datavorous/entrance-exam-dataset huggingface dataset:
A block mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ( $g$ is acceleration due to gravity) will be:

Options:
A $m g \cos \theta$
B $m g \sin \theta$
C $m g$
D $\mathrm{mg} / \cos \theta$

The correct answer is:
$\mathrm{mg} / \cos \theta$

Solution:
Here $N \cos \theta=m g$
$\Rightarrow N=\frac{m g}{\cos \theta}$

Topic: Physics, Laws of Motion, JEE Main
A block moving horizontally on a smooth surface with a speed of 40 m s-1 splits into two equal parts. If one of the parts moves at 60 m s-1 in the same direction, then the fractional change in the kinetic energy will be x: 4 where x= ________.

Options:

The correct answer is:
1

Solution:
2m×40=60×m+mv
V=20 m/s
Now ΔKE=KEf-KEi=12 m×(20)2+12 m×(60)2-122 m×(40)2
ΔKE=12 m(800)⇒KEi=122 m×(40)2=12 m×3200
Fractional change =ΔKEKEi=14

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A block moving horizontally on a smooth surface with a speed of 40 m s-1 splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m s-1 in the same direction, then the fractional change in kinetic energy is :

Options:
A 13
B 23
C 14
D 18

The correct answer is:
18

Solution:
m×40=23mv+m3×60
120=2v+60
v=30 m s-1
Now Ki=12×m×402=800m
Kf=12×2 m3(30)2+12×m3×(60)2
=16{ m(1800+3600)}=m(5400)6=900m
ΔKEKi=900m-800m800m=18

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A block of $1 \mathrm{~kg}$ is stopped against a wall by applying a force $F$ perpendicular to the wall. If $\mu=0.2$ then minimum value of $F$ will be

Options:
A $980 \mathrm{~N}$
B $49 \mathrm{~N}$
C $98 \mathrm{~N}$
D $490 \mathrm{~N}$

The correct answer is:
$49 \mathrm{~N}$

Solution:
$F=\frac{W}{\mu}=\frac{1 \times 9.8}{0.2}=49 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
A block of 3 kg is attached to a string whose other end is attached to the wall. An unknown force F is applied so that the string makes an angle of 30° with the wall. The tension T is : (Given g =10 m s–2)

Options:
A 20 N
B 25 N
C 10 N
D 15 N

The correct answer is:
20 N

Solution:
From geometry, θ=30°.

For equilibrium,

Tcosθ=mg

⇒T=mgcos30°⇒T=3×1032

⇒T=20 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is:

Options:
A 0.0314 N
B 9.859×10-2 N
C 6.28×10-3 N
D 9.859×10-4 N

The correct answer is:
9.859×10-4 N

Solution:
N=mω2R

Given m=0.2 kg,T=40 S,R=0.2 m

Put values in equation 1

N=9.859×10-4 N

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A block of ice at \( -10^{\circ} \mathrm{C} \) slowly heated and converted to steam at \( 100^{\circ} \mathrm{C} \). Which of the following curves represents this phenomenon qualitatively?

Options:
A
B
C
D

The correct answer is:

Solution:
The temperature of ice will increases from -10℃ to 0℃ .
Heat supplied in this process will be
Q=msi(10)
Here, m=mass of ice
si=specific heat of ice
Then ice starts melting. Temperature during melting will remain constant (0℃). Heat supplied in this process will be
Q2=mL where, L=latent heat of melting
Now, the temperature of water will increase from 0℃ to 100℃ . Heat supplied will be Q3=msw(100)
where sw=specific heat of water.
Finally water at 100℃ will be converted into steam at 100℃ and during this process temperature again remains constant. Temperature versus heat supplied graph will be as follows.

Topic: Physics, Thermal Properties of Matter, JEE Main
A block of ice of mass 120 g at temperature 0°C is put in 300 g of water at 25°C. The x g of ice melts as the temperature of the water reaches 0°C. The value of x is

[Use: Specific heat capacity of water =4200 J kg-1 K-1, Latent heat of ice =3.5×105 J kg-1]

Options:

The correct answer is:
90

Solution:
According to the principle of calorimetry, heat lost by hot body is equal to the heat gained by cold body.

The heat released by water

∆Q=ms∆T=0.3×4200×25=31500 J

let m kg ice melts then,

∆Q=mL=m×3.5×105=31500 J

⇒m=31500×10-53.5=9000×10-5

m=0.09 kg=90 g

⇒x=90

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A block of mass 0.1 kg is attached to a cord passing through a hole in a horizontal frictionless horizontal surface as shown in the figure. The block is originally revolving at a distance of 0.4 m from the hole, with an angular velocity of 2 rad s-1 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.2 m. Considering the block to be point mass, calculate the new angular velocity (in rad s-1).

Options:

The correct answer is:
8

Solution:
By conservation of angular momentum

I1ω1=I2ω2

⇒ MR12ω1=MR22ω2

⇒ ω2=R1R22ω1

=0.40.22×2

=8 rad s-1

Topic: Physics, Rotational Motion, JEE Main
A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 N m-1 and oscillates in a damping medium of damping constant 10-2 kg s-1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to-

Options:
A 2 s
B 5 s
C 3 s
D 7 s

The correct answer is:
7 s

Solution:
Given,

E=Eo2
⇒A=Ao2
Amplitude of damped oscillation=A=Aoe-bt2m
⇒Ao2=Aoe-bt2m
2=ebt2m
ln⁡2=bt2m=10-22×0.1t=0.1t2
t2=10ln⁡2=10×12×0.693
t=10×0.693=6.93
t≈7 s.

Topic: Physics, Oscillations, JEE Main, JEE Main 2017 (09 Apr Online)
A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m s-1 in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m s-2. Assume there is no rotational motion and loss of energy after the collision is negligible.]

Options:
A 21 J
B 20 J
C 19 J
D 23 J

The correct answer is:
21 J

Solution:
Conservation of linear momentum

0.1×20=0.1+1.9×v

v=1 m s-1

Using work energy theorem

Wg=∆k

2×g×1=k-12×2×12

∴ k=21 J

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take ln2=0.693 )

Options:
A 1.16×10-2 kg s-1
B 3.3×102 kg s-1
C 1.16×102 kg s-1
D 5.7×10-3 kg s-1

The correct answer is:
1.16×10-2 kg s-1

Solution:
The amplitude of damped oscillation with time t and damping constant b is given by,

A=A0e-bt/2m

⇒lnA0A=bt2m

⇒ln2=b2m×120; here t=2 min=120 s

⇒0.693×2×1120=b

⇒b=1.16×10-2 kg s-1.

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A block of mass 1 kg is connected with a smooth plank of the same mass is performing oscillations. The value of the spring constant is 200 N m-1. The block and the plank are free to move and there is no friction anywhere. The angular frequency of the oscillation is Ï rad s-1. Find the value of Ï.

Options:

The correct answer is:
20

Solution:
Using the concept of reduced mass, we obtain:-
ω=kmreduced=2km ∵mreduced=m×mm+m=m2

⇒ω=2×2001=20 rad s-1

Topic: Physics, Oscillations, JEE Main
A block of mass 1 kg is dropped on a spring-mass system as shown in the figure. The block travels 100 meters in the air before striking the 3 kg mass. Calculate maximum compression in the spring, if both the blocks move together after the collision. Spring constant of the string k=1.25Ã106.

Options:
A 2 cm
B 4 cm
C 8 cm
D 16 cm

The correct answer is:
2 cm

Solution:
The velocity of 1 kg block just before striking
V=2×10×100=205 m s-1
Applying conservation of momentum, the blocks stick after the collision and move with velocity v
Or 1×205=4v
Final velocity v=55 ,
Using energy conservation
KE+PE= const
12×4×552+12k3gk2
=0+12k.x2
∴x2=4×25×51.25×106+30×301.25×1062
x=201000 m=2 cm

Topic: Physics, Oscillations, JEE Main
A block of mass 1 kg is pushed towards another block of mass 2 kg from a 6 m distance as shown in the figure. If just after the collision, the velocity of 2 kg block is 4 m s-1, then

Options:
A Coefficient of restitution between the blocks is 1
B Coefficient of restitution between two blocks is 12
C Coefficient of restitution between two blocks is 13
D Coefficient of restitution between two blocks is 14

The correct answer is:
Coefficient of restitution between the blocks is 1

Solution:
From conservation of linear momentum we can see that velocity of 1 kg block just after collision is 2 m/s leftwards.
Now relative velocity of approach = 6 m/s
and relative velocity of separation = 6 m/s
e = Relative velocity of separation Relative velocity of approach = 1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : g=10 m s-2

Options:
A 53 J
B 5 J
C 5×103 J
D 10 J

The correct answer is:
5 J

Solution:
Work done against frictional force

=μN×10=μmgcos60°×10

=0.1×5×10=5 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A block of mass 1 kg is released from the top of a rough incline having Î¼ = 1 3. The initial speed of the block is 2 m s-1. The inclined plane has unknown length and has a spring of spring constant k = 1 N m-1 connected at the base as in the figure. Find the maximum compression of spring. (answer in meter).

Options:

The correct answer is:
2

Solution:
ma = mg sin 3 0 o - 1 3 mg cos 6 0 o
â a = 0
by energy conservation
wmg + wfr + wspring = Î K.E.
wmg=âwfr
-12kx2=0-12mv2âx=mkv=2 m

Topic: Physics, Work Power Energy, JEE Main
A block of mass $1 \mathrm{kg}$ starts from rest at $x=0$ and moves along the $X$ -axis under the action of a force $F=k t,$ where $t$ is time and
$k=1 \mathrm{Ns}^{-1}$. The distance the block will trave] in 6 seconds is

Options:
A $36 \mathrm{m}$
B $72 \mathrm{m}$
C $108 \mathrm{m}$
D $18 \mathrm{m}$

The correct answer is:
$36 \mathrm{m}$

Solution:
From Newton's second law
$$
\begin{array}{l}
F=m a \\
m=1 \mathrm{kg} \\
F=1 \times a=\frac{d v}{d t}
\end{array}
$$
According to the question,
$$
F=k t=\frac{d v}{d t}
$$
$$
d v=k t \cdot d t
$$
Integration gives
$$
\begin{aligned}
v &=\frac{k t^{2}}{2}=\frac{d x}{d t} \\
x &=\frac{k t^{3}}{6}=\frac{1 \times 6 \times 6 \times 6}{6} \\
&=36 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
A block of mass 2 kg is kept at the origin at t=0 and is having velocity 45 m s-1 in the positive x -direction. The only force on it is a conservative and its potential energy is defined as U=- x3+6x2+15. Its velocity when the force acting on it is minimum (after the time t=0 ) is -

Options:

The correct answer is:
8

Solution:
At x=0
K=12×2×80=80 J and U= 15 J
∴ E= K+U= 95 J
F=-dUdx⇒F=3x2-12x
For F=minimum, dFdx=0
⇒ x=2 m
At x =2 m
E= K+U
⇒ 95=12×2×v2+-8+24+15
⇒v= 8 m s-1

Topic: Physics, Work Power Energy, JEE Main
A block of mass 2 kg is attached with two identical springs of spring constant 20 N m-1 each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is πX in SI unit. The value of X is______.

Options:

The correct answer is:
5

Solution:
When the block is displaced by a distance x, force acting on block due to both springs will be in the same direction. Therefore,

F=-2kx⇒a=-2kmx,

Comparing it witha=-ω2x , we get

ω=2km=2×202

=20 rad s-1

Now, T=2πω=2π20=π5

Hence, required value of X=5.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A block of mass $2 \mathrm{~kg}$ is kept on the floor. The coefficient of static friction is 0.4 . If a force $F$ of 2.5 Newtons is applied on the block as shown in the figure, the frictional force between the block and the floor will be

Options:
A $2.5 \mathrm{~N}$
B $5 \mathrm{~N}$
C $7.84 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$2.5 \mathrm{~N}$

Solution:
Applied force $=2.5 \mathrm{~N}$
Limiting friction $=\mu \mathrm{mg}=0.4 \times 2 \times 9.8=7.84 \mathrm{~N}$
For the given condition applied force is very smaller than limiting friction.
$\therefore$ Static friction on a body $=$ Applied force $=$
$2.5 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
A block of mass 2 kg moving on a horizontal surface with speed of 4 m s-1 enters a rough surface ranging from x=0.5 m to x=1.5 m. The retarding force in this range of rough surface is related to distance by F=-kx where k=12 N m-1. The speed of the block as it just crosses the rough surface will be

Options:
A 2 m s-1
B 2.5 m s-1
C 1.5 m s-1
D zero

The correct answer is:
2 m s-1

Solution:
Acceleration of the block a=fm=-12x2=-6x (Here the block is retarding)

Now, using, a=vdvdx
⇒v dv = a dx
On integrating, we get

∫4vvdv=-6∫0.51.5xdx

⇒v2-422=-61.52-0.522

⇒v2=16-12⇒v=2 m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A block of mass 2 kg slides on an inclined plane which makes an angle of 30Â° with the horizontal. The coefficient of friction between the block and the surface is 32. The force along the plane that needs to be applied to the block so that it moves up without any acceleration is ( take g = 10 m s-2 )

Options:
A 10 N
B 30 N
C 20 N
D 25 N

The correct answer is:
25 N

Solution:
For acceleration to be zero, the net force should be zero.

F=mgsinθ+μmgcosθ⇒F=2×10×12+32×2×10×32⇒F=25 N

Topic: Physics, Laws of Motion, JEE Main
A block of mass $4 \mathrm{~kg}$ is suspended through two light spring balances $A$ and $B$. Then $A$ and $B$ will read respectively

Options:
A $4 \mathrm{~kg}$ and zero $\mathrm{kg}$
B Zero $\mathrm{kg}$ and $4 \mathrm{~kg}$
C $4 \mathrm{~kg}$ and $4 \mathrm{~kg}$
D $2 \mathrm{~kg}$ and $2 \mathrm{~kg}$

The correct answer is:
$4 \mathrm{~kg}$ and $4 \mathrm{~kg}$

Solution:
As the spring balances are massless therefore the reading of both balance should be equal

Topic: Physics, Laws of Motion, JEE Main
A block of mass 5 kg is (i) pushed in case A and (ii) pulled in case B, by a force F=20 N, making an angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is μ=0.2. The difference between the accelerations of the block, in case B and case A will be:
g=10 m s-2

Options:
A 3.2 m s-2
B 0 m s-2
C 0.8 m s-2
D 0.4 m s-2

The correct answer is:
0.8 m s-2

Solution:
For Block A:
N=mg+Fsin⁡30°
=5×10+20×12
=60 N
fr=0.260= 12 N
Fcos⁡30° –fr= ma
2032 -12=5×a
⇒aA=103-125 ……(i)
For Block B:
Fsin⁡30°+N=mg
N=5×10 –20×12
=40 N
fr=0.2×40=8 N
Fcos30° –fr=ma
2032-8=5×a
⇒aB=103-85
aB-aA=45=0.8 m/s2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased until the block just begins to slide down. The sin of angle made by the plane with horizontal is 35 when the block start sliding. The coefficient of friction between the block and the plane is (g = 10 m s-2)

Options:
A 3 5
B 3 4
C 4 5
D 2 3

The correct answer is:
3 4

Solution:
The block will just start sliding when the angle θ made by the inclined plane with the horizontal will be equal to the angle of repose.

θ=θrepose=tan-1μ

⇒tanθ=μ
μ=34

Topic: Physics, Laws of Motion, JEE Main
A block of mass 5 kg starting from rest pulled up on a smooth incline plane making an angle of 30° with horizontal with an effective acceleration of 1 m s-2. The power delivered by the puling force at t=10 s from the start is _____ W.
[Use g=10 m s-2]
(Calculate the nearest integer value)

Options:

The correct answer is:
300

Solution:
Let's consider the following free-body diagram:

From the above diagram, it can be written that

F-mgsin30°=ma ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the force applied on the block.

F-50 sin30°=5×1⇒F=5+25 N=30 N

The velocity of the block v at t=10 s can be written as

v=0+1 m s-2×10 s=10 m s-1

Hence, the required power is given by

P=Fv= 30 N×10 m s-1= 300 W

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis will be g=10  m s-2,

Options:
A 10 rad s-1
B 102π rad s-1
C 10  rad s-1
D 10 π  rad s-1

The correct answer is:
10  rad s-1

Solution:
âââââââ
N=mω2R
f=mg
For minimum angular velocity,
f=μN
mg=μ·mω2R
ω=gμR=100.1×1=10 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g=10 ms-2 )

Options:
A 23 N
B 25 N
C 18 N
D 32 N

The correct answer is:
32 N

Solution:
Block will be at rest
mgsin⁡θ+3N=P+f
10×10×12+3=P+μmgcos⁡θ
( ∵ To get P minimum friction must be minimum)
Pmin=502+3-0.6×10×10×12
=202+3=32N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A block of mass 10 kg is moving along x-axis under the action of force F=5x N. The work done by the force in moving the block from x=2 m to 4 m will be _______ J.

Options:

The correct answer is:
30

Solution:
Work done is W=∫Fdx

It is given that

F=5x N

So, work done is

W=∫245xdx=5x2224⇒W=5(16-4)2=30 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A block of mass $10 \mathrm{~kg}$ is placed on a horizontal frictionless surface and is attached to a cord which passes over two light frictionless pulleys as shown in the figure. The hanging block tied to the other end of the cord is initially at rest $2 \mathrm{~m}$ above the horizontal floor.If the hanging block strikes the floor $2 \mathrm{~s}$ after the system is released, then weight of the hanging block is ....... $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A 22.22 N
B 11.11 N
C 1.11 N
D 2.22 N

The correct answer is:
11.11 N

Solution:
For hanging block,
$$
\begin{aligned}
& \quad s=u t+\frac{1}{2} a t^2[\because u=0] \Rightarrow 2=\frac{1}{2} \times a \times 2^2 \\
& \Rightarrow \quad a=1 \mathrm{~ms}^{-2}
\end{aligned}
$$
So, velocity of hanging block $=$ velocity of $10 \mathrm{~kg}$ block after $2 \mathrm{~s}$
$$
\begin{array}{ll}
\Rightarrow & v=u+a t \\
\Rightarrow & v=0+1 \times 2 \text { or } v=2 \mathrm{~ms}^{-1}
\end{array}
$$
From work - energy theorem,
$\mathrm{KE}$ of $10 \mathrm{~kg}$ block $=$ Work done by gravity on hanging block $=\frac{1}{2} \times 10 \times 2^2=m \times 10 \times 2$
$$
\Rightarrow \quad m=1 \mathrm{~kg}
$$
Weight of hanging block $=1 \times 10=10 \mathrm{~kg} \approx 11.1 \mathrm{lN}$

Topic: Physics, Laws of Motion, JEE Main
A block of mass $10 \mathrm{~kg}$ placed on the rough horizontal surface having coefficient of friction $\mu=0.5$, if a horizontal force of $100 \mathrm{~N}$ acting on it then acceleration of the block will be:

Options:
A $10 \mathrm{~ms}^{-1}$
B $5 \mathrm{~ms}^{-2}$
C $15 \mathrm{~ms}^{-2}$
D $0.5 \mathrm{~ms}^{-2}$

The correct answer is:
$5 \mathrm{~ms}^{-2}$

Solution:
Here $f_{\max }=\mu N=\mu m g$ $=0.5 \times 10 \times 10=50 \mathrm{~N}$ $\Rightarrow a=\frac{\text { net force }}{\text { mass }}$ $=\frac{100-50}{10}=5 \mathrm{~m} / \mathrm{sec}^2$

Topic: Physics, Laws of Motion, JEE Main
A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms-1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :[use g=9.8 ms-2]

Options:
A 9.8 m
B 4.9 m
C 12.5 m
D 19.6 m

The correct answer is:
9.8 m

Solution:
Given initial velocity u=9.8 m s-1.

Acceleration a=μg=0.5×9.8=4.9 m s-2

Using, v2=u2+2as, here, final velocity v=0.

So, 0=9.82+24.9s⇒s=9.8 m

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A block of mass $0.50 \mathrm{~kg}$ is moving with a speed of $2.00 \mathrm{~m} / \mathrm{s}$ on a smooth surface. It strikes another mass of $1.00 \mathrm{~kg}$ and then they move together as a single body. The energy loss during the collision is

Options:
A
$0.16 \mathrm{~J}$

B
$1.00 \mathrm{~J}$

C
$0.67 \mathrm{~J}$

D
$0.34 \mathrm{~J}$

The correct answer is:
$0.67 \mathrm{~J}$

Solution:
$$
\begin{aligned}
& m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\
& v=2 / 3 \mathrm{~m} / \mathrm{s} \\
& \text { Energy loss }=\frac{1}{2}(0.5) \times(2)^2-\frac{1}{2}(1.5) \times\left(\frac{2}{3}\right)^2=0.67 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2008
A block of mass $90 \mathrm{~kg}$ is suspended by three strings $A, B$ and $C$ as shown in figure. Tensions in the strings $A, B$ and $C$ respectively are $\left(g=10 \mathrm{~ms}^{-2}, \sin 37^{\circ}=0.6, \cos 37^{\circ}=0.8\right)$

Options:
A $400 \mathrm{~N}, 500 \mathrm{~N}$ and $300 \mathrm{~N}$
B $500 \mathrm{~N}, 300 \mathrm{~N}$ and $900 \mathrm{~N}$
C $300 \mathrm{~N}, 600 \mathrm{~N}$ and $900 \mathrm{~N}$
D $1200 \mathrm{~N}, 1500 \mathrm{~N}$ and $900 \mathrm{~N}$

The correct answer is:
$1200 \mathrm{~N}, 1500 \mathrm{~N}$ and $900 \mathrm{~N}$

Solution:
Given, mass of block, $m=90 \mathrm{~kg}$
Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ $T_A, T_B$ and $T_C$ are tensions in strings $A, B$ and $C$.
According to given diagram,
$$
\begin{aligned}
\text { Weight } & =m g \\
& =90 \times 10 \\
& =900 \mathrm{~N} \\
\therefore \quad T_C & =900 \mathrm{~N}
\end{aligned}
$$
By using Lami's theorem,
$$
\begin{array}{rlrl}
& & \frac{T_A}{\sin \left(90^{\circ}+37^{\circ}\right)} & =\frac{T_B}{\sin 90^{\circ}}=\frac{T_C}{\sin \left(90^{\circ}+53^{\circ}\right)} \\
\Rightarrow & \frac{T_A}{\cos 37^{\circ}} & =T_B=\frac{900}{\cos 53^{\circ}} \\
\therefore & & T_B & =\frac{900}{\cos 53^{\circ}}=1500 \mathrm{~N} \\
& \text { and } & T_A & =\frac{900}{\cos 53^{\circ}} \cos 37^{\circ} \\
& & =1197=1200 \mathrm{~N}
\end{array}
$$

Topic: Physics, Laws of Motion, JEE Main
A block of mass $100 \mathrm{~g}$ is suspended vertically from a massless spring system of spring constant, $k=1 \mathrm{~N} / \mathrm{m}$ each. The block is hit from above to impart an impulse of 2 Ns. Calculate the maximum displacement from the Equilibrium position of the block. (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $2 \mathrm{~m}$
B $4 m$
C $5 \mathrm{~m}$
D $9 \mathrm{~m}$

The correct answer is:
$4 m$

Solution:
Given, mass $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$,
spring constant, $k=1 \mathrm{~N} / \mathrm{m}$ and impulse, $I=2 \mathrm{~N} \mathrm{~s}$ The given system has parallel connection of the springs.
So,
time period, $T=2 \pi \sqrt{\frac{m}{k_p}}=2 \pi \sqrt{\frac{0.1}{2}}\left(\because k_p=k+k\right)$
$=\frac{2 \pi}{\sqrt{20}}=\frac{\pi}{\sqrt{5}}=1.40$
Impulse acts at equilibrium position, so it acts only $\frac{T}{4}$ seconds.
Now, force act on the spring system,
$\begin{array}{rlrl} & & k_p x & =m g+\frac{I}{T / 4} \\ \Rightarrow & & 2 x=0.1 \times 10+\frac{2 \times 4}{1.4} \\ \Rightarrow & & x=3.35 \mathrm{~m}=4 \mathrm{~m}\end{array}$
Hence, the maximum displacement from the equilibrium position of the block is $4 \mathrm{~m}$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main