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 en one question of two marks and two questions of three marks You have to attempt only one of the choices in such questions
- You may use log tables if necessary but use of calculator is not allowed.

Section – A

Question 1.

Show that maximum intensity in interference pattern is four times the intensity due to each slit. Hence show that interference involves only redistribution of energy.

Answer:

Suppose the amplitude of waves from each slit is a. Therefore, intensity due to each slit = a^{2}.

When interference is destructive,

∴ Minimum intensity =(a – a)^{2} = 0

When interference is constructive,

∴ Maximum intensity = (a + a)^{2} = (2a)^{2} = 4a^{2} = 4 times the intensity due to each slit, which was to be proved.

Also average intensity in the interference pattern = \(\frac{0+4 a^{2}}{2}\) = 2a^{2} = sum of intensities due to two slits .

Hence we conclude that interference involves only redistribution of energy keeping the total energy fixed.

Question 2.

(a) According to de-Broglie, when is the wave associated with matter?

(b) Write the expression for the de-Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V.

OR

If the radius of second electron orbit in hydrogen atom be r then what will be the radius of the third orbit will be?

Answer:

(a) When it is in motion with any velocity.

(b) de-Broglie wavelength associated with the particle.

λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}\)

OR

r_{n} ∝ n^{2}

⇒ \(\frac{r_{2}}{r_{3}}\) = \(\frac{4}{9}\)\

⇒ r_{3} = \(\frac{9}{4}\)r

⇒ r_{3} = 2.25r

Question 3.

Classify the dopants used in the doping process.

Answer:

There are two types of dopants:

(i) Pentavalent dopant: The elements whose each atom has five valence electrons are known as pentavalent dopants. For example, Arsenic (As), Antimony (Sb), Phosphorus (P), etc.

(ii) Trivalent dopant: The elements whose each atom has three valence electrons are known as trivalent dopants. For example, Indium (In), Gallium (Ga), Aluminium (Al), etc.

Section – B

Question 4.

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.

Answer:

The distance of the nth 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_{1} = 3 × 10^{-3} m

D = 1.5 m

λ = 450 × 10^{-9} m

3 × 10^{-3} = \(\frac{1 \times(1 \cdot 5) \times\left(450 \times 10^{-9}\right)}{a}\)

a = \(\frac{1 \times(1.5) \times\left(450 \times 10^{-9}\right)}{3 \times 10^{-3}}\)

a = 0.225 mm

Question 5.

Find the de-Broglie wavelength associated with an electron moving with a velocity of 0.5 c and rest mass 9.1 × 10^{-31} kg. [Given h = 6.6 × 10^{-34} Js and c = 3 × 10^{8} ms^{-1}]

Answer:

Given,

Velocity of electron, υ = 0.5 c = 0.5 × 3 × 10^{8} = 1.5 × 10^{8} ms^{-1}.

Mass of electron = \(\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

= \(\frac{9.1 \times 10^{-31}}{\sqrt{1-(0.5)^{2}}}\)

\(\frac{9.1 \times 10^{-31}}{\sqrt{0.75}}\) = 10.51 × 10^{-31} kg

de-Broglie wavelength, λ = \(\frac{h}{m v}\)

= \(\frac{6.6 \times 10^{-34}}{10.51 \times 10^{-31} \times 1.5 \times 10^{8}}\)

= 0.4187 × 10^{-11} m

= 0.04187 Å.

Question 6.

Following data was recorded for values of object distance and the corresponding values of image distance, in the study of real image formation by a convex lens of power +5 D. One of these observations is incorrect. Identify and give reason:

Answer:

Here, P = + 5D

f = \(\frac{100}{P}=\frac{100}{5}\)

= 20 cm

In observation 3, where u = 35 cm, object lies between f and 2f. The image must be beyond 2f, i.e., υ > 40 cm. But υ = 37 cm, i.e., image also lies between/and 2f which is wrong.

Question 7.

In a typical nuclear reaction, e.g.,

\({ }_{1}^{2} \mathrm{H}\) + \({ }_{1}^{2} \mathrm{H}\) → \({ }_{2}^{3} \mathrm{He}\) + \({ }_{0}^{1} n\) + 3.27 MeV,

although number of nucleons is conserved, yet energy is released. How? Explain.

Answer:

In a given nuclear reaction, the sum of the masses of the target nucleus (\({ }_{1}^{2} \mathrm{H}\)) and the bombarding particle (\({ }_{1}^{2} \mathrm{H}\)) may be greater than the product nucleus (\({ }_{2}^{3} \mathrm{He}\)) and the outgoing neutron \({ }_{0}^{1} n\). So from the law of conservation of mass – energy some energy (3.27 MeV) is released due to mass defect in the , nuclear reaction. This energy is called Q – value of the nuclear reaction.

Question 8.

Deduce the relation μ = \(\frac{\text { Real depth }}{\text { Apparent depth }}\).

OR

(a) A liquid of refractive index 1.5 is poured into a cylindrical jar of radius 20 cm, upto a height of 20 cm. A small bulb is lighted at the centre of the bottom of the jar. Find the area of the liquid surface through which the light of the bulb passes into air.

(b) The above figure shows a solid glass sphere of radius 5 cm that has a small air bubble B trapped at a distance 2 cm from the centre C. The refractive index of the material of glass is 1.5. Find the apparent position of the bubble where it will appear, when seen through the surface of the sphere from an outside point E.

Answer:

Consider an object O placed in a liquid. A ray of light OA is incident along the normal to the surface of the liquid. Another ray OB, on entering from air to water bends away from the normal. The two refracting rays appear to come from point I, therefore, object though lying at depth OA, appears to be at depth AI. Accordingly, AO is called ‘real depth’ and AI is called ‘apparent depth’ of the object O.

The refractive index of the medium air V with respect to liquid ‘l’ is given by

OR

(a) InΔAOL, tan C = \(\frac{\mathrm{AO}}{\mathrm{OL}}\)

⇒ AO = OL tan C

⇒ r = 20 tan C

Also, μ = \(\frac{1}{\sin C}\)

or sin C \(\frac{1}{\mu}=\frac{1}{1 \cdot 5}=\frac{2}{3}\)

∴ tan C = \(\frac{2}{\sqrt{5}}\)

Thus, r = 20 × \(\frac{2}{\sqrt{5}}\)

Therefore, the required surface area,

A = πr^{2}

or A = π × \(\frac{40 \times 40}{5}\) = 320π cm^{2} = 1005.71 cm^{2}.

(b) Since, \(\frac{n}{v}-\frac{1}{u}=\frac{n-1}{\mathrm{R}}\)

Where, n = _{g}µ_{g} = \(\frac{1}{{ }_{a} \mu_{g}}=\frac{1}{1 \cdot 5}=\frac{2}{3}\)

Also u = -(5 – 2)

= -3 cm

and R = -5 cm

Thus \(\frac{2 / 3}{v}-\frac{1}{-3}=\frac{\frac{2}{3}-1}{-5}\)

On solving υ = – 2.5 cm.

Thus, the apparent position of the bubble will be 2-5 cm inside the sphere from the centre.

Question 9.

Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Answer:

The Rydberg formula is

The wavelengths of the first four lines in the Lyman series correspond to transitions from n_{i} = 2,3, n_{f} = 1. We know that

Substituting n_{i} = 2, 3, 4, 5, we get λ_{21} = 1218 Å, λ_{31} = 1028 Å, λ_{41} = 974.3 Å and λ_{51} = 951.4 Å.

Question 10.

Consider a two slit interference arrangement, (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of X such that the first minima on the screen falls at a distance D from the centre O.

Answer:

As it is clear from figure,

T_{2}P = TO + OP = S_{2}C + OP = D + x

T_{1}P =OT_{1} – OP = CS_{1} – OP = D – x

Now, S_{1}P = \(\sqrt{S_{1} \mathrm{~T}_{1}^{2}+\mathrm{T}_{1} \mathrm{P}^{2}}\) = [D^{2} + (D – x)^{2}]^{1/2}

and S_{2}P = \(\sqrt{\mathrm{S}_{2} \mathrm{~T}_{2}^{2}+\mathrm{T}_{2} \mathrm{P}^{2}}\) = [D^{2} + (D + x^{2})]^{1/2}

Path difference between the waves reaching P from S_{1} and S_{2}

= S_{2}P- S_{1}P

= [D^{2} + (D + x)^{2}]^{1/2} – [D^{2} + (D – x)^{2}]^{1/2}

For first minimum to fall at P, n = 1

Path duff. = S_{2}P – S_{1}P = \(\frac{1}{2}\)λ

or [D^{2} + (D + x)^{2}]^{1/2} – [D^{2} + (D – x)^{2}]^{1/2} = \(\frac{\lambda}{2}\)

if x = D

[D^{2} + 4D^{2}]1^{1/2} – D = \(\frac{\lambda}{2}\)

D(√5 – 1) = \(\frac{\lambda}{2}\)

D(2.236 – 1) = \(\frac{\lambda}{2}\)

or D = \(\frac{\lambda}{2.472}\).

Question 11.

(a) What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves?

(b) Give four basic properties of electromagnetic waves.

OR

In a plane electromagnetic wave, the electric field oscillates with a frequency of 2 * 1010 s_1 and amplitude of 40 Vm^{-1}.

(a) What is the wavelength of the wave?

(b) What is the energy density due to electric field?

Answer:

(a) Since electromagnetic waves are transverse in nature. We have electric and magnetic fields associated with an electromagnetic wave perpendicular to each other and perpendicular to the direction of propagation of electromagnetic waves.

Let the direction of electric field and magnetic field is along Y-axis and Z-axis then the direction of propagation of EM waves will be along positive X-axis.

(b)

(i) All electromagnetic waves travel with the speed of light.

(ii) \(\overrightarrow{\mathrm{E}}\) is perpendicular to \(\overrightarrow{\mathrm{B}}\) and both are perpendicular to the direction of propagation.

(iii) \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) oscillate in same phase.

(iv) The ratio of their amplitudes is equal to the velocity of waves in free space i.e., \(\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}\) = c.

OR

Wavelength. λ = \(\frac{C}{v}\)

= \(\) = 1.5 × 10^{-2}m

(b) Given: E_{0} = 40 Vm^{-1}

Energy density due to electric field = \(\frac{1}{2}\)ε_{0}\(\mathrm{E}_{\mathrm{rms}}^{2}\)

\(\frac{1}{2}\)ε_{0}(\(\frac{\mathrm{E}_{0}}{\sqrt{2}}\))^{2} = \(\frac{1}{4}\)ε_{0}\(\)

= \(\frac{1}{4}\) × 8.85 × 10^{-12} (40)^{2}

= 3.5 × 10^{-9} J/m^{3}.

Section – C

Question 12.

Light Emitting Diode (LED): LED is the photoelectronic device which converts electrical energy into the light energy. It has heavily doped p-n junction diode and it gives spontaneous radiation when it is connected in forward bias. In this the upper layer is of p-type semiconductor and lower layer is of n-type.

To control the brightness of light emitted by LED, a resistance is connected in the circuit with battery.

The specific materials used for making LED’s are Gallium-Arsenide-Phosphide (GaAsP) for yellow or red light, Gallium-phosphide (GaP) for red or green light etc.

LED’s are used in making calculators digital watches, burglar alarms, computers, picture phones, remote control and traffic light etc.

(a) LED is a:

(i) Photo device

(ii) Electrical device

(iii) Mechanical device

(iv) Photo electronic device

(b) p-n junction diode used in LED’s is:

(i) Lightly doped

(ii) Moderately doped

(iii) Heavily doped

(c) The brightness of light is controlled by using …………………. in the circuit.

(i) Inductor

(ii) Resistance

(iii) Capacitor

(d) The lower layer of LED’s is of ………………… semiconductor.

(i) p-type

(ii) n-type

(iii) q-type

(iv) Both (i) and (ii)

(e) What kind of material is used in making LED’s:

(i) GaAsP

(ii) GaP

(iii) GaAs

(iv) All of these

Answer:

(a) (iv) Photo electronic device

(b) (iii) Heavily doped

(c) (ii) Resistance

(d) (ii) n-type

(e) (iv) All of these