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JEE Main Practice Test- 10 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Practice Test- 10

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JEE Main Practice Test- 10 - Question 1

A solid ball rolls down a parabolic path ABC from a height h as shown in figure. Portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

Detailed Solution for JEE Main Practice Test- 10 - Question 1

At B , total kinetic energy = mgh

Here,

m = mass of ball

The ratio of rotational to kinetic energy would be , Kr/Kt = 2/5

where, Kr = 2/7mgh and Kt = 5/7 mgh

In portion BC, friction is absent . Therefore, rotational K.E will remain constant and Translational K.E will convert into potential energy.

Hence, if H be the height to which ball climbs in BC, then

mgH = Kt

mgH = 5/7mgh

H = 5/7h

JEE Main Practice Test- 10 - Question 2

A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.

Detailed Solution for JEE Main Practice Test- 10 - Question 2


where T1 and T2 are the temperatures of a source and sink respectively.
When T1 and T2 both are decreased by 100 K each, (T1 - T2) stays constant. T1 decreases.
∴  η increases.

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JEE Main Practice Test- 10 - Question 3

A particle is moving in a uniform circular motion on a horizontal surface. Particle position and velocity at time t = 0 are shown in the figure in the coordinate system. Which of the indicated variable on the vertical axis is incorrectly matched by the graph shown alongside for particle's motion-

Detailed Solution for JEE Main Practice Test- 10 - Question 3



JEE Main Practice Test- 10 - Question 4

In Coolidge tube experiment, if applied voltage is increased to three times, the short wavelength limit of continuous X - ray spectrum shifts by 20 pm. What is the initial voltage applied to the tube ?

Detailed Solution for JEE Main Practice Test- 10 - Question 4

Cut-off wavelength for continuous x-ray is given as : 
∴ hc eV0 = λ  &  hc 3 eV0 = λ - Δλ

⇒   V0 = 2 hc 3 e Δ λ = 41 kV

JEE Main Practice Test- 10 - Question 5

A washer is made of metal having resistivity 10–7 Ωm. The washer has inner radius 1 cm, outer radius 3 cm and thickness 1 mm. A magnetic field, oriented normal to the plane of the washer, has the time dependent magnitude B = (2t) tesla/sec. Find the current (in ampere) around the washer

Detailed Solution for JEE Main Practice Test- 10 - Question 5

Electric field at a general radial distance is E
E = rN/c
J = σE
Current in circular element di = j(dr)t
So net current in washer i = 

JEE Main Practice Test- 10 - Question 6

Ends of two wires A and B having resistivity  and  of same cross section area joined in series together to from a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths  given that temperature coefficient of resistivity of wire A and B is 

αA= 4 ×10–5/ °C and αB = –6 × 10–6/°C. Assume that mechanical dimensions do not change with temperature

Detailed Solution for JEE Main Practice Test- 10 - Question 6

As net resistance does not changes on change temp. so

JEE Main Practice Test- 10 - Question 7

A cone of radius = height = r is under a liquid of density d. Its base is parallel to the free surface of the liquid at a depth H from it as shown in the figure. What is the net force due to liquid on its curved surface? (neglect atmospheric pressure)

Detailed Solution for JEE Main Practice Test- 10 - Question 7

Force due to liquid on curved surface = F

F = weight of liquid above the cone

JEE Main Practice Test- 10 - Question 8

A spherical black body has a radius R and steady surface temperature T, heat sources ensure the heat evolution at a constant rate and distributed uniformly over its volume. What would be the new steady surface temperature of the object if the radius is decreased by half? Assume surrounding to be at absolute zero and heat evolution rate through unit volume remain same.

Detailed Solution for JEE Main Practice Test- 10 - Question 8

Net heat getting generated in complete volume of sphere = rate of heat radiated by its surface

JEE Main Practice Test- 10 - Question 9

The smallest length scale known in physics is the Planck length. It is an important ingredient in some current cosmological theories. Which of the following expressions could represent this Planck length? (Symbols has usual meaning)

Detailed Solution for JEE Main Practice Test- 10 - Question 9

Dimensional formula of only option D matches with length.

JEE Main Practice Test- 10 - Question 10

In the figure shown two motors P1 & P2 fixed on a plank which is hanging with light string passing over fixed Pulley P. If the motors start winding the thread with angular velocity ω1 & ω2 then velocity of plank V is (here R1 & R2 are the radii of motor rotor respectively) [Given: ω1 = 2 rad/s, R1 = 2m, ω2 = 2 rad/s, R2 = 3m]

Detailed Solution for JEE Main Practice Test- 10 - Question 10

JEE Main Practice Test- 10 - Question 11

Five identical balls each of mass m and radius r are strung like beads at random and at rest along a smooth, rigid horizontal thin road of length L, mounted between immovable supports. Assume 10r < L and that the collision between balls or between balls and supports are elastic. If one ball is struck horizontally so as to acquire a speed v, the average force felt by the support is

Detailed Solution for JEE Main Practice Test- 10 - Question 11

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 12

A double star is a system of two stars rotating about their centre of mass only under their mutual gravitational attraction. Let the stars have masses m and 2m and let their separation be l. Their time period of rotation about their centre of mass will be proportional to

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

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 13

If white light is used in a Young’s double – slit experiment. Point C represents centre of a screen

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 14

Consider a thermodynamic cycle in a PV diagram shown in the figure performed by one mole of a monoatomic gas. The temperature at A is T0 and volume at A and B are related as VB = VC = 2VA. Choose the correct option(s) form the following

Detailed Solution for JEE Main Practice Test- 10 - Question 14

temprature at state 'B' is maximum

Net work done by gas in cyclic process

Heat capacity for process 

*Multiple options can be correct
JEE Main Practice Test- 10 - Question 15

If the plank’s constant would be double the present value, in the Bohr’s model for hydrogen atom

Detailed Solution for JEE Main Practice Test- 10 - Question 15

JEE Main Practice Test- 10 - Question 16

In 1906, Robert Millikan devised an experiment that allowed him to determine the charge of an electron. A schematic of Millikan’s set – up is shown below:
Two metal plates are connected by a series of batteries to form a capacitor. There is an electric field between the plates. The metal plates are inside an insulated cylindrical container.
Oil drops are introduced into the container through a small hole in the top. The oil drops acquire a negative charge as they pass through the nozzle of the oil can. Some of the drops fall through a hole in the upper plate. By adjusting the voltage between the plates, certain drops can be suspended between them. The relationship between the electric field between the plates and the voltage across the plates is ∆V = EL
Where E is the electric field and L is the plate separation.
Millikan chose oil because of its relatively low vapour pressure and high charge holding ability. (To answer the following question assume oil drops as to be non-conducting tiny spheres)

In order for an oil drop of mass m, radius r and volume charge density ρ, to be suspended between the plates, the magnitude and direction of the electric field must be:

Detailed Solution for JEE Main Practice Test- 10 - Question 16

JEE Main Practice Test- 10 - Question 17

In 1906, Robert Millikan devised an experiment that allowed him to determine the charge of an electron. A schematic of Millikan’s set – up is shown below:
Two metal plates are connected by a series of batteries to form a capacitor. There is an electric field between the plates. The metal plates are inside an insulated cylindrical container.
Oil drops are introduced into the container through a small hole in the top. The oil drops acquire a negative charge as they pass through the nozzle of the oil can. Some of the drops fall through a hole in the upper plate. By adjusting the voltage between the plates, certain drops can be suspended between them. The relationship between the electric field between the plates and the voltage across the plates is ∆V = EL
Where E is the electric field and L is the plate separation.
Millikan chose oil because of its relatively low vapour pressure and high charge holding ability. (To answer the following question assume oil drops as to be non-conducting tiny spheres)

Suppose the original oil droplet were replaced with a positively charged one that had twice the charge and three times the mass of the original droplet, how would the magnitude of the electric field have to be changed in order for the drop to remain suspended?

JEE Main Practice Test- 10 - Question 18

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.
Speed of the revolving particle is, in the first Bohr orbit.

Detailed Solution for JEE Main Practice Test- 10 - Question 18

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

JEE Main Practice Test- 10 - Question 19

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.
Radius of hypothetical atom is

Detailed Solution for JEE Main Practice Test- 10 - Question 19

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

JEE Main Practice Test- 10 - Question 20

In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is  It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is r0 speed of electron is V0, and total energy is –E0. Now answer the following questions.
The momentum of an emitted photon when it makes a transition from the second excited state to ground state, is

Detailed Solution for JEE Main Practice Test- 10 - Question 20

If mass of revolving particle is m, and change q. change at nucleus Q

Energy of nth orbit

Radius of first orbit of this atom

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 21

The given figure shows a plot of the time dependent force x F acting on a particle in motion along the x-axis. What is the total impulse (in kg-m/s) delivered by this force to the particle from time t = 0 to t = 2 second?


Detailed Solution for JEE Main Practice Test- 10 - Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 22

In the figure shown a small block B of mass m is released from the top of a smooth movable wedge A of the same mass m. The height of wedge A shown in figure is h = 16 cm. B ascends another movable smooth wedge C of the same mass. Neglecting friction anywhere find the maximum height (in cm) attained by block B on wedge C.


Detailed Solution for JEE Main Practice Test- 10 - Question 22

Let u and v be the speed of wedge A and block B at just after the Block B get off the wedge A. Applying conservation of momentum
in horizontal direction, we get
mu = mv   ...(1)
Applying conservation of energy between initial and final state as shown in figure (1), we get

At the instant block B reaches amximum height h' on the wedge C (figure 2) the speed of block B and wedge C are v'
Applying conservation of momentum in horizontal direction, we get
mv=(m+m)v'  ...(4)

Applying conservation of energy between initial and final state

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 23

Mass 2m is kept on a smooth circular track (R = 9 meters) of mass m which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity  towards left and released. Find the maximum height reached by 2m in meters.


Detailed Solution for JEE Main Practice Test- 10 - Question 23

Let v be the final speed of block when it is at maximum height. At that instant the speed of circular track shall also be v

From conservation of momentum

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 24

Two blocks of masses m1 and m2 are connected by spring of constant K such that m2/m1=9. The spring is initially compressed and the system is released from rest at t = 0 second. The work done by spring on the blocks m1 and m2 be ω1 and ω2 respectively by time t. The speeds of both the blocks at time ‘t’ are non-zero. Then find the value of ω12.


Detailed Solution for JEE Main Practice Test- 10 - Question 24

Here in this question when we released the system this will start moving

So ω1 =  work done by spring on the block 1 = change in kinetic energy of block 1

ω2 = work done by spring on the block 2 = change in kinetic energy of block 2

So only spring will do the work

So 

and  

so  

And we know there is no external force on this system so momentum will be conserved

So           m1v1 = m2v2

so     v1/v2 = m1/m2 

So  now put this in the ratio

We get  ω12  = m2/m1 = 9

*Answer can only contain numeric values
JEE Main Practice Test- 10 - Question 25

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.


Detailed Solution for JEE Main Practice Test- 10 - Question 25

 If t is the time taken by pendulums to come in same phase again first time after t=0.

and NS= Number of oscillations made by shorter length pendulum with time period TS.

NL= Number of oscillations made by longer length pendulum with time period TL

Then t=NSTS=NLTL

 ⇒ NS=2NL i.e. if NL=1
⇒ NS=2

JEE Main Practice Test- 10 - Question 26

Copper reduces  into NO and NO2 depending upon concentration of HNO3 in solution. Assuming  [Cu2+] = 0.1M, and PNO = PNO2 = 10–3 bar. At which concentration of HNO3, thermodynamic tendency for reduction of  into NO and NO2 by copper is same?

Detailed Solution for JEE Main Practice Test- 10 - Question 26

JEE Main Practice Test- 10 - Question 27

A radioactive material (t1/2 = 30 days) gets spilled over the floor of a room. If initial activity is ten times the permissible value, after how many days will it be safe to enter the room

Detailed Solution for JEE Main Practice Test- 10 - Question 27

It will be safe to enter the room after the nuclei decreases to 1/10 of its initial amount.

→ Half life = 30 days

→ Using integration law,

 ∴ After 100 days, it will be safe to enter the room.

JEE Main Practice Test- 10 - Question 28

Detailed Solution for JEE Main Practice Test- 10 - Question 28

Trinitro benzene diagenium ionis strong electrophile type and show coupling even with mesitylene

JEE Main Practice Test- 10 - Question 29

 A mixture of all possible stereoisomers from the above structure is subjected to fractional distillation, which of the following statements is correct

JEE Main Practice Test- 10 - Question 30

Which of the following about SF4, SOF4 and COF2 molecules is correct?

Detailed Solution for JEE Main Practice Test- 10 - Question 30

Equatorial FSF bond angle is less in SFthan in SOF4 since lone pair repulsion is more than two electron pairs in double bond
In both SF4 and SOF4 the hybridisation state of S is same sp3d
The OCF bond angle on COF2 is more than 120° since two electron pairs in double bond repel more than one electron pair in C-F bonds.
Due to repulsion by lone pair the axial FSF bond angle is less than 180°

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