiatimes.com/ business/india-business/india-vs- pakistan-a-tale-of-two-economies/ articleshow/7945005 l.cms
Answer:
Though it is widely understood that infrastructure is the foundation of development, India is yet to wake up to the call. India invests only 30 percent of its GDP on infrastructure, which is far below that of China. So it can be concluded that India is investing in infrastructural facilities less than China and this gap is too large. Some economists have projected that India will become the third biggest economy in the world a few decades from now. Considering this fact, this is a relatively lower proportion in this direction. For that to happen, India will have to boost its infrastructure investment.

Question 8.
“India and Pakistan started their economic journey at the same time but now India is far ahead of Pakistan.” In the light of above statement, give any two reasons. (3)
Answer:
India is ahead of Pakistan as following have affected Pakistan’s economy:

(i) Unstable Government: In 1950, Pakistan’s per person GDP was US $1268, which was almost 50 per cent greater than India that year. However, in the backdrop of sustained political uncertainty and upheaval, Pakistan stagnated throughout the 1950s while a politically stable India grew. As a result, by 1960, India had almost caught up with Pakistan in per capita GDP terms with the per capita income gap having shrunk to 15 percent.

(ii) Homegrown Terror Groups or Ailing Economy: Pakistan’s economy has suffered a direct and indirect cost linked to terrorist activities of almost $126.79 billion, which is equal to ₹10762.14 billion. Pakistan today is facing global criticism for failing to rein in homegrown terror groups even as it grapples with an ailing economy and mounting global debt.

Question 9.
Compare and analyze India and Pakistan economies in the terms of GDP growth. (3)
Answer:
India has surpassed Pakistan’s per capita GDP over the years. From 2007, India’s per capita GDP has consistently stayed higher than that of Pakistan’s. India is today the fastest-growing trillion-dollar economy in the world and the fifth Largest overall, according to data compiled by IMF’s World Economic Outlook. It jumped from the 9th spot in 2010 to the 5th spot in a span of just 9 years. The GDP of India is almost 10 times that of Pakistan, which is pLaced at the 45th position. Since 2008, India’s GDP has risen almost 140 percent as against an increase of about 63 percent for Pakistan.

Question 10.
Suppose a ban is imposed on the consumption of tobacco. Examine its likely effects on:
(A) gross domestic product, and (B) welfare. (3)
Answer:
(A) Ban on the consumption of tobacco will bring down the production of tobacco. Since it is counted in GDP, GDP will fall.
(B) The ban will improve health in general. It will thus increase welfare.

Question 11.
“To boost the falling demand in the economy, Reserve Bank of India recently reduced Repo Rate and Reverse Repo Rate.” Elaborate the rationale behind the steps taken by the Central Bank. (5)
Answer:
The steps taken by the Central Bank to boost the falling demand in the economy are justified as the reduction in the Repo rate and Reverse Repo Rate will increase the availability of funds in the market through the commercial banks. Rationale: A decrease in Repo/Reverse Repo Rate will push the commercial banks to reduce the lending rate and will eventually make the borrowings cheaper for the general public. As a result, the consumption demand in the economy may increase.

Question 12.
(A) Define ‘net factor income from abroad’. How is it different from ‘net exports’?
(B) Calculate the value of “Rent” from the following data:

Particulars	Amount (In ₹ crores)
Gross Domestic Capital Formation	18,000
Mixed-Income of Self Employed	7,000
Subsidies	250
Interest	800
Rent	?
Profit	975
Compensation of Employees	6,000
Consumption of Fixed Capital	1,000
Indirect Tax	2,000

OR

(A) State any two precautions that must be taken into consideration while estimating national income by the value-added method. (5)

(B) In an economy, the following transactions took place. Calculate the value of output and value-added by Firm B: i. Firm A sold to firm B goods of 80 crores; to firm C 50 crore; to househoLd 30 crore and goods of vaLue, 10 crore remains unsold ii. Firm B sold to firm C goods of 70 crores; to firm D 40 crore; goods of vaLue 30 crores were exported and goods of value 5 crores were sold to the government. (5)
Answer:
(A) Net Factor Income from Abroad is the excess of factor incomes (rent, wages, interest, profit) earned from abroad over factor incomes (rent, wages, interest, profit) paid to abroad, whereas; net export refers to the excess of the value of exports over the value of imports of a country in an accounting year.

(B) GDP_MP = NDP_FC + Depreciation + Net indirect tax
(i) = (vii) + (ii)+ [(iv) + (vi) + Rent] + (viii) + [(ix)-(iii)]
18000 = (6000 + 7000 + (800 + 975 + Rent) + 1000 + (2000 – 250) 18000
= 17525 + Rent Rent
= ₹ 475 crore
OR
(A) Precautions of value added method are:

1. Value of sale and purchase of second hand goods is not considered while estimating value added as the value of second hand goods is already accounted during the year they were produced.
2. Value of intermediate goods is not included in the estimation of value added because value of intermediate goods is reflected in the value of final goods.

(B) Value of output of firm B = Sales of firm B to firm C + Sales of firm B to firm D + Exports + Sales of firm B to Government = 70 + 40 + 30 + 5 = 145 crores Value Added by Firm B = Value of output by Firm B – Purchases by Firm B from firm A = 145 – 80 = 65 crores

Question 13.
(A) “If the rate of resource extraction exceeds the rate of regeneration, it leads to a reduction in carrying capacity of the environment.” Discuss the rationale of the given statement with valid reasons.

(B) Over 60% of India’s arable and is estimated to suffer from environmental degradation. This has been caused by both by a rapidly growing poor population seeking subsistence and by the misappropriation of natural resources by the wealthy for Luxury consumption. In the Light of these Unes, state and discuss any two principaL causes of environmental degradation. (5)
Answer:
(A) The environment is able to perform its functions uninterruptedly so long as the demand for these functions is within the carrying capacity of the environment. This means that the resources are not extracted beyond the rate of their regeneration. If there is a disequilibrium (demand being more than supply), the environment fails to replenish itself and it will lead to environmental crisis. Thus, to maintain a healthy environment, the carrying capacity of the environment must be valued and respected.

(B) Two principal causes behind environmental degradation:

1. Population Rise: Rising population is one of the major causes for degradation of the environment as it adds to the burden on natural resources, leading to environmental crisis.
2. Consumption Levels: It has been observed that the developing and developed nations have witnessed affluent consumption levels in the past, bringing disequilibrium in the demand and supply of the natural resources, pushing the world to the threshold of the environmental crisis.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 6 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Name the reagents used in the following reactions: (Any two) (2)

Answer:
(A)
LiAlH₄

Explanation:

(B) KMnO₄/KOH, heat

Explanation:
When ethyl benzene is oxidised with alkaline KMnO₄ or acidic Na₂Cr₂O₇, the entire side chain (in benzene homologues) with at least one H at a carbon, regardless of length is oxidised to -COOH

(C) H₂/Pd – BaSO₄

Explanation:
Rosenmund reduction is a hydrogenation process in which an acyl chloride is selectively reduced to an aldehyde. The reaction was named after Karl Wilhelm Rosenmund who first reported it.

So, benzoyl chloride is reduced to benzaldehyde in presence of Pd-BaSO₄ and H₂.

Question 2.
Give reason
(A) Boiling point of methyl amine is 60°C while that of trimethyl amine is 30°c. State reason for this difference in boiling point.
Answer:
Trimethyl amine is a tertiary amines and it does not have hydrogen atoms bonded to the nitrogen atom and therefore are not hydrogen bond donors. Thus tertiary amines cannot form intermolecular hydrogen bonds. As a result., they have lower boiling points than primary amine like methyl amine.

(B) Why does Aniline not react with dilNaOH? (2)
Answer:
Aniline does not react with dil. NaOH because in basic conditions, the molecule is uncharged and only very slightly polar. Its polarity is so low that interactions between water and the NH₂^– group are not sufficient to make the molecule, souble in water.

Question 3.
Identify which of the following AX, AY or AZ indicates a strong .weak or a non electrolyte. (2)

Answer:
AX – fig (a) indicates a non electrolyte as the particles do not dissociate at all.
Examples:

AY -fig (b) indicates a Weak Electrolytes: as the particles are not completely dissociated into ions in solution.
Examples: CH₃COOH, NH₄OH, HCN etc.

AZ -fig (c) indicates a Strong Electrolytes: In these electrolytes particles completely dissociateinto ions in solution.
Examples: HCl, KCl, NaOH, NaCl

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
(A) Identify the Coordination complex which contain Cyanide as the ligand and Iron as the central metal atom. This complex is used to identify both Cu²⁺ and Fe³⁺ cation during salt analysis.
Answer:
The coordination complex is K4[Fe(CN)6]

Related Theory:
Both Cu²⁺ and Fe³⁺ are detected with potassium fer- rocyanide in the following manner

(B) While NH₂ is a strong ligand but NH₄⁺ is not. Explain.
Answer:
Ammonia is converted into ammonium ion in the following manner:
NH₃ + H⁺ ⇌ NH₄⁺

Complexes are formed by donation of a pair of electrons from ligand to metal. In ammonia, N atom has one lone pair of electrons. So it can form complexes. Nitrogen donates this lone pair of electrons to proton to form ammonium ion. NH₄⁺ ion does not possess any lone pair of electrons which it can donate to central metal ion hence it does not form complexes.

(C) Select bidentate of didentate ligand from the following? (3)
CO, SCN^–, CH₃COO^–, C₂O₄^2-
Answer:
Bidentate Ligand are the one which donates 2 pair of electrons to metal atom. Among the options provided, C2C>42_ is the only Ligand which is bidentate ligand. It donates electrons from its 2 oxygen atoms.

Question 5.
Answer the following question:
(A) Bleeding due to a small cut is stopped by applying alum.
Answer:
Alum is used to stop blood from a cut. It allows the blood clot to form. The high ionic strength promotes flocculation of the blood . Thus, blood is positive sol and sol particles are coagulated more efficiently by S042~ ions

(B) You must have observed at sunset an orange colour develops in the sky .Why is it so?
Answer:
At the time of sunset, the sun is at horizontal. The Light emitted by the sun has to travel a relatively longer distance through the atmosphere. As a result, blue part of light is scattered away by the particulate (colloidal particles)in the atmosphere causing red part to be visible.

(C) Why is Chemisorption called activated absorption? (3)
Answer:
Chemisorption is also called as activated adsorption because there is an involvement of chemical bond formation between the reactants and the adsorbent. This formation of chemical bonds requires a high activation energy.

Related Theory:
Chemisorption is highly specific in nature as the molecules are held on the solid surface through a chemical bond. For eg., oxygen is adsorbed on metals due to oxide formation and also hydrogen gets adsorbed on metals due to hydride formation

Question 6.
(A) Name the cell that drives a non-spontaneous redox reaction through the application of electrical energy with the help of a suitable example
Answer:
An electrolytic cell is an electrochemical cell that drives a nonspontaneous redox reaction through the application of electrical energy.

Example: The decomposition of water into hydrogen and oxygen, and bauxite into aluminum and other chemicals.

(B) How much electricity in Faraday is required to produce
(i) 80g of Ca from molten CaCl₂?
Answer:
From given data,
Ca²⁺ + 2e^– → Ca

Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce
80 g of calcium = \(\frac{2 \times 80}{40}\)F
= 4 F

(ii) 54g of Al from molten Al₂O₃?
OR
A voltaic cell is set up at 25°C with the following half cells:
\(\frac{\mathrm{Al}}{\mathrm{Al}^{3+}}\) (0.001M) and \(\frac{\mathrm{Ni}}{\mathrm{Ni}^{2+}}\) (0.50M).
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. (3)
Answer:
From given data
Al³⁺ + 3e^– → Al

Electricity required to produce 27 g of Al
= 3 F

Therefore, electricity required to produce 54g of Al = \(\frac{3 \times 54}{27}\)F
= 6 F
OR
E₀(Al³⁺/Al) = – 1.66V, E₀(Ni²⁺/Ni) = -0.25V

Anode: [Al → Al³⁺ + 3e^–] × 2
Cathode: [Ni²⁺ + 2e^– → Ni] × 3

Net cell reaction: 2Al +3Ni²⁺ → 2Al³⁺+ 3Ni
n = 6

Caution:
Students should not forget to put powers of stoichiometric coefficients on the concentrations of products and reactants in Nernst equation.

Question 7.
Account for the following:
(A) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequency and why?
Answer:
Copper exhibits + 1 oxidation state more frequently i.e., Cu⁺¹ because of its electronic configuration 3d¹⁰4s¹. It can easily lose 4s¹ electron to give stable 3d¹⁰ configuration.

(B) Which of the following cations are coloured in aqueous solutions and why? SC³⁺, V³⁺, Ti⁴⁺, Mn²⁺.
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer:
Sc³⁺ = 4S° 3d³ = no unpaired electron
V³⁺ = 3d² 4s° = 2 unpaired electron
Ti⁴⁺ = 3d° 4s° = no unpaired electron
Mn²⁺ = 3d⁵ 4s° = 5 unpaired electron
Thus V³⁺ and Mn²⁺ are coloured in their aqueous solution due to presence of unpaired electron.

(C) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points. In Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, the metallic bonds present in them are weak (3)
Answer:
The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points because of stronger metallic bonding and high enthalpies of atomization.

Question 8.
Answer the following questions:
(A) Give the product for:

Answer:
NaBH₄ does not reduce ester group. It is mild reducing reagent which reduces aldehyde and ketone into their respective alcohols.

(B) Which among the following will undergo cannizaro reaction? Benzaldehyde, acetone .formaldehyde
Answer:
Benzaldehyde and formaldehyde undergo cannizaro reaction as these compounds do not contain alpha hydrogen

Related Theory:
An alpha carbon is the first carbon that is joined to the functional group. In the case of aldehydes and ketones, a functional group is a carbonyl group. The functional group is responsible for the formation of alpha hydrogen.

(C) Iodoform reaction is obtained when methyl ketones react with hypoiodite not with iodine. Why? (3)
Answer:
Hypoiodite ion is very strong oxidising agent .hence it can oxidise methyl ketone to iodoform, iodide ion is a reducing agent and hence it cannot oxidise methyl ketone to iodoform.

Related Theory:
Reaction of Iodine with NaOH gives sodium hypoiodite

The Haloform Reaction;

Formation of carboxylic acids only happens with methyl ketones
Iodoform test (NaOH/I₂) is historically valuable for identifying methyl ketones

Question 9.
Write the IUPAC names of the foLlowing coordination compounds
(A) [Cr(NH₃)₃Cl₃]
(B) K₄[Fe(CN)₆]
(C) [CoBr₂(en)₂]⁺, (en = ethylenediamine)
OR
(A) Write the formulae for the following coordination compounds:
(i) Tetraammineaquachloridocobalt (III) chloride
(ii) Potassium tetracyanonickelate (II)
(B) Write the hybridization of the complex
[NiCl₄]^2-. (Atomic number of Ni = 28) (3)
Answer:
(A) Triamminetrichloridochromium (III)

(B) Potassiumhexacyanidooferrate (III)

(C) Dibromidobis (ethane 1, 2-diamine) cobalt (III)
OR
(A) (i) [Co(NH₃)₄(H₂O)Cl]Cl₂
(ii) K₂[Ni(CN)₄]

(B) Ni(28) : [Ar] 4s²3d⁸
Ni²⁺: [Ar] 4s°3d⁸
Since Cl^– is a weak field ligand, the pairing of electrons will not take place.

It has sp³ hybridization. It has a tetrahedral shape and complex is paramagnetic in nature due to presence of unpaired electrons.

Question 10.
Write the structures of the main product when acetone (CH₃—CO—CH₃) reacts with the following reagents.
(A) Zn—Hg/Conc. HCl
(B) H₂N—NHCONH₂/H⁺
(C) CH₃MgBr and then H₃O⁺
OR
An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid also produced (B). On dehydration (C) gives but-l-ene. Write the equations for the reactions involved. (3)
Answer:

Related Theory:
Clemmensen reduction is used to reduce aldehydes and ketones to hydrocarbons. Zinc amalgam and concentrated hydrochloric acid are used as Clemmensen’s reagent Clemmensen reduction is a reaction which involves the reduction of a carbonyl compound to form a simple hydrocarbon.

Related Theory:
Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide. Semicarbazones are useful for identification and characterization of aldehydes and ketones.

Explanation:
Organic compound A is an ester as on acid hydrolysis it gives a mixture of an acid and an alcohol.
Oxidation of alcohol (C) gives acid (B). Hence, the number of carbon atoms in (B) and (C) are same.
Ester (compound A) has eight C atoms. Hence, both carboxylic acid B and alcohol C must contain 4 C atoms each.
Dehydration of alcohol C gives but-l-ene. Hence, C must be a straight chain alcohol, i.e butan-l-ol Oxidation of (C) gives (B). Hence, (B) is butanoic acid.

Question 11.
Write the chemical equations involved when aniline is treated with the following reagents:
(i) Br₂ water
(ii) CHCl₃ + KOH
(iii) HCl
OR
Explain giving suitable reason:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) p-methylaniline is more basic than p nitroaniline.
Answer:

OR
(i) Since aniline is basic, it reacts with Lewis acid AlCl₃ to form salt. .
(ii) It is because methyl group in p methylaniline is electron releasing, it donates electrons for donation whereas nitrogroup in p^– nitroaniline is electron withdrawing.
(iii) Since -NH₂ group is ortho -para directing hence acetylation is done so as to decrease the reactivity of Aniline towards electrophilic substitution reactions.

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the questions that follow:
Reaction 1 Initiation
CH₃CHO → CH₃ + CHO
Propagation
CH₃ + CH₃CHO → CH₃CO + CH₄
CH₃CO → CH₃ + CO

Termination
2 .CH₃ → C₂H₆

Reaction 2
decomposition of acetaldehyde is an example of fractional order reaction because the order of the reaction is a fractional value.
It can be a whole number or a fraction. It can be noted that when the order of reaction is a fraction, the reaction is generally a chain reaction or follows some other complex mechanism. An example of a chemical reaction with a fractional reaction order is the pyrolysis of acetaldehyde.

The rate of decomposition of acetaldehyde into methane and CO in the presence of 12 at 800 K follows the rate law:
Rate = k[CH₃CHO][I₂l
The decomposition is believed to go by a two- step mechanism:
CH₃CHO + I₂ → CH₃I + CO + HI
CH₃I + HI → CH₄ + I₂
(A) Which of the two steps is a slower one?
(B) Explain why order cannot be determined by a balanced chemical equation
(C) Give one example of fractional order reaction.
(D) A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when:
(i) the concentration of B alone is increased to three times.
(ii) the concentrations of A as well as B are doubled?
OR
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? (5)
Answer:
(A) Rate law is obtained from the slowest step Hence step 1: CH₃CHO + I₂ → CH₃I + CO + HI this step is the slower one Here I₂ is the catalyst as it is consumed in first step and regenerated in the next step.

(B) The sum of powers of the concentration of the reactants in the rate law expression is .called the order of that chemical reaction. Order cannot be determined by a balanced chemical equation it may or may not be equal to the sum of stoichiometric coefficients as it is an experimental property.

(C) Decomposition of acetaldehyde Explanation- The order of the decomposition of acetaldehyde is 3/2 or 1.5.

(D) r = k[A]¹ [B]²
(i) When concentration of B increases to 3 times, the rate of reaction becomes 9 times
r = kA(3B)² ∴ r = 9kAB² = 9 times

(ii) r = k(2A) (2B)² ∴ r = 8kAB² = 8 times
OR
Rate = k [A]² = ka²
If [A] = \(\frac{1}{2 a}\); Rate = k \(\left(\frac{a}{2}\right)^{2}=\frac{1}{4}\)ka²

∴ Rate = 1/4th (one fourth of original rate)
The unit of rate constant is L mol^-1 s^-1

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 6 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks.
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the chokes in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

SECTION – A
(Section A contains 3 questions of 2 marks each.)

Question 1.
How is an n-type semiconductor formed?
Draw the energy band diagram n-type semiconductor. Name the majority charge carriers in it. (2)
Answer:
If pentavalent impurity atoms of Bi, Sb or P are doped in a tetravalent crystal of silicon or Germanium, we get a n-type semiconductor. Electrons are the majority charge carriers in a n-type semiconductor.

Question 2.
Name the characteristics of electromagnetic waves that:
(A) increases
(B) remains constant in the electromagnetic spectrum, as one moves from radioactive region towards ultraviolet region. (2)
Answer:
As one moves from radioactive region towards ultraviolet region,
(A) the frequency of the electromagnetic waves increases.
(B) The speed of the electromagnetic waves remains constant.

Related Theory
Speed of electromagnetic wave remains constant in the vacuum. Every massless object will travel with the speed of light.

Question 3.
Explain with the help of Einstein’s photoelectric equation any two observed features in photoelectric effect which cannot be explained by wave theory.
OR
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV. (2)
Answer:
Features of photoelectric equation which cannot be explained by wave theory :
(A) The wave theory could not explain the instantaneous process of photoelectric effect.
(B) ‘Maximum kinetic energy’ of the emitted photo electrons is independent of intensity of incident light.

Related Theory
Einstein gives the correction in photoelectric effect and give Einstein’s photoelectric equation.
OR
Let the maximum energies of emitted electrons are K₁ and K₂ when 600 nm and 400 nm visible light are used according to question
K₁ = 2K₁

SECTION – B
(Section B consists of 8 questions of 3 marks each.)

Question 4.
In compound microscope, length of microscope is 21.5 cm, focal length of objective is 1.6 cm, focal length of eye piece is 2.1 cm and if final image is formed at infinity, then find the distance of the object from objective lens? (3)
Answer:
Here,
v = 21.5 – 2.1
= 19.4 cm
because the image formed by the objective should lie at the focus of the eyepiece, from relation,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\)
= \(\frac{f-v}{v f}\)
⇒ u = \(\frac{v f}{f-v}\)
= \(\frac{1.6 \times 19.4}{1.6-19.4}\)
= -1.74 cm

Question 5.
In a single slit diffraction pattern, how does the angular width of central maxima change when
(A) Silt width is decreased
(B) Distance between the slit and screen is increased and
(C) Light of short visible wavelength is used? Justify your answer. (3)
Answer:
(A) Angular width of centrol ,maximum
2θ = \(\frac{2 \lambda}{d}\)
where d is slit width.
if slit width decreases, angular width increases.

(B) It does not depend upon D. So, it will not have any effect.

(C) Angular width
2θ = \(\frac{2 \lambda}{d}\)
∴ If Light of smaller wavelength is used then angular width decreases.

Question 6.
Spectra are produced due to electronic transitions from higher energy level to lower energy level.
(A) Name the series spectra produced by hydrogen.
(B) Which series lies in the visible region.
(C) Write down the Balmer formula for the wavelength of H_α line. (3)
Answer:
(A) Lyman, Balmer, Paschen, Brackett and Pfund series.
(B) Balmer series.
(C) Wavelength of Ha line \(\frac{1}{\lambda}\) = R \(\frac{1}{2^{2}}-\frac{1}{n^{2}}\)

Question 7.
Two convex lenses of an astronomical telescope of focal length 5 cm and 20 cm respectively, are shown in the given figure.

(A) Which of the given lenses will be used as an objective lens?
(B) How should the lenses adjusted to have the telescope in its normal adjustment position?
(C) What will be the magnifying power of the telescope in its normal adjustment position? (3)
Answer:
(A) Because of the larger focal length and larger radius, the lens B should be used as an objective Lens.
(B) In normal adjustment, the distance between objective and eyepiece will be
f_o + f_e = 20 + 5
= 25 cm
The distance required to be increased between the two lenses.
= 25 – 15 = 10 cm

(C) Magnifying power of the telescope in normal adjustment will be,
m = \(\frac{f_{0}}{f_{e}}\) = \(\frac{20 \mathrm{~cm}}{5 \mathrm{~cm}}\) = 4

Question 8.
A projectile of mass m, charge Ze, initial speed v and impact parameter b is scattered by a heavy nucleus of charge Ze. Obtain a relation between minimum distance s of particle from nucleus an impact parameter using conservation of angular momentum and energy. Show that if b = 0, s becomes distance of closest approach. (Ignoring size of nucleus and recoil motion).
OR
Considering that the protons and neutrons have equal masses, calculate how many times nuclear matter is denser than water. Take the value of nuclear radius R = 1.2 × 10^-15 A^1/3 metre and mass of the nucleus = 1.67 × 10^-27 kg. (3)
Answer:

Angular momentum of the projectile at α from the nucleus = mvb
At point of minimum distance (s) from nucleus velocity v’ of projectile is normal to radius vectors.
Angular momentum of projectile at minimum nucleus = mv’s
By Law of Conservation of angular momentum, mv’s = mvb
⇒ V’ \(\frac{v b}{s}\)
By Conservation of Energy
(KE + PE) act as = (KE + PE) at distance s

Dividing by \(\frac{1}{2}\) mv² equation (i)
b² + \(\frac{1}{2 \pi \varepsilon_{0}} \frac{Z Z^{\prime} e^{2}}{m v^{2}}\) = s²
s = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{Z Z^{\prime} e^{2}}{m v^{2}}\)
which is distance of closest approach.
OR
Density pf nucleus
ρ_n = \(\frac{m}{\frac{4}{3} \pi R^{3}}\) = \(\frac{3 m}{4 \pi R^{3}}\)
ρ_n = \(\frac{3 \times 1.67 \times 10^{-27}}{4 \times 3.14 \times\left(1.20 \times 10^{-15}\right)^{3}}\)
p_n = 2.307 × 10¹⁷ kg m^-3
Density of water ρ_w = 10³ kg m^-3
> \(\frac{\rho_{n}}{\rho_{w}}\) = \(\frac{2.307 \times 10^{17}}{10^{3}}\) = 2.307 × 10¹⁴

Question 9.
Draw a circuit diagram showing the biasing of an LED and state the factors which control wavelength of light and intensity of light emitted by a diode. (3)
Answer:

Wavelength of light is controlled by the band gap of semiconductor material.
Intensity of ammeter by the diode concentration of impurity in the junction diode.

Related Theroy
The p-n junctions can be designed so that current through them changes either by causing electron excitation by light photons or conversely through electron excitation by a suitable bias voltage resulting in emission of light photons.

Question 10.
You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.
(A) Which lenses should be used as objective and eyepiece? Justify your answer.
(B) Why is the aperture of the objective preferred to be large?
OR
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (3)
Answer:
(A) Magnification,
m = \(\frac{f_{0}}{f_{e}}\)
= \(\frac{P_{e}}{P_{0}}\)
Therefore, the lens of 0.5 D should be used as objective and the lens of 10 D should be used as eye-piece in order to achieve higher magnification.

(B) The aperture of the objective lens is made larger so, that it receives as much light as coming from the distant object and the resolving power of the telescope increases.
OR
Given:
λ₁ = 800 nm
= 800 × 10^-9 m
λ₂ = 600 nm
= 600 × 10^-9 m
D = 1.4 m
d = 0.28 mm
= 0.28 × 10^-3m
Suppose n₁^thaximum corresponds to wavelength λ₁ and it coincides with n₂^th maximum corresponding to wavelength λ₂.
∴ n₁\(\frac{\lambda_{1} D}{d}\) = n₂\(\frac{\lambda_{2} \mathrm{D}}{d}\)
or \(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{2}}\) = \(\frac{600}{800}=\frac{3}{4}\)

Thus, 3^rd maximum corresponding to wavelength 800 nm coincides with 4^th maximum corresponding to wavelength 600 nm.
And the minimum distance is given by,
X_min = n₁\(\frac{\lambda_{1} D}{d}\)
\(\frac{3 \times 800 \times 10^{-9} \times 1.4}{0.28 \times 10^{-3}}[ /latex]
X_min = 12mm

Question 11.
Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only by using Einstein’s equation. (3)
Answer:
(A) Existence of threshold frequency: According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of threshold frequency.

(B) Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it dependence on frequency and not on intensity of the incident light.

(C) Instantaneous emission of electrons: According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.

Related Theory
Some features cannot be explained in photoelectric current. So, Einstein gave his correction to explain the effect.

SECTION – C
(Section C consists one case study-based question of 5 marks.)

Question 12.
The wave theory of light was first put forward by Christian Huygen. During that period, everyone believed in Newton’s corpuscular theory, which had satisfactorily explained the phenomenon of reflection, refraction, the rectilinear propagation of light and the fact that light could propagate through vacuum. Huygen’s model was accepted when Thomas Young performed the interference experiment.

(A) Which one of the following phenomena is not explained by Huygen’s construction of wave front?
(i) Refraction
(ii) Reflection
(iii) Diffraction
(iv) Origin of spectra

(B) When light waves suffer reflection at the interface from air to glass, the change in phase of reflected waves is equal to:
(i) Zero
(ii) [latex]\frac{\pi}{2}\)
(iii) π
(iv) 2π

(C) In the given diagram a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through the glass slab is shown also along with normal drawn at A and D.

The refractive intex of glass will be equal
(i) \(\frac{\mathrm{BD}}{\mathrm{AC}}\)
(ii) \(\frac{A B}{C D}\)
(iii) \(\frac{B D}{A D}\)
(iv) \(\frac{A C}{A D}\)

(D) Select the right options in the following:
(i) Christian Huygen’s a contemporary
of Newton established the wave theory of light by assuming that light waves were transverse.
(ii) Maxwell provided the compelling theoretical evidence that light is transverse wave.
(iii) Thomas Young experimentally proved the wave behaviour of light and Huygen’s assumption.
(iv) All the statements given above, correctly answers the question.

(E) The diagram below shows two sources, A and B, vibrating in phase in the same uniform medium and producing circular wave fronts.

Which phenomenon occurs at point P?
(i) Destructive interference
(ii) Constructive interference
(iii) Reflection
(iv) Refraction (5)
Answer:
(A) (d) Origin of spectra
Explanation: The Huygen’s construction of wavefront does not explain the phenomena of origin of spectra.

(B) (c) π
Explanation: When a light wave suffers reflection at the interface from air to glass, it experiences a change in phase because the refractive index of glass is more than air, as we know that air is a rarer medium and glass is denser medium. So, the change in phase of the reflected wave is π.

Caution
Students are often confused about the phase change in reflection. When a light wave travels from rarer to denser than the phase of the incident wave change with angle it and when it travels from denser to rare then the phase of the incident wave does not change.

(C)(a) \(\frac{B D}{A C}\)
Explanation: BD = V₁t,
AC = V₂t
\(\frac{\sin \theta}{\sin \phi}\) = \(\frac{\left(\frac{B D}{A D}\right)}{\left(\frac{A C}{A D}\right)}\)
\(\frac{\mathrm{BD}}{\mathrm{AC}}\) = \(\frac{v_{1} t}{v_{2} t}\)
= \(\frac{v_{1}}{v_{2}}=\frac{\mu_{2}}{\mu_{1}}\)
= \(\frac{\mu_{2}}{\mu_{1}}=\frac{\mathrm{BD}}{\mathrm{AC}}\)

Caution
Students are often confused in the geometry of the diagram. They should observe the angles and the common arm of the triangle.

(D) (b)
Maxwell provided the compelling theoretical evidence that light is transverse wave
Explanation: In Huygens principle the wave theory was established by assuming that the waves were longitudinal. So, (a) is wrong.

In Maxwell’s theory of light, he stated that light is a transverse wave and conducted multiple experiments to prove. So, (b) is correct.

Thomas Young conducted his double slit experiment and tried to prove the wave nature of light by proving that light undergoes interference which is a phenomenon showed by waves. He never proved the type of wave light is, i.e., transverse or longitudinal. So, (c) is wrong.

These statements mostly highlight the wave nature of light. They do not comment on the particle nature of light or the dual nature of light established. Hence, they clearly do not answer the question “what is light” correctly. So, option (D) is also wrong.

(E) (b) Constructive interference Explanation: Wave fronts for a source represents locus of points at same phase. A and B are vibrating in phase. Hence, all the wave fronts shown in the diagram are in the same phase. At P, there are wave fronts of both A and B. Hence, the waves of A and B are at same phase at P. Hence, constructive interference occurs at P.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 12 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 12 for Practice

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Arrange the following compounds in increasing order of their property as indicated – (Any two) (2)
(A) CH₃CHO, C₆H₅CHO, HCHO (reactivity towards nucleophilic addition reaction).
(B) 2, 4-dinitrobenzoic acid, 4-methoxybenzoic acid, 4 nitrobenzoic acid(acidic character).
(C) CH₃CH₂CH₂CH₃, CH₃OCH₂CH₃,
CH₃CH₂CHO, CH₃COCH₃, CH₃CH₂CH2₂OH
(boiling point)

Question 2.
Why do amines act as nucleophiles? Give example of a reaction in which methylamine acts as a nucleophile. (2)

Question 3.
(A) In the expression of rate of reaction in terms of reactants, what is the significance of negative sign?
(B) What is the use of integrated rate equation? (2)

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
(A) What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³ and 3d⁴?
(B) What are interstitial compounds? Why are such compounds well known for transition metals? (3)

Question 5.
When Liquid ‘A’ is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. liquid ‘B’ also forms a white crystalline solid with sodium hydrogen sulphite but it does not give test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
OR
Complete the following equation and write the structures of A, B, C, D, E, and F.

Question 6.
(A) Differentiate between the following with the help of one example of each
(i) Homoleptic and Heterolytic Complexes
(ii) Double salt and a complex
(B) What is the coordination number of Fe in [Fe(EDTA)]³⁺ ? (3)

Question 7.
(A) How does the precipitation of colloidal smoke takes place in Cottrell precipitator?
(B) Observe the following diagram and answer the questions that follow –

(i) Name the property which is shown by the colloidal solution.
(ii) Explain the reason for the property shown by colloidal solution. (3)

Question 8.
Write the structures of the following compounds.
(A) α-Methoxypropionaldehyde
(B) 3-Hydroxybutanal
(C) 2-Hydroxycyclopentanecarbaldehyde (3)

Question 9.
(A) Why does acetylation of -NH₂ group of aniline reduce its activating effect?
(B) What is the product when C₆H₅CH₂NH₂ reacts with HNO₂ ?
(C) What is the structure and IUPAC name of the compound, allyl amine?
OR
How will you convert the following:
(A) Nitrobenzene into aniline
(B) Ethanoic acid into methanamine
(C) Aniline into N-phenylethanamide (3)

Question 10.
For the complex [Fe(en)₂ Cl₂ ], Cl, (en = ethylene diamine), identify:
(A) the oxidation number of iron,
(B) the hybrid orbitals and the shape of the complex,
(C) the magnetic behaviour of the complex,
OR
(A) What type of isomerism is shown by the complex [Co(NH₃ )₆ ][Cr(CN)₆ ]?
(B) Why a solution of [Ni(H₂O)₆ ]²⁺ is green while a solution of [Ni(CN)₄]^2- is colourless? (At. no. of Ni = 28)
(C) Write the IUPAC name of the following complex: [Co(NH₃ )₅ (CO₃ )]Cl. (3)

Question 11.
For the reaction A + B → products, the following initial rates were obtained at various given initial concentrations:

[A] mol / L	[B] mol / L	Initial rate M/s
1) 0.1	0.1	0.05
2) 0.2	0.1	0.10
3) 0.1	0.2	0.05

Determine the half-life period.
OR
The following data were obtained during the first order thermal decomposition of
SO₂Cl₂ at a constant volume:
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

Experiment	Time/s-1	Total Pressure/atm
1	0	0.4
2	100	0.7

Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010) (3)

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the following questions:
ALl chemical reactions involve interaction of atoms and molecules. A large number of atoms/ molecules are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCI is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.
(A) How many moles of chlorine gas will be evolved from the reaction given in the above passage? .
(B) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which will the products formed at cathode and anode?
(C) What will be number of moles of electrons exchanged during electrolysis of aqueous solution of NaCl?
(D) Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5A was passed through the solution of AgNO₃. (Molar mass of Ag = 108 g mol- 1,1 F = 96500 C mol^-1). ”
OR
How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: IF = 96,500 C mol^-1) (5)