CBSE Class 12 Physics Question Paper 2019 (Series: BVM/1) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Physics with Solutions and CBSE Class 12 Physics Question Paper 2023 (Series: BVM/1) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Physics Question Paper 2019 (Series: BVM/1) with Solutions

General Instructions:

Read the following instructions very carefully and follow them:

1. This question paper contains 35 questions. All questions are compulsory.
2. Question paper is divided into FIVE sections – Section A, B, C, D and E.
3. In Section-A: question number 1 to 18 are Multiple Choice (MCQ) type questions carrying 1 mark each.
4. In Section-B: question number 19 to 25 are Short Answer-1 (SA-1) type questions carrying 2 marks each.
5. In Section-C: question number 26 to 30 are Short Answer-2 (SA-2) type questions carrying 3 marks each.
6. In Section-D: question number 31 to 33 are Long Answer (LA) type questions carrying 5 marks each.
7. In Section-E: question number 34 and 35 are case-based questions carrying 4 marks each.
8. There is no overall choice. However, an internal choice has been provided in 2 questions in Section-B, 2 questions in Section-C, 3 questions in Section-D and 2 questions in Section-E.
9. Use of calculators is NOT allowed.
   c = 3 × 10 ⁸ m/s; h = 6.63 × 10 ^-34 Js
   e = 1.6 × 10 ^-19 ; µ ₀ = 4π × 10 ^-7 T m A ^-1
   \(\varepsilon_0\) = 8.854 × 10 ^-12 C ² N ^-1 m ^-2
   \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10 ⁹ N m ² C ^-2
   Mass of electron (m _e ) = 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 × 1o ²³ per gram mole
   Boltzmann constant = 1.38 × 10 ^-23 JK ^-1

Section A

Question 1.
Draw the pattern of electric field lines, when a point charge – Q is kept near an uncharged conducting plate. [1]
Answer:
Pattern of Electric field lines:

Question 2.
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:
No change will occur in the mobility of electrons when the potential difference applied is doubled.

Question 3.
Define the term “threshold frequency”, in the context of photoelectric emission. [1]
Or
Define the term “Intensity” in phcfton picture of electromagnetic radiation.
Answer:
Threshold frequency equals the minimum frequency of incident radiation (light) that can cause photoemission from a given photosensitive surface.
Or
Intensity of radiation is proportional to (/equal to) the number of energy quanta (photons) per unit area per unit time.

Question 4.
Differentiate between a ray and a wave front. [1]
Answer:
Ray defines the path of light.
Wave front is the locus of points in the light wave having the same phase of oscillation at any instant.

Question 5.
Define the term ‘wavefront’.
Answer:
The wavefront is defined as the locus of all particles of a medium, which are vibrating in the same phase.

Section B

Question 6.
Two bulbs are rated (P ₁ , V) are (P ₂ , V). If they are connected (i) in series and (ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P ₁ and P ₂ . [2]
Answer:

Question 7.
Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D? [2]
Or,
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index \(4\sqrt{2}\) / 5 .
Answer:
Given, μ ₂ = 1.5, μ ₁ = 1.4, Power (P) = -5 D, R = ?

Or,

μ ₂ = 1.6, μ ₁ = \(\frac{4 \sqrt{2}}{5}\), D _m = ?

Therefore, the angle of minimum deviation is 30°.

Question 8.
An α-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field \(\vec{B}\) , acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. [2]
Answer:
As We known, \(\frac{m v^2}{r}\) = \(\frac{q v}{B}\) and k = \(\frac{1}{2}\)mv²

Question 9.
Show that the radius of the orbit in hydrogen atom varies as n², where n is the principal quantum number of the atom. [2]
Answer:
When an electron moves around hydrogen nucleus, the electrostatic force between electron and hydrogen nucleus provides necessary centrepetal force.
\(\frac{m v^2}{r}\) = \(\frac{1}{4 \pi \epsilon_0}\) \(\frac{e^2}{r^2}\)

or mv²r = \(\frac{e^2}{4 \pi \epsilon_0}\)

Also we know from Bohr’s postulate,
mvr = \(\frac{ n h }{2 \pi}\)

m²v²r² = \(\frac{ n^2 h^2 }{4 \pi^2}\)

Dividing (ii) by (i), we have
mr = \(\frac{ n^2 h^2 }{4 \pi^2}\) × \(\frac{4 \pi \epsilon_0}{e^2}\)

∴ r = \(\frac{ n^2 h^2 }{4 \pi^2 me^2}\) • 4π∈ ₀

∴ r ∝ n²

Question 10.
The oscillating electric field of an electromagnetic wave is given by :
E _y = 30 sin [2 × 10 ¹¹ t + 300 π x] Vm ^-1
(a) Obtain the value of the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field. [2]
Answer:
(a) We compare the given expression with

Question 11.
Calculate the orbital period of the electron in the first excited state of hydrogen atom. [2]
Answer:
We know, Radius (r _n ) = \(\frac{h^2 \varepsilon_0}{\pi m e^2}\)n²

Question 12.
Plot a graph showing variation of de Broglie wavelength (λ) associated with a charged particle of mass m, versus 1/\(\sqrt{V}\), where V is the potential difference through which the particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle? [2]
Answer:
(i) A graph between λ and 1/\(\sqrt{V}\) :

(ii) λ = \(\frac{h}{\sqrt{2 m q V}}\)

Slope of the curve
\(\frac{\lambda}{\left(\frac{1}{\sqrt{V}}\right)}\) = \(\frac{h}{\sqrt{2 m q }}\)
Squaring both sides of equation (i), we have

Section C

Question 13.
(a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole. [3]
Answer:
(a) Equipotential surface in a constant electric field is shown in the adjoining diagram.

Equipotential surfaces are not equidistant, because V ∝ \(\frac{1}{r}\)

(b) Expression for magnetic field due to dipole on its axial lane :

Electric field intensity at point P due to charge -q,

Question 14.
Using Kirchhoff s rules, calculate the current through the D 40 Ω and 20 Ω resistors in the given circuit: [3]

Answer:
The given circuit diagram is modified as shown below:
To Find I ₁ = ?, I ₂ = ?

Let the current be I ₁ through 40 Ω and I ₂ through 20 Ω.
In loop ABCFA,
+80 – 20 I ₂ + 40 I ₁ = 0
-20I ₂ + 40 I ₁ = -80 or I ₂ – 2 I ₁ = 4 … (i)

In loop FCDEF,
-40 I ₁ – 10(I ₁ + I ₂ ) + 40 = 0
-50 I ₁ – 10 I ₂ + 40 = 0
5 I ₁ + I ₂ = 4 …(ii)
Solving these two equations (i) and (it), we have
I ₁ = 0A and I ₂ = 4A

Question 15.
(a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range.
(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field. [3]
Answer:
(a) In Radar: Mircrowaves : Frequency range (~10 ¹⁰ to 10 ¹² hz)
In Eye surgery: Ultraviolet rays : Frequency range (~10 ¹⁵ to 10 ¹⁷ )

(b) Average energy density of the electric field = \(\frac{1}{2}\)∈ ₀ E²

Question 16.
Using Huygen’s wave theory, verify the laws of reflection. [3]
Or,
Define the term, “refractive index” of a medium. Verify Snell’s laws of refraction when a plane wavefront is propagating from a denser to a rarer medium.
Answer:
(t) The wavefront is a locus of points which oscillate in phase.

(ii) Diagram of a plane wave front for Reflection:

Since time taken by waves from point B to C and from A to E is same
∴ BC = AE = vτ

In ∆ABC and ∆AEC,
AC = AC (common)
∠ABC = ∠AEC (90° each)
AE = BC
∴ ∆ABC = ∆AEC
Hence ∠BAC = ∠ECA
∠i = ∠r
i. e., Angle of incidence = Angle of reflection

Question 17.
(a) Define mutual inductance and write its S.I. unit.
(b) A square loop of side V carrying a current I ₂ is kept at distance x from an infinitely long straight wire carrying a current I ₁ , as shown in the figure. Obtain the expression for the resultant force acting on the loop. [3]
Answer:
(a) Mutual inductance equals the induced emf in a coil when the rate of change of current in its neighbouring coil is one ampere/second.
S.I. Unit: henry (H) or weber/ampere

(b) Force per unit length between two parallel straight conductors is given by
F = \(\frac{\mu_0}{4 \pi}\) \(\frac{2 \mathrm{I}_1 \mathrm{I}_2}{d}\)

∴ Force on the part of the loop which is parallel to infinite straight wire and at a distance x from it.
F ₁ = \(\frac{\mu_0}{2 \pi}\) \(\frac{\mathrm{I}_1 \mathrm{I}_2 a}{x}\) …………(i)

Force on the part of the loop which is at a distance (x + a) from it.
F ₂ = \(\frac{\mu_0}{2 \pi}\) \(\frac{\mathrm{I}_1 \mathrm{I}_2 a}{(x+a)}\) …………(i)

Net Force, F = F ₁ – F ₂

F = \(\frac{\mu_0}{2 \pi}\) I ₁ I ₂ a [latex]\frac{1}{x}-\frac{1}{x+a}[/latex] (away from the infinite straight wire)

Question 18.
(a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.
(b) Explain the significance of a radial magnetic field when a current carrying coil is kept in it. [3]
Answer:
(a) Let I = current through the coil
a, b = sides of the rectangular loop
A = ab = area of the loop
n = Number of turns in the loop
B = Magnetic field
θ = angle between magnetic field
B and area vector A

Force exerted on the arm DA inward
F ₁ = I b B
Force exerted on the arm BC outzvard
F ₂ = l b B
∴ F ₂ = F ₁
Thus net force on the loop is zero
∴ Two equal and opposite forces form a couple which exerts a torque
∴ Magnitude of the torque on the loop is,

(b) In a radial magnetic field two sides of the rectangular coil remain parallel to the magnetic field lines while its other two sides remain perpendicular to the magnetic field lines. This holds for all positions of the coil.

Question 19.
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. [3]
A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 × 10 ⁶ m, and the radius of lunar orbit is 3.8 × 10 ⁸ m.
Answer:
(i)

Reflecting telescope in normal adjustment

Drawbacks:

• Chromatic aberration is there.
• Lesser resolving power.

Expression: M = \(\frac{\tan \beta}{\tan \alpha}\) = \(\frac{\beta}{\alpha}\),

M = \(\frac{h}{f_e}\) × \(\frac{f_0}{h}\) = \(\frac{f_0}{f_e}\)

(ii) Angular magnification of the telescope = \(\frac{f_0}{f_e}\) = \(\frac{15}{0.01}\) = 1500 ………(i)

For objective lens, tan α = \(\frac{\text { Diameter of moon }}{\text { Radius of lunar orbit }}\) = \(\frac{3.48 \times 10^6}{3.8 \times 10^8}\) ………….(ii)

For eyepiece, tan ß = \(\frac{h_i}{f_e}\) = \(\frac{h_i}{f_e}\) ………..(iii)
Where hi,- = diameter of moon of image

Magnifying Power = \(\frac{\beta}{\alpha}\) = \(\frac{\tan \beta}{\tan \alpha}\)
= \(\frac{\frac{h_i}{10^{-2}}}{\frac{3.48 \times 10^6}{3.8 \times 10^8}}\)

= \(\frac{h_i \times\left(3.8 \times 10^8\right)}{\left(3.48 \times 10^6\right) \times 10^{-2}}\)

= 1500
On calculating, h = 13.73 cm …(From equation (i))

Question 20.
(a) State Gauss’s law for magnetism. Explain its significance.
(b) Write the four important properties of the magnetic field lines due to a bar magnet. [3]
Or,
Write three points of differences between para-, dia- and ferro- magnetic materials, giving one example for each.
Answer:
(a) Gauss’s law for magnetism states that “The total flux of the magnetic field, through any closed surface, is always zero.”
= \(\oint_S \vec{B} \cdot \vec{d} \cdot s\) = 0
This law implies its significance that “magnetic monopoles do not exist” and thus magnetic field lines form closed loops.

(b) Four properties of magnetic field lines are:
(i) Magnetic field lines always form continuous closed loops.
(ii) The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at the point.
(iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
(iv) Magnetic field lines do not intersect.

Or

Examples:
Diamagnetic materials: Bi, Cu, Pb, Si, water, NaCl, Nitrogen (at STP)
Paramagnetic materials: Al, Na, Ca, Oxygen (at STP), Copper chloride
Ferromagnetic materials: Fe, Ni, Co, AlniCo

Question 21.
Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is
(i) attractive and
(ii) repulsive. Write any two characteristic features of nuclear forces. [3]
Answer:
The graph indicates that the attractive force between the two nucleons is strongest at a separation r0 = 1 fm. For a separation greater than r0, the force is attractive and for separation less than r0, the force is strongly repulsive.

Two characteristic features of nuclear forces :
1. Strongest interaction
2. Short-range force
3. Charge independent character

Question 22.
(a) Three photo diodes D ₁ D ₂ and D ₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?
(b) Why photodiodes are required to operate in reverse bias? Explain. [3]
Answer:
(a) Given:
D ₁ = 2.5 eV, D ₂ = 2 eV, D ₃ = 3 eV, λ = 600 nm or 600 × 10 ^-19

The band gap energy of diode D ₂ (= 2eV) is less than the energy of the photon.
Hence, diode D ₂ will not be able to detect light of wavelength 600 nm.

(b) A photodiode when operated in reverse bias, can measure the fractional change in minority carrier dominated reverse bias current with greater ease.

Question 23.
A proton and an alpha particle are accelerated through the same potential. Which one of tin- two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy? Justify your answers. [3]
Or,
Draw the circuit diagram of a full wave rectifier and explain its working. Also, give the input and output waveforms.
Answer:
As a charge q is accelerated by a potential V
We have W = qV = \(\frac{1}{2}\)mv² = \(\frac{p^2}{2m}\)

As the work done on the charge becomes K.E.
p = \(\sqrt{2 m q V}\)

(i) Since A =\(\frac{h}{p}\)

⇒ \(\frac{\lambda_p}{\lambda_\alpha}\) = \(\sqrt{\frac{2 \times 4 m_p \times 2 e \times V}{2 \times m_p \times e \times V}}\) = \(2 \sqrt{2}\) > 1

∴ Proton will have greater de-Broglie wavelength

(ii) Energy E = qV.
So one proton having lesser charge in coulomb will have less K.E.

Or

Diagram of full wave rectifier :

Working :
The diode D ₁ is forwardbiased during one half cycle and current flows through the resistor, but diode D ₂ is reverse-biased and no current flows through it. During the other half of the signal, D ₁ gets reverse-biased and no current passes through it, D ₂ gets forward-biased and current flows through it. In both half cycles, current through the resistor flows in the same direction.

Question 24.
Derive the expression for the current density of a conductor in terms of the conductivity and applied electric field. Explain, with reason how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant. [3]
Answer:
(i) Derivation of expression for current density-
Using Ohm’s law,
V = IR = \(\frac{\mathrm{I} \rho l}{\mathrm{~A}}\) = I(\(\frac{\rho l}{\mathrm{~A}}\)) ……..(i)

Potential difference (V), across the ends of a conductor of length ‘/’ where field ‘E’ is applied, is given by
V = El ………..(ii)

From equations (i) and (ii),
El = I(\(\frac{\rho l}{\mathrm{~A}}\))

But current density
J = \(\frac{I}{A}\)
El = J ρ l = \(\frac{JI}{\sigma}\)

⇒ J = σE

(ii) Mobility, μ = \(\frac{v_d}{\mathrm{E}}\) = \(\frac{v_d}{\frac{V}{l}}\)

So, as potential is doubled, drift velocity also gets doubled, therefore, there will be no change in mobility.