CBSE Sample Papers For Class 12 Physics Mock Paper 1

Q1. The electric dipole moment of an electron and a proton 4 nm apart is

Opt:

6.4 × 10^-24 C m

6.88 × 10^-24 C m

6.4 × 10^-28 C m

8.4 × 10^-28 C m

Ans:

6.4 × 10^-28 C m

Q2. Force acting upon a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed then the force acting on the same particle will become

Opt:

zero

F/2

F

2F

Ans:

F/2

Q3. A battery of internal resistance r having no load resistance has an e.m.f E volt. The observed e.m.f across the terminals of the battery when a load resistance R(=r) is connected to its terminals is

Opt:

E

2E

E/2

E/4

Ans:

E/2

Q4. Two identical cells are first connected in series then in parallel. The ratio of power consumed in the two cells
Opt:

1:1.

1:4.

2:1.

4:1.
Ans:

1:1.

Q5. An electron is revolving at 5 × 10¹¹ cycles/sec, if the radius is 3.14 × 10^-15 m. then the magnetic field produced at the center is

Opt:

4 Tesla

8 Tesla

16 Tesla

32 Tesla

Ans:

16 Tesla

Q6. Magnetic field strength due to a short bar magnet on its axial line at a distance r is B. Magnetic field strength at the same distance on equatorial line is

Opt:

B/2

B

2B

4B

Ans:

B/2

Q7. Complete the analogy.

Electric generator: Electromagnetic induction:: Electromagnetic brake

Opt:

Eddy current

Superconduction

Hydraulic resistance

Viscous drag

Ans:

Eddy current

Q8. In an LCR the capacitance is changed from C to 5C. For the same resonant frequency , the inductance should be changed from L to

Opt:

2L

L/5

L/4

4L

Ans:

L/4

Q9. The electromagnetic waves having the highest penetrating power are

Opt:

Gamma rays

U-V rays

Infrared rays

X-rays

Ans:

Gamma rays

Q10. The transition from state n = 4 to n =3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

Opt:

2 → 1

3 → 2

4 → 2

5 → 4

Ans:

5 → 4

Q11. The momentum of a photon of a radiation of wavelength 330 nm is

Opt:

5.0 × 10^–28 kg m/s

2.0 × 10^–28 kg m/s

2.0 × 10^–27 kg m/s

5.0 × 10^–27 kg m/s

Ans:

2.0 × 10^–27 kg m/s

Q12.The half-life of a radioactive substance is 10 days. It means that
Opt: The whole of the substance will disintegrate in 20 days. half of the substance will disintegrate in 20 days. one-fourth of the substance will disintegrate in 5 days. three-fourth of the substance will disintegrate in 20 days.
Ans: three-fourth of the substance will disintegrate in 20 days.

Q13. The focal length of magnifying lens is 0.05 m. If distance of distinct vision is 0.25 m then its magnifying power is

Opt:

3

5

6

9

Ans:

6

Q14. For silicon, the width of the depletion region is of the order of 10^-6 m and the potential barrier is 0.7 V. The magnitude of barrier electric field for a silicon junction is

Opt:

5 × 10³ V/m

7 × 10³ V/m

6 × 10⁴ V/m

7 × 10⁵ V/m

Ans:

7 × 10⁵ V/m

Q15. For a given medium, the polarising angle is 60^o. The critical angle for this medium will be

Opt:

35º16’

53º16’

55º16’

75º16’

Ans:

35º16’

Q16. The near point of a long-sighted person is 100 cm. The power of the lens, which enables him to see the objects at a distance 20 cm or more than that distinctly, is

Opt:

+1 D

+2 D

+3 D

+4 D

Ans:

+4 D

Q17. In an LCR circuit 10 Ω resistance, 10 mF capacitor and a 5 mH coil are connected in series. When a alternating current source of suitable frequency is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency

Opt:

gets doubled

gets halved

gets quadrupled

remains unchanged

Ans:

remains unchanged

Q18. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Ans:

Here, applied magnetic field, B = 0.25 T
Magnetic moment, M = 0.6 T⁻¹
Angle between the axis of the solenoid and the direction of the applied field, θ=30°

Torque acting on the solenoid is given as,τ=MBsinθ∴τ=0.6×0.25sin30o=0.075J.

Q19. Plutonium decays with a half cycle of 24000 years. If the plutonium is stored for 72000 years, what fraction of it remains?

OR

A stream of electrons traveling with a speed ‘v’ m/s at right angles to a uniform electric field ‘E’ is deflected in a circular path of radius ‘r’.
Prove that,

em=v2rE \frac{\mathrm{e}}{\mathrm{m}}=\frac{{\mathrm{v}}^2}{\mathrm{rE}}

me​=rEv2​
Ans:

We have the relation,

NNo=(12)tT1/2 \frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{T}}_{1/2}}}

No​N​=(21​)T1/2​t​

Substituting the given values in the above equation we get,

NNo=(12)7200024000=18. \frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}={\left(\frac{1}{2}\right)}^{\frac{72000}{24000}}=\frac{1}{8}.

No​N​=(21​)2400072000​=81​.
Therefore after 72000 years plutonium that will remain will be 1/8 of its original value.

OR

We know that,

Force on electron due to electric field = Centripetal force.
F_E = F_C
Substituting the required values on L.H.S and R.H.S, we get,

eE=mv2ror,em=v2rE \mathrm{eE}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}\phantom{\rule{0ex}{0ex}}\mathrm{or},\quad \frac{\mathrm{e}}{\mathrm{m}}=\frac{{\mathrm{v}}^2}{\mathrm{rE}}

eE=rmv2​or,me​=rEv2​

Q20. A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?

Ans:

Here f_o = 144 cm and f_e = 6.0 cm
∴ Magnifying power =

–fofe=–1446=–24

and Separation between the two lenses = f_o + f_e = (144 + 6) cm = 150 cm or 1.5 m.

Q21. If the voltage in an a.c. circuit is given by v = 220√2 sin(314t – θ) then find the
a) peak and rms value of voltage, and
b) frequency of a.c.

Ans:

Given

Q22. In a single slit diffraction experiment, if the width of slit is doubled, how does the
1. intensity of light and
2. width of the central maximum change?
Give reason for your answer.

Ans:

1. If the width of slit is doubled then intensity of light also gets doubled. This is because intensity of light is directly proportional to slit width.
2. The width of central maximum will become half. Central maximum is given by the relation,

βo=λDd

, here d=width of slit.

Q23. Four metallic plates each with surface area ‘A’ and inter plate separation ‘d’ are placed as shown. Alternate plates are connected to points A and B. What is the equivalent capacitance of the system?

Ans:

The given arrangement equivalent to three capacitors.
Ist capacitor is formed joining plate 1 and plate 2,
2nd capacitor is formed by joining plate 2 and plate 3,
3rd capacitor is formed plate 3 and plate 4.
The three capacitors, each of capacity,
are joined in parallel.

Q24. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 ms⁻²

Ans:

\begin{array}{l}\text{Here,}l\text{ength of rod},\text{l}=0.\text{45 m}\\ \text{Mass suspended}\text{by}\text{vertical}\text{wires},\text{m}=\text{6}0\text{g}\\ =\text{6}0\times \text{1}{0}^{-\text{3}}\text{kg}\\ \text{Acceleration due to gravity},\text{g}=\text{9}.{\text{8 ms}}^{\text{-2}}\\ \text{Current in the rod passing through the wire},\text{I}=\text{5 A}\\ \left(\text{a}\right)\text{Magnetic field strength}\text{is equal and opposite to the}\\ \text{weight}\text{of the wire i}.\text{e}.,\\ BIl=mg\\ \therefore B=\frac{mg}{Il}=\frac{\text{6}0\times \text{1}{0}^{-\text{3}}\times \text{9}.\text{8}}{5\times 0.\text{45}}=0.26T\\ \text{A horizontal magnetic field of}0.\text{26 T perpendicular to}\\ \text{the length of the conductor should be set}\text{up in order}\\ \text{to get zero tension in the wire}.\\ \text{The magnetic field must be such that}the\text{Fleming}’\text{s left}\\ \text{hand rule gives an upward magnetic force}.\\ \left(\text{b}\right)\text{If}\text{we reverse}\text{the direction of current},\text{then the}\\ \text{force due to magnetic field and}\text{the weight of the wire}\\ \text{act in the vertically downward direction}.\\ \therefore \text{Total tension in the wire}=\text{BIl}+\text{mg}\\ \text{=}0.\text{26}\times 5\times 0.\text{45}+\left(\text{6}0\times \text{1}{0}^{-\text{3}}\right)\times 9.8\\ =1.176N\end{array}

Q25. Calculate the rate at which the flux linked with the generated area changes with time when a rod of length l is (a)translated (b) rotated in a uniform magnetic field B as shown in the figur

Ans:

Q26. For photoelectric effect in a metal, the figure shows the plot of cut-off voltage versus frequency of incident radiation.

Calculate:

(a) The threshold frequency.

(b) The work function for metal.

OR

An electron and proton have the same kinetic energy.

(a) Which of the two has greater wavelength? Justify your answer.

(b) What is the momentum of a photon of frequency

$\mathrm{\nu }$

ν ?

Opt:

For photoelectric effect in a metal, the figure shows the plot of cut-off voltage versus frequency of incident radiation.

Calculate:

(a) The threshold frequency.

(b) The work function for metal.

OR

An electron and proton have the same kinetic energy.

(a) Which of the two has greater wavelength? Justify your answer.

(b) What is the momentum of a photon of frequency

$\mathrm{\nu }$

ν ?

Ans:

(a) For the threshold frequency the cut off potential is zero. From the graph given above it follows that for frequency of 2.5 x 10¹⁴ Hz the cut off voltage is zero.

Hence the value of cut off frequency,

νo=2.5×1014Hz. {\mathrm{\nu }}_{\mathrm{o}}=2.5\quad \times {10}^{14}\mathrm{Hz}.

νo​=2.5×1014Hz.

(b) The work function of any metal is given by relation,

W=hνoeeV=6.63×10−34×2.5×10141.6×1019 \mathrm{W}=\frac{{\mathrm{h\nu }}_{\mathrm{o}}}{\mathrm{e}}\mathrm{eV}=\frac{6.63\times {10}^{-34}\times 2.5\times {10}^{14}}{1.6\times {10}^{19}}

W=ehνo​​eV=1.6×10196.63×10−34×2.5×1014​
W=1.0359 eV

OR

(a) The de Broglie wavelength for a particle of kinetic energy K and mass m is given by the relation,

λ=h2mK \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mK}}}

λ=2mK​h​
Given that the kinetic energy of the particles are same, hence the wavelength of particle will only depend on mass of particle.

∴λ∝1m \therefore \mathrm{\lambda }\propto \frac{1}{\sqrt{\mathrm{m}}}

∴λ∝m​1​
The mass of proton is greater than that of electron, hence electron will have greater wavelength.
(b) The momentum of a photon of frequency

ν \mathrm{\nu }

ν is given by,

P =

Ec=hνc \frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{h\nu }}{\mathrm{c}}

cE​=chν​

Q27. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and n = 3 orbits?

Ans:

As radius of orbit of hydrogen atom is given byrn∝​ n2rn = kn2where k = 4πε0h24π2me2hence, for n = 1r1 = k(1)2 or k = r1 = 5.3×10-11 mwhere r1 is the radius of innermost electron orbit.similarly for n = 2r2 = k(2)2 = 4 kwhere, k = 5.3×10-11 m∴r2 = 4×5.3×10-11 m = 2.12×10-10 msimilarly for n = 3r3 = k(3)2 = 9kr3 = 9×5.3×10-11 m = 4.77×10-10 m

Q28. What is choke coil? Why is it preferred to resistance in a.c. circuit? If the frequency of emf is increased, then what will be the effect on the currents flowing through the circuits given below?

OR

A 100 μF capacitor in series with a 50 resistance is connected to a 150 V, 30 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the phase difference between current and voltage?

Ans:

A choke coil is made up of copper wire. It has got very high inductance with very small resistance. In such a coil, the current will lag behind the voltage by almost by

\pi

π​ /2,

thereby reducing the power consumption in the coil to a negligible value.

Thus a choke coil can reduce the current flowing in an a.c. circuit without consuming much power.

The same purpose can be achieved through resistance too but there will be a huge loss of power in this case which is given mathematically by,

I²_rms

.R.

Figure (a) contains resistance only hence current flowing through the circuit does not depend upon frequency.

Figure (b) contains inductor only; here current flowing through the circuit will be given by,

From the above equation it follows that as frequency in the circuit is increased then current in the circuit decreases.

Figure (c) contains capacitor only; here current flowing through the circuit will be given by,

From the above equation it follows that as frequency in the circuit is increased the current in the circuit increases.

OR

Q29. Give an expression for Torque experienced by the dipole in a uniform electric field.

What is the potential energy of a dipole placed in an external field?
What will be the torque experienced by the dipole if it is in the direction of the electric field?

OR

A 300pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 300pF capacitor. How much energy is lost in the process?
Ans:

Consider a dipole as shown in fig below with charges q₁ = +q and q₂ = -q placed in an uniform electric field E

Then the torque experienced by the dipole,

τ = p x E.
Suppose an external torque
τ_ext
applied in such a manner that it just neutralizes this torque and rotates it in the plane of paper from θ

_o to θ

₁ without angular acceleration.

Amount of work done by external torque,

The torque experienced by the dipole if it is in the direction of the electric field = 0

OR

Q30. (a) What is the effect on the relaxation time of electrons in a metal if temperature is increased?

(b) An electric cable of copper has just one wire of radius 9 mm. Its resistance is 5 ohm. Five different well-insulated copper wires of radius 3mm each replace this single wire of cable. What will be the equivalent resistance of cable?

Or

(a) A circuit shown in the fig. has potential 6V, 3V, and 2V at points A, B and C respectively. Find the potential at point O.

(b) A battery of emf 6V and internal resistance 1 ohm is connected to a resistor. If the current in the resistor is 0.5A, find the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Ans:

(a) The relaxation time of electrons in a metal decreases with the increase in temperature.

(b)

OR

(a)

(b)

Q31. (a) The focal length of objective and eyepiece of a telescope are 20 cm and 2 cm respectively. An object at a distance of 200 cm from the objective is focused by the telescope. The final image is formed at a distance of 40 cm from the eyepiece. Calculate the length of the telescope.

(b) A double concave lens of glass of refractive index 1.6 has radii of curvature of 40 cm and 60 cm. Calculate its focal length in air.

OR

(a) Define critical angle with reference to total internal reflection. Calculate the critical angle for glass-air surface, if a ray of light which is incident in air on the glass surface is deviated through 15^o, when angle of incidence is 45^o.

(b) What are optical fibres? Give their one use.
Ans:

(a)

(b) Here n = 1.6, for a double concave lens R₁ is –ve but R₂ = +ve,
Hence R₁ = -40 cm and R₂ = 60 cm
From lens maker’s formula

OR

(a) The critical angle for a pair of media may be defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90^o

and beyond which the light is totally internally reflected back in the denser medium itself.
In the given problem angle of incidence in air i = 45^o

As the ray deviates through 15^o

and glass is optically denser,
hence r = 45^o

– 15^o

= 30^o

∴ Refractive index of glass

(b) Optical fibres consist of many, long, high quality composite glass/quartz fibres. Each fibre consists of a central core of higher refractive index and a cladding of lesser refractive index. Optical fibres work on the principle of total internal reflection and are used for optical signal transmission.

Q32. Case Study:

Read the following paragraph and answer the questions.

According to Huygen, light is a wave. It is proved experimentally by Young’s double slit experiment (YDSE).
In YDSE setup, S is a narrow slit illuminated by a monochromatic source of light sends wave fronts in all directions. Slits S1 and S2 become the source of secondary wavelets which are in phase and of same frequency. These waves are superimposed on each other which give rise to interference.

Alternate dark and bright bands are obtained on a screen (called interference fringes) placed certain distance from the plane of slit S1 and S2.

Following figure shows an experimental set up similar to Young’s double slit experiment to observe interference of light.

Here SS₂-SS₁=(λ/4), Write the condition of (i) Constructive (ii) Destructive, interference at any point P is in terms of path difference.

(iii) Does the central fringe observed in the above set up lie above or below O? Give reason in support of your answer.

OR

(iii) In a single slit diffraction experiment, how does the intensity of light and width of the central maximum change if the width of slit is doubled? Give reason for your answer.
Ans:

(i) Constructive interference is given by,

∆total=∆initial+∆=nλ {∆}_{\mathrm{total}}={∆}_{\mathrm{initial}}+∆=\mathrm{n\lambda }

∆total​=∆initial​+∆=nλ , where n = 0, 1, 2….

Given that,

∆initial=λ4 {∆}_{\mathrm{initial}}=\frac{\mathrm{\lambda }}{4}

∆initial​=4λ​
.
Substituting the value Δ_initial in the above equation we get,

(ii) Similarly destructive interference is given by,

∆total=∆initial+∆=(2n−1)λ2, {∆}_{\mathrm{total}}={∆}_{\mathrm{initial}}+∆=(2\mathrm{n}-1)\frac{\mathrm{\lambda }}{2},

∆total​=∆initial​+∆=(2n−1)2λ​, where n = 1,2,3….

Substituting value of

∆initial=λ4 {∆}_{\mathrm{initial}}=\frac{\mathrm{\lambda }}{4}

∆initial​=4λ​

in the above equation we get,

(iii) The value of central fringe is given by substituting the value of n = 0 in the equation of constructive interference i.e.

∆=(n−14)λ ∆=\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{\lambda }

∆=(n−41​)λ , here substituting n = 0 we get,

∆=−λ4 ∆=-\frac{\mathrm{\lambda }}{4}

∆=−4λ​

The negative sign shows that the central fringe shifts below point O.

OR

If the width of slit is doubled then intensity of light also gets doubled. This is because intensity of light is directly proportional to slit width.

So, the width of central maximum will become half. Central maximum is given by the relation,

βo=λDd {\mathrm{\beta }}_{\mathrm{o}}=\frac{\mathrm{\lambda D}}{\mathrm{d}}

βo​=dλD​ , here d=width of slit.