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JEE Main Mock Test - 17 - JEE MCQ


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JEE Main Mock Test - 17 - Question 1

A man swims from a point A on one bank of a river of width 100 m. When he swims perpendicular to the water current, he reaches the other bank 50 m downstream. The angle to the bank at which he should swim, to reach the directly opposite point B on the other bank is

Detailed Solution for JEE Main Mock Test - 17 - Question 1

Refer to Fig.

Refer to Fig.

So, it is 60 upstream.

JEE Main Mock Test - 17 - Question 2

A cubical block of wood of side l floats at the interface between oil of relative density n and water with its lower face x = l/4 below the interfaces as shown in Fig. The relative density of wood is

Detailed Solution for JEE Main Mock Test - 17 - Question 2

Total upward buoyant force on the block is
U = buoyant force due to oil + buoyant
force due to water
= ρ0(l − x)l2g + ρwxl2g
For floating U = weight of the block = ρbl3g
∴ ρbl3g = ρ0(l − x)l2g + ρwxl2g

So the correct choice is (c).

JEE Main Mock Test - 17 - Question 3

If a particle of charge 10−12 coulomb moving along the with a velocity 105 m/s experiences a force of 10−10 newton in  due to magnetic field, then the minimum magnetic field is

Detailed Solution for JEE Main Mock Test - 17 - Question 3

JEE Main Mock Test - 17 - Question 4

Young's double slit expetiment is performed using light of wavelength λ. One of the slits is covered by a thin glass sheet of refractive index μ at this wavelength. The smallest thickness of the sheet to bring the adjacent minimum to the centre of the screen is

Detailed Solution for JEE Main Mock Test - 17 - Question 4

If t is the thickness of the glass sheet, the fringes are displaced by an amount given by

In order to bring the adjacent minimum to the centre of the screen (i.e. to bring the first dark fringe the central bright fringe), the fringes must be displaced by half the fringe width, i.e.

JEE Main Mock Test - 17 - Question 5

The current I drawn from a 5 volt source will be

Detailed Solution for JEE Main Mock Test - 17 - Question 5
The equivalent circuit is a balanced Wheatstone's bridge. Hence, no current flows through arm BD.

AB and BC are in series

AD and DC are in series

ABC and ADC are in parallel

or

Hence,

JEE Main Mock Test - 17 - Question 6

In the circuit shown in figure, A and B are two cells of the same emf, E and of internal resistances rA and rB, respectively. L is an ideal inductor and C is an ideal capacitor. The key K is closed. When the current in the circuit becomes steady, what should be the value of R so that the potential difference across the terminals of cell A is zero?

Detailed Solution for JEE Main Mock Test - 17 - Question 6

When the key, K is inserted, the current starts growing and after some time it acquires a steady value. At this stage, no current flows through the capacitor (because an ideal capacitor offers an infinite resistance to a steady current). All the current flows through the inductor (because an ideal inductor offers zero resistance to a steady current). Now, the network of resistors is a balanced wheatstone bridge. Hence, no current flows through the resistance 2R. Therefore, this resistance can be ignored. The net resistance between points X and Y = resistance of the parallel combination of 2 R and 2 R = = R

Hence, the current in the circuit is

I =

Now, the terminal voltage of cell A is

2rA = R + rA + rB or R = rA – rB

So, the correct choice is (a).

JEE Main Mock Test - 17 - Question 7

A coil of inductance 0.20 H is connected in series with a switch and a cell of emf 1.6 V. The total resistance of the current is 4.0Ω. What is the initial rate of growth of the current when the switch is closed?

Detailed Solution for JEE Main Mock Test - 17 - Question 7
When the switch is closed at(t = 0s), no current flows, voltage drop across the inductor is the same as the supply voltage of 1.6 V.

Voltage equation for the circuit, we have

Where i is the current drawn from the source , at t = 0s, i = 0, we thus have

JEE Main Mock Test - 17 - Question 8

Mass of the earth is 81 times the mass of the moon and the distance between the earth and moon is 60 times the radius of the earth. If R is the radius of the earth, then the distance between the moon and the point on the line joining the moon and earth where the gravitational force becomes zero is

Detailed Solution for JEE Main Mock Test - 17 - Question 8

Let d be the distance between the moon and the point at force on mass m is zero then

JEE Main Mock Test - 17 - Question 9

A magnet of magnetic moment M and length 2l is bent at its mid-point such that the angle of bending is 60°. Now, the magnetic moment is

Detailed Solution for JEE Main Mock Test - 17 - Question 9

In new situation we have

As the length of magnet is halved , Magnetic moment M’= m(l) = M/2

Resultant magnetic Moment

JEE Main Mock Test - 17 - Question 10

Three identical uniform rods of the same mass M and length L are arranged in xy plane as shown in the figure. A fourth uniform rod of mass 3 M has been placed as shown in the xy plane. What should be the value of the length of the fourth rod such that the centre of mass of all the four rods lie at the origin?

Detailed Solution for JEE Main Mock Test - 17 - Question 10

Let 'x' be the length of the 4th rod and the centre of the mass of all the rods lies at origin then Xcm = 0

JEE Main Mock Test - 17 - Question 11

A metallic wire with tension T and at temperature 30°C vibrates with its fundamental frequency of 1 kHz. The same wire with the same tension but at 10°C temperature vibrates with a fundamental frequency of 1.001 kHz. The coefficient of linear expansion of the wire is

Detailed Solution for JEE Main Mock Test - 17 - Question 11

The frequency at which a wire vibrates is given by:

Here, g is the acceleration due to gravity and l is the length of the wire.

Let’s express the fundamental frequency of the vibration at T1​=30C as follows,

Let’s express the fundamental frequency of the vibration at T2​=10C as follows

Dividing equation (2) by equation (1), we get,

We can express the linear expansion of the wire due to change in temperature as,

Here, α is the linear expansion coefficient and ΔT is the change in temperature.

Using equation (3), we can rewrite the above equation as,

Substituting 1KHz for f1​, 1.001KHz for f2​ and −20C for ΔT in the above equation, we get,

JEE Main Mock Test - 17 - Question 12

For a gas, the difference between the two specific heats at constant pressure and constant volume is 4150 J kg-1 K-1 and their ratio is 1.4. What is the specific heat of the gas at constant volume in units of J kg-1 k-1?

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

Given: Cp - Cv = 4150 Jkg-1K-1 and Cp/Cv = 1.4 Or Cp = 1.4 Cv.

Therefore,

1.4 Cv - Cv = 4150

Or CV = 4150/0.4 = 10375 J kg-1 K-1

JEE Main Mock Test - 17 - Question 13

Among the following pairs, which one does not have identical dimensions?

Detailed Solution for JEE Main Mock Test - 17 - Question 13

Moment of inertia (I) = mr2

[I] = [ML2]

Moment of force (C) = rF

[C] = [r][F] = [L][MLT-2] or [C] = [ML2T-2]

Moment of inertia and moment of a force do not have identical dimensions.

JEE Main Mock Test - 17 - Question 14

Which of the following pairs of physical quantities does not have same dimensional formula ?

Detailed Solution for JEE Main Mock Test - 17 - Question 14

Tension=[MLT−2]

Surface Tension = [ML0T−2]

Clearly these two have different dimension.

JEE Main Mock Test - 17 - Question 15

A plank is held at an angle αα to the horizontal on two fixed supports A and B. The plank can slide against the supports (without friction) because of its weight MgMg. Acceleration and direction in which a man of mass mm should move so that the plank does not move is:

Detailed Solution for JEE Main Mock Test - 17 - Question 15

F.B.D. of man and plank are

For plank to be at rest, applying Newton's second law to plank along the incline

Mgsinα = f

and applying Newton's second law to man along the incline.

mgsinα + f = ma

a = gsinα(1 + M/m) down the incline

JEE Main Mock Test - 17 - Question 16

A force is acting at a point . The value of αα for which angular momentum about origin is conserved is:

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

If = constant then

So should be parallel to so coefficient should be in same ratio. So

So α = −1

JEE Main Mock Test - 17 - Question 17

A simple pendulum of length ll is suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination θ. What will be the time period of the pendulum?

Detailed Solution for JEE Main Mock Test - 17 - Question 17
Here, the point of suspension has an acceleration. (down the plane). Further, can be resolved into two components gsinθ (along the plane) and gcosθ (perpendicular to the plane).

= gcosθ (perpendicular to plane)

JEE Main Mock Test - 17 - Question 18

The frequency of a sonometer wire is f but when the weights producing tension are completely immersed in water the frequency becomes f/2 and on immersing the weights in certain liquid the frequency becomes f/3. The specific gravity of the liquid is

Detailed Solution for JEE Main Mock Test - 17 - Question 18

Let ρ the density of weight.

σ be the density of liquid.

and V be the volume of the weights.

From (1) and (2)

From (1) and (3)

JEE Main Mock Test - 17 - Question 19

Which of these is correct in regard to a magnet?

Detailed Solution for JEE Main Mock Test - 17 - Question 19
Magnetic length is the distance between the two equal and opposite poles of the magnet.

Poles are not actually at the end of the magnet they are little inside from the end this makes the magnetic length of the magnet little less than the Geometric length.

Geometric length is the actual length of the magnet.

So when the relationship between these two measured it comes to be,

Magnetic length = 0.8× Geometric length

JEE Main Mock Test - 17 - Question 20

The mean lives of radioactive substances are 1620 y and 405 y for α- emission and β- emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α- emission and β- emission simultaneously.

Detailed Solution for JEE Main Mock Test - 17 - Question 20
Let at some instant of time tt, number of atoms of the radioactive substance are N. It may decay either by α - emission on by β- emission. So, we can write,

If the effective decay constant is λ, then

Now,

*Answer can only contain numeric values
JEE Main Mock Test - 17 - Question 21

Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of 100 kg and they are 100 m apart. They are initially at rest relative to each other. How long (in number of days) will it be before the gravitational attraction brings them 1 cm closer together? (Nearest integer)


Detailed Solution for JEE Main Mock Test - 17 - Question 21

a1 = acceleration of first

a2 = acceleration of second

Net acceleration for approach

Now

Solving , we get t = 1.41 days

≈ 1 day

*Answer can only contain numeric values
JEE Main Mock Test - 17 - Question 22

An object O is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The plane surface of the lens is silvered as shown in the figure below:

The image is formed at a distance of ___ cm to the left of the lens. Fill in the blank with appropriate integer.


Detailed Solution for JEE Main Mock Test - 17 - Question 22

The effective focal length of the silvered lens is given by:

, which gives F = 15/2 cm.

The silvered lens behaves like a concave mirror.

Using the spherical mirror formula , we have

, which gives v = -12 cm. The negative sign indicates that the image is formed to the left of the lens.

*Answer can only contain numeric values
JEE Main Mock Test - 17 - Question 23

Water is drawn from a well into a 5 kg drum of capacity 55 litres, by two ropes connected to the top of the drum. The linear mass density of each rope is 0.5 kg m-1. The work done in lifting water to the ground from the surface of water in the well 20 m below is c × 104 J. The value of c, up to one decimal place, is [g = 10 ms-2]

(Nearest integer)


Detailed Solution for JEE Main Mock Test - 17 - Question 23

Work done in lifting water and drum = (60 x 10 x 20) J = 12000 J

Total mass of ropes = 40 x 0.5 kg = 20 kg

As the mass is uniformly distributed over the length, centre of the mass lies at the centre of the rope and while lifting the rope, CM rises by 10 m.

Work done in the case of ropes = (20 x 10 x 10) J = 2000 J

Total work done = 14000 J

Total work done ≈ 1 × 104 J

*Answer can only contain numeric values
JEE Main Mock Test - 17 - Question 24

An iron bar of length 10 m is heated from 0 °C to 100 °C. If the coefficient of linear thermal expansion of iron is 10×10−6 °C−1, the increase in the length of the bar in cm is


Detailed Solution for JEE Main Mock Test - 17 - Question 24

The change in length ∆l is proportional to ll and ∆T. Stated mathematically

Δl = αlΔT

Where, α is called the coefficient of linear thermal expansion for the material.

Given, ΔT = 100 °

l = 10 m

∴ Δl=10 × 100 × 10 × 10−6

= 10−2 m = 1 cm

*Answer can only contain numeric values
JEE Main Mock Test - 17 - Question 25

An uncharged capacitor of capacitance 4 μF and a resistance of 2.5 MΩ are connected in series with 12 V battery at t = 0. Find the time in seconds (approximate to the nearest integer) after which the potential difference across the capacitor is three times the potential difference across the resistor. [Given, ln(2) = 0.693]


Detailed Solution for JEE Main Mock Test - 17 - Question 25

During charging voltage across capacitor with time is given by

Vc =  V(1 − e−t/T)

V = VC + VR

Here, V is the applied potential.

Given that VC = 3VR = 3(V−VC)

Here τ = CR =10 s

⇒ t = 2 × 10 × 0.693 = 13.86 s ≃ 14 s

JEE Main Mock Test - 17 - Question 26

The reaction, N2(g) + 3H2(g) ⇌ 2NH3(g) is exothermic and reversible. A mixture of N2(g),H2(g) and NH3(g) is at equilibrium in a closed container. When a certain quantity of extra H2(g) is introduced into the container, while keeping the volume constant, then which statement among the following is true?

Detailed Solution for JEE Main Mock Test - 17 - Question 26

Since, the reaction is exothermic and volume is kept constant. So, the temperature will increase.

JEE Main Mock Test - 17 - Question 27

What is the pH of 0.01M glycine solution? For glycine Ka1 = 4.5 × 10−3 and Ka2 = 1.7 × 10−10 at 298 K

Detailed Solution for JEE Main Mock Test - 17 - Question 27

JEE Main Mock Test - 17 - Question 28

The precipitate of Al(OH)3 dissolves in NaOH solution. It is due to the formation of

Detailed Solution for JEE Main Mock Test - 17 - Question 28

Dissolution of Al(OH)3 by a solution of NaOH produces a complex compound shown as below:
Al(OH)+ NaOH ⟶ [Al(H2O)2(OH)4]

JEE Main Mock Test - 17 - Question 29

Sulphuric acid reacts with PCl5 to give

Detailed Solution for JEE Main Mock Test - 17 - Question 29

HO − SO− OH + PCl→ Cl − SO− O + POCl+ HCl

JEE Main Mock Test - 17 - Question 30

For all x, y ∈ R, the maximum value of sin2x + sin2y + sin2(x + y) equals

Detailed Solution for JEE Main Mock Test - 17 - Question 30


LHS   (equality occurs if x = y = π/3)

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