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JEE Main Mock Test - 12 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Mock Test - 12

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JEE Main Mock Test - 12 - Question 1

In an LCR circuit as shown below, both switches are open initially. Later, switch S1 is closed and S2 is kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statements is correct?

Detailed Solution for JEE Main Mock Test - 12 - Question 1

Switch S1 is closed and switch S2 is kept open. Now, capacitor is charging through a resistor R.

Charge on a capacitor at any time t,

q = q0(1 - e-t/τ)

q = CV(1 - e-t/τ) [As q0 = CV]

At t = τ/2, q = CV(1 - e-τ/2τ) = CV(1 - e-1/2)

At t = τ, q = CV(1 - e-τ/τ) = CV(1 - e-1)

At t = 2τ, q = CV(1 - e-2τ/τ) = CV(1 - e-2)

JEE Main Mock Test - 12 - Question 2

In an ac generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is

Detailed Solution for JEE Main Mock Test - 12 - Question 2

In an ac generator, emf induced, e = NABω sinωt

Hence, amplitude of induced emf or maximum emf = NABω

JEE Main Mock Test - 12 - Question 3

A rectangular loop has a sliding connector PQ of length ℓ and resistance RΩ and is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are

Detailed Solution for JEE Main Mock Test - 12 - Question 3
Emf induced across PQ is ε = Bℓv. The equivalent circuit diagram is as shown in the figure.

Applying Kirchhoff's first law at junction Q, we get I = I1 + I2 ... (i)

Applying Kirchhoff's second law for the closed loop PLMQP, we get

-I1R - IR + ε = 0

I1R + IR = Bv ... (ii)

Again, applying Kirchhoff's second law for the closed loop PONQP, we get

-I2R - IR + ε = 0

I2R + IR = Bℓv ... (iii)

Adding equations (ii) and (iii), we get

2IR + I1R + I2R = 2Bℓv

2IR + R(I1 + I2) = 2Bℓv

2IR + IR = 2Bℓv (Using (i))

3IR = 2Bℓv

Substituting this value of I in equation (ii), we get I1 = Bℓv/3R

Substituting the value of I in equation (iii), we get I2 = Bℓv/3R

JEE Main Mock Test - 12 - Question 4

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if Q is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 4
Since the hypotenuse of the triangle is √2a, the net electrostatic energy is

For U = 0, we require

Hence, option 2 is the correct answer.

JEE Main Mock Test - 12 - Question 5

A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration 'a' to keep the block stationary. Then, 'a' is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 5

The incline is given an acceleration a.

Acceleration of the block is to the right.

Pseudo acceleration 'a' acts on block to the left.

Equate resolved parts of 'a' and 'g' along the incline.

ma cos α = mg sin α

or, a = g tan α

JEE Main Mock Test - 12 - Question 6

The diameter of a drop of liquid fuel changes with time due to combustion according to the following relationship.

While burning, the drop falls at its terminal velocity under Stokes flow regime. Find the distance that it will travel before complete combustion.

Detailed Solution for JEE Main Mock Test - 12 - Question 6
Given, diameter of the liquid drop,

At complete combustion

D = 0

∴ t = tb

And at t = 0, D = D0

Thus, the initial diameter of the drop is D0 and changes according to given equation and the time for compelete combustion is tb.

From Stokes law, the velocity of drop,

Also

JEE Main Mock Test - 12 - Question 7

The half-life period of a radioactive element X is the same as the mean life time of another radioactive element Y. Initially, they have the same number of atoms. Then,

Detailed Solution for JEE Main Mock Test - 12 - Question 7
T1/2 (half life of X) = τY (mean life of Y)

⇒ λx = λY(0.693)

Y will decay faster than X.

JEE Main Mock Test - 12 - Question 8

An object is 1 metre in front of the curved surface of a plano-convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?

Detailed Solution for JEE Main Mock Test - 12 - Question 8

Here,

u = 100 cm

v = 120 cm

Here,

FL - focal length of the lens

FM - focal length of the mirror

JEE Main Mock Test - 12 - Question 9

Consider a two particle system with particles having masses m1 and m2. The first particle is pushed towards the centre of mass through a distance d. By what distance should the second particle be moved, so as to keep the centre of mass at the same position?

Detailed Solution for JEE Main Mock Test - 12 - Question 9
Let m2 be moved by x, so as to keep the centre of mass at the same position.

∴ m1d + m2 (- x) = 0

or, m1d = m2x

JEE Main Mock Test - 12 - Question 10

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, then the final equilibrium temperature of the gas in the container will be

Detailed Solution for JEE Main Mock Test - 12 - Question 10

As this is a simple mixing of gas, PV = nRT for adiabatic as well as isothermal changes.

The total number of molecules is conserved.

Final state = (n1 + n2)RT

JEE Main Mock Test - 12 - Question 11

In a hydrogen atom, the electron moves around the nucleus in a circular orbit of radius 5 × 10−11 m. Its time period is 1.5 × 10−16s. The current associated with the electron motion is (charge of electron is 1.6 × 10−19 C)

Detailed Solution for JEE Main Mock Test - 12 - Question 11

Time taken by charge to cross a point on the orbit = 1.5 × 10−16 s

JEE Main Mock Test - 12 - Question 12

A cubical vessel open from top of side L is filled with a liquid of density ρ then the torque of hydrostatic force on a side wall about an axis passing through one of bottom edges is-

Detailed Solution for JEE Main Mock Test - 12 - Question 12
Force by hydrostatic pressure

and centre of pressure is at height L/3.

JEE Main Mock Test - 12 - Question 13

A certain ideal gas undergoes a polytropic process PVn = constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be γ, then the range of values of n will be

Detailed Solution for JEE Main Mock Test - 12 - Question 13
The molar specific heat capacity in a polytropic process PVn = constant is given by

From this equation, we see that C will be negative when n < γ="" n="" />< γ="" />

n > 1, simultaneously, i.e., 1 < n="" />< γ.="" since="" γ="" for="" all="" ideal="" gases="" is="" greater="" than="" 1,="" if="" n="" /> γ or n < 1,="" then="" />v will be positive.

JEE Main Mock Test - 12 - Question 14

A ball is projected from the bottom of an inclined plane of inclination 30o, with a velocity of 30 m s−1, at an angle of 30o with the inclined plane. If g = 10 ms−2, then the range of the ball on given inclined plane is

Detailed Solution for JEE Main Mock Test - 12 - Question 14

Range on incline plane,

JEE Main Mock Test - 12 - Question 15

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

Detailed Solution for JEE Main Mock Test - 12 - Question 15
Frequency of LC oscillation

JEE Main Mock Test - 12 - Question 16

If the radius of the earth were to shrink by one percent and its mass remains the same, the acceleration due to gravity on the earth's surface would

Detailed Solution for JEE Main Mock Test - 12 - Question 16

g will increase if R decreases.

JEE Main Mock Test - 12 - Question 17

A black body, at 200 K, is found to have maximum energy at a wavelength of 14 μm. When its temperature is raised to 1000 K, the wavelength at which maximum energy is emitted is

Detailed Solution for JEE Main Mock Test - 12 - Question 17

JEE Main Mock Test - 12 - Question 18

The time constant of charging of the capacitor shown in the diagram is

Detailed Solution for JEE Main Mock Test - 12 - Question 18
The given circuit is equivalent to two simple circuits connected in parallel with the battery as shown below.

τc = 2RC

JEE Main Mock Test - 12 - Question 19

Spherical wavefronts shown in figure, strike a plane mirror. Reflected wavefronts will be as shown in

Detailed Solution for JEE Main Mock Test - 12 - Question 19
According to Huygens principle, ray of light will be normal to the wavefront.

JEE Main Mock Test - 12 - Question 20

A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height 2R above the earth's surface where R is the radius of the earth. The value of

Detailed Solution for JEE Main Mock Test - 12 - Question 20
The periodic time of a simple pendulum is given by,

When taken to height 2R,

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 21

The magnetic field due to a current-carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. Its value at the centre of the loop is CμT. What is the value of C?


Detailed Solution for JEE Main Mock Test - 12 - Question 21
Magnetic field at any point on the axis of coil,

At the centre of coil,

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 22

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell, in ohm, is


Detailed Solution for JEE Main Mock Test - 12 - Question 22
The internal resistance of a cell is given by

Hence:

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 23

The height, in terms of R (radius of Earth), at which the acceleration due to gravity becomes g/9 (where g = acceleration due to gravity on the surface of Earth), is nR. What is the value of n?


Detailed Solution for JEE Main Mock Test - 12 - Question 23
The acceleration due to gravity at a height h from the ground is given as g/9.

At the surface of Earth, the acceleration due to gravity is

At r = R + h:

Acceleration due to gravity

∴ r = 3R

The height above the ground = h = r - R = 2R

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 24

A particle executes S.H.M. given by x = 0.24cos(400t − 0.5) in SI units. Find amplitude in cmcm.


Detailed Solution for JEE Main Mock Test - 12 - Question 24
Here, x = 0.24 cos (400 t - 0.5) ...(i)

The standard equation for S.H.M. is

x = r cos(2 π v t − ϕ)…(ii)⁡

Comparing the equations (i) and (ii),we have

r = 0.24 m

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 25

The apparent depth of needle lying at the bottom of the tank which is filled with water to a height of 15.5 cm is measured by a microscope to be 8.5 cm. If water is replaced by a liquid of refractive index 1.94 to the same height as earlier, then the displacement of the microscope needed to establish the focus on the needle again is n μmn μm. The value of n is


Detailed Solution for JEE Main Mock Test - 12 - Question 25

Actual depth of the needle in water,

h1 = 15.5 cm

Apparent depth, H = 15.5/1.94 = 7.99 cm

Here, H is less than h2. Thus to focus the needle again, the microscope should be moved up by the distance 8.5 − 7.99 = 0.51 cm

JEE Main Mock Test - 12 - Question 26

The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnetic field in one direction, are

Detailed Solution for JEE Main Mock Test - 12 - Question 26
Deflection in the magnetic field confirms that the particles are charged. It does not confirm the charge on the particles.

Hence, option (4) is correct.

α-rays consist of positively charged particles (He++) and β-rays consist of negatively charged particles . Since they are oppositely charged, so they get deflected in opposite directions. γ-rays are neutral (carry no charge). So, they remain undeflected.

JEE Main Mock Test - 12 - Question 27

Both the molecules possess dipole moment in which of the following pairs?

Detailed Solution for JEE Main Mock Test - 12 - Question 27
H2O and SO2 both possess dipole moment due to bent structure.

JEE Main Mock Test - 12 - Question 28

Which of the following solutions will have pH close to 1.0?

Detailed Solution for JEE Main Mock Test - 12 - Question 28
In option (4),

pH = 1

JEE Main Mock Test - 12 - Question 29

Out of H2S2O3, H2S4O6, H2SO5 and H2S2O8, peroxy acids are

Detailed Solution for JEE Main Mock Test - 12 - Question 29
A peroxy acid (sometimes called peracid) is an acid which contains an acidic -OOH group. H2SO5 and H2S2O8 are peroxy acids as shown in their structures.

JEE Main Mock Test - 12 - Question 30

Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl-, CN- and H2O, respectively, are

Detailed Solution for JEE Main Mock Test - 12 - Question 30
Ni2+ + 4Cl- → [NiCl4]2- - tetrahedral due to sp3 hybridisation.

Ni2+ + 4CN- → [Ni(CN)4]2- - dsp2 hybridisation, it is square planar.

Ni2+ + 6H2O → [Ni(H2O)6]2+ - sp3d2 hybridisation, it is octahedral.

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