## JEE Main 2020 Physics Paper With Solutions Jan 7 Shift 1

**1.**A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just 10 % of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero is

a. 45

^{o}

b. 90

^{o}

c. 71.6

^{o}

d. 18.4

^{0}

Intensity after polarisation through polaroid = I_{o}cos^{2}φ

So, 0.1I_{o} = I_{o}cos^{2}φ

⇒cosφ = √0.1

⇒cosφ = 0.316

Since, cos φ < cos 45^{o} therefore, φ > 45^{o} If the light is passing at 90^{o} from the plane of polaroid, than its intensity will be zero.

Then, θ = 90^{o} − φ therefore, θ will be less than 45^{o}. So, the only option matching is option d which is 18.4^{o}

**Answer: (d)**

**2.**Which of the following gives reversible operation?

Since, there is only one input hence the operation is reversible.

Answer: (c)

**3.**A 60 HP electric motor lifts an elevator with a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (Given 1 HP = 746 W, g = 10 m/s

^{2})

a. 1.5 m/s

b. 1.7 m/s

c. 2.0 m/s

d. 1.9 m/s

P = Fv + Mgv

Applied power = 4000 × v + 20000 v

60 × 246 = 4000 v + 20000 v

v = 1.865

v = 1.9 m/s Rounding it off.

**4**. A long solenoid of radius R carries a time (t) dependent current I(t) = I

_{0}t(1 − t). A ring of radius 2R is placed coaxially near its middle. During the time instant 0 ≤ t ≤ 1, the induced current (IR) and the induced EMF (V

_{R}) in the ring changes as:

a. Direction of IR remains unchanged and V

_{R}is maximum at t = 0.5

b. Direction of IR remains unchanged and V

_{R}is zero at t = 0.25

c. At t = 0.5 direction of IR reverses and V

_{R}is zero

d. At t = 0.25 direction of IR reverses and V

_{R}is maximum

Field due to solenoid near the middle = µ_{o}NI

Flux, φ = BA

Where (A = πR^{2})

= µ_{o}NI_{o}t(1 − t)πR^{2}

E = − dφ/dt [By Lenz’s law]

E = −πµ_{o}I_{o}NR^{2}(1 − 2t)

Current will change its direction when EMF will be zero

= (1- 2t) = 0

So, t = 0.5 sec

Answer: (c)

**5.**Two moles of an ideal gas with Cp/Cv = 5/3 are mixed with 3 moles of another ideal gas with Cp/Cv = 4/3. The value of Cp/Cv for the mixture is

a. 1.47

b. 1.4

c. 1.42

d. 1.50

For γ = Cp/Cv = 5/3

Using formula Cp = Rγ/γ-1

Cv = R/γ-1

Cp = 5R/2

Cv = 3R/2

Similarly for 2nd gas having γ = Cp/Cv =4/3

Cp = 4R

Cv = 3R

Now γ of mixture =

Given that n_{1} = 2 and n_{2} = 3

= 17/12 = 1.42

**Answer: (c)**

**6.**Consider a circular coil of wire carrying current

*I*, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by

*φ*The magnetic flux through the area of the circular coil area is given by

_{i}*φ*

_{0}.

Which of the following option is correct?

a. φ

_{i}= −φo

b. φ

_{i}>φ

_{o}

c. φ

_{i}<φ

_{o}

d. φ

_{i}=φ

_{o}

As magnetic field line of ring will form close loop.

∴ φ_{1} = −φ_{0} (because the magnetic field lines going inside is equal to the magnetic field lines coming out.)

**Answer: (a)**

**7.**The current (in A) flowing through 1 Ω resistor in the following circuit is

a. 0.40 A

b. 0.25 A

c. 0.20 A

d. 0.5 A

Net resistance across CD = 1/2 Ω

Net resistance across BE = 2 + (1/2) = (5/2) Ω

Net resistance across BE = [(5/2) x 2] / [(5/2) x 2] = (10/9) Ω

Total current in circuit = V/R = 9/10 A

In the given circuit,voltage across BE = voltage across BF = 1 V

Current across BE = V_{BE}/R = (2/5) A

Current across CD and DE will be same which is 2/5 A.

Now, current across any 1 Ω resistor will be same and given by = I = (1/2) × (2/5) = 1/5 = 0.20 A

**Answer: (c)**

**8.**Two infinite planes each with uniform surface charge density +σ C/m

^{2}are kept in such a way that the angle between them is 30

^{o}. The electric field in the region shown between them is given by:

a.

b.

c.

d.

Field due to single plate =

**Answer: (a)**

**9.**If the magnetic field in a plane electromagnetic wave is given by B˙ = 3 × 10

^{−8 }sin(1.6 × 10

^{3}x + 48 × 10

^{10 }t)ˆj T then what will be expression for electric field?

a. E˙ = 3 × 10

^{−8}sin(1.6 × 10

^{3}x + 48 × 10

^{10}t)ˆi V/m

b. E˙ = 3 × 10

^{−8}sin(1.6 × 10

^{3}x + 48 × 10

^{10}t)ˆj V/m

c. E˙ = 60 sin(1.6 × 10

^{3}x + 48 × 10

^{10}t)kˆ V/m

d. E˙ = 9 sin(1.6 × 10

^{3}x + 48 × 10

^{10}t)kˆ V/m

We know that,

B_{0 }= 3 × 10^{−8}

=⇒ E_{0} = B_{0} × c = 3 × 10^{−8} × 3 × 10^{8}

= 9 N/C

∴ E = E_{o}sin(ωt − kx + φ)kˆ = 9 sin(ωt − kx + φ)kˆ

E˙ = 9 sin (1.6 × 10^{3}x + 48 × 10^{10}t)kˆ V/m

**Answer: (d)**

**10.**The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10

^{−16}s. The frequency of revolution of the electron in its first excited state (in s

^{−1}) is

a. 6.2 × 10

^{15}

b. 7.8 × 10

^{14}

c. 1.6 × 10

^{14}

d. 5.6 × 10

^{12}

Time period is proportional to n^{3} /Z^{2}

Let T_{1 }be the time period in ground state and T_{2} be the time period in its first excited state.

T_{1 = }n^{3} /2^{2}

(Where, n = excitation level and 2 is atomic no.)

(T_{1}/T_{2}) = (n_{1}/n_{2})^{3}

Given,

T_{1} = 1.6 × 10^{−16}s

So,

(1.6 × 10^{−16}) / T_{2}= (1/2)^{3}

T_{2} = 12.8 x 10^{-16} s

Frequency is given by f = 1/T

f = 1/ (12.8 x 10^{-16}) Hz

f = 7.8128 × 10^{14} Hz

**Answer: (b)**

**11.**A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ’b’, the correct equivalence will be

a. L ↔1/b, C ↔1/m, R ↔1/k

b. L ↔ k, C ↔ b, R ↔ m

c. L ↔ m, C ↔ k, R ↔ b

d. L ↔ m, C ↔ 1/k , R ↔ b

For damped oscillator by Newton’s second law

− kx − bv = ma

kx + bv + ma = 0

kx + b(dx/dt) + m(d^{2}x/dt^{2}) =0

For LCR circuit by KVL

− IR − L (dI/dt)–q/c = 0

⇒ IR + L (dI/dt) + q/c = 0

⇒ q/c + R(dq/dt) + L(d^{2}q/dt^{2}) = 0

By comparing

R ⇒ b

c ⇒ 1/k

m ⇒ L

**Answer: (d)**

**12.**Visible light of wavelength 6000 × 10

^{−8 }cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minima is at 60

^{o}from the central maxima. If the first minimum is produced at θ

_{1}, then θ

_{1}is close to

a. 20

^{o}

b. 45

^{o}

c. 30

^{o}

d. 25

^{o}

For single slit diffraction experiment: Angle of minima are given by

sin θ_{n} = nλ/d (sin θ_{n} ≠ θ_{n} as θ is large )

sin θ_{2} = sin 60^{◦} = √3/2 = 2λ/ d = (2 × 6000 × 10^{−10})/d ……1

sin θ_{1} = λ/d= (6000 x 10^{−10})/d ……2

Dividing (1) and (2)

⇒

⇒ sin θ_{1} = √3/4 = 0.43

As, the value is coming less than 30^{o} the only available option are 20^{o} and 25^{o} but by using approximation we get θ_{1} = 25^{◦}

**Answer: (d)**

**13.**The radius of gyration of a uniform rod of length

*l*about an axis passing through a point

*l/4*away from the center of the rod, and perpendicular to it, is

a. √ (7/48)

*l*

b. √ (3/8)

*l*

c. (1/4)

*l*

d. (1/8)

*l*

Moment of inertia of rod about axis perpendicular to it passing through its centre is given by

I = Ml^{2}/12 + M(l/4)^{2}

I = (3Ml^{2 }+ 4Ml^{2})/48

Now, comparing with I = Mk^{2} where k is the radius of gyration

k=√7l^{2}/48

k=l√7/48

**Answer: (a)**

**14.**A satellite of mass m is launched vertically upward with an initial speed u from the surface of the earth. After it reaches height R(R = radius of earth), it ejects a rocket of mass m/10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G = gravitational constant; M is the mass of earth)

a.

b.

c.

d.

T.E_{ground}= T.ER

(1/2)mu^{2} + (-GMm/R) = (1/2)mv^{2} +(-GMm/2R)

(1/2)mv^{2 }= (1/2)mu^{2}+(-GMm/2R)

v^{2 }= u^{2}+(-GMm/R)

The rocket splits at height R. Since, separation of rocket is impulsive therefore conservation of momentum in both radial and tangential direction can be applied.

=

**Answer: (a)**

**15.**Three point particles of mass 1 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at the point:

a. 0.9 cm right and 2.0 cm above 1 kg mass

b. 2.0 cm right and 0.9 cm above 1 kg mass

c. 1.5 cm right and 1.2 cm above 1 kg mass

d. 0.6 cm right and 2.0 cm above 1 kg mass

Taking 1 kg as the origin

x_{com}= (m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3})/(m_{1}+m_{2}+m_{3})

= (1 × 0 + 1.5 × 3 + 2.5 × 0)/5

x_{com} = 0.9

y_{com}= (m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3})/ (m_{1} + m_{2}+m_{3})

y_{com}= (1 × 0 + 1.5 × 0 + 2.5 × 4)/5

ycom = 2

Centre of mass is at (0.9, 2)

**Answer: (a)**

**16.**If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece should be close to:

a. 22 mm

b. 12 mm

c. 2 mm

d. 33 mm

Magnification of compound microscope for least distance of distinct vision setting (strained eye)

M = (L/f_{0})(1 + (D/f_{e}))

where L is the tube length

f_{0} is the focal length of objective

D is the least distance of distinct vision = 25 cm

∴ fe ≈ 21.7 × 10^{−3}m = 22 mm

**Answer: (a)**

**17.**Speed of transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm

^{2}) is 90m/s. If the Young’s modulus of wire is 16 × 10

^{11}Nm

^{−2}, the extension of wire over its natural length is

a. 0.03 mm

b. 0.04 mm

c. 0.02 mm

d. 0.01 mm

Given, M = 6 grams = 6 × 10^{−3} kg

L = 60 cm = 0.6 m

A = 1 mm^{2} = 1 × 10^{−6}m^{2}

Using the relation, v^{2} = T/ µ

⇒ T = µv^{2}

= V^{2} x M/L

As Young’s modulus, Y = stress/strain

strain = stress/Y= T/AY

strain =∆L/L = (V^{2}(M/L))/AY = V^{2}(M/AYL)

∆L = V^{2}M/AY

∆L =(8100 × 6 × 10^{−3})/(1 × 10^{−6} × 16 × 10^{11})

∆L = 0.03 mm

**Answer: (a)**

**18.**1 liter of dry air at STP expands adiabatically to a volume of 3 litres. If γ = 1.4, the work done by air is (31.4 = 4.655) (take air to be an ideal gas)

a. 48 J

b. 100.8 J

c. 90.5 J

d. 60.7 J

Given, P_{1} = 1 atm, T_{1} = 273 K (At STP)

In adiabatic process,

P_{1}V_{1}^{γ }= P_{2}V_{2}^{γ}

P_{2} = 1 × (1/3)^{1.4}

P_{2} = 0.2164 atm

Work done in adiabatic process is given by

W =( P_{1}V_{1} − P_{2}V_{2})/(γ – 1)

W = ((1 × 1 − 3 × .2164)/ 0.4) x 101.325

Since, 1 atm = 101.325 kPa and 1 Liter = 10^{−3} m^{3}

W=89.87J

Closest answer is 90.5 J

**Answer: (c)**

**19.**A bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from the rest, the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:

a. r√3/4gh

b. (1/r)√4gh/3

c. ( r√3)/2gh

d. (1/r)√2gh/3

By energy conservation,

mgh = (1/2) mv^{2} + (1/2)Iω

⇒gh = v^{2}/2 +ω^{2}r^{2}/4 ————(1)

Since the rope is inextensible and also it is not slipping,

∴ v = rω——–(2)

from eq. (1) and (2)

gh = ω^{2}r^{2}/2 +ω^{2}r^{2}/4

gh = (3/4) ω^{2}r^{2}

⇒ ω^{2} = 4gh/3r^{2}

ω= (1/r) )√4gh/3

**Answer: (b)**

**20.**A parallel plate capacitor has plates of area A separated by distance ‘d’ between them. It is filled with a dielectric which has a dielectric constant varies as k (x) = k(1 + αx), where ‘x’ is the distance measured from one of the plates. If (αd <<1), the total capacitance of the system is best given by the expression:

a.

b.

c.

d.

Given, k(x) = k (1 + αx)

dC = Aε_{0}k/dx

Since all capacitance are in series, we can apply

On putting the limits from 0 to d

= ln (1 + αd)/ kε_{0}Aα

Using expression ln(1 + x) = x – x^{2}/2 +—–

And putting x = αd where, x approaches to 0.

**Answer: (b)**

**21.**A non- isotropic solid metal cube has coefficient of linear expansion as 5 × 10

^{−5}/

^{◦}C along the x-axis and 5 × 10

^{−6}/

^{◦}C along y-axis and z-axis. If the coefficient of volumetric expansion of the solid is C × 10

^{−6}/

^{◦}C then the value of C is

We know that, V = xyz

∆v/v = (∆x/x) + (∆y/y) + (∆z/z)

(1/T)∆v/v = (1/T)((∆x/x)+(∆y/y)+(∆z/z))

y = α_{n} + α_{y} + α_{z}

y = (50 × 10^{−6}/^{◦}C) + (5 × 10^{−6}/^{◦}C) + (5 × 10^{−6}/^{◦}C)

y = 60 × 10^{−6}/^{◦}C

∴ C = 60

**22.**A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0), C(5,5,0), D(0,5,0), E(0,5,5), F(0,0,5). The magnetic field in this region is B˙ = (3ˆi + 4kˆ) T. The quantity of flux through the loop ABCDEFA (in Wb) is ———

As we know, magnetic flux =

⇒ (B_{x} + B_{z}) • (A_{x} + A_{z})

⇒ (3ˆi + 4kˆ) • (25ˆi + 25kˆ)

⇒ (75 + 100) Wb

⇒ 175 Wb

**23.**A carnot engine operates between two reservoirs of temperature 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy in(J ) delivered by the engine to the low temperature reservoir, in a cycle, is

η = 1 – (T_{2}/T_{1}) = 1- (300/900) =2/3

Given, W = 1200 J

From conservation of energy

Q_{1} − Q_{2} = W

η = Q_{1} − Q_{2} = W

⇒ Q_{1}= 1800 J

⇒ Q_{2} = Q_{1} − W = 600 J

**24.**A particle of mass 1 kg slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P(height 1 m) the kinetic energy of the particle (in J ) is:

(Figure drawn is schematic and not to scale; take g = 10 m/s

^{2})

As the particle starts from rest the total energy at point A = mgh = T.EA( where h = 2 m) After reaching point P

T.Ec = K.E. + mgh

By conservation of energy

T.E_{A} = T.Ep

⇒ K.E. = mgh = 10 J

**25.**A beam of electromagnetic radiation of intensity 6.4× 10

^{−5}W/cm

^{2}is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function φ = 2 eV) of surface area 1cm

^{2}. If one in 103 photons ejects an electron, total number of electrons ejected in 1s is 10x (hc = 1240 eV − nm, 1 eV = 1.6 × 10

^{−19 }J ), then x is

P = Intensity × Area

= 6.4 × 10^{−5}Wcm^{−2} × 1 cm^{2}

= 6.4 × 10^{−5} W

For photoelectric effect to take place, energy should be greater than work function Now,

E = (1240/310) = 4eV>2eV

Therefore, photoelectric effect takes place

Here n is the number of photons emitted.

n × E = I × A

n = IA/E = (6.4 × 10^{−5)}/(6.4 × 10^{−19})

n = 10^{14}

Where, n is number of incident photon

Since, 1 out of every 1000 photons are successful in ejecting 1 photoelectron Therefore, the number of photoelectrons emitted is

= 10^{14}/10^{3}

x = 10^{11}

**Answer:** (10^{11})

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