JEE Advanced 2021 was conducted by IIT Kharagpur. JEE Advanced is the second level of examination for admission into the Indian Institute of Technology. JEE Advanced is one of the toughest entrance examinations conducted in the country. Solving previous yearsβ question papers will help the candidates top the examination. The JEE Advanced Question Paper 2021 Physics Paper 2 solved by the subject experts at BYJUβS is available on this page. The PDF format of the solutions is also available here for download. The candidates can download the question paper with solutions and keep it for future reference.

JEE Advanced 2019 Question Paper

Physics Paper 1

**Question 1: One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q, at a height L above the hinge at point O. A block of weight αW is attached at point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of (2√2) W Which of the following statement(s) is(are) correct?**

- a. The vertical component of the reaction force at O does not depend on α
- b. The horizontal component of the reaction force at O is equal to W for α = 0.5
- c. The tension in the rope is 2W for α = 0.5
- d. The rope breaks if α > 1.5

Solution:

Answer: (a, b, d)

W x (L/2) + αW x L = T x (1/√2) x L

$\Rightarrow T = \sqrt{2}\left ( \frac{1}{2}+\alpha \right )W$ $T\times \frac{1}{\sqrt{2}}+F_{v}= W+\alpha W$ $\frac{W}{2}+\alpha W+F_{v}= W+\alpha W$ F

_{v}= W/2At α = ½,

$T = \sqrt{2}\left ( \frac{1}{2}+\frac{1}{2} \right )W$ = √2WF

_{H }(at α = ½) = √2W x ½ = WAt α = 1.5, T = √2 x (½ + 3/2) W = 2√2W

**Question 2: A source, approaching with speed u towards the open end of a stationary pipe of length L, is emitting a sound of frequency f _{s}. The farther end of the pipe is closed. The speed of sound in air is v and f_{0} is the fundamental frequency of the pipe. For which of the following combination(s) of u and f s, will the sound reaching the pipe lead to a resonance?**

- a. u = 0.8v and f
_{s}= f_{0} - b. u = 0.8v and f
_{s}= 2f_{0 } - c. u = 0.8v and f
_{s}= 0.5f_{0} - d. u = 0.5v and f
_{s}= 1.5f_{0}

Solution:

Answer: (a, d)

$\frac{v}{v-u}\times f_{s}= (odd)\times f_{0}$ (a)

$\frac{v}{v-0.8v}\times f_{0}= 5 f_{0}$ (b)

$\frac{v}{v-0.8v}\times 2f_{0}= 10 f_{0}$ (c)

$\frac{v}{v-0.8v}\times \frac{f_{0}}{2}= (5/2)f_{0}$ (d)

$\frac{v}{v-0.5v}\times \frac{3f_{0}}{2}= 3f_{0}$

**Question 3: For a prism of prism angle θ = 60º, the refractive indices of the left half and the right half are, respectively, n _{1} and n_{2} (n_{2 }≥ n_{1}) as shown in the figure. The angle of incidence is chosen such that the incident light rays will have minimum deviation if n_{1} = n_{2 }= n = 1.5. For the case of unequal refractive indices, n_{1} = n and n_{2} = n + Δn (where Δn << n), the angle of emergence e = i + Δe. Which of the following statement(s) is(are) correct?**

- a. The value of Δe (in radians) is greater than that of Δn
- b. Δe is proportional to Δn
- c. Δe lies between 2.0 and 3.0 milliradians if Δn = 2.8 × 10
^{–3} - d. Δe lies between 1.0 and 1.6 milliradians if Δn = 2.8 × 10
^{–3}

Solution:

Answer: (b, c)

For n

_{1}= n_{2}= n = 1.5,r = 30º

Therefore, sin i = 1.5 × sin (30º) = 3/4

⇒ sin e = 3/4 for n

_{1}= n_{2}Now, r′ = 30º and n

_{2}= n + Δnn

_{2}x sin (r’) = 1 x sin e⇒ Δn

_{2 }x sin 30^{0}= cos e x Δe$\Rightarrow \Delta e = \frac{(\Delta n)\times \frac{1}{2}}{\sqrt{1-\frac{9}{16}}}= \frac{2}{\sqrt{7}}\Delta n$ ⇒ Δe < Δn and, Δe ∝ Δn

At Δn = 2.8 × 10

^{–3}, Δe = 2.12 × 10^{–3}rad

**Question 4: A physical quantity $\vec{S}$ is defined as $\vec{S} = (\vec{E}\times \vec{B})/\mu _{0}$, where $\vec{E}$ is electric field, $\vec{B}$ is magnetic field and μ0 is the permeability of free space. The dimensions of $\vec{S}$ are the same as the dimensions of which of the following quantity(ies)?**

- a. Energy /(Charge x Current)
- b. Force/ (Length x Time)
- c. Energy/Volume
- d. Power/Area

Solution:

Answer: (b, d)

$\vec{S} = (\vec{E}\times \vec{B})/\mu _{0}$ $\vec{S}$ is known as poynting vector and represents intensity of electromagnetic waves$\left [ \vec{S} \right ]=\left [ MT^{-3} \right ]=\left [ \frac{Power}{Area} \right ]= \left [ \frac{Force}{Length \times Time} \right ]$

**Question 5: A heavy nucleus N, at rest, undergoes fission N → P + Q, where P and Q are two lighter nuclei. Let δ = M _{N} – M_{P} – M_{Q}, where M_{P}, M_{Q} and M_{N} are the masses of P, Q and N, respectively. E_{P} and E_{Q }are the kinetic energies of P and Q, respectively. The speeds of P and Q are V_{P} and V_{Q}, respectively. If c is the speed of light, which of the following statement(s) is(are) correct?**

- a. E
_{P }+ E_{Q}= c^{2}δ - b.
$E_{p}=\left ( \frac{M_{P}}{M_{P}+M_{Q}} \right )c^{2}\delta$ - c.
$\frac{V_{P}}{V_{Q}}=\frac{M_{Q}}{M_{P}}$ - d. The magnitude of momentum for P as well as Q is
$c\sqrt{2\mu\delta }$ , where$\mu =\frac{M_{P}M_{Q}}{M_{P}+M_{Q}}$

Solution:

Answer: (a, c, d)

E

_{P}+E_{Q}= δc^{2}(Q-value of nuclear reaction)$\sqrt{2M_{P}E_{P}}=\sqrt{2M_{Q}E_{Q}}$ M

_{P}V_{P }= M_{Q}V_{Q}$\Rightarrow \frac{E_{P}}{E_{Q}}=\frac{M_{Q}}{M_{P}}$ $\Rightarrow E_{P}= \frac{M_{Q}}{M_{P}+M_{Q}}\delta c^{2}$ ⇒ Momentum of P or

$Q = \sqrt{\frac{2M_{P}M_{Q}}{M_{P}+M_{Q}}\delta c^{2}}$

**Question 6: Two concentric circular loops, one of radius R and the other of radius 2R lie in the xy-plane with the origin as their common centre, as shown in the figure. The smaller loop carries current I _{1 }in the anti-clockwise direction and the larger loop carries current I_{2} in the clockwise direction, with I_{2 }> 2I_{1}. **

- a.
$\vec{B}(x,y)$ is perpendicular to the xy-plane at any point in the plane - b.
$\left | \vec{B}(x,y) \right |$ depends on x and y only through the radial distance$r =\sqrt{x^{2}+y^{2}}$ - c.
$\left | \vec{B}(x,y) \right |$ is non-zero at all points for r < R - d.
$\vec{B}(x,y)$ points normally outward from the xy-plane for all the points between the two loops

Solution:

Answer: (a, b)

A magnetic field due to a circular loop at any point in its plane will be perpendicular to the plane. Due to symmetry, it will depend only on the distance from the centre. The field will be in opposite direction inside and outside the loop. The field may be non-zero for r < R, as it is in opposite direction due to both the loops.

**Question Stem for Question Nos. 7 and 8 **

**A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test tube with some air (ideal gas) trapped as shown in the figure. The test tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially, the bottle is sealed at atmospheric pressure p _{0} = 10^{5 }Pa so that the volume of the trapped air is V_{0} = 3.3 cc. When the bottle is squeezed from outside at a constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p_{0} + Δp without changing its orientation. At this pressure, the volume of the trapped air is V_{0} – ΔV. **

**Let ΔV = X cc and Δp = Y × 10 ^{3} Pa.**

**Question 7: The value of X is __________. **

Solution:

Answer: (0.30)

**Question 8: The value of Y is __________. **

Solution:

Answer: (10.00)

**Solution of Q. Nos. 7 & 8**When buoyant force on (tube + air) system will become equal to the weight of the tube then the tube will start sinking. (Here we can neglect weight of air as compared to weight of tube)

Now, Let volume of air in this case = V

_{air }F

_{B }= mgSo, δ

_{w}(V_{tube}+ V_{air}) g = mg⇒

$1\left ( \frac{5}{2.5}cm^{3}+V_{air} \right ) =5$ ⇒ 2 + V

_{air}= 5V

_{air}= 3 cm^{3}As initial volume of air = 3.3 cm

^{3}So, ΔV = 0.3 cc

So, X = 0.30

As temperature of air is constant

So, PV = constant

P

_{0}3.3 = P_{f}3, P_{f}is final pressure of air⇒ P

_{f}= 1.1 P_{0 }= P_{0 }+ 0.1 P_{0}So, ΔP = 10

^{4}PaSo, Y = 10

So, X = 0.30

Y = 10.00

**Question Stem for Question Nos. 9 and 10 **

**A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L = 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg-m ^{2}/s. The kinetic energy of the pendulum just after the lift-off is K Joules. **

**Question 9: The value of J is __________. **

Solution:

Answer: (0.18)

**Question 10: The value of K is __________. **

Solution:

Answer: (0.16)

**Solution of Q. Nos. 9 and 10**V

_{0 }=0.2/0.1 =2 m/s9. L = P × H

= 0.2 × 0.9 = 0.18 kg m

^{2}/s10. V

_{1}=V_{0}cos θ = 2 x (0.9/1)K = (½) x (0.1) x (2 x 0.9)

^{2}= 0.162 Joules

**Question Stem for Question Nos. 11 and 12 **

**In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase angle (in degrees) between the current and the supply voltage is ϕ. Assume, π√3 = 5. **

**Question 11: The value of C is ___. **

Solution:

Answer: (100)

**Question 12: The value of ϕ is ___. **

Solution:

Answer: (60)

**Solution of Q. Nos. 11 & 12**P = V

^{2}/2⇒ 500 = 100

^{2}/R⇒ R = 20 Ω

Now across resistance 500 = I × 100

⇒ I

_{rms}= 5 AV

_{rms }= 200 V,V

_{rms/rea}l = 100 Vcos φ = 100/200 = ½ ⇒ Φ = 60

^{0}tan Φ = X

_{C}/ R = 1/ωRC√3 = 1/100π(20)C

C = 1/(20π√3 x 100)

= 10

^{-4}F= 100 μF

**Question Stem for Question Nos. 13 and 14 **

**A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its centre at the origin. A magnetic dipole of moment m is brought along the axis of this loop from infinity to a point at distance r (>> a) from the centre of the loop with its north pole always facing the loop, as shown in the figure below.**

**The magnitude of the magnetic field of a dipole m, at a point on its axis at distance r, is $\frac{\mu _{0}m}{2\Pi r^{3}}$, where μ _{0} is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m_{1} and m_{2}, separated by a distance r on the common axis, with their north poles facing each other, is km_{1}m_{2}/r^{4}, where k is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.**

**Question 13: When the dipole m is placed at a distance r from the centre of the loop (as shown in the figure), the current induced in the loop will be proportional to?**

- a. m/r
^{3} - b. m
^{2}/r^{2} - c. m/r
^{2} - d. m
^{2 }/r

Solution:

Answer: (a)

Magnetic flux due to dipole through ring =

$\frac{\mu _{0}}{2\Pi }\times \frac{m}{r^{3}}\times \Pi a^{2}$ for net magnetic flux through the loop to be zero.Magnetic flux due to dipole = Magnetic flux due to induced current

⇒

$\frac{\mu _{0}}{2\Pi }\times \Pi a^{2}\times \frac{m}{r^{3}}=l\times \Pi a^{2}\times \frac{k}{a}$ , where k is proportionality constant.⇒ l∝ m/r

^{3}

**Question 14: The work done in bringing the dipole from infinity to a distance r from the centre of the loop by the given process is proportional to**?

- a. m/r
^{5} - b. m
^{2}/r^{5} - c. m
^{2}/r^{6} - d. m
^{2}/r^{7}

Solution:

Answer: (c)

$F = \frac{km_{1}m_{2}}{r^{4}}=k(l\pi a^{2})\left ( \frac{m}{r^{4}} \right )$ F = C(m

^{2}/r^{7}) where C is a constant obtained by substituting the value of I from Q.13$\left | W \right |=\int_{\infty }^{r}Fdr=Cm^{2}\int_{\infty }^{r}\frac{dr}{r^{7}}=\frac{C'm^{2}}{r^{6}}$ where C′ is a constant$\left | W \right |\propto \frac{m^{2}}{r^{6}}$

**Question Stem for Question Nos. 15 and 16 **

**A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, C _{V} = 2R. Here, R is the gas constant. Initially, each side has a volume V_{0} and temperature T_{0}. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result, the partition moves slowly towards the right, reducing the right side volume to V_{0} /2. Consequently, the gas temperatures on the left and the right sides become T_{L} and T_{R}, respectively. Ignore the changes in the temperatures of the cylinder, heater and partition.**

**Question 15: The value of T _{R}/T_{0} is**

- a. √2
- b. √3
- c. 2
- d. 3

Solution:

Answer: (a)

PV

^{ץ}= C⇒

$TV^{\gamma -1}=C$ $\Rightarrow T_{0}V_{0}^{\gamma -1}= T_{R}(\frac{V_{0}}{2})^{\gamma -1}$ $C_{V}=\frac{R}{\gamma -1}$ $\Rightarrow 2R=\frac{R}{\gamma -1}$ $\gamma -1=\frac{1}{2}$ $\gamma=\frac{3}{2}$ $\Rightarrow \frac{T_{R}}{T_{0}}=2^{\gamma -1}=\sqrt{2}$

**Question 16: The value of Q/RT _{0} is**

- a. 4(2√2 +1)
- b. 4(2√2 -1)
- c. (5√2 +1)
- d. (5√2 -1)

Solution:

Answer: (b)

Q = ΔU

_{1}+ ΔU_{2}ΔU

_{1 }= C_{V}ΔT_{1}= 2R(T_{L}– T_{0})ΔU

_{2}= C_{V}ΔT_{2 }= 2R(T_{R}– T_{0})T

_{L}= 3√2T_{0}, T_{R }= √2T_{0}Q = 2R[3√2 -1]T

_{0}+ 2R(√2 -1]T_{0}Q = 4RT

_{0}[2√2 -1]⇒ Q/RT

_{0}= 4[2√2 -1]

**Question 17:****In order to measure the internal resistance r _{1} of a cell of emf E, a meter bridge of wire resistance R_{0} = 50 Ω, a resistance R_{0}/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r _{1} = ___ Ω.**

Solution:

Answer: (3)

Current will flow in the main circuit

$I = \frac{E}{r_{1}+\frac{3R_{0}}{2}}$ $+E- IR_{0}\times 0.72-Ir_{1}-\frac{E}{2}=0$ $\frac{E}{2}=\frac{2E}{2r_{1}+3R_{0}}\times [0.72R_{0}+r_{1}]$ $2r_{1}+3R_{0}= 4[0.72R_{0}+r_{1}]$ 0.12R

_{0}= 2r_{1}r

_{1}= 3Ω

**Question 18:****The distance between two stars of masses 3M _{S }and 6M_{S} is 9R. Here R is the mean distance between the centres of the Earth and the Sun, and MS is the mass of the Sun. The two stars orbit around their common centre of mass in circular orbits with period nT, where T is the period of Earth’s revolution around the Sun. The value of n is ___.**

Solution:

Answer: (9)

Centre of mass of system lies at 6R from lighter mass

$\left [ 3M_{s}\omega ^{2}\times 6R \right ]=\frac{G(18M^{2}_{s})}{81R^{2}}$ $\omega ^{2}R = \frac{GM}{81R^{2}}$ $T^{'}=\sqrt{\frac{81R^{3}}{GM_{s}}}$ T’ = 9T

n = 09

**Question 19: In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are E _{P}, E_{Q} and E_{R}, respectively, and they are related by E_{P }= 2E_{Q} = 2E_{R}. In this experiment, the same source of monochromatic light is used for metals P and Q while a different source of monochromatic light is used for metal R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is ___.**

Solution:

Answer: (6)

$\frac{hc}{\lambda _{1}}=\phi _{p}+E_{p}$ $\frac{hc}{\lambda _{1}}=\phi _{Q}+E_{Q}$ E

_{p}= 2E_{Q}E

_{P}- E_{Q }= 0.5⇒ E

_{p}= 1.0 eV, E_{Q}= 0.5 eVE

_{R }= 0.5 eVEnergy of incident photon on R = φ

_{R}+ E_{R}= 6 eV

JEE Advanced 2021 Physics Paper 2 Solutions