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JEE Main 2019 January 9 Shift 2 Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2019 January 9 Shift 2 Paper & Solutions

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JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 1

Two plane mirrors arc inclined to each other such that a ray of light incident on the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 1


Assuming angles between two mirrors be θ as per geometry,
sum of anlges of Δ
3θ = 180°
θ = 60°

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 2

In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30°≤θ≤30°is:

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 2


Pam difference dsinθ = nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32 × 10–3 sin 30 = n × 500 × 10–9
n = 320
Hence total no. of maximas observed in angular range –30°≤ θ ≤ 30° is
maximas = 320 + 1 + 320 = 641

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JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 3

At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio RB/RA  of their activities after time t itself decays with time t as e-3t [f the half-life of A is m2, the half-life of B is :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 3

Half life of A = ℓn2

at t = 0  RA = RB

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 4

Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of Vo changes by : (assume that the Ge diode has large breakdown voltage)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 4

Initially Ge & Si are both forward biased so current will effectivily pass through Ge diode with a drop of 0.3 V if "Ge" is revesed then current will flow through "Si" diode hence an effective drop of (0.7 – 0.3) = 0.4 V is observed.

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 5

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to 

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 5

Frequency of torsonal oscillations is given by


f2 = 0.8 f1

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 6

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature of 27°C. Amount of heat transferred to the gas, so that RMS velocity of molecules is doubled, is about :
[Take R = 8.3 J/ K mole]

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 6

Q = nCvΔT as gas in closed vessel

Q = 10000 J = 10 kJ

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 7

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 7



According to the question, U = k


∴ Correct answer is (3)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 8

A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 8

Frequency of the sound produced by flute,

Velocity of observer, 

∴ frequency detected by observer, f' =

= 335.56 × 2 = 671.12
∴ closest answer is (4)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 9

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108m/s,h = 6.6 × 10–34 J-s)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 9


= 3.75 × 1014 Hz
1% of f = 0.0375 × 1014 Hz
= 3.75 × 1012 Hz = 3.75 × 106 MHz
number of channels = 
∴ correct answer is (4)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 10

Two point charges qand  are placed on the x-axis at x = l m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 10


Let are the vaues of electric field dueto q1 & q2 respectively magnitude of


E2 = 9 × 103 V/m










∴ correct answer is (3)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 11

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 11





in the same way we get,



on comparing equation (i) to equation (ii), we get

This does not match with any of the options so probably they have assumed the wrong combination




However this is one of the four options.
It must be a "Bonus" logically but of the given options probably they might go with (4)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 12

A rod of length 50cm is pivoted at one end. It is raised such that if makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s–1) will be (g = 10ms–2)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 12


Work done by gravity from initial to final position is,


According to work energy theorem




∴ correct answer is (1)

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 13

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL) to that at the centre of the coil (BC), i.e. R   will be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 13


L = 2πR L = N × 2πr
R = Nr


JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 14

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 x 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is:

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 14

Usurface + E1 = Uh
KE of satelite is zero at earth surface & at height h



Gravitational attraction




JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 15

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 15

Average energy density of magnetic field,

is maximum value of magnetic field.
Average energy density of electric field,

now,



uE = uB
since energy density of electric & magnetic field is same, energy associated with equal volume will be equal.
uE = uB

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 16

A series AC circuit containing an inductor (20 mH), a capacitor (120 μF) and a resistor (60Ω) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 16

R = 60Ω f = 50Hz, ω = 2πf = 100 π




Q = P.t = 8.64 × 60 = 5.18 × 102

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 17

Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 17




on comparing the powers of M, L, T
– x + y = 0 ⇒  x = y
3x + 2y + z = 0 ⇒  5x + z = 0 .. ..(i)
–2x – y – z = 1 ⇒ 3x + z = –1 .. .(ii)
on solving (i) & (ii) 

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 18

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(c = 3 × 108ms–1, h = 6.6 × 10–34 J-s)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 18

B = B0sin (π × 107C)t + B0sin (2π × 107C)t
since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
B1 = B0sin(π × 107C)t

B2 = B0sin(2π × 107C)t v2 = 107C
where C is speed of light  C = 3 × 108 m/s v2 > v1
so KE of photoelectron will be maximum for photon of higher energy.
v2 = 107C Hz
hv  = φ + KEmax
energy of photon
Eph = hn = 6.6 × 10-34 × 107 × 3 × 109
Eph = 6.6 × 3 × 10–19J

KEmax = Eph – φ
= 12.375 – 4.7 =  7.675 eV ≈ 7.7 eV

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 19

Charge is distributed within a sphere of radius R with a volume charge density  where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 19






JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 20

Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1(= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3(= 400 K). Calculate the temperature T2 if the work outputs of the two engines are equal :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 20


w1 = w2
Δu1 = Δu2
T3 – T2 = T2 – T1
2T2 = T1 + T3
T2 = 500 K

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 21

A carbon resistance has a following colour code. What is the value of the resistance ?

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 21


R = 53 × 104 ± 5% = 530 kΩ ± 5%

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 22

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 22

x = 3t2 + 5

v = 6t + 0
at t = 0 v = 0
t = 5 sec v = 30 m/s
W.D. = ΔKE

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 23

The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cosωt y = a sinωt and z = aωt The speed of the particle is :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 23

vx = –aωsinωt   ⇒  vy = aωcosωt
vz = aω  

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 24

In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 Ω, R3 = 100 Ω and R4 = 500 Ω and the reading of an ideal voltmeter across R4 is 5V, then the value R2 will be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 24


V4 = 5V

V3 = i1R3 = 1V
V3 + V4 = 6V = V2
V1 + V3 + V4 = 18V
V1 = 12 V

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 25

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 25


at equation

F = 100 N

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 26

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then 'v' is equal to

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 26

For A & B let time taken by A is t0
from ques.
vA – vB = v = (a1 – a2)t0 – a2t .. ..(i)


 .. ..(ii)
putting t0 in equation


JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 27

A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 27

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 28

The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 28

In flow volume = outflow volume





JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 29

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 29


LC = 0.5 × 10–2 mm
+ve error = 3 × 0.5 × 10–2 mm
= 1.5 × 10–2 mm = 0.015 mm
Reading = MSR + CSR – (+ve error)
= 5.5 mm + (48 × 0.5 × 10–2) – 0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm

JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 30

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is
(Given charge of electron =1.6 × 10–19C)

Detailed Solution for JEE Main 2019 January 9 Shift 2 Paper & Solutions - Question 30


mv = qBR .. .. (i)
Path is straight line
it qE = qvB
E = vB .. ..(ii)
From equation (i) & (ii)

m = 2.0 × 10–24 kg

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