RBSE 12th Physics Model Paper Set 3 with Answers in English 

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RBSE Class 12 Physics Model Paper Set 3 with Answers in English

Time . 2 Hours 45 Min.
Max Marks: 56

General Instruction to the Examinees:

• Candidate must write first his/her Roll No. on the question paper compulsorily.
• All the questions are compulsory.
• Write the answer to each question in the given answer book only.
• For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A

Question 1.
Write the Correct Answer from multiple choice question 1 (i to ix) and write in given answer book:

(i) A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium, if pis equal to: [1]
(a) –\(\frac{Q}{2}\)
(b) –\(\frac{Q}{4}\)
(c) +\(\frac{Q}{4}\)
(d) +\(\frac{Q}{2}\)
Answer:
(b) –\(\frac{Q}{4}\)

(ii) Figure shows the field lines of a positive pointU charge. The work done by the field in moving a small positive charge from Q to P is: [1]

(a) Zero
(b) Positive
(c) Negative
(d) Data insufficient
Answer:
(c) Negative

(iii) A wire of resistance 12 ohms per metre is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points, A and B as shown in the figure below is: [1]

(a) 6 π Ω
(b) 3 Ω
(c) 6 Ω
(d) 0.6 π Ω
Answer:
(b) 3 Ω

(iv) A charged particle would continue to move with a constant velocity in a region wherein, which of the fo1lowing conditions is not correct? [1]
(a) E = 0,B ≠ 0
(b) E ≠ 0 B ≠ 0
(c) E ≠ 0,B = 0
(d) E = 0,B = 0
Answer:
(c) E ≠ 0,B = 0

(v) Direction of current induced in wire moving in a magnetic field is found using: [1]
(a) Fleming’s left hand rule
(b) Fleming’s right hand rule
(c) Ampere’s rule
(d) Right hand clasp rule
Answer:
(b) Fleming’s right hand rule

(vi) The figure given below shows stopping potential V0 and frequency v for two different metallic surfaces A and B. The work function of A, as compared to that of B is: [1]

(a) less
(b) more
(c) equal
(d) can be said
Answer:
(a) less

(vii) The average binding energy per nucleon is maximum for the nucleus: [1]
(a) ₂He⁴
(b) ₈O¹⁶
(c) ₂₆Fe⁵⁶
(d) ₉₂U²³⁸
Answer:
(c) ₂₆Fe⁵⁶

(viii) Bonding in a germanium crystal is: [1]
(a) Metallic
(b) Ionic
(c) Vander Waal’s type
(d) Covalent
Answer:
(d) Covalent

(ix) In an insulator, the forbidden energy gap between the valence bond and conduction band is of the order of: [1]
(a) 1 MeV
(b) 0.1 MeV
(c) 1 eV
(d) 5 eV
Answer:
(d) 5 eV

Question 2.
Fill in the blanks:

(i) The induction of electric dipole moment in a dielectric substance in the presence of external electric field is called the ……………….. .[1]
(ii) Unit of mobility is ………………….. .[1]
(iii) The mathematical form of Gauss law for magnetism is ………………………. . [1]
(iv) …………………….. circuits process a signal in the form of continuous, time varying voltage or current. [1]
Answer:
(i) polarisation of dielectric substance
(ii) m²s^-1 volt^-1
(iii) \(\oint_{s} \overrightarrow{\mathrm{B}} \cdot d \vec{A}\) = 0
(iv) Analog.

Question 3.
Give the answer of the following questions in one line:

(i) How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant? [1]
Answer:
The mobility of electrons in a conductor is given by
μ = \(\frac{e \tau}{m}\)
where, e = charge on electron, m = mass of electrons and τ = relaxation time.
Also, τ ∝ T.
But here temperature (T) is kept constant. As mobility is independent of potential difference, so there is no change in it.

(ii) A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency? [1]
Answer:
As we know that in a circular path, frequency of a charged particle is given by υ = \(\frac{q B}{2 \pi m}\) or υ ∝ \(\frac{1}{m}\) Since, m > m_e, therefore electron will move in circular path with higher frequency.

(iii) Why is the core of a transformer laminated? [1]
Answer:
The core of a transformer is laminated to prevent eddy current being produced in the core.

(iv) State de-Broglie hypothesis. [1]
Answer:
de-Broglie hypothesis : A moving object sometimes acts as a wave and sometimes as a particle or a wave is associated with the moving particle which control this particle in very respect. This wave associated with the moving particle is called matter wave or de-Broglie wave. Its wavelength is given by
λ = \(\frac{h}{m v}\)
where, h = Planck’s constant, m = mass of object, υ = velocity of the object.

(v) Write the relationship of de-Broglie wavelength X associated with a particle of mass m in terms of its kinetic energy E. [1]
Answer:
Kinetic energy, K = \(\frac{p^{2}}{2 m}\)
p = momentum, m = mass and
K = kinetic energy
⇒ p = \(\sqrt{2 m K}\)
de-Broglie wavelength, λ = \(\frac{h}{p}\)
where, p = \(\sqrt{2 m K}\)
λ = \(\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m E}}\) [K = E]

(vi) A nucleus undergoes (1-decay. How does its: [1]
(i) mass number, and
(ii) atomic number change?
Answer:
During β-decay,
(a) no change in mass number.
(b) atomic number increases by 1.

(vii) Write the relationship between the size and the atomic mass number of a nucleus. [1]
Answer:
Relationship between the size and the atomic mass nubmer of a nucleus is
\(\frac{R}{R_{o}}\) A^1/3 ⇒ R = R_oA^1/3

(viii) The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Spacing Identify the region, if any over which the semiconductor has a negative resistance. [1]

Answer:
Resistance of a material can be found out by the slope of the curve V versus I. Part BC of the curve shows the negative resistance as in this region current decreases by increasing the voltage.

Section – B

Question 4.
If two similar large plates, each of area A having surface charge densities + σ and -σ are separated by a distance d in air, find the expression for:
(a) Field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) The potential difference between the plates.
(c) The capacitance of the capacitor so formed. [1½]
Answer:
According to questions,

(a) Electric field due to a plate of positive charge at point, P = σ/2ε₀ Electric field due to other plate = σ/2ε₀
Since, they have same direction, so
E_net = \(\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\)
Outside the plate, electric field will he zero because of opposite direction.

(b) Potential difference between the plates is given by
V + Ed – σd/ε₀
( E = σ/ε₀)

(c) Capacitance of the capacitor is given by ( Q = CV)
C = Q = \(\frac{Q}{V}=\frac{\sigma A}{\sigma d}\) ε₀ = \(\frac{\varepsilon_{0} A}{d}\)

Question 5.
Answer carefully: (a) Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \(\frac{Q_{1} Q_{2}}{4 \pi \varepsilon_{0} r^{2}}\) where r is distance between their centres?
(b) If Coulomb law involved 1/r³ dependence (instead of 1/r²,) would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line of force passing through that point? [1½]
Answer:
(a) When the charged spheres are brought close together, the charge distributions on the spheres become non-uniform. Therefore. Coulomb’s law is not valid. Hence the magnitude of force is not given exactly by this formula.
(b) No, Gauss’s law will not be true if Coulomb’s law involved 1/r³ dependence instead of 1/r² dependence.
(c) The line of force gives the direction of acceleration of charge. If the electric line of force is linear, the test charge will move along the line. If the line of force is nonlinear, the charge will not go along the line.

Question 6.
Describe briefly with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell. [1½]
Answer:
Measurement of internal resistance of a cell using potentiometer

The cell of emf, E (internal resistance r) is connected across a resistance box (R) through key K₂.
E = Φl₁ …………….. (i)
when K₂ is open balance length is obtained at length,
AN₁ = l₁
∴ v = Φl₂ ……………….. (ii)
From Eqs. (i) and (ii), we get
\(\frac{E}{V}=\frac{l_{1}}{l_{2}}\) ……………… (iii)
E = I(r + K) and V = IR
\(\frac{E}{V}=\frac{r+R}{R}\) ……………. (iv)
From Eqs. (iii) and (iv), we get
\(\frac{R+r}{R}=\frac{l_{1}}{l_{2}}\) ⇒ \(\frac{r}{R}=\frac{l_{1}}{l_{2}}\) – 1
r = R(\(\frac{l_{1}}{l_{2}}\) – 1)
We known l₁; l₂ and R, so we can calculate r.

Question 7.
Three cells of emf E, 2E and 5E having internal resistances r, 2r and 3r respectively are connected across a variable resistance R as shown in the figure. Find the expression for the current. Plot a graph for variation of current with R. [1½]

Answer:
Here, E₁ = E, E₂ = – 2E and E₃ = 5E, r₁ = r,r₂ = 2r and r₃ = 3r Equivalent emf of the cell is E = E₁ + E₂ + E₃
= E – 2E + 5E = 4E
Equivalent resistance

= r₁ + r₂ + r₃ + R
= r + 2r + 3r + R = 6r + R
∴ Current, I = \(\frac{4 E}{6 r+R}\)
The graph for variation of current I with resistance R is shown above.

Question 8.
Define mutual inductance between two long coaxil solenoids. Find out the expression for the mutual inductance of inner solenoid of lenght l having the radius r₁ and the number of turns n₁ per unit length due to the second outer soldenoid of same length and n₂ number of turns per unit length. [1½]
Answer:
Mutual Inductance of Two Concentric Coils of Different Radii
Let two co-axial coils S₁ and S₂ of radii r₁ and r₂ (such that r₁ < r₂) be placed in air. If the current I flows through the S₂, then magnetic flux linked with the coil S₁ is given by
Φ_B = B₂A = (\(\frac{\mu_{0} I}{2 r_{2}}\))πr₁² ……….. (1)

But by the definition of mutual inductance
Φ_B = MI …………… (2)
From the equation (1) and (2) we get
MI = \(\frac{\mu_{0} I}{2 r_{2}}\)πr₁²
∴ M = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)
Since M₁₂ = M₂₁ = M = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)
= \(\frac{\mu_{0}}{4 \pi}\left[\frac{2 \pi^{2} r_{1}^{2}}{r_{2}}\right]\)

Question 9.
A wheel with 10 metallic spokes, each 0.50 m long, is rotated with a speed of 120 rev/min in a plane normal to the earth’s magnetic field B_e at the place. If B_e = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1G = 10^-4 T. [1½]
Answer:
We are given that radius of the wheel, r = 0.50 m angular frequency of rotation of the wheel,
f = 120/60 = 2rps
Earth’s magnetic field,
B_e = 0.40 G = 0.40 × 10^-4 T
Let ε be induced emf across the ends of a spoke due to its motion in the magnetic field.
Area swept by the spoke in one second = (πr²f)
Thus, ε = Bπr²f = [(0.40 × 10^-4)
(3.14) (0.50)² (2)] V
or ε = 6.28 × 10^-5 V

Question 10.
How is the working of a telescope different from that of a microscope? [1½]
Answer:
Differences between telescope and microscope are given as below :

Characteristics	Telescope	Microscope
Position of object	At infinity	Near objective at a distance lying between fo and 2fo
Position of image	Focal plane of objective	Beyond 2fo when fo is the focal length of objective.

Question 11.
Use the mirror equation to deduce that an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [1½]
Answer:
From mirror formula
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{u f}\)
υ = \(\frac{u f}{u-f}\)
For concave mirror
υ = \(\frac{(-u)(-f)}{(-u)-(-f)}\)
υ = \(\frac{u f}{f-u}\)
Since, the object is between pole and the focus, therefore,
0 < u < f (f – u) > 0
and υ = \(\frac{u f}{(f-v)}\) = positive quantity.
i. e., the image will form erect, virtual and behind the mirror.
Magnification of the image, m = \(\)
m = \(\frac{\frac{u f}{(f-u)}}{u}=\frac{f}{f-u}\)
(f – u) < f ∴ m > 1
i. e., the size of the image will be larger than that of the object.

Question 12.
The objective of an astronomical telescope has a diameter of 150 mm and a focal length of 4.00 m. The eyepiece has a focal length of 25.00 mm. Calculate the magnifying and resolving power of telescope. (λ. = 6000 Å for yellow colour). [1½]
Answer:
The diameter of objective of the telescope
= 150 × 10^-3 m
f_o = 4m
f_e = 25 × 10^-3 m and D = 0.25 m
Magnifying power,

Question 13.
What are the advantages of a pair of eyes instead of a single eye? [1½]
Answer:
There are following advantages of pair of eye instead of a single eye.

• There is a correct estimation of distance by both eyes instead of one eye. This is why it is difficult to fill ink in a pen or string in the hole of needle by one eye.
• Instead of one eye, both eyes increase range of vision.
• Instead of one eye, both eyeis help us to see 3-D shape of and object.

Question 14.
Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei 20 < A < 240. How do you explain the constancy of binding energy per nucleon in the range of 30 < A < 170 using the property that nuclear force is short-ranged? [1½]
Answer:
The binding energy per nucleon curve is shown as below:

The binding energy per nucleon in the range of 30 < A < 170 has average binding energy per nucleon = 8.5 MeV. The higher value of binding energy per nucleon is due to stability of these nucleons. Neutron-proton ratio is higher in this range of mass number which leads to stability of the nuclei. Also, the nuclear force is strongly attractive enough to overcome the coulombian repulsive force acting between positively charged protons.

Question 15.
The half-life of \({ }_{92}^{238} \mathrm{U}\) undergoing α-decay is 4.5 × 10⁹ years. What is the activity
of lg sample of \({ }_{92}^{238} \mathrm{U}\) ? [1½]
Answer:
T_1/2 = 4.5 × 10⁹ years
= 1.42 × 10¹⁷ s
( 1 years = 3.16 × 10⁷ s)
Number of atoms,
N = \(\frac{6.023 \times 10^{23}}{238}\)
= 25.3 × 10²⁰ atoms
Activity (\(\frac{d N}{d t}\)) = λN
\(\frac{0.693}{T_{1 / 2}}\) × 25.3 × 10²⁰
\(\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}\)s^-1
= 1.23 × 10⁴ s^-1
= 1.23 × 10⁴ Bq ( 1S^-1 = 1Bq)

Section – C

Question 16.
Drive the expression for force per unit length between two long straight parallel current carrying conductors. Hence, define one ampere. [3]
Or
Find the magnetic field at a point on the axis of a circular coil carrying current and hence find the magnetic field at the centre of the circular coil carrying current. [3]
Answer:
Suppose two long straight current carrying wires AB and CD having current I₁ and I₂ respectively are placed parallel to each other at a distance of r as shown in the diagram. Each wire is situated in magnetic field of other wire, therefore both of them will experience magnetic force.

When the current in both wires is in same direction, then they will experience force of attraction between them and when the currents are in opposite directions, then they experience a force of repulsion between them. The directions of force can be decided by right hand rule (for magnetic field produced) and Fleming’s left hand rule (for force experienced). This force was calculated by applying Biot-Savart’s law and Lorentz’s force therefore this is also known as ‘Ampere’s law’. It can be explained as under : According to Biot-Savart’s law, the magnetic field produced by current Ix in the position of wire CD,
B₁ = \(\frac{\mu_{0}}{2 \pi} \frac{I_{1}}{r}=\frac{\mu_{0}}{4 \pi} \frac{2 I_{1}}{r}\) NA^-1m^-1

According to Right Hand Palm Rule rio.l, the direction of this magnetic field will be normal to plane of the paper inwards. Since wire CD is situated in this field at right angles to field, therefore force acting on CD,
F₂ = I₂B₁l sin 90° = I₂B₁l
= I₂\(\frac{\mu_{0}}{2 \pi} \frac{I_{1}}{r}\)l
or F₂ = \(\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2} l}{r}\)N ……………. (1)
or F₂ = 2 × 10^-7 \(\frac{I_{1} I_{2} l}{r}\)
The force acting on unit length, similarly, the force acting on wire AB
F₁ = \(\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2} l}{r}\)N ……………… (2)
F₁ = 2 × 10^-7\(\frac{I_{1} I_{2} l}{r}\)N
From equations (1) and (2), it is clear that
F₁ = F₂ = F
Thus, we can write, the magnetic force between two parallel current carrying wires,
F = \(\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2} l}{r}\)
or F = 2 × 10^-7\(\frac{I_{1} I_{2} l}{r}\)
If l = 1 m, then
F = 2 × 10^-7\(\frac{I_{1} I_{2}}{r}\)N.m^-1
If I₁ = I₂ = 1A; r = 1m; then F = 2 × 10^-7Nm^-1
i.e., “If two wires are placed at lm apart parallel to each other and the force (attraction or repulsion) acting between them is 2 × 10^-7Nm^-1 then the current flow in a current carrying conductors are one ampere.

Question 17.
(i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working.
(ii) Why must both the objective and they eyepiece of a compound microscope have short focal lengths? [3]
Or
Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eyepiece. [3]
Answer:
(i) Adjustment and Formation of Image: In adjustment process, eye lens or eye pieces is adjusted first. For this, eye piece is moved forward or backward up to such extent that cross wire should be seen clearly. Now the object is placed infront of field lens and it is moved to adjust in such a way that the clear image of the object would be seen. In this situation inverted, larger and virtual image of the object forms on cross-wire. The ray diagram of formation of image is given in the figure below.

AB is a small object and its inverted, real and larger image A’B’is formed ‘ by objective lens. This image acts as virtual object for eye lens, therefore it is moved forward or backward such that the image A’B’ may lie within the focal length f_e of the eye piece. The image A’B’ acts as an object for – the eye piece which essentially acts like a simple microscope. The eye piece forms a virtual and magnified final image A”B” of the object AB. Clearly, the final image A”B” is inverted with respect to the object A.
(ii) f_o and f_e of compound microscope must be small so as to have large magnifying power as
M – \(\frac{L}{f_{o}}\) (1 + \(\frac{D}{f_{e}}\))

Question 18.
Explain with the help of Einstein’s photoelectric equation any three observed features in photoelectric effect which cannot be explained by wave theory. [3]
Or
Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions Φ₁ and (Φ₁ > Φ₂). On what factors does the:
(i) slope, and
(ii) intercept of the lines depend? [3]
Answer:
Photo-electric effect can be explained with the help of photoelectric equation:
1. \(v_{\max }^{2}\) ≥ 0
(v – v₀)>0 or v ≥ v₀
Thus, it is clear that for photo-electric emission, the frequency of light should be either equal to threshold freuency (v₀) or greater than v₀.

2. It is clear from equation eυ₀ = \(\frac{1}{2} m v_{\max }^{2}\) = h(υ – υ₀) that on increasing the frequency of incident light, kinetic energy of photo- electrons (\(\frac{1}{2} m v_{\max }^{2}\)) will increase and vice-versa.

3. On increasing the intensity of incident light, the number of photons falling on the surface of metal will increase and hence the number of photo-electrons will also increase. In other works, the rate of emission of electrons will be dierctly proportional to intensity of incident light.

Section – D

Question 19.
(i) Derive an expression for the electric field E due to a dipole of length 21 at a point distant r from the centre of the dipole on the axial line.
(ii) Draw a graph of E versus r for r >> l.
(iii) If this dipole is kept in a uniform external electric field E₀, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases. [4]
Or
(i) A sphere S₁ of radius r₁ encloses a net charge Q. If there is another concentric sphere S₂ of radius r₂ (r₂ > r₁) enclosing charge Q, find the ratio of the electric flux through S₁ and S₂. How will the electric flux through sphere S₁ change. If a medium of dielectric constant K is introduced in the space inside S₂ in a place of air?
(ii) A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. [4]
Answer:
(i) Electric field of a dipole at a point on the axis

(ii) E ∝ \(\frac{1}{r^{3}}\) As r increase, E will sharply decrease. The shape of the graph will be as given in the figure.

(iii) When the dipole were kept in a uniform electric field E₀. The torque acting on dipole,
τ = p × E = pE sinθ

(a) If θ = 0°,then τ = 0, p||E
(b) If θ = 180°, then τ = 0, p||-E

Question 20.
(i) How is a depletion region formed in p-n junction?
(ii) With the help of a labelled circuit diagram, explain how a junction diode is used as a full wave rectifier. Draw its input, output waveforms. [4]
Or
(i) Describe the working of light emitting diode (LEDs).
(ii) Which semiconductors are preferred to make LED’s and why?
(iii) Give two advantages of using LED’s over conventional incandescent low power lamps.
(iv) Why are Si and Ga as preferred materials for solar cells? [4]
Answer:
(i)

As the P-N junction starts, due to the concentration gradient across P-and N-sides, holes diffuse form P-side to N side (P → N) and electrons diffuse form N-side to P-side (N → P). This motion of charge gives rise to diffusion current across the junction. When an electron diffuses from N P, it leaves behind an ionised donor on N-side. This ionised donor (positive) charge is immobile as it is bounded to the surrounding atoms. As the electrons continue to defuse form N P, a level of positive charge (or positive space-charge-region) on n side of the junction is developed. Similarly when a hole diffuses form P → N due to concentration gradient, it leaves being an ionised acceptor (negative charge) which is impossible. As the holes continue to diffuse,a layer of negative charge (or negative space-charge region) on the P-side of the junction is developed. This space-charge region of either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges.

(ii) Full wave Rectifier

Here, the P-side of the diodes are connected to the ends of the secondary of the transformer. The N-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer. So for a full wave rectifier the secondary of the transformer is provided with a center tapping end so it is called center tap transformer. Suppose, the input voltage to A with respect to the centre tap at any instant is positive. It is clear, at that instant, voltage at B being out of phase will be negative.

So diode D₁ gets forward biased and conducts (while D₂ being reverse biased is not conducting). Hence, during the positive half cycle we get an output current (and a output voltage across the load resistor R_L). In the case of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D₁ would not conduct but diode D₂ would giving an output current and output voltage (across R) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the half wave rectifier.