CBSE Class 12 Physics Question Paper 2020 (Series: HMJ/4) with Solutions

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CBSE Class 12 Physics Question Paper 2020 (Series: HMJ/4) with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

1. This questions paper comprises four sections-A, B, C and D.
2. There are 37 questions in the question paper. All questions are compulsory.
3. Section A : Q. no. 1 to 20 are very short-answer type questions carrying 1 mark each.
4. Section B : Q. no. 21 to 27 are short-answer type questions carrying 2 marks each.
5. Section C : Q. no. 28 to 34 are long-answer type questions carrying 3 marks each.
6. Section D : Q. no. 35 to 37 are also long answer type questions carrying 5 marks each.
7. There is no overall choice in the question paper. However, an internal choice has been provided in two questions of one mark, one question of two marks. You have to attempt only one of the choices in such questions.
8. However, separate instructions are given with each section and question, wherever necessary.
9. Use of calculators and log tables is not permitted.
10. You may use the following values of physical constants wherever necessary:
    c = 3 × 10 ⁸ m/s, h = 6.63 × 10 ^-34 Js, e = 1.6 × 10-19 C, µ0 = 4π × 10 ^-7 TmA ^-1
    ε ₀ = 8.854 × 10 ^-12 C ² N ^-1 m ^-2 , \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10 ⁹ Nm ² C ^-2
    Mass of electron (m) = 9.1 × 10 ^-31 kg, Mass of neutron = 1.675 × 10 ^-27 kg
    Mass of proton = 1.673 × 10 ^-27 Kg, Avogadro’s number = 6.023 × 10 ²³ per gram mole
    Boltzmann constant = 1.38 × 10 ^-23 JK ^-1
    * Modified as per Latest CBSE Curriculum.

Set I Code No. 55/4/1
Section A

Select the most appropriate option from those given below each question.

Question 1.
A cell of emf (E) and internal resistance r is connected across a variable external resistance R. The graph of terminal potential difference V as a function of R is—

Answer:
(b)

[Hint : V = \(\frac{E}{\left(1+\begin{array}{l}
r \\
R
\end{array}\right)}\)

Question 2.
Which statement is true for Gauss law?
(a) All the charges whether inside ot outside the Gaussian surface contribute to the electric flux.
(b) Electric flux depends upon the geometry of the Gaussian surface.
(c) Gauss theorem can be applied to non-uniform electric field.
(d) The electric field over the Gaussian surface remains continuous and uniform at every point.
Answer:
(d) The electric field over the Gaussian surface remains continuous and uniform at every point.

Question 3.
A current I flows through a long straight conductor which is bent into a circular loop of radius R in the middle as shown in the figure.
The magnitude of the net magnetic field at point O will be
(a) zero
(b) \(\frac{\mu_0 I}{2 R}(1+\pi)\)
(c) \(\frac{\mu_0 I}{4 \pi R}\)
(d) \(\frac{\mu_0 I}{2 R}\left(1-\frac{1}{\pi}\right)\)
Answer:
(d) \(\frac{\mu_0 I}{2 R}\left(1-\frac{1}{\pi}\right)\)

\(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\left(1-\frac{1}{\pi}\right)\) [Hint: \(\left(\frac{\mu_0 I}{2 R}-\frac{\mu_0 I}{2 \pi R}\right)\)

Question 4.
A circular loop of radius r, carrying a current I lies in y-z plane with its centre at tin origin. The net magnetic flux through the loop is: [1]
(a) directly proportional to r
(b) zero
(c) inversely proportional to r
(d) directly proportional to I
Answer:
(b) zero

Question 5.
The kinetic energy of a proton and that of an α-particle is 4 eV and 1 eV, respectively. The ratio of the de-Broglie wavelengths associated with them, will be
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 4 : 1
Answer:
(b) 1 : 1
[Hint: λ = \(\frac{h}{\sqrt{2 m E_k}}\)

Question 6.
A photocell connected in an electrical circuit is placed at a distance ‘d’ from a source of light. As a result, current I flows in the circuit. What will be the current in the circular when the distance is reduced to ‘d/2’? [1]
(a) I
(b) 2I
(c) 4I
(d) I/2
Answer:
(c) 4I
[Hint: I ∝ \(\frac{1}{d^2}\)

Question 7.
A current of 10A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due north on the table is:
(a) 2 × 10 ^-5 T, acting downwards
(b) 2 × 10 ^-5 T, acting upwards
(c) 4 × 10 ^-5 T, acting downwards
(d) 4 × 10 ^-5 T, acting upwards
Answer:
(a) 2 × 10 ^-5 T, acting downwards

Question 8.
When a wave undergoes reflection at an interface from rarer to denser medium, adhe change in its phase is:
(a) π/2
(b) 0
(c) π
(d) π/4
Answer:
(c) π

Question 9.
An alternating voltage source of variable angular frequency ‘ω’ and fixed amplitude is connected in series with a capacitance C and electric bulb of resistance R (inductance zero). When ‘ω’ is increased—
(a) The bulb glows dimmer.
(b) The bulb glows brighter.
(c) Net impedance of the circuit remains unchanged.
(d) Total impedance of the circuit increases.
Answer:
(b) The bulb glows brighter.

As angular frequency, ω↑ (increase)

∴ X _C = \(\frac{1}{2 \pi f_c}\) = \(\frac{1}{\omega c}\) ↓(decrease) i.e., X _C ↓ ⇒ I↑

Question 10.
In the α-particle scattering experiment, the shape of the trajectory of the scattered α-particles depends upon: [1]
(a) only on impact parameter.
(b) only on the source of α-particles.
(c) both impact parameter and source of α-particles.
(d) impact parameter and the screen material of the detector.
Answer:
(a) only on impact parameter.

Note: Fill in the blanks with appropriate answers:

Question 11.
Torque acting on an electric dipole placed in an electric field is maximum when the angle between the electric field and the dipole moment is ____. [1]
Answer:
90° or π/2 radian [Hint: T = pE Sin θ

Question 12.
A proton released from rest in an electric field, will start moving towards a region of ____ potential in the field. [1]
Answer:
Lower

Question 13.
Two wires of equal length, one of copper and the other of manganin have the same resistance. The Thicker wire is ____. [1]
Answer:
As R = \(\frac{\rho l}{\mathrm{~A}}\) ∴ A = \(\frac{\rho l}{R}\)
For both wires R and I are sanie and ρ copper < ρ manganin
∴ A copper < A manganin
i.e., Manganin wire is thicker than copper wire.

Question 14.
Why is the terminal voltage of a cell is less than its emf because ____.
Answer:
Terminal voltage of a cell is less than emf because some curent, however small, may be is drawn to measure terminal voltage due to internal resistance of the cell.

Question 15.
Relationship b/w the radius of a nucleus related to its mass number A is _____. [1]
Answer:
R ∝ (A) ^1/3

Note: Answer the following:

Question 16.
The work done in moving a charge particle between two points in a uniform electric field, does not depend on the path followed by the particle. Why? [1]
Answer:
Because the electrostatic force is conservative in nature and hence work done by or against the electric field does not depend upon the path followed.

Question 17.
Define ‘magnetic declination’ at a place on earth. [1]
Answer:
Magnetic Declination is defined as the angle between the magnetic meridian and the geographic meridian at a place on the earth.

Question 18.
An A.C. source with variable frequency is connected to a parallel plate capacitor. How will the displacement current be affected with the decrease in frequency of the source?
Answer:
The displacement current will decrease with the decrease in the frequency of the A.C. source, when connected to a parallel plate capacitor. [Hint: I _C = I _D = \(\frac{V}{X_c}\) = \(\frac{V}{1 / \omega \mathrm{C}}\) = ωCV

Question 19.
An astronomical telescope may be a refracting type or a reflecting type. Which of the two produces image of better quality? Justify your answer. [1]
Answer:
Reflecting type telescope produces image of better quality, because it has a mirror of large aperture, high resolving power and is free from chromatic and spherical aberrations.

Question 20.
Can a slab of y-type semi-conductor be physically joined to another n-type semi-conductor slab to form p-n junction? Justify your answer. [1]
Answer:
No, A slab of p-type semi-conductor cannot be physically joined to another n-type semiconductor slab to form a p-n junction.
Justification. As such there will be discontinuity for the flow of charge carriers and hence no contact at atomic level.

Or

In a p-n junction diode the forward bias resistance is low as compared to the reverse bias resistance. Give reason.
Answer:
The forward current is large due to majority charge carriers which are very large in number. Hence resistance in forward bias is low.

Section – B

Question 21.
Find the total charge stored in the network of capacitors connected between A and B as shown in the figure. [2]

Answer:
In the given circuit diagram, the positive terminal of the battery is connected at the point A, when it splits and flows in capacitors 4 μF and 6 μF, which are in parallel combination.
∴ C’ = 4 + 6 = 10 μF …….[∵ C _p = C ₁ + C ₂
Similarly, two capacitors 2 μF and 3 μF are connected in parallel
∴ C” = 2 + 3 = 5μF ……[∵ C _p = C ₃ + C ₄
Now, C’ and C” are connected in series, hence
\(\frac{1}{C_{\text {net }}}\) = \(\frac{1}{C^{\prime}}+\frac{1}{C^{\prime \prime}}\) = \(\frac{1}{10}+\frac{1}{5}\) = \(\frac{1+2}{10}\) = \(\frac{3}{10}\)
Net Capacitance of the combination, C _net = \(\frac{10}{3}\) μF
∴ Tota Charge (Q) = C _net × V = \(\frac{10}{3}\) × 3 = 10 μC

Question 22.
A wire of length L ₀ has a resistance R ₀ . It is gradually stretched till its length becomes 2L ₀ .
(a) Plot a graph showing variation of its resistance R with its length l during stretching.
(b) What will be its resistance when its length becomes 2L ₀ ?
Answer:
Given. Initial length of wire (L ₁ ) = and its Resistance = R ₀
Final length after stretching (L ₂ ) = 2L ₀ ; Resistance after stretching (R ₂ ) = ?
We know that, R = \(\rho \frac{L}{A}\) = \(\frac{\rho^2}{\mathrm{AL}}\) = \(\frac{\mathrm{pL}^2}{\mathrm{~V}}\) …. where [V is volume
∴ R ∝ L ²

(a)

(b) Resistance becomes 4 R ₀ when length becomes 2L ₀ , because R ∝ L ²

Question 23.
A resistor R and an inductor L are connected in series to a source of voltage V = V ₀ sin ωt. The voltage is found to lead current in phase by π/4. If the inductor is replaced by a capacitor C, the voltage lags behind current in phase by π/4. When L, C and R are connected in series with the same source, [2]
Find the:
(i) average power dissipated and
(ii) instantaneous current in the circuit.
Answer:
With inductor (L) connected to Resistance (R), voltage is found to lead current in phase by π/4. Similarly, with capacitor (C) connected to resistance (R), voltage lags behind current in phase by π/4. When both inductor (L) and capacitor (C) are connected in series with resistance (R), then the phase difference (π/4 – π/4) will be zero.

Question 24.
Light of same wavelength is incident on three photo-sensitive surfaces A, B and C. The following observations are recorded.
(i) From surface A, photo electrons are not emitted.
(ii) From surface B, photo electrons are just emitted.
(iii) From surface C, photo electrons with some kinetic energy are emitted.
Compare the threshold frequencies of the three surfaces and justify your answer. [2]
Or.
If the frequency of light incident on the cathode of a photo-cell is increased, how will the following be affected? Justify your answer.
(i) Energy of the photo electrons,
(ii) Photo current.
Answer:
Let the wavelength of incident light be = λ
Let the frequency of incident light = v Let v _0A , v _0B and v _oC be the threshold frequency of 3 metallic surfaces A, B and C respectively.
(i) From Surface A, photo electrones are not emitted, which means v _0A > v.
(ii) From Surface B, photo electrons are just emitted, which means v _0B ≈ v.
(iii) From Surface C, photo electrons with some kinetic energy are emitted, which means
v _oC < v.
From the above, it is concluded that,
V _0A > V _0B > V _oC
Justification: If the frequency of incident light of photon is v, then hv = hv ₀ + E _k and therforce v _0A > v, v _0B = v and v _oc < v.

Or

(i) The energy (E _k ) of the emitted photo electrons increases as the frequency (v) of light incident on the cathode of a photo-cell increases, because E _k = (hv – φ ₀ )

(ii) Photo current will not be affected with the increase in frequency of incident light, because with increase of v, E _k will increase, but not the number of photo electrons. Also, we can say the photo current depends upon the intensity of light and not on frequency of incident light.

Question 25.
Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron. [2]
Answer:
From Bohr’s theory, the frequency f of the radiation emitted when an electron de-excites from level n ₂ to level n ₁ is given as

Question 26.
Distinguish between an intrinsic semiconductor and p-type semiconductor. Give reason, why, a p-type semiconductor crystal is electrically neutral although n _h >> n _e ? [2]
Answer:
(i)
Intrinsic Semiconductor and p-type semiconductor

Intrinsic Semiconductor	p-type semiconductor
1. The pure semiconductors (Ge or Si) in which the electrical conduct-ivity is totally governed by elect-rons thermally excited from the valence bond to the conduction bond are called intrinsic semiconductors.	1. A tetravent semi-conductor of Si or Ge doped with trivalent impurity atoms of B, Al or In is called a p-type semiconductor.
2. They have equal number of densities of free electrons and holes i.e., n e = n h .	2. It has’ more density of holes than density of free electrons i.e., n h >> n e .

(ii) In a p-type semiconductor, the trivalent impurity atom shares its three valence electrons with the three tetravalent host atoms while the fourth bond remains unbounded. The impurity atom as a whole is electrical neutral. Hence the p-type semiconductor is also neutral.

Question 27.
Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagram. [2]
Answer:
Distinguishing features between conductors, semiconductors and insulators :

(i) Insulator. In insulator, the valence band is completely filled. The conduction band is empty and forbidden energy gap is quite large. So no electron is able to go from valence band to conduction band even if electric field is applied. Hence electrical conduction is impossible. The solid/substance is an insulator.

(ii) Conductors (Metals). In metals, either the conduction band is partially filled or the conduction and valence band partly overlap each other. If small electric field is applied across the metal, the free electrons start moving in a direction opposite to the direction of electric field. Hence, metal behaves as a conductor.

(iii) Semiconductors. At absolute zero kelvin, the conduction band is empty and the valence band is filled. The material is insulator at low temperature. However the energy gap between valence band and conduction band is small. At room temperature, some valence electrons acquire thermal energy and jump to conduction band where they can conduct electricity. The holes left behind in valence band act as a positive charge carrier.

Section – C

Question 28.
A hollow conducting sphere of inner radius r ₁ and outer radius r ₂ has a charge Q on its surface. A point charge +q is also placed at the centre of the sphere. [3]
(a) What is the surface charge density on the
(i) inner and
(ii) outer surface of the sphere?
(b) Use Gauss’ law of electrostatics to obtain the expression for the electric field at a point lying outside the sphere.
Or,
(a) An infinitely long thin straight wire has a uniform linear charge density λ. Obtain the expression for the electric field (E), using Gauss’ law.
(b) Show graphically the variation of this electric field E as a function of distance x from the wire.
Answer:
(a) Surface charge density on the:

(i) Inner surface of the shell, σ _in = \(\frac{-q}{4 \pi r_1^2}\)
(ii) Outer surface of the shell, σ _out = \(\frac{Q+q}{4 \pi r_2^2}\)
(b) Derivation of expression of electric field for a spherical Gaussian surface when x > r ₂ . As per Gauss’ law,

Or

(a) Gauss’s law in electrostatics. It states that “the total electric flux over the surface S in vaccum is \(\frac{1}{\varepsilon_0}\) times the total charge (q)”.
∴ φ _E = \(\oint \vec{E} \cdot d \vec{S}\) = \(\frac{q}{\varepsilon_0}\)

Electric field due to an infinitely long straight wire: Consider an infinitely long straight line charge having linear charge density λ to determine its electric field. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S ₁ and is directed radially outward.

Total flux through the cylindrical surface,

(b)

(b) Variation of electric field (E) as a function of distance (x). [Hint: E ∝ \(\frac{1}{x}\)

Question 29.
(a) Derive the relation between current density ‘\(\overrightarrow{\mathrm{J}}\)‘ and potential difference ‘V’ across a current carrying conductor of length ‘l’, area of cross-section ‘A’ and the number density ‘n’ of free electrons.
(b) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10 ^-7 m ² carrying a current of 1.5 A. [Assume that the number density of conduction electrons is 9 × 10 ²⁸ m ^-3 ] [3]
Answer:
(a) Suppose a potential difference V is applied across a conductor of length l and of uniform cross-section A. The electric field E set up inside the conductor is given by
E = \(\frac{\mathrm{V}}{\mathrm{l}}\)
Under the influence of field \(\overrightarrow{\mathrm{E}}\), the free electrons begin to drift in the opposite direction \(\vec{E}\) with an average drift velocity v _d .

Let the number of electrons per unit volume or electron density = n
Charge on an electron = e
No. of electrons in length l of the conductor
= n × volume of the conductor = n × Al
Total charge contained in length l of the conductor is
q = enAl ……….. (i)
All the electrons which enter the conductor at the right end will pass through the conductor at the left end in time,
t \(=\frac{\text { distance }}{\text { velocity }}\) = \(\frac{l}{v_d}\) …… (ii)
Using equations (i) and (ii), we get

Current density ‘J’ is given by
J = \(\frac{\mathrm{I}}{\mathrm{A}}\) = \(\frac{n e \mathrm{~A} v_d}{\mathrm{~A}}\) = nev _d ∴ J ∝ v _d
Hence the current density of a metallic conductor is directly proportional to the drift speed of electrons.

(b) Since I = neAv _d ⇒ v _d = \(\frac{1}{n e A}\)

Question 30.
(a) Differentiate between self inductance and mutual inductance.
(b) The mutual inductance of two coaxial coils is 2H. The current in one coil is changed uniformly from zero to 0.5A in 100 ms. Find the:
(i) change in magnetic flux through the other coil.
(ii) emf induced in the other coil during the change. [3]
Answer:
(a) Self Inductance is the property of the coil when induced emf is produced in the coil
itself with the change of flux associated with it.
Mutual Inductance is the property of the coil, when induced emf is produced in a neighbouring coil with the change of flux associated with the pair of coils.
Both have same unit as Henry (H).

(b) Given. M = 2H, ∆I = (0.5 – 0) A = 0.5 A; ∆t = 100ms = 100 × 10 ^-3 s

(i) Change in magnetic flux (∆φ) = M × ∆I = 2 × 0.5 = 1 Wb
(ii) emf induced in the other coil (e) = \(\frac{\Delta \phi}{\Delta t}\) = \(\frac{1}{\left(100 \times 10^{-3}\right)}\) = 10 V

Question 31.
Explain with the help of a diagram, the working of a step-down transformer. Why is a laminated iron core used in a transformer? [3]
Answer:
Principle. It is a device which converts high voltage a.c. into low voltage a.c. and vice versa. It is based upon the principle of mutual induction. When alternating current passes through a coil, an induced emf is set up in the neighbouring coil.

Construction. A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as Primary ‘Y is connected to A.C. supply. The other coil known as Secondary ‘S’ is connected to the load.

Working. When an alternating current passes through the primary, the magnetic flux through the iron core changes which does two things. It produces emf in the primary and an induced emf is also set up in the secondary. If we assume that the resistance of primary is negligible, the back emf will be equal to the voltage applied to the primary.
∴ V _P = N _P \(\frac{d \phi}{d t}\)

In a step-up transformer: N _S > N _P ∴ V _S > V _P
In a step-down transformer: N _S < N _P ∴ V _S < V _P
Use of laminated core. Laminated core is used in transformers to minimise production of eddy currents and hence to minimise the energy loss.

Question 32.
Name the eletro-magnetic waves with their frequency range, produced in [3]
(a) some radioactive decay;
(b) sparks during electric welding;
(c) TV remote
Answer:
(a) Gamma rays produced in radioactive decay with frequency range 10 ¹⁹ to 1024 Hz.
(b) UV rays produced during electric welding with frequency range 10 ¹⁵ to 10 ¹⁷ Hz.
(c) Infrared rays produced in TV remote with frequency range 10 ¹² to 10 ¹⁴ Hz.

Question 33.
Two coherent light waves of intensity 5 x 10 ^-2 Wm ^-2 each super-impose and produce the interference pattern on a screen. At a point where the path difference between the waves is λ/ 6, λ being wavelength of the wave, find the [3]
(a) phase difference between the waves.
(b) resultant intensity at the point.
(c) resultant intensity in terms of the intensity at the maximum.
Answer:

Question 34.
Two objects P and Q when placed at different positions in front of a concave mirror of focal length 20 cm, form real images of-equal size. Size of object P is three times size of object Q. If the distance of P is 50 cm from the mirror, find the distance of Q from the mirror. [3]
Answer:

Putting the value of f and u ₁ , we have
∴ u ₁ = \(\frac{2(-20)+(-50)}{3}\) = \(\frac{-40-50}{3}\) = \(\frac{-90}{3}\) = -30 cm
Hence the distance of object Q from the mirrors is 30 cm.

Section – D

Question 35.
(a) Show that a current carrying solenoid behaves like a small bar magnet. Obtain the expression for the magnetic field on its axis.
(b) A steady current of 2A flows through a circular coil having 5 turns of radius 7 cm. The coil lies in X-Y plane with its centre at the origin. Find the magnitude and direction of the magnetic dipole moment of the coil. [5]
Or
(a) Derive the expression for the force acting between two long parallel current carrying conductors. Hence, define 1A current.
(b) A bar magnet of dipole moment 3 Am ² rests with its centre on a frictionless pivot. A force F is applied at right angles to the axis of the magnet, 10 cm from the pivot. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of 30° with the field. Calculate the value of F.
How will the equilibrium be effected if F is withdrawn?
Answer:
(a)
(i) A solenoid may be regarded as a combination of large number of identical circular current loops, in which each behaves like a magnetic dipole. Hence, the current carrying solenoid will behave like a bar magnet.
(ii) Megnetic field due to Solenoid
Let length of solenoid = L

Total number of turns in solenoid = N
∴ Number of turns per unit length = \(\frac{\mathrm{N}}{\mathrm{L}}\) = n
ABCD is an Ampere’s loop
AD is a long axis, Length of AD = x and Current in one turn = I ₀

Number of turns in x length = nx; Current in turns nx, I = nx I _o
According to Ampere’s circuital law —
Bx = μ ₀ I; Bx = μ ₀ nxI ₀ ∴ B = μ ₀ nI ₀

(b) Given. I = 2A, N = 5, a = 7 cm = (7 × 10 ^-2 )m
To Find. \(\overrightarrow{\mathrm{M}}\) = ? and direction =?
Magnetic dipole moment (\(\overrightarrow{\mathrm{M}}\)) = NIA = NI (πa ² )
= 5 × 2 × {\(\frac{22}{7}\) × 7 × 10 ^-2 ) ² } = 0.154 Am ²
Therefore, the direction of \(\overrightarrow{\mathrm{M}}\) will be perpendicular to x—y plane or parallel to z-axis.

Or,

(a) Consider two infinitely long parallel conductors carrying current I ₁ and I ₂ in the same direction. Let r be the distance of separation between these two conductors.
B ₁ = \(\frac{\mu_0 \mathrm{I}_1}{2 \pi d}\)
F ₂ = I ₂ × I ₂ × B ₁ sin θ {sin θ = 1}
⇒ F ₂ = I ₂ × I ₂ × \(\frac{\mu_0 \times \mathrm{I}_1}{2 \pi d}\)

Hence, force is attractive in nature.
Ampere: Ampere is that current which if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10 ^-7 N on each metre of the other wire.

Then current flowing is 1A
F = \(\frac{\mu_0 \times 1 \times 1}{2 \pi \times 1}\)
∴ F = \(\frac{\mu_0}{2 \pi}\) = 2 × 10 ^-7 N

(b) Given: M = 3 Am ² , B = 0.25 T, θ = 30°, r = 10 cm = (10 × 10 ^-2 )m
To find. F = ?
Restoring Torque (RT) = F × r; Deflecting Torque (DT) = MB sinθ
In equilibrium position, these two torques are equal and opposite
∴ F × r = MB sin θ
or F = \(\frac{M B \sin \theta}{r}\) = \(\frac{(3) \times(0.25)\left(\sin 30^{\circ}\right)}{\left(10 \times 10^{-2}\right)}\)
⇒ F = \(\frac{(3) \times 0.25}{10 \times 10^{-2}} \times \frac{1}{2}\)
∴ F = 3.75 N ….. [sin 30° = \(\frac{1}{2}\)
When force is withdrawn, the magnet oscillates for some time, but finally aligns along the original direction of external magnetic field.

Question 36.
(a) Draw the ray diagram showing refraction of ray of light through a glass prism. Derive the expression for the refractive index μ of the material of prism in terms of the angle of prism A and angle of minimum deviation δ _m .
(b) A ray of light PQ enters an isosceles right angled prism ABC of refractive index 1.5 as shown in figure.
(i) Trace the path of the ray through the prism.
(ii) What will be the effect on the path of the ray if the refractive index of the prism is 1.4? [5]

Or
(a) Two thin lenses are placed coaxially in contact. Obtain the expression for the focal length of this combination in terms of the focal lengths of the two lenses.
(b) A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length.
Answer:
(a) Ray diagram: The minimum deviation D _m , the refracted ray A inside the prism becomes parallel to its base, we have

For a small angle prism, i.e., a thin prism, D _m is also very small and we get

(b)
(i)

(ii) If the refractive index of the prism becomes 1.4, total internal reflection will not occur as shown in the ray diagram. The ray will emerge out of prism after refraction.
For μ = 1.5, the critical angle is 42° less than 45°, while for μ = 1.4, the critical angle is 46°, more than 45°.

Or

(a) Power of a lens is defined as the ability to converge a beam of light facing on the lens
Since P = \(\frac{1}{f}\)
Its S.I. unit is dioptre (D).
Let C ₁ , C ₂ be the optical centres of two thin convex lenses L ₁ and L ₂ held co-axially in contact with each other in air.

Suppose f ₁ and f ₂ are their respective focal lengths. Let a point object O be placed on the common principal axis at a distance OC ₁ = u
The lens L ₁ alone would form its image at I’ where C ₁ I’ = v’.
From the lens formula for L ₁ ,
\(\frac{1}{v^{\prime}}-\frac{1}{u}\) = \(\frac{1}{f_1}\)
I’ would serve as a virtual object for lens L ₂ , which forms a final image I at a distance of
C ₁ I = v.
As the lenses are thin, therefore, for the lens L ₂
u = I ₂ I’ ≈ C ₁ I = v’
From the lens formula for L ₂ ,
\(\frac{1}{v}-\frac{1}{v^{\prime}}\) = \(\frac{1}{f_2}\) ……. (ii)
Adding equations (i) and (ii), we get
\(\frac{1}{f_1}+\frac{1}{f_2}\) = \(\frac{1}{v}-\frac{1}{u}\)
Let the two lenses be replaced by a single lens of focal length f, which forms image I at distance v, of an object at distance u from the lens.
For this lens, \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{f}\) = \(\frac{1}{f_1}+\frac{1}{f_2}\) ⇒ \(\frac{1}{f}\) = \(\frac{f_2+f_1}{f_1 f_2}\) ⇒ f = \(\frac{f_1 \times f_2}{f_2+f_1}\)

(b) According to lens maker formula:

Question 37.
(a) State the postulates of Bohr’s model of hydrogen atom and derive the expression for Bohr radius.
(b) Find the ratio of the longest and the shortest wavelengths amongst the spectral lines of Balmer series in the spectrum of hydrogen atom.
Answer:
(a) Basic postulates of Bohr’s atomic model:
(i) Every atom consists of a central core called nucleus in which entire positive charge and mass of the atom are concentrated. A suitable number of electrons revolve around the nucleus in circular orbit. The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

(ii) Electron can resolve only in certain discrete non-radiating orbits, called stationary orbit. Total angular momentum of the revolving electron in an integral multiple of h/2π. …where [h is plank constant]
mvr = \(\frac{n h}{2 \pi}\)

(iii) The radiatìon of energy occurs only when an electron jumps from one permitted orbit to another. The difference in the total energy of electron in the two permitted orbit is absorbed when the electron jumps from inner to the outer orbit and emitted when electron jumps from outer to inner orbit.

Radii of Bohr’s stationare orbits. According to Bohr’s postulates, angular momentum of electron for any permitted orbit is,
mvr = \(\frac{n h}{2 \pi}\) or v = \(\frac{n h}{2 \pi m r}\) ….. (i)
Also, according to Bohr’s postulates, the centripetal force is equal to electrostatic force between the electron and nucleus.

It is the expression for Bohr’s radius, which is about 0.5 A
For first orbit of hydrogen atom, n = 1, Z = 1 then we have r = 0.529Å

(b) The shortest wavelength in Balmer series is obtained when an electron jumps to the second stationary orbit (n _i ) from n _f = infinity.

The longest wavelength in Balmer series is obtained when an electron jumps to the second stationary orbit (n _i = 2) from next higher stationary orbit (n _f = 3).

Set II Code No. 55/4/2

Note: Except for the following questions, all the remaining questions have been asked in Set-I.

Question 1.
An electron and a proton are moving along the same direction with the same kinetic energy. They enter a uniform magnetic field acting perpendicular to their velocities. The dependence of radius of their paths on their masses is: [1]
(a) r ∝ m
(b) r ∝ \(\sqrt{m}\)
(c) r ∝ \(\frac{1}{m}\)
(d) r ∝ \(\frac{1}{\sqrt{m}}\)
Answer:
(b) r ∝ \(\sqrt{m}\)

r ∝ \(\sqrt{m}\) [Hint: r = \(\frac{\sqrt{2 m \mathrm{E}_k}}{q \mathrm{~B}}\)

Question 4.
A photocell connected in an electrical circuit is placed at a distance ‘d’ from a source of light. As a result, current I flows in the circuit. What will be the current in the circuit, when the distance is reduced to ‘d/3’? [1]
(a) I
(b) 6 I
(c) 9 I
(d) 1/3I
Answer:
(c) 9 I

Question 6.
A current of 5 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due south on the table is: [1]
(a) 1 × 10 ^-5 T acting downwards
(b) 1 × 10 ^-5 T acting upwards
(c) 2 × 10 ^-5 T acting downwards
(d) 2 × 10 ^-5 T acting upwards
Answer:
(b) 1 × 10 ^-5 T acting upwards

Question 11.
Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of their nuclear radii? [1]
Answer:
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left(\frac{A_1}{A_2}\right)^{1 / 3}\) = \(\left(\frac{27}{125}\right)^{1 / 3}\) = \(\frac{3}{5}\)
∴ Ratio of their nuclear radii = 3 : 5

Question 12.
Draw the shape of the wavefront coming out of a concave mirror when a plane wave is incident on it. [1]
Answer:

Question 19.
Depict equipotential surfaces due to an electric dipole. [1]
Answer:

Question 20.
Define ‘angle of dip’ at a place on earth. [1]
Answer:
Angle of dip is defined as the “angle subtended by the resultant magnetic field of earth (at a given place) to the horizontal”.

Question 21.
Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’. [2]
Answer:
Definition : Drift velocity is defined as the velocity with which free electrons in a conductor get drifted in a direction opposite to the direction of the applied field. Its unit is m ^-1 s and dimensions [LT ^-1 ]
Expression : The magnitude of electric field set up across the conductor is given by E = \(\frac{\mathrm{V}}{l}\)
Let n be the number of free electrons per unit volume of the conductor.

Then, total number of free electrons in the conductor = n × Volume of the conductor
Hence, Q = (nAl)e
Time taken by the charge to cross the conductor length l is given by
t = \(\frac{l}{v_d}\) …where [v _d is drift velocity of electrons
I = \(\frac{\mathrm{Q}}{t}\) = \(\frac{n \mathrm{Ale}}{\frac{l}{v_d}}\) = neAvd I = neAv _d ∴ I = neAv _d

Question 31.
(a) We feel the warmth of the sunlight but not the pressure on our hands. Explain.
(b) Which out of wavelength, frequency and speed of an electro-magnetic wave does not change on passing from one medium to another?
(c) A thin ozone layer in the upper atmosphere is crucial for human survival on earth, why? [3]
Answer:
(a) We feel the warmth of the sunlight but not the pressure on our hands, because the momentum transferred to our hands is extremely small. Hence we do not feel any pressure on our hands.
(b) When an electro-magnetic wave passes from one medium to another, frequency does not change.
(c) A thin ozone layer in the upper atmosphere is crucial for human survival, because it prevents harmful ultra-violet (UV) rays to penetrate the earth’s atmosphere.

Question 33.
An object is placed in front of a concave mirror of focal length of 12 cm. There are two possible positions of the object for which the image formed is three times the size of the object. [3]
(a) Draw the ray diagram for each case, and
(b) Find the distance between the two positions of the object.
Answer:
Given.f = 12 cm and m = 3
(a) There are two possibilities to get enlarged image, when
(i) the object is kept between ‘F’ & ‘C’ to get real and inverted image;
(ii) the object is kept between the pole and focus to get virtual and erect image.

(b) When the image is real and inverted, then m = -3 and v = 3u f = 12

Therefore, the distance between two positions of the object is 8 cm (16 cm – 8 cm).

Set III Code No. 55/4/3

Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set-II.

Question 1.
A photo-cell connected in an electrical circuit is placed at a distance ‘d from a source of light. As a result current I flows in the circuit. What will be the current in the circuit when the distance is increased to ‘2d’? [1]
(a) 21
(b) I/4
(c) I/2
(d) 4I
Answer:
(b) I/4

Question 2.
There are uniform electric and magnetic fields in a region pointing along X-axis. An α-particle is projected along Y-axis with a velocity v. The shape of the trajectory will be [1]
(a) circular n XZ plane
(b) circular in YZ plane
(c) helical with its axis parallel to X-axis
(d) helical with its axis parallel to Y-axis.
Answer:
(c) helical with its axis parallel to X-axis
[Hint: r = \(\frac{m v \sin \theta}{q B}\)

Question 5.
A current of 10 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance 10 cm vertically above
the wire is: 1
(a) 1.2 × 10 ^-5 T, acting towards south
(b) 2 × 10 ^-5 T, acting towards north
(c) 3 × 10 ^-5 T, acting downwards
(d) 2 × 10 ^-5 T, acting upwards
Answer:
(b) 2 × 10 ^-5 T, acting towards north

Question 6.
When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? 1
Answer:
No, only the drift vek’ ‘ ties of the electrons are superposed over their random (haphazard) thermal velocities. Tlu solid line shows the random path followed by a free electron in the absence of an external. ield. The electron proceeds from A to B, making six collisions on its path. The dotted curve shows how the random motion of the same electron gets modified when an electric field is applied.

Question 13.
A neutron is bombarded on a \({ }_5^{10} B\) nucleus and an alpha particle is emitted. The nuclear reaction involved is \({ }_0^1 \mathrm{n}+{ }_5^{10} \mathrm{~B} \longrightarrow{ }_2^4 \mathrm{He}+\) + ……..
Answer:
\({ }_3^7 \mathrm{Li}\) [Hint: \({ }_0^1 \lambda+{ }_5^{10} B\) → \({ }_2^4 \mathrm{He}+{ }_3^7 \mathrm{Li}\), because sum of mass numbers and atomic numbers of reactants and products must be equal.

Question 19.
What is the change in the value of angle of dip when one goes from the equator to the north pole of earth? [1]
Answer:
When we go from the equator to the north pole, the value of angle of dip increases from 0 to 90°.

Question 20.
A charged particle (+q) moves in a uniform electric field (\(\vec{E}\)) in the direction opposite to \(\vec{E}\). What will be the effect on its electrostatic potential energy during its motion? [1]
Answer:
When a charged particle (+q) moves in a uniform electric field (\(\overrightarrow{\mathrm{E}}\)) in the direction opposite to (\(\overrightarrow{\mathrm{E}}\)), electrostatic potential energy increases during motion.

Question 23.
A wire of length L ₀ has a resistance R ₀ . It is gradually stretched till its length becomes 1.5 L ₀ .
(a) Plot the graph showing variation of its resistance R with its length l during stretching.
(b) What will be its resistance when its length becomes 1.5 L ₀ ?
Answer:
Given: Initial length (L ₁ ) = L ₀ and Final length (L ₂ ) = 1.5 L ₀

(a) As we know, R = \(\rho \frac{l}{\mathrm{~A}}\) = \(\rho \frac{l \times l}{(\mathrm{~A} \times l)}\) = \(\rho \frac{l^2}{\mathrm{~V}}\) …[Here V is the volume
For a given wire, V is constant and also resistivity ρ is constant for the material of the given wire.
∴ R ∝ l ²
Graph will be

(b)

Resistance would become 2.25 times of initial resistance when length of wire is stretched to 1.5 times its initial length.

Question 24.
You are given three capacitors of 2 μF, 3 μF and 4 μF, respectively.
(a) Form a combination of all these capacitors of equivalent capacitance 13/3 μF.
(b) What is the maximum and minimum value of the equivalent capacitance that can be obtained by connecting these capacitors? [2]
Answer:
Given: C ₁ = 2µF, C ₂ = 3µF, C ₃ = 4µF and C _R = 13/3 µF, Combination = ?
(a) If these capacitors are connected in series, then C _R would be less than the minimum value of these 3 capacitors, i.e,
C _R < 2µF[\(\frac{1}{\mathrm{C}}\) = \(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)]
If these capacitors are connected in parallel, then C _R would become 9 µF (C = C ₁ + C ₂ + C ₃ ). Therefore, these 3 capacitors have to be joined in mixed combination. The most appropriate combination would be to get C _R = \(\frac{13}{3}\).

(b) C _max = 2 + 3 + 4 = 9μF when connected in parallel. (C = C ₁ + C ₂ + C ₃ )
When connected in series: \(\frac{1}{\mathrm{C}_{\min }}\) = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\) = \(\frac{6+4+3}{12}\) = \(\frac{13}{12}\) ∴ C _min = \(\frac{12}{13}\)µF

Question 29.
(a) How are electro-magnetic waves produced? Depict an electro-magnetic wave propagating in Z-direction with its magnetic field B oscillating along X-direction.
(b) Write two characteristics of electro-magnetic waves. [3]
Answer:
(a) Electromagnetic Waves. Accelerating electric charge produces electromagnetic waves.
Sketch of a plane electromagnetic wave propagating along the z-direction with oscillating electric field E
along the x-direction and the oscillating magnetic field B along the y-direction.

(b) Characteristics of electro-magnetic wàves:

(i) The electromagnetic waves are transverse in nature.
(ii) The velocity of electromagnetic waves in free space is given by
c = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\) = 3 × 10 ⁸ ms ^-1
(iii) The ratio of the magnitudes of electric and magnetic fields is constant and is equal to the velocity of electromagnetic waves in free space
∴ c = \(\frac{\mathrm{E}_0}{\mathrm{~B}_0}\)
(iv) The equation for the oscillating electric field is, E ₂ = E ₀ sin \(\left(\frac{2 \pi x}{\lambda}+\frac{2 \pi t}{T}\right)\) and the equation of magnetic field varying sinusoidally with x and fis
B _y = B ₀ sin \(\left(\frac{2 \pi x}{\lambda}+\frac{2 \pi t}{T}\right)\) (any two)

Question 30.
A concave mirror forms a real image of an object kept at a distance 9 cm from it. If the object is taken away from the mirror by 6 cm, the image size reduces to \(\frac{1}{4}\)th of its previous size. Find the focal length of the mirror. [3]
Hence the focal length of the given concave mirror is 7 cm.
Answer:

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