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JEE Main 2019 January 10 Shift 2 Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2019 January 10 Shift 2 Paper & Solutions

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JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 1

Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 1

4F2 + 9F2 + 12F2 cos θ = R2
4F2 + 36 F2 + 24 F2 cos θ = 4R2
4F2 + 36 F2 + 24 F2 cos θ
= 4(13F2 + 12F2cosθ) = 52 F2 + 48F2cosθ

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

The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R = V/1 , where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is:

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


0.95 × 30 = 0.05 Rυ
Rυ = 19 × 30 = 570 Ω

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JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 3

An unknown metal of mass 192 g heated to a temperature of 100ºC was immersed into a brass calorimeter of mass 128 g containing 240 g of water a temperature of 8.4ºC Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5ºC (Specific heat of brass is 394 J kg–1 K–1)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 3

192 × S × (100 – 21.5)
= 128 × 394 × (21.5 – 8.4)
+ 240 × 4200 × (21.5 – 8.4)
⇒ S = 916

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 4

A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s?

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 4

S = Area under graph
1/2× 2 × 2 + 2 × 2 + 3 × 1 = 9 m

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 5

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is:

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 5




JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 6

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 6

P = I2R
4.4 = 4 × 10–6 R
R = 1.1 × 106 Ω

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 7

The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 7

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 8

At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 8

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

The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?

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


∴ Range of signal = 2250 Hz to 2750 Hz
Now check all options : for 2000 KHZ
fmod = 200 Hz
∴ Range = 1800 KHZ to 2200 KHZ

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

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then :

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

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 11

The electric field of a plane polarized electromagnetic wave in free space at time t= 0 is given by an expression

The magnetic field : (cis the velocity of light)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 11


i.e. direction of 'c'.



JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 12

Condiser the nuclear fission
Ne20 →  2He4 + C12
Given that the binding energy/nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 12

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 13

Two vectors have equal magnitudes. The magnitude of  is 'n' times the magnitude of  The angle between   is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 13


JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 14

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 14


JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 15

Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ?

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 15

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 16

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 16

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 17

Half mole  of an ideal monoatomic gas is heated at constant pressure of 1atm from 20ºC to 90ºC. Work done by gas is close to :
(Gas constant R = 8.31 J /mol.K)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 17

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 18

A metal plate of area 1 × 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2.The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
[1 eV = 1.6 × 10–19J]

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 18


JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 19

Charges -q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences and electrostatic force F. If O is now moved alons the equatorial line to P' such that OP = (y/3), the force on Q will be close to (y/3 >> 2a)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 19

Electric field of equitorial plane of dipole

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 20

Two stars of masses 3 × 1031 kg each, and at distance 2 × 1011m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 ×10–11 Nm2 kg–2)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 20

By energy convervation between 0 & ∞.

[M is mass of star m is mass of meteroite)

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 21

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 21

For closed organ pipe, reson ate frequency is odd multiple of fundamental frequency.
∴ (2n + 1) f0 ≤ 20,000
(fo is fundamental frequency = 1.5 KHz)
∴ n = 6

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 22

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 22


Applying torque equation about point P.

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 23

For the circuit shown below, the current through the Zener diode is:

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 23

Assuming zener diode doesnot undergo breakdown, current in circuit = 120/15000 = 8mA
∴ Voltage drop across diode = 80 V > 50 V.
The diode undergo breakdown.

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 24

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 24



(Potential at ∞  = 0)

∴Work required to put a fifth charge Q at origin is equal to 

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 25

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5 cm then ω close to : (density of water = 103 kg / m3)

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 25


Extra Boyant force = δAxg
B0 + B × mg = ma
B = ma

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 26

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates the work done by the capacitor on the slab is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 26


JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 27

Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2 . The density of the gas is 8 kg /m3. What is the order of energy of the gas due to its thermal motion ?

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 27

Thermal energy of N molecule



order will 104

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 28

A particle which is experiencing a force, given by  undergoes a displacement of If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 28

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 29

The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code ( Orange, Red, Brown). The resistors R2 and R4 are 80Ω and 40Ω, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 29

JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 30

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

Detailed Solution for JEE Main 2019 January 10 Shift 2 Paper & Solutions - Question 30

For Ball
using parallel axis theorem.



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