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JEE Main Mock Test - 16 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Mock Test - 16

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JEE Main Mock Test - 16 - Question 1

The dimensions of 1/2 ε0E2= permittivity of free space and E = electric field) are

Detailed Solution for JEE Main Mock Test - 16 - Question 1

1/2 ε0E2 is the expression for electrostatic energy density, i.e., the energy stored per unit volume in a parallel plate capacitor.

JEE Main Mock Test - 16 - Question 2

A pendulum has a length l. Its bob is pulled aside from its equilibrium position through an angle α and then released. The speed of the bob when it passes through the equilibrium position is given by

Detailed Solution for JEE Main Mock Test - 16 - Question 2

As shown in Fig., the height attained by the bob when the string subtends an angle α with the vertical is
h = l − lcos α = l(1 − cos α)

Its potential energy at the highest point A = mgh where m is the mass of the bob. Let v be the speed of the bob when it passes through O. Its kinetic energy at O = 1/2mv2. From the principle of conservation of energy, we have

Hence the correct choice is (c).

JEE Main Mock Test - 16 - Question 3

Three +ve charges of equal magnitude 'q' are placed at the vertices of an equilateral triangle of side l. How can the system of charges be placed in equilibrium ?

Detailed Solution for JEE Main Mock Test - 16 - Question 3

To keep the system in equilibrium, net force experienced by charges at 'A′,′B′ and 'C′ should be zero. For this another charge of the opposite sign should be placed at the centroid of the triangle. Let this charge be '−Q′.

JEE Main Mock Test - 16 - Question 4

A particle of charge q and mass m moves in a circular orbit of radius r with the angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Detailed Solution for JEE Main Mock Test - 16 - Question 4

The effective current is i =  and the area is A = πr2
The magnetic moment is
The angular momentum is

JEE Main Mock Test - 16 - Question 5

If a wire is stretched to make it 0.1% longer, its resistance will

Detailed Solution for JEE Main Mock Test - 16 - Question 5
Resistance of the wire is given by

On stretching, volume (V) remains constant.

Percentage change in the resistance of the wire is:

Hence, when the wire is stretched by 0.1%, its resistance will increase by 0.2%.

JEE Main Mock Test - 16 - Question 6

Directions: The following question contains statement-1 and statement-2. Of the four choices given, choose the one that best describes the two statements.

Statement-1: For a mass M kept at the centre of a cube of side a, the flux of gravitational field passing through its sides is 4πGM.

Statement-2: If the direction of a field due to a point source is radial and its dependence on the distance r from the source is given as 1/r2, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

Detailed Solution for JEE Main Mock Test - 16 - Question 6
Let A be the Gaussian surface enclosing a spherical charge Q.

Electric field at any point at a distance 'r' from the charge Q is:

Flux through the Gaussian surface,

Every line passing through A has to pass through B, whether B is a cube or any surface. It is only for Gaussian surface that the lines of field should be normal, assuming the mass to be a point mass.

Gravitational field,

Flux,

Here, B is a cube. As explained earlier, whatever be the shape, all the lines passing through A are passing through B, although all the lines are not normal.

Statement-2 is correct because when the shape of the earth is spherical, area of this Gaussian surface is 4πr2. This ensures inverse square law.

JEE Main Mock Test - 16 - Question 7

When the current in the portion of the circuit, shown in the figure, is 2 A and is increasing at the rate of 1 A/s, then the measured potential difference Vab is 8 V. However, when the current is 2 A and is decreasing at the rate of 1 A/s, the measured potential difference Vab is 4V. The values of R and L are

Detailed Solution for JEE Main Mock Test - 16 - Question 7

For the first case, when the current is increasing di/dt = 1A / s

8 = 2R + L …..(1)

For the first case, when the current is increasing di/dt = -1A / s

4 = 2R - L …..(2)

From 1 and 2, R = 3Ω, L = 2H

JEE Main Mock Test - 16 - Question 8

A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e-bt in the x direction.

Which of the following curves depicts its speed v(t)?

Detailed Solution for JEE Main Mock Test - 16 - Question 8
F(t) = F0e-bt (Given)

ma = F0e-bt

Integrating both sides, we get

JEE Main Mock Test - 16 - Question 9

A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)

Detailed Solution for JEE Main Mock Test - 16 - Question 9
For the wire having length L, the elongation in the wire is given by l.

Then,

Therefore,

In the second case, due to pulley arrangement,

The length of wire is L/2 on each side.

So, the elongation on either side will be:

Hence, total increase in length on both sides is l.

JEE Main Mock Test - 16 - Question 10

Two particles, A and B, of equal masses are suspended from two massless springs of spring constants k1 and k2, respectively. If the maximum velocities, during oscillations, are equal, then the ratio of amplitudes of A and B is

Detailed Solution for JEE Main Mock Test - 16 - Question 10
Maximum velocity under simple harmonic motion, vm = aω.

JEE Main Mock Test - 16 - Question 11

The amplitude of wave disturbance propagating in the positive x-axis is given by at t = 0 and at t = 2s, where x and y are in metres. The shape of the disturbance does not change during the propagation. The velocity of the wave is

Detailed Solution for JEE Main Mock Test - 16 - Question 11

The given pulse is of the form

Here, v is the wave velocity.

The given equation is:

Comparing equations (1) and (2), we get

JEE Main Mock Test - 16 - Question 12

Two cylindrical rods of uniform cross-sectional area A and 2A, having free electrons per unit volume 2n and n respectively are joined in series. A constant current I flows through them in steady state. The ratio of the drift velocity of free electrons in the left rod to that of the right rod is

Detailed Solution for JEE Main Mock Test - 16 - Question 12
Since current, I = ne Avd through both rods is the same.

⇒ (2n)eAvL = ne(2A)vR

⇒ vL/vR = 1

JEE Main Mock Test - 16 - Question 13

Water is filled up to a height h in a cylindrical vessel. It takes time t to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height 4h then the time it takes to completely drain the vessel is

Detailed Solution for JEE Main Mock Test - 16 - Question 13
Time required to empty the tank,

∴ t2 = 2t

JEE Main Mock Test - 16 - Question 14

A refrigerator is to maintain the eatables, kept inside it, at 9°C. The coefficient of performance of the refrigerator, if the room temperature is 36°C, is

Detailed Solution for JEE Main Mock Test - 16 - Question 14
Here, T1 = 36℃ = 36 + 273 = 309 K

T2 = 9℃ = 9 + 273 = 282 K

The coefficient of performance of a refrigerator is given by

JEE Main Mock Test - 16 - Question 15

An elevator is going upward with an acceleration a = g/4 and a ball is released from rest relative to the elevator at a distance h1 above the floor. The speed of the elevator at the time of ball release is v0. Then the bounce height h2 of the ball with respect to the elevator is (the coefficient of restitution for the impact is e)

Detailed Solution for JEE Main Mock Test - 16 - Question 15

JEE Main Mock Test - 16 - Question 16

The magnetic flux through a circuit of resistance R changes by an amount Δϕ in a time Δt. The total magnitude of electric charge Q, that passes through any point in the circuit during this time is

Detailed Solution for JEE Main Mock Test - 16 - Question 16
Induced emf is given by

Current, [where Q is total charge passed in time Δt]

JEE Main Mock Test - 16 - Question 17

To get maximum current through a resistance of 2.5 Ω, one can use mm rows of cells, each row having n cells. The internal resistance of each cell is 0.5 Ω. What are the values of n & m, if the total number of cells is 45?.

Detailed Solution for JEE Main Mock Test - 16 - Question 17

For maximum current :

Net external resistance = net internal resistance.

JEE Main Mock Test - 16 - Question 18

An iron wire AB of length 3 m at 0°C is stretched between the opposite walls of a brass casing at 0°C0°C as shown in the figure below. The diameter of the wire is m0.6 mm. What extra tension will be set up in the wire when the temperature of the system is raised to 40°C?

Given αbrass = 18 × 10−6 K−1, αiron= 12 × 10−6 K−1, Yiron = 21 × 1010 N m−2

Detailed Solution for JEE Main Mock Test - 16 - Question 18
Increase in length of the iron wire = LαΔt = L × 12 × 10−6 × 40 = 48L × 10−5 m

Increase in length of the brass tube = LαΔt = L × 18 × 10−6 × 40 = 72L × 10−5 m

Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then,

T = 21 × 1010 × 24 × 10−5 × 9π × 10−8 = 14.2 N

JEE Main Mock Test - 16 - Question 19

The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used is

Detailed Solution for JEE Main Mock Test - 16 - Question 19
For the first diffraction minimum, dsinθ=λ

And if the angle is small, sinθ = θ

dθ = λ

i.e., Half angular width, θ = λ/d

Full angular width w = 2θ = 2λ/d

= 6000 × 0.7 = 4200 Å

JEE Main Mock Test - 16 - Question 20

What will be the displacement equation of the simple harmonic motion obtained by combining the motions?

(Note: ϕ is an acute angle)

Detailed Solution for JEE Main Mock Test - 16 - Question 20
The resultant equation is

x = A sin(ωt + ϕ)

∑Ax = 2 + 4cos30° + 6 cos 60° = 8.46

Thus, the displacement equation of combined motion is

x = 11.25sin(ωt + ϕ)

Where, ϕ = 40.4°

*Answer can only contain numeric values
JEE Main Mock Test - 16 - Question 21

The minimum frequency of a γ-ray that causes a deuteron to disintegrate into a proton and a neutron is p × 1020 Hz. The value of p is (Nearest integer)

(Given, md = 2.0141 a.m.u., mp = 1.0078 a.m.u. and mn = 1.0087 a.m.u.)


Detailed Solution for JEE Main Mock Test - 16 - Question 21
Total mass of the products = 2.0165 a.m.u., which is greater than the mass of the deuteron by 0.0024 a.m.u.

The extra must be provided by the energy of the photon so that minimum possible frequency must be given by

*Answer can only contain numeric values
JEE Main Mock Test - 16 - Question 22

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength (in V/m) is


Detailed Solution for JEE Main Mock Test - 16 - Question 22
In electromagnetic wave, the peak value of electric field (E0) and peak value of magnetic field (B0) are related by

E0 = B0c

E0 = (20 × 10-9 T)(3 × 108 ms-1)

= 6 V/m

*Answer can only contain numeric values
JEE Main Mock Test - 16 - Question 23

Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C, whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm. respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units. respectively. Rate of heat flow (in cal/s) through copper rod is

(Nearest integer)


Detailed Solution for JEE Main Mock Test - 16 - Question 23

*Answer can only contain numeric values
JEE Main Mock Test - 16 - Question 24

The forward-bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. What is the value of the forward bias resistance?


Detailed Solution for JEE Main Mock Test - 16 - Question 24

The forward biased resistance of a diode is

JEE Main Mock Test - 16 - Question 25

The precipitate of Ag2CrO4(Ksp = 1.9 × 10−12) is obtained when equal volumes of the following are mixed.

Detailed Solution for JEE Main Mock Test - 16 - Question 25

Precipitation occurs when the ionic product exceeds Ksp value. When equal volumes of two solutions are mixed, the concentration of each is reduced to half. Therefore, In first case, Ionic product,


∴ No precipitation occurs.
In fourth case,

JEE Main Mock Test - 16 - Question 26

Identify the least stable ion.

Detailed Solution for JEE Main Mock Test - 16 - Question 26
This option is correct because the electronic configuration of Be = 1s2 2s2.

Therefore, the electronic configuration of Be- = 1s2 2s2 2p1.

So, the stable electronic configuration of Be has broken up and it is highly unstable.

JEE Main Mock Test - 16 - Question 27

The reaction, 2SO2(g) + O2(g) → 2SO3(g) is carried out in a 1 dm3 vessel and a 2 dm3 vessel, separately. The ratio of the reaction velocities, respectively, will be

Detailed Solution for JEE Main Mock Test - 16 - Question 27
2SO2(g) + O2(g) → 2SO3(g)

For 1dm3, R = k[SO2]2[O2]

JEE Main Mock Test - 16 - Question 28

Cyanide process is used for the extraction of

Detailed Solution for JEE Main Mock Test - 16 - Question 28
Gold and silver are extracted from their native ores by MacArthur-Forrest cyanide process.

4Au(s) + 8NaCN(aq) + O2(g) + 2H2O(l) → 4Na[Au(CN)2](aq) + 4NaOH(aq)

2Na[Au(CN)2](aq) + Zn(s) → Na2[Zn(CN)4] + 2Au

JEE Main Mock Test - 16 - Question 29

KO2 (potassium superoxide) is used in oxygen cylinders in space and submarines because it

Detailed Solution for JEE Main Mock Test - 16 - Question 29
Potassium superoxide is widely used as a CO2 scrubber and O2 generator in spacecrafts and submarines.

The following reaction takes place:

4KO2 + 2CO2 → 2K2CO3 + 3O2

JEE Main Mock Test - 16 - Question 30

Which of the following dissolves in water but does not give any oxyacid solution?

Detailed Solution for JEE Main Mock Test - 16 - Question 30
In OF2, fluorine has negative oxidation state due to high electronegativity. Therefore, OF2 dissolves in water but does not give any oxyacid solution because unlike other halogens, fluorine cannot acquire a positive oxidation state. In oxyacids of halogens, all halogen atoms have positive oxidation states.

SO2, SCI4 and SO4 give oxyacid solutions in water and acquire positive oxidation states as shown below.

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