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

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 2 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 a p-type semiconductor formed? Draw energy band of p-type semiconductors and name the majority charge carriers in it. (2)
Answer:
1. If trivalent impurity atoms of B, Al, or In are doped in a pure semiconductor of Silicon or Germanium, we get a p-type semiconductor. Holes are the majority charge carriers in a p-type semiconductor. The diagram of energy band of p-type semiconductors is given below.

Related Theory
There are two types of dopants used for doping the tetravaLent Silicon or Germanium.
1. Pentavalent dopants which have five valence electrons like Arsenic, Antimony, and Phosphorus.
2. Trivalent dopants have three valence electrons like Indium, Boron, and Aluminium.

Question 2.
Photons, with a continuous range of frequencies, are made to pass through a sample of rarefied hydrogen.
Transitions of electrons are shown in figure:

(A) Identify the spectral series, of the hydrogen emission spectrum corresponding to these lines.
(B) Which of these lines corresponding to absorption of radiation of maximum wavelength?
OR
Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions W₂ and W₂ (W₁ > W₂).
On what factors does the (i) slope and (ii) intercept of the lines depend? (2)
Answer:
(A) Transition I → Lyman series
Transition II → Balmer series
Transition III → Paschen series of spectral absorption lines in the continuous spectrum.

(B) Energy of electron in hydrogen atom
E_n = \(-\frac{13.6}{n^{2}}\) eV
E₁ = -13.6 eV
E₂ = – 3. 4 eV and similarly we can find E₃, E₄, ……………..
For transition III energy absorbed is:
ΔE = E₄ – E₃ = 0.66 eV

which is least out of three transitions. Transition III corresponds to the absorption of radiation of maximum wavelength.
OR
The graph of stopping potential Vs and frequency (v) for two metals 1 and 2 is shown in fig.

(A) Slope of graph tan θ= \(\frac{h}{e}\) and depends on h and e.
(B) Intersect of lines depend on the work function.

Question 3.
Distinguish between the metals and insulators on the basis of energy band theory. Why does a diamond behave as an insulator? (2)
Answer:
For metals, the valence band is completely filled and the conduction band have the possibilities,
either it is partially filled within extremely small energy gap between valence band and conduction band or it is empty, with two bands overlapping each other.

For insulators, the energy gap between the conduction and valence band are very large, also the conduction band is partially empty. When an electric field is applied across such a solid, the electrons find it difficult to acquire sufficient energy to jump to conduction band. There is a Large forbidden band of 6eV in diamond. It is difficult to excite the electrons from valence band to the conduction band. Due to the absence of free charges, diamond behaves as an insulator.

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

Question 4.
(A) State Bohr’s postuLate to define stable orbits in hydrogen atoms. How does Louis de BrogUe’s hypothesis explain the stability of these orbits?
(B) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 LeveL Estimate the frequency of the photon. (3)
Answer:
(A) Bohr’s postulate for stable orbits in hydrogen atom: An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of \(\frac{h}{2 \pi}\). where h is Planck’s constant.

If n is the principal quantum number of orbit, then an electron can revolve only in a certain orbits of definite radii. These are called stable orbits. Louis de Broglie’s explanation of stabiLity of orbits:

According to de Broglie. orbiting electron around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if the standing wave does not close upon itself only those Louis de Brog lie waves exist for which the circumference of circular orbit Contains a whole number of wavelengths i.e., for orbit circumference of nth orbit as 2nπr_n
2πr_n = nλ
2πr_n = n \(\left(\frac{h}{m v}\right)\)
mvr_n = n \(\left(\frac{h}{2 \pi}\right)\)
which is the quantum condition proposed by Bohr.

(B) In ground state, n = 1
In excited state, n = 4
\(\frac{1}{\lambda}=R\left[\frac{1}{(1)^{2}}-\frac{1}{(4)^{2}}\right]\) where R is Rydberg constant
\(\frac{1}{\lambda}\) = 1.097 x 10⁷ x \(\frac{15}{16}\)
= 1.028 x 10⁷ m^-1
Frequency, v = \(\frac{c}{\lambda}\) = 3 x 10⁸ x 1.028 x 10⁷
= 3.09 x 10¹⁵ Hz.

Question 5.
(A) When p-n junction Is said to be reverse-biased, what will be the direction of flow of majority charge carriers?
(B) Which semiconductors are preferred to make LED and why?
(C) Give two advantages of using LED. (3)
Answer:
(A) A p-n junctión ¡s set to be reverse biased when its p-region is maintained at lower potential with respect to its n-region. i.e, p-end is connected with negative terminal, and n-end is connected with positive terminal In a reverse-biased p-n junction.
majority charge carriers move away from the junction.
(B) MateriaLs Like gaLLium phosphide (Gap) Gallium arsenide (GaAs), etc., are used.
They emit the maximum amount of energy in form of tight radiation.
(C) The following are the advantages of LEDs:
(i) They are operational at two voltages.
(ii) They are quick in action and their power consumption is Low.

Question 6.
What is meant by the size of nucleus? A nucleus of mass number 225 splits into two fresh nucLei having radius R = 1.1 x 10^-15 A^1/3 m, find the radii of the new nuclei formed. (3)
Answer:
The extremely small central core of the atom in which whole of the positive charge and practically the entire mass is confined is known as nucleus. Different nuclei differ in size. The radius of nuclei is directly proportional to A^1/3.
where A is its mass number.
Here A = 225, R = 1.1 x 10^-15 A^1/3 m
Let A₁ and A₂ be the respective mass number of two new nuclei formed.
A₁= \(\frac{3}{(3+2)}\) A
= \(\frac{3}{5}\) x 225
= 135
And,
A₂ = \(\frac{3}{(3+2)}\) A
=\(\frac{2}{5}\) x 225 = 90

R₁= 1.1 x 10^-15 A^1/3
= 1.1 x 10^-15 x 135^1/3
= 5.643 x 10^-15 m

R₂ = 1.1 x 10^-15 A^1/3
= 1.1 x 10^-15 x 90^1/3
= 4.93 x 10^-15 m.

Question 7.
State Huygens’ principLe. Using this principle draw a diagram to show how a plane wavefront incident at the interface of the two media gets refracted when It propagates from a rarer to a denser medium. Hence, verify Snell’s Law of refraction. (3)
Answer:
(A) Huygens’ Principle:
(i) Each point on the primary wavefront acts as a source of secondary wavelets, transferring out a disturbance in all directions in the same way as the original source of Light does.
(ii) The new position of the wavefront at any instant is the envelope of the secondary wavelets at that instant.

(B) Refraction on the basis of wave theory:
(i) Consider any point Q on the incident wavefront
(ii) Suppose when disturbance from point P on incident wavefront reaches point P’ on the refracted wavefront, the disturbance from point Q reaches Q’ on the refracting surface XY.

(iii) Since P’A’ represents the refracted wavefront, the time taken by light to travel from o point on incident wavefront to the corresponding point on refracted wavefront should always be the same. Now, time taken by Light to go from Q to Q will be

t = \(\frac{Q K}{c}+\frac{Q K}{v}\) ………………………… (i)

In right-angled ΔAQK,
∠QAK = i
∴ QK=AK sin i …………………………………. (ii)
In right-angled ΔP’Q’K.
∠Q’P’K = r
and Q’K = KP’ sin r …………………………. (iii)
Substituting (ii) and (iii) in equation (i).

The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e., given equation (iv) is independent of AK. It will happen so. if
\(\frac{\sin i}{c}-\frac{\sin r}{v}\) = 0
⇒ \(\frac{\sin i}{\sin r}=\frac{c}{v}\)
⇒ n = \(\frac{\sin i}{\sin r}\)
This is the Snells low for refraction of light.

Retated Theory
Wave theory explained Lows of reflection and refraction both.

Question 8.
Draw a labelled ray diagram of an astronomical telescope in the near-point adjustment position. A giant refracting telescope at an observatory has an objective Lens of Focal Length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens.
The diameter of the Moon is 3.48 x 10⁶ m and the radius of lunar orbit is 3.8 x 10⁶ m.
OR
ExpLain the refraction of Light through a glass-slab with a neat ray diagram. (3)
Answer:
Astronomical teLescope in near points adjustment

Here, f₀ = 15 m,
f_e=1 cm = 0.1 m
Diameter of moon,
d = 3.48 x 10⁶ m

Radius of lunar orbit
μ₀ = 3.8 x 10⁸ m
size of the image of moon I = ?
The angle subtended by the moon at the objective lens,
α = \(\frac{d}{\left|\mu_{0}\right|}\) ………………………….. (i)
And the angle subtended by the image of the moon at the objective lens,
β = \(\frac{I}{f_{0}}\) ……………………… (ii)

Both these angles in equations (i) and (ii) wift be equal.
∴ \(\frac{d}{\left|\mu_{0}\right|}=\frac{1}{f_{0}}\)
⇒ I = \(\frac{d \times f_{0}}{\left|\mu_{0}\right|}\)
= \(\frac{3.48 \times 10^{6} \times 15}{3.8 \times 10^{8}}\) m
= 13.73 x 10^-2 m
= 13.73 cm.
OR

On entering into the glass medium-light ray bends towards the normal that is tight ray gets refracted on entering the gLass medium. After getting refracted this ray now travels through the glass slab and comes Out of the glass slab by refraction from the other interface boundary. Since ray goes from glass medium to air it again gets refracted and bends away from normal.

The incident ray and the emergent ray are parallel to each other. i₁ is the angLe of incidence, r₁ is the angLe of
refraction and r₂ is the angle of emergence. The angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other. When a Lightray is incident normally to the interface of two media then there is no bending of tight ray and it goes straight through the medium.

Question 9.
In the study of a photoeLectric effect the graph between the stopping potential V and frequency y of the incident radiation on two different metals P and Q is shown below:
(A) Which one of the two metals has higher threshold frequency?
(B) Determine the work function of the metal which has greater value.
(C) Find the maximum kinetic energy of electron emitted by light of frequency 8 x 10¹⁴ Hz for this metal. (3)
Answer:
(A) Q has higher threshoLd frequency.
(B) Work function of Q
Φ_o = hv₀
hv0 = (6.6 x 10^-34) x \(\frac{6 \times 10^{14}}{1.6 \times 10^{-19}}\) eV
= 2.5 eV

(C) K_max = h(v – v₀)
= \(\frac{6.6 \times 10^{-34}\left(8 \times 10^{14}-6 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}\)
= \(\frac{6.6 \times 10^{-34} \times 2 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV}\)
K_max=0.83eV.

Question 10.
(A) Monochromatic Light of wavelength 589 nm is incident from air on a water surface. If μ for water is 1.33, find the wavelength, frequency, and speed of the refracted light.
(B) A doubLe convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature.
Find the radius of curvature required, if the focal length is 20 cm. (3)
Answer:
(A) Given: m= 1.33
λ_a= 589 nm
We know that,
μ = \(\frac{C}{V}\)
v = \(\frac{c}{\mu}\)
= \(\frac{3 \times 10^{8}}{1.33}\)
Speed, v = 2.26 x 10⁸ m/s

Frequency remains same.
∴ v = \(\frac{c}{\lambda_{a}}=\frac{3 \times 10^{8}}{589 \times 10^{-9}} \)
Frequency, v = 5.09 x 10¹⁴ Hz
Wavelength.
λ_w = \(\frac{\lambda_{a}}{\mu} \)
= \(\frac{589 n m}{1.33}\)
= 442.8 nm

(B) Given: μ= 1.55.
20 cm
We know that,

R = 20 x 0.55 x 2
R= 22 cm.

Question 11.
Which constituent radiation of the electromagnetic spectrum is used
(A) in radar
(B) to photograph internal part of a human body, and
(C) for taking photographs of the sky during night and foggy conditions?
Give one reason for your answer in each case.
OR
(A) When a convex Lens of Focal Length 30 cm is in contact with concave tens of focal length 20 cm, find out if the system is converging or diverging.
(B) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of
refracting angle A so that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is μ.)
Answer:
(A) Microwaves are used .in the operation of radar because microwaves have small wavelengths and are not bent by objects coming in its path. Due to it, these waves can be sent in a particular direction as a beam signal.

(B) X-rays are used to photograph internal parts of a human body because X-rays are a quite short wavelengths. They can easily pass-through flesh but not bones.
(i) Infrared rays are used for taking photographs of the sky during night and foggy conditions because these
can easily pass-through fog and dark.
OR
(A) f₁= f_convex= 30 cm
f₂ = f_concave = – 20 cm
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
= \(\frac{1}{30}-\frac{1}{20} \)
= \(\frac{2-3}{60}\)
\(\frac{1}{f}=-\frac{1}{60}\)
[∴ concave (diverging) lens)]

(B) μ = \(\frac{\sin \left(\frac{\delta_{\min }+A}{2}\right)}{\sin \frac{A}{2}}\)
Let C is the critica angle,
sin C = \(\frac{1}{\mu}\)

r₁+r₂ = A
r₁+C = A
r₁ = A – C
μ = \(\frac{\sin i}{\sin r_{1}}\)
⇒ sin i = μ sinr₁
sin i = μ sin(A-C)
i = sin^-1μ sin(A-C)

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

Question 12.
If the setup of Young’s doubLe slit experiment is immersed in a liquid then the Wavelength of the Light gets changed. Since an immersion frequency remains unchanged, velocity decreases hence, waveLength decreases. Now, alt those parameters which are dependent on wave Length will be affected. If the fringe width is air in \(\frac{\lambda D}{d}\) then in Light it will be \(\frac{\lambda D}{\mu d}\) in the same way, the angular fringe width also changes and the position of cringes on-screen aIs changes. (3)

(A) In Young’s experiment when sodium Light of waveLength 5893A° is used, then 62 fringes are seen in the field of view. Instead, if violet Light of waveLength 4358A° is used, then the number of fringes that will be seen in the field of view will be:
(a) 54
(b) 64
(c) 74
(d) 84
at x = 0. The third maximum (taking the central maximum as zero maximum) will be at x equal to:
(a) 1.67 x 10^-2 m
(b) 1.5 x 10^-2 m
(c)O.5 10^-2m
(d) 5.0 10^-2 m

(C) In Young’s experiment, the ratio of maximum to minimum intensities of the fringe system is 4:1. The amplitudes of the coherent sources are in the ratio:
(a) 4:1
(b) 3:1
(c) 2:1
(d) 1:1

(D) The displacement of two Interfering Light waves are y₁ = 4sinωt and Y₂ = 3sinω \(\left(\frac{\omega t+\pi}{2}\right)\) .The ampLitude of the resultant wave is:
(a) 5
(b) 7
(c) 1
(d) 0
(E) In Young’s experiment, tight of waveLength 4000Å is used to produce bright fringes of width 0.6 mm, at a
dipped in a Liquid of refractive index 1.5, then the fringe width will be:
(a) 0.2 mm
(b) 0.3 mm
(c) 0.4 mm
(d) 1.2 mm (5)
Answer:
(A) (d) 84
Explanation: From Young’s double slit experiment,
λ₁ = 5893 Å
λ₂= 4359 Å
\(\frac{n_{1} \lambda_{1} D}{d}\) = \(\frac{n_{2} \lambda_{2} D}{d}\)
The above condition is total extent of fringes is constant for both wavelengths.
\(\frac{62 \times 5893 \times 10^{-10} \times \mathrm{D}}{d}=\frac{n_{2} \times 4359 \times 10^{-10} \times \mathrm{D}}{d}\)
n₂ = \(\frac{62 \times 5893}{4359}\)
= \(\frac{365366}{4359}\) = 83.8
n₂ = 84

Related Theory:
In Young’s double-slit, experiment the number of fringes ate inversely proportional to the fringe width.
(B) (b) 1.5 x 10^-2 m
Explanation: x = (n) \(\lambda \frac{\mathrm{D}}{d} \)
= 3 x 5000 x 10^-10 x \(\frac{2}{0.2 \times 10^{-3}}\)
= 1.5 x 10^-2 m = 1.5 cm

Caution:
Students are often confused about the number of slits. In Young’s double-slit experiment two slits are there.
(C) (b) 3:1

Explanation: \(\frac{I_{(m a x)}}{I_{(m i n)}}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)
⇒ \(\frac{4}{1}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)

Related Theory:
In Young’s experiment, intensity of tight is directly proportional to the square of the amplitude.
(D) (a) S
Explanation: Here,
Φ = \(\frac{\pi}{2}\)
Amplitude of first wave = A₁ = 4
Amplitude of second wave = A₂ = 3
Resultant = \(\sqrt{\left[(4)^{2}+(3)^{2}\right]}\)
= 16 + 9 = 5cm.

(E) (c) 0.4 mm
Explanation: Fringe width.

Caution:
Students are often confused in the relation of fringe width and wavelength of Light. Here, fringe width is
directly proportional of a wavelength of Light.

Students can access the CBSE Sample Papers for Class 12 Maths 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 Maths Term 2 Set 5 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper contains three sections-A. B and C. Each part is compulsory.
• Section-A has 6 short answer type (SA1) questions of 2 marks each.
• Section-B has 4 short answer type (SA2) questions of 3 marks each.
• Section-C has 4 long answer type questions (LA) of 4 marks each.
• There is an Internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A
(Section – A has 6 short answer type (SA-1) questions of 2 marks each.)

Question 1.

Question 2.
A die is rolled. If the outcome is an odd number, what is the probability that it is a prime number? (2)

Question 3.
A line makes the same angle θ with x and z-axes. If the angle β which it makes with y-axis is such that sin²β = 3 sin²θ, then find the value of cos²θ.
OR
If the distance of the point (1,1,1) from the origin is half its distance from the plane x + y + z + k = 0, then find k. (2)

Question 4.
Write the probability distribution of the random variable “number of heads” when one coin is tossed. (2)

Question 5.
Find the vector projection of \(\vec{a}\) on \(\vec{b}\), where \(\vec{a}\) = Sj – 3k and \(\vec{b}\) = î + ĵ + k̂. (2)

Question 6.
Show that xdy – (y + 2x²) dx = 0 is a linear differential equation. (2)
Answer:
We have,
x dy – (y + 2x²) dx = 0
or \(\frac{d y}{d x}\) = \(\frac{y+2 x^{2}}{x}=\frac{y}{x}+2 x\)
⇒ \(\frac{d y}{d x}-\frac{1}{x} y\) = 2x
This is of the type \(\frac{d y}{d x}\) + P(x)y = Q(x), where P(x)
= \(\frac{1}{x}\) and Q(x) = 2x. x
∴ The given differential equation is a Linear differential equation.
Hence proved.

SECTION – B
(Section – B has 4 short answer type (SA-2) questions of 3 marks each.)

Question 7.
Evaluate \(\int \frac{1}{3 \sin x+4 \cos x}\) dx. (3)
OR

Answer:
Let I = \(\int \frac{1}{3 \sin x+4 \cos x}\) dx
Let r cos α = 3 and r sin α = 4.
Then, r² = 25, i.e., r = 5 and α = tan^-1\(\left(\frac{4}{3}\right)\)
∴ 3 sin x + 4 cos x
= 5 cos α sin x + 5 sin α cos x
= 5 sin (x + α)
Now, I = \(\int \frac{1}{5 \sin (x+\alpha)}\) dx
= \(\frac{1}{5}\) ∫cosec (x + α) dx
= \(\frac{1}{5}\) log |cosec (x + α) – cot (x + α)| + C
= \(\frac{1}{5}\) log |cosec (x + tan^-1\(\frac{4}{3}\) – cot (x + tan^-1\(\frac{4}{3}\)| + C
OR

Question 8.
Show that the points A(- 2, 3, 5), B (1, 2, 3) and C (7, 0, -1) are collinear. (3)
Answer:

Question 9.
Prove that \(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})\) = 0. (3)
Answer:

Question 10.
Sketch the bounded region and find the area of the region bounded by the curve y = |x – 1| and the line y = 1.
OR
Find the area of the region bounded by the curve y² = 4x, the lines x = 1 and x = 4 and the x-axis in the first quadrant. (3)
Answer:
We have, y = |x – 1|

Now, required area = ar(rect OBCD) – ar(∆OAD) – ar(∆ABC)

OR
y² = 4x represents a parabola whose vertex is at the origin, axis of symmetry is x-axis, focus on the positive direction of x-axis and it opens to the right.

SECTION – C
(Section – C has 4 long answer type questions (LA) of 4 marks each.)

Question 11.
Show that the points (î – ĵ + 3k̂) and 3 (î + ĵ + k̂)are equidistant from the plane \(\vec{r}\). (5î + 2ĵ – 7k̂) + 9 = 0 and lies on opposite side of it.
OR
Find the point(s) on the line through the points A(l, 2, 3) and B (3, 5, 9) which are at a distance of 14 units from the mid-point of AB. (4)
Answer:
Mid-point of vectors (î – ĵ + 3k̂) and (3î + 3ĵ + 3k̂)

If the vectors (î – ĵ + 3k̂) and (3î + 3ĵ + 3k̂) are equidistant and lies on opposite side of the
plane \(\vec{r}\) . (5î + 2ĵ – 7k̂)+ 9 = 0, then their mid-point must satisfy the equation of the plane.
Since, (2î + ĵ + 3k̂) . (5î + 2ĵ – 7k̂) + 9
= 10 + 2 – 21 + 9
= 0 = R.H.S.
Hence, the vectors (î – ĵ + 3k̂) and (3î + 3ĵ + 3k̂) are equidistant and lies on opposite side of the given plane.
Hence, proved.
The equation of the line AB joining A(1, 2, 3) and B(3, 5, 9) is
\(\frac{x-1}{3-1}=\frac{y-2}{5-2}=\frac{z-3}{9-3}\)
⇒ \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{6}\) = λ(say)
∴ Any point on this line is
(2λ + 1, 3λ + 2, 6λ + 3) …(i)
Since, this point is at a distance of 14 units from the mid-point of AB i.e., \(\left(2, \frac{7}{2}, 6\right)\).
∴ \(\sqrt{(2 \lambda+1-2)^{2}+\left(3 \lambda+2-\frac{7}{2}\right)^{2}+(6 \lambda+3-6)^{2}}\) [Using distance formula]
i.e., (2λ – 1)² + (3λ – \(\frac{1}{2}\))² + (6λ – 3)² = 196 [squaring both sides]
⇒ 49λ – 49λ – \(\frac{735}{4}\) = 0
or 4λ – 4λ – 15 = 0
⇒ (2λ + 3) (2λ – 5) = 0
⇒ λ = \(\frac{5}{2}\), – \(\frac{3}{2}\)
Substituting the values of λ in (i), we get the
requirec points as (6, \(\frac{19}{2}\), 18) and (- 2, –\(\frac{5}{2}\), – 6).

Question 12.
Solve the differential equation: \(\frac{d y}{d x} \sqrt{1+x+y}\) = x + y – 1 (4)
Answer:
Given differential equation is:
\(\frac{d y}{d x} \sqrt{1+x+y}\) = x + y – 1 ……. (i)
Put \(\sqrt{1+x+y}\) = z
or 1 + x + y = z²
Differentiating w.r.t x, we get
0 + 1 + \(\frac{d y}{d x}\) = 2z\(\frac{d z}{d x}\)
Substituting these values in (i), we get

⇒ 4 – 2z = A(z – 1) + B(z + 2)
⇒ 4 – 2z = (A + B)z – A + 2B
Comparing coefficients of z and constant terms both sides, we get
A + B = – 2
and – A + 2B = 4
Solving these equations, we get
B = \(\frac{2}{3}\), A = – \(\frac{8}{3}\)
∴ \(\frac{2 z^{2}}{z^{2}+z-2}\) = 2 + \(\left(-\frac{8}{3}\right) \frac{1}{z+2}+\frac{2}{3} \frac{1}{z-1}\)
Substituting this value in (ii), we get
⇒ \(2\left\{1+\frac{1}{3(z-1)}-\frac{4}{3(z+2)}\right\}\)dz = dx
Integrating on both sides, we have

= x + C
Which is the required solution of the given differential equation.

Question 13.
Evaluate (4)

Answer:

CASE-BASED/DATA-BASED

Question 14.
There are two shops in a village market named Vijay General Store and Anand General Store. In Vijay General Store, 30 tin of pure mustard oil, 40 tin of adulterated mustard oil, while in Anand General Store, 50 tin of pure mustard oil 60 tin of adulterated mustard oil are there. Mr. Gautam wants to buy one tin of mustard oil from any store selecting at random.
Based on the above information, answer the following two questions:
(A) Find the conditional probability than an
adulterated mustard oil tin is bought, given that (i) Vijay General Store is selected at random (ii) Anand General Store is selected at random. (2)
(B) Mr. Gautam wants to know quality of mustard oil Before purchasing, he selected a store at random and then selected a tin of mustard oil at random. If the tin selected at random has adulterated oil, then find the probability that the selected tin is from Vijay General Store. (2)
Answer:
Let E₁, E₂ and A be the events defined as follows:
E₁: Selection of Vijay General Store
E₂: Selection of Anond General Store
A: Purchasing of a tin having adulterated mustard oil.

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

CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 2 with Solutions

Maximum Marks : 35
Time : 2 hours

Instructions

• The question paper is divided into 3 sections -A, B and C.
• Section A, consists of 7 questions (1-7). Each question carries 2 marks
• Section B, consists of 3 questions (8-10). Each question carries 3 marks.
• Section C, consists of 3 questions (11-13). Each question carries 4 marks.
• Internal choices have been given for question numbers -1, 3, 8 and 12.

Section-A
(Each question carries 2 Marks)

Question 1.
The branches of Digital Cinemas were operating without any network between them. Looking at the advancement of digital technologies, they want to implement networks in their branches . Help them understand the advantages they would enjoy with networks. Also, inform about the different types of networks that can be implemented.
Or
While studying protocols, Mr. Thakur discovered existence of VoIP . Help him to understand the protocol and also explain him the advantages of it.
Answer:
There are some advantages of computer networking, which are as follows

• It enhances communication and availability of information.
• It allows for more convenient resource sharing.
• It makes file sharing easier.
• It is highly flexible.
• It is an inexpensive system.
• It increases cost efficiency.
• It boosts storage capacity.
• It lacks independence.

Categories of networks on the basis of distance, they support

• LAN (Local Area Network) A network of computers within a small area such as : a room, a building or a campus.
• MAN (Metropolitan Area Network) A network spread across a city.
• WAN (Wide Area Network) A network spread across countries, states and continents, i.e. a larger geographical area.

OR

VoIP is an IP telephony term for a set of facilities used to manage the delivery of voice information over Internet. It enables a user to make cheap telephone calls over a broadband Internet connection, instead of using a regular telephone service.

A major advantage of VoIP is that avoids the tolls charged by ordinary telephone service. A user can make a call locally or in other parts of US or Canada, or anywhere else in the world, eliminating long distance fees by using a VoIP service.

The concept of VoIP is used in wireless LAN networks and sometimes referred to as WVoIP, VoFI, VoWi-Fi and Wi-Fi VoIP.
Advantages of VoIP

• The biggest single advantage of VoIP has over standard telephone systems is low cost.
• Using services such as true VoIP, subscribers can call one another at no cost to other party.
• Routing phone calls over existing data networks eliminate the need for separate voice and data networks.
• The ability to transmit more than one telephone call over a single broadband connection.
• VoIP consists advance telephone features, e.g. call routing, screen POP and IVR.

Question 2.
(i) Anish is a web developer and can write web pages but he is not conversant with how to host the web pages to a server. Explain him the concept of web hosting.
Answer:
Web hosting is a service that makes our site or web application accessible on the Internet. It’s also one of the most essential elements to consider when building a website. Web hosting is typically provided by web hosts, which are businesses that maintain, configure and run physical servers that house websites.

(ii) Rishita finds the word www in front of all the URLs and website addresses. She is curious about what is www? Explain her the concept.
Answer:
WWW stands for World Wide Web, which is also known as the Web. It is a collection of websites or web pages stored in web servers and connected to local computers through the Internet. These websites contain text pages, digital images, audios, videos, etc. Users can access the content of these sites from any part of the world over the Internet using their devices such as computers, laptops, cell phones, etc. The WWW, along with Internet, enables the retrieval and display of text and media to your device.

Question 3.
Some of the MySQL numeric functions are not properly understood by Niharika . Help her with an explanation and example of each of the functions given below
(i) POWER()
(ii) MOD()
OR
Write the uses of following MySQL functions with one example of each.
(i) MID()
(ii) LEFT()
Answer:
(i) POWER()/POW() function returns the power of a number.
e.g. SELECT POWERt (2, 5);

(ii) MOD() function returns the remainder of a number dividing by another number.
e.g. SELECT MOD(12, 2);

Or

(i) MID() function returns a substring of the specified length starting from the specified position.
e.g. MID( ‘HELLO’ ,3);
Output

(ii) LEFT() function returns the left most number of characters as specified,
e.g. LEFT(‘HELLO’,1);
Output

Question 4.
Rishabh, a beginner in web surfing came to know that for surfing the web a computer/ smartphone equipped with a modem, an internet connection and a web browser is required . He understands the other requirements but has less knowledge about web browsers.
Explain him the concept of web browser.
Answer:
A web browser or simply browser, is an application used to access and view websites. Common web browsers include Microsoft Edge, Internet Explorer, Google Chrome, Mozilla Firefox and Apple Safari.

A web browser takes you anywhere on the Internet. It retrieves information from other parts of the web and displays it on your desktop or mobile device. The information is transferred using the HyperText Transfer Protocol, which defines how text, image and video are transmitted on the web.

Question 5.
Samadrita wants to find the outputs of the following queries using SUBSTRQ and MOD() functions. Help her to find the outputs.
(i) SELECT SUBSTR(“testcases”,-5.3);
Answer:

The SUBSTR() function extracts certain number of characters from a string . Here the function extracts characters starting from position – 5 and then 3 characters from the right.

(ii) SELECT M0D( 17.7,3);
Answer:

The MOD() function returns the remainder of division.

Question 6.
Ramesh had learnt the ORDER BY and GROUP BY clauses of SQL a few time back. Now, he is not very comfortable about the terms . Help him to recall back the differences.
Answer:

ORDER BY clause	GROUP BY clause
It is used to arrange records of a table.	It is used to group records on common values of a column.
It works on individual records.	It works on groups of records.
The attribute can be under aggregate function under ORDER BY statement.	The attribute cannot be under aggregate function under GROUP BY statement.
It has two types-Ascending and Descending.	No such variations.

Question 7.
Priya is not able to find the outputs of the following SQL queries on the Drinks table. Her manager had instructed her to find out some aggregate results based on the table. Help her in finding the outputs.

Write the output of the following SQL commands.
(i) SELECT COUNT(DISTINCT Brand) FROM Drinks;
(ii) SELECT SUM(Price) FROM Drinks WHERE Brand=“Coca – Col a” ;
Or Write the output of the following SQL commands.
(i) SELECT MAX(Price).MIN(Price) FROM Drinks;
(ii) SELECT COUNT(Price) FROM Drinks;
Answer:

Section B
(Each question carries 3 Marks)

Question 8.
Mr. Siladitya wants to find outputs of some MySQL functions, although he is well conversant with some functions, but not all and specially when working with multiple functions together. Help him in finding the outputs in the following cases.
Write the output of the following SQL commands.
(i) SELECT UCASE(LEFT(‘school ’ ,4));
(ii) SELECT C0NCAT( LEFT(‘ Happy ’ , 2), RIGHT( ‘ Days ’, 2));
(iii) SELECT M0NTH(‘2021-09-08’);
Or
(i) SELECT SUBSTR(‘Christmas’,2,4);
(ii) SELECT SUBSTR( ‘ Wi ntervacati on ’ ,-5,3);
(ii) SELECT MIDC ’SCIENTIFIC’ ,3)
Answer:

Question 9.
Ankita is writing the following queries but not getting the proper outputs. Help her in correcting the errors.

(i) SELECT MAX(Salary) Employee;
Answer:
SELECT MAX(Salary) FROM Employee;

(ii) SELECT MEAN(Salary) FROM Employee;
Answer:
SELECT AVG(Salary) FROM Employee;

(iii) SELECT L0WEST(Sal ary) FROM Employee;
Answer:
SELECT MIN(Salary) FROM Employee;

Question 10.
Vijay is not clear about how he can use the following MySQL functions, explain him the uses of the following functions.
(i) TRIM()
Answer:
TRIM() It removes any extra spaces from right and left of a string,but not from the middle,
e.g.
SELECT TRIM(‘ Zebra crossing ’);

(ii) SUBSTR()
Answer:
SUBSTR() It extracts certain number of characters from any part of a string,
e.g. SELECT SUBSTR
(‘Elephantsinafrica’ ,2,4);

Output

(iii) RIGHT()
Answer:
RIGHT() It extracts characters from right of a string.
e.g. SELECT RIGHT(‘Writersbui1ding’,5);

Output

Section-C
(Each question carries 4 Marks)

Question 11.
Akhilesh a clerical staff in a Doctor’s Dispensary. He maintains records of visiting doctors in a table Doctor. He wants to analyse some data and find some results such as total number of doctors of certain departments , average charges in some departments etc. Help him in writing proper queries to get the results.

(i) To display the maximum charges among the Ortho doctors.
Answer:
SELECT MAX(Charges) FROM Doctor WHERE Dept=“0rtho”;

(ii) To display the doctor names in uppercase along with their department names concatenated.
Answer:
SELECT CONCAT(UCASE(Dname),Dept) FROM Doctor;

(iii) To display each department and the total number of doctors in them.
Answer:
SELECT Dept,COUNT(*) FROM Doctor GROUP BY Dept;

(iv) To display each department and the average charges of each.
Answer:
SELECT Dept, AVG(Charges) FROM Doctor GROUP BY Dept;

Question 12.
Simran has prepared a table storing details of some contractual job applicants. Her colleague who left the organisation, had written some queries to generate certain outputs from the table. She was told by her manager to find the outputs of the queries, Help her in finding the outputs.

(i) SELECT Name, JoinYear FROM Applicants WHERE Gender=‘F’ AND C_ID=‘A02’ ;
(ii) SELECT MIN(JoinYear) FROM Applicants WHERE l 3ender=‘M’;
(iii) SELECT AVG(Fee) FROM Applicants WHERE C_ID= ‘A01’ OR c_: [D=‘AO 5 ’ ;
(iv) SELECT SUM(Fee), C_ID FROM Applicants GROUP BY C_ID HAVING COUNT(*)=2;

OR
(i) SELECT MAX(Fee) FROM Applicants WHERE Name LIKE “A%”;
(ii) SELECT COUNT(DISTINCT Gender) FROM Applicants;
(iii) SELECT SUM(Fee) FROM Applicants WHERE Gender=“M”;
(iv) SELECT COUNT(*) FROM Applicants;
Answer:

Question 13.
Green Valley Public School has 4 buildings in its campus . Distance between the buildings and the number of computers in each is given below

Building	Number of Computers
A	150
B	10
C	25
D	30

 

Building	Distance
A-B	10 m
A-C	1250 m
A-D	25 m
B-C	30 m
B-D	2000 m

(i) Which building is best suitable for placement of server ?
Answer:
Building A, as it has maximum number of computers.

(ii) If building A to D is to be connected, which device will be required for strong signals ?
Answer:
A repeater would be required, as signals become weak in long distances which can be amplified by a repeater.

(iii) Which building would need a switch/hub ?
Answer:
Hub/switch would be required in all the buildings as they connect multiple computers.

(iv) Which topology would you suggest for connecting computers in each building ?
Answer:
Star topology would be the best as it has multiple positive features.