datavorous/entrance-exam-dataset huggingface dataset:
A boy of mass $20 \mathrm{~kg}$ is standing on a $80 \mathrm{~kg}$ free to move long cart. There is negligible friction between cart and ground. Initially, the boy is standing $25 \mathrm{~m}$ from a wall. If he walks $10 \mathrm{~m}$ on the cart towards the wall, then the final distance of the boy from the wall will be

Options:
A
$15 \mathrm{~m}$

B
$12.5 \mathrm{~m}$

C
$15.5 \mathrm{~m}$

D
$17 \mathrm{~m}$

The correct answer is:
$17 \mathrm{~m}$

Solution:
No Solution Available

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2013 (23 Apr Online)
A boy playing on the roof of a $10 \mathrm{~m}$ high building throws a ball with a speed of $10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of $10 \mathrm{~m}$ from the ground?
$$
\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]
$$

Options:
A
$5.20 \mathrm{~m}$

B
$4.33 \mathrm{~m}$

C
$2.60 \mathrm{~m}$

D
$8.66 \mathrm{~m}$

The correct answer is:
$8.66 \mathrm{~m}$

Solution:
From, the question if the horizontal distance is none other than the horizontal range on the level of the roof of building
$$
\text { Range }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{(10)^2 \sin (2 \times 30)}{\mathrm{g}}=\frac{10 \times 10 \times \sqrt{3}}{2 \times 10}=8.66
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2003
A boy pushes a box of mass 2 kg with a force F→=20i^+10j^N on a frictionless surface. If the box was initially at rest, then _______ m is displacement along the x-axis after 10 s

Options:

The correct answer is:
500

Solution:
F→=20i^+10j^

a→=F→m=20i^+10j^2⇒10i^+5j^

∴ s→=12a→t2=1210i^+5j^×102

⇒5010i^+5j^m

∴ Displacement along x-axis

⇒50×10⇒500 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A boy pushes a toy box $2.0 \mathrm{~m}$ along the floor by means of a force of $10 \mathrm{~N}$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

Options:
A $6 \mathrm{~J}$
B $8 \mathrm{~J}$
C $10 \mathrm{~J}$
D $12 \mathrm{~J}$

The correct answer is:
$10 \mathrm{~J}$

Solution:
$\mathrm{W}=\mathrm{F} \mathrm{s} \cos \theta=10 \times 2 \cos 60^{\circ}=10 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main
A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be:

Options:
A t1t2t2-t1
B t1+t22
C t1t2t2+t1
D t2-t1

The correct answer is:
t1t2t2+t1

Solution:
L= Length of escalator

Vb/ecc=Lt1

When only escalator is moving,

Vesc=Lt2

when both are moving,

Vb/g=Vb/esc+Vesc

Vb/g=Lt1+Lt2⇒t=LVb/g=t1t2t1+t2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A boy throws a ball at an angle Î¸ with the vertical. If the vertical component of the initial velocity is 20 m s-1 and the wind imparts a horizontal acceleration of 8 m s-2 to the left, the angle at which the ball must be thrown so that the ball returns to the boy's hand is Î¸. What is the value of 10tanâ¡Î¸?

Options:

The correct answer is:
8

Solution:
Taking the upward vertical as positive y-direction
uy=20 m s-1
ay=-10 m s-2
sy=uyt+12ayt2
t=2uyay=2×2010=4 s
Taking the right-side horizontal as positive x-direction
sx=0=uxt+12ax×t2
0=ux4-8242
ux=16
tan⁡θ=uxuy=1620
10tan⁡θ=10×1620= 8

Topic: Physics, Motion In Two Dimensions, JEE Main
A boy throws a cricket ball from the boundary to the wicket keeper. If the frictional force due to air fa cannot be ignored, the forces acting on the ball at the position X are represented by

Options:
A
B
C
D

The correct answer is:

Solution:
The fiction force due to air on the ball fa will always act in the opposite direction of the velocity of the ball, and the velocity of the ball at any point on the path is always in the tangential direction of the path of the projectile.

The weight of the body always acts downward perpendicular to the surface of the earth and along the centre of the earth.

Topic: Physics, Laws of Motion, JEE Main
A boy ties a stone of mass $100 \mathrm{~g}$ to the end of a $2 \mathrm{~m}$ long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of $80 \mathrm{~N}$. If the maximum speed with which the stone can revolve is $K / \pi r e v . / m i n$. The value of $K$ is :
(Assume the string is massless and unstretchable)

Options:
A 400
B 300
C 600
D 800

The correct answer is:
600

Solution:
$\begin{aligned} & \mathrm{T}=\mathrm{m} \omega^2 \mathrm{r} \\ & \Rightarrow 80=0.1 \times(2 \pi \times \mathrm{K} / \pi \times 1 / 60)^2 \times 2 \\ & \Rightarrow 800 / 2=\mathrm{K}^2 / 900 \\ & \Rightarrow \mathrm{K}=30 \times 20=600\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is Kπ rev min-1. The value of K is :
(Assume the string is massless and un-stretchable)

Options:
A 400
B 300
C 600
D 800

The correct answer is:
600

Solution:
The maximum angular velocity of the stone is ω=rpm×2π60=K30 rad s-1

The tension in the string will provide the required centripetal force. Therefore,

T=mω2l⇒80=0.1×K302×2⇒K2=360000⇒K=600

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A boy walks to his school at a distance of $6 \mathrm{~km}$ with constant speed of 2.5 $\mathrm{km} /$ hour and walks back with a constant speed of $4 \mathrm{~km} / \mathrm{hr}$. His average speed for round trip expressed in $\mathrm{km} /$ hour, is

Options:
A 24/13
B 40/13
C 3
D 1/2

The correct answer is:
40/13

Solution:
Distance average speed $=\frac{2 v_1 v_2}{v_1+v_2}=\frac{2 \times 2.5 \times 4}{2.5+4}$
$=\frac{200}{65}=\frac{40}{13} \mathrm{~km} / \mathrm{hr}$

Topic: Physics, Motion In One Dimension, JEE Main
A boy weighing 50 kg eat some bananas. The energy content of the bananas is 1000 cal. If the boy uses all of this energy to jump upwards from the ground, then the height up to which his centre of mass will rise is [mechanical equivalent of heat = 4.2 J cal-1, g=10 m s-2]

Options:

The correct answer is:
8.4

Solution:
Energy used by the boy is
E=1000Ã4.2 J

â4200=50Ã10Ãh

âh=8.4 m

Topic: Physics, Thermodynamics, JEE Main
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms-1 . Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to:

Options:
A 106N m-2
B 104N m-2
C 108N m-2
D 103N m-2

The correct answer is:
106N m-2

Solution:
Elastic potential energy =12.Y. ∆lL2AL
=12Y. 0.20.422.π×32×10-6×0.42
=1.35×10-6Y
Kinetic energy =12mv2=12×(0.02×202)
1.35×10-6 Y=4
Y=3×106 N m-2

Topic: Physics, Mechanical Properties of Solids, JEE Main
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms-1 . Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to:

Options:
A 106N m-2
B 104N m-2
C 108N m-2
D 103N m-2

The correct answer is:
106N m-2

Solution:
Elastic potential energy =12.Y. ∆lL2AL
=12Y. 0.20.422.π×32×10-6×0.42
=1.35×10-6Y
Kinetic energy =12mv2=12×(0.02×202)
1.35×10-6 Y=4
Y=3×106 N m-2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A brown coloured mixture of two gases is obtained by the reduction of 6N nitric acid with metallic copper. This mixture on cooling condenses to a blue liquid which on freezing at -30oC gives a blue solid. The correct choice for blue liquid or solid is.

Options:
A It is referred to as the anhydride of nitrous acid.
B It is an acidic oxide and hence dissolves in alkalies producing nitrites.
C It can also be prepared by the action of 50% HNO3 on arsenious oxide and then cooling to 250 K.
D All of these.

The correct answer is:
All of these.

Solution:
(i) It is an acidic oxide and anhydride of HNO 2
N2O3+H2O→2HNO2
(ii) With alkali it forms nitrites
N2O3+2NaOH→2NaNO2+H2O
(iii) It can also be prepared by the action of 50% HNO3 on arsenious oxide and then cooling to 250 K
All are correct here.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A brown ring appears in the test for

Options:
A Nitrate
B Nitrite
C Bromide
D Iron

The correct answer is:
Nitrate

Solution:
At the junction of salt solution and $\mathrm{FeSO}_4$ solution with conc. $\mathrm{H}_2 \mathrm{SO}_4$ a brown ring is obtained
$\left(\mathrm{FeSO}_4 \cdot \mathrm{NO}\right)$
Brown ring

Topic: Chemistry, Practical Chemistry, JEE Main
A bubble of air is underwater at temperature $15^{\circ} \mathrm{C}$ and the pressure $1.5 \mathrm{~bar}$. If the bubble rises to the surface where the temperature is $25^{\circ} \mathrm{C}$ and the pressure is $1.0 \mathrm{~bar}$, what will happen to the volume of the bubble?

Options:
A Volume will become greater by a factor of 1.6
B Volume will become greater by a factor of 1.1
C Volume will become smaller by a factor of 0.70
D Volume will become greater by a factor of 2.9

The correct answer is:
Volume will become greater by a factor of 1.6

Solution:
$\because \quad \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
(By ideal gas equation)
or $\frac{1.5 \times V_1}{288}=\frac{1 \times V_2}{298}$
$\therefore \quad V_2=1.55 V_1$
i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.

Topic: Chemistry, States of Matter, NEET, NEET 2011 (Mains)
A bucket contains water filled upto a height= 15 cm. The bucket is tied to a rope which is passed over a frictionless light pulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket+water). The water pressure above the atmospheric pressure at the bottom is

Options:
A 0.5 kPa
B 1 kPa
C 5 kPa
D 20kPa

The correct answer is:
1 kPa

Solution:
P = h ρ g eff ⇒ P = h ρ g - a ....(1)
a = M - M / 2 g M + M / 2
a = g 3 ....(2)
From (1) & (2)
P = 1 5 × 1 0 - 2 × 1 0 3 × g - g / 3 ⇒ P = 1 5 0 × 2 g 3
P = 1 0 0 0 P a ⇒ P = 1 kPa

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A bucket full of water tied with the help of a $2 \mathrm{~m}$ long string performs a vertical circular motion. The minimum angular velocity of the bucket at the uppermost point so that water will not fall will be:

Options:
A $2 \sqrt{5} \mathrm{rad} / \mathrm{s}$
B $\sqrt{5} \mathrm{rad} / \mathrm{s}$
C $5 \mathrm{rad} / \mathrm{s}$
D $10 \mathrm{rad} / \mathrm{s}$

The correct answer is:
$\sqrt{5} \mathrm{rad} / \mathrm{s}$

Solution:
as water should not fall
$w w^2 r=m g$
$\omega=\sqrt{\frac{8}{\gamma}}=\sqrt{\frac{10}{2}}=\sqrt{5}$

Topic: Physics, Laws of Motion, JEE Main
A bucket full of water weighs 5 kg, it is pulled from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate. If it is observed that for every metre the bucket loses 0.2 kg mass of water, then the total work done in pulling the bucket up from the well is [g=10 m s-2]

Options:
A 600 J
B 400 J
C 100 J
D 500 J

The correct answer is:
600 J

Solution:
The weight of bucket when it has been pulled up a distance x is (5-0.2x).
Hence, the required work is
W=â«x=20x=0-5-0.2xÃ10Ãdx
=50xx=0x=20-2x22x=0x=20
W=50Ã20-202=600 J

Topic: Physics, Work Power Energy, JEE Main
A buffer solution is a mixture of

Options:
A Strong acid and strong base
B Weak acid and weak base
C Weak acid and conjugate acid
D Weak acid and conjugate base

The correct answer is:
Weak acid and conjugate base

Solution:
A buffer solution (more precisely, $\mathrm{pH}$ buffer or hydrogen ion buffer) is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Buffer solutions are used as a means of keeping $\mathrm{pH}$ at a nearly constant value in a wide variety of chemical applications.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A buffer solution is prepared by mixing $0.01 \mathrm{M}$ weak acid and $0.05 \mathrm{M}$ solution of a salt of weak acid and strong base. What is the $\mathrm{pH}$ of buffer solution? $(\mathrm{pKa}=4.74)$

Options:
A $3.34$
B $4.80$
C $5.44$
D $6.93$

The correct answer is:
$5.44$

Solution:
$\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { salt }]}{[\text { acid }]} \\ & =4.74+\log _{10} \frac{0.05}{0.01} \\ & =4.74+0.7=5.44\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
A buffer solution with $\mathrm{pH}_9$ is to be prepared by mixing $\mathrm{NH}_4 \mathrm{Cl}$ and. $\mathrm{NH}_4 \mathrm{OH}$. Calculate the number of moles of $\mathrm{NH}_4 \mathrm{Cl}$ that should be added to one litre of $1.0 \mathrm{M} \mathrm{NH}_4 \mathrm{OH} .\left[K_b=1.8 \times 10^{-5}\right]$

Options:
A 3.4
B 2.6
C 1.5
D 1.8

The correct answer is:
1.8

Solution:
$p H=-\log K_b+\log \frac{[\text { Salt }]}{[\text { Acid] }]}$
$\begin{aligned} & p H=-\log \left[1.8 \times 10^{-s}\right]+\log \frac{[\text { Salt }]}{1.0} \\ & 9=4.7+\log \frac{[\text { Salt }]}{1.0} ; \log \frac{[\text { Salt }]}{1.0}=4.7-9=-4.3 \\ & \frac{[\text { Salt }]}{1.0}=\text { Antilog } \frac{1}{4.3} ;[\text { Salt }]=1.8\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

Options:
A increases
B decreases
C remains same
D becomes zero

The correct answer is:
increases

Solution:
$Z=\sqrt{R^2+X_C^2} \& X_C=\frac{1}{W C}$
due to dielectric
$\mathrm{C} \uparrow \Rightarrow \mathrm{X}_{\mathrm{C}} \downarrow \Rightarrow \mathrm{Z} \downarrow$
So, current increases $\&$ thus bulb will glow more brighter.

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance between parallel walls is ' $d$ ' then the required focal length of the lens placed in between the walls is:

Options:
A only $\frac{d}{4}$
B only $\frac{d}{2}$
C more than $\frac{d}{4}$ but less than $\frac{d}{2}$
D less than or equal to $\frac{d}{4}$

The correct answer is:
less than or equal to $\frac{d}{4}$

Solution:
Since $v+u=d \Rightarrow v=d-u$
From lens formula
$$
\begin{aligned}
\frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{f} & =\frac{1}{d-u}-\frac{1}{-u} \\
& =\frac{d}{(d-u) u} \\
\Rightarrow \quad f & =\frac{d u-u^2}{d} \ldots(\mathrm{i})
\end{aligned}
$$
Differentiating eq. (i) we have
$$
\frac{d f}{d u}=\frac{1}{d}(d-2 u)
$$
for maximum value of $f_1=\frac{d f}{d u}=0$
$$
\begin{aligned}
& \Rightarrow \frac{1}{d}(d-2 u)=0 \Rightarrow d=24 \\
& \Rightarrow u=\frac{d}{2}
\end{aligned}
$$
putting these value in eq. (i)
$$
f=\frac{d}{4}
$$

Topic: Physics, Ray Optics, JEE Main
A bulb is rated at 100 V, 100 W, it can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 Hz.

Options:
A π3 H
B 100Â H
C 2π H
D 3Ï H

The correct answer is:
3Ï H

Solution:
From the rating of the bulb, the resistance of the bulb can be calculated.
R=Vrms2P=(100)2100=100âÎ©
For the bulb to be operated at its rated value the rms current through it should be 1 A
Also,
I rms=VrmsZ
â´ââ1=2001002+(2Ï50L)2ââââLâ=3Ï H

Topic: Physics, Alternating Current, JEE Main
A bullet emerges from a barrel of length 1.2 m with a speed of 640 m s-1. Assuming constant acceleration, the approximate time that is spent in the barrel after the gun is fired is

Options:
A 4 ms
B 40 ms
C 400 μs
D 1 s

The correct answer is:
4 ms

Solution:
Given, s=1.2 m, v=640 ms-1, a=?, u=0;t=?
We have the third equation of motion
2as=v2-u2
2a×1.2=640×640
Or a=8 × 64 × 1033
And by first equation of motion
v=u+at
Or t=va=154×10-3
=3.75×10-3s ≈4 ms

Topic: Physics, Motion In One Dimension, JEE Main
A bullet fired into a fixed target loses half of its velocity after penetrating $3 \mathrm{~cm}$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Options:
A
$3.0 \mathrm{~cm}$

B
$2.0 \mathrm{~cm}$

C
$1.5 \mathrm{~cm}$

D
$1.0 \mathrm{~cm}$

The correct answer is:
$1.0 \mathrm{~cm}$

Solution:
$F .3=\frac{1}{2} m v^2-\frac{1}{2} m \frac{v^2}{4}$
$F(3+x)=\frac{1}{2} m v^2$
$x=1 \mathrm{~cm}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2005
A bullet is fired from a gun at the speed of 280 m s-1 in the direction 30° above the horizontal. The maximum height attained by the bullet is (g=9.8 m s-2, sin30°=0.5):

Options:
A 3000 m
B 2800 m
C 2000 m
D 1000 m

The correct answer is:
1000 m

Solution:
Maximum height of the projectile is given by,

H=u2sin2θ2g=2802×sin230°2×9.8=1000 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A bullet is fired from a gun. The force on the bullet is given by F=600-2×105t where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Options:
A 9 Ns
B 1.8 Ns
C 0.9 Ns
D 0.3 Ns

The correct answer is:
0.9 Ns

Solution:
Given, F=600-2Ã105 t=0
â¹ t=3Ã10-3 s
Impulese I=â«0tFâdt
=â«03Ã10-3600-2Ã105 tdt
=[600 t-105 t2]03Ã10-3=0.9 Ns

Topic: Physics, Laws of Motion, JEE Main
A bullet is fired from a gun. The force on the bullet is given by F=600â2Ã105t, where F is in newton and t is in seconds. If the force on the bullet becomes zero as soon as it leaves the barrel, then the average impulse imparted to the bullet is

Options:
A 1.8 N s
B 18 N s
C 9 N s
D 0.9 N s

The correct answer is:
0.9 N s

Solution:
Given F=600−(2× 10 5 t)

The force is zero at time, t, given by

0=600−2× 10 5 t

⇒t= 600 2× 10 5 =3× 10 −3  seconds

∴  Impulse= ∫ 0 t Fdt = ∫ 0 3× 10 −3 (600−2× 10 5 t) dt

= [ 600t− 2× 10 5 t 2 2 ] 0 3× 10 −3 =600×3× 10 −3 − 10 5 (3× 10 −3 ) 2

=1.8 – 0.9 = 0.9 N s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bullet is fired with a velocity u making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component o the velocity of the bullet when it reaches the maximum height is

Options:
A $\mathrm{u}$
B $0$
C $\frac{\sqrt{3 u}}{2}$
D $\frac{\mathrm{u}}{2}$

The correct answer is:
$\frac{\mathrm{u}}{2}$

Solution:
Horizontal velocity would be constant so the value of velocity at the highest point will be $u / 2$

Topic: Physics, Motion In Two Dimensions, JEE Main
A bullet loses 1nth of its velocity passing through one plank. Considering uniform retardation, the number of such planks that are required to stop the bullet can be:

Options:
A Infinite
B n
C n22n-1
D 2n2n-1

The correct answer is:
n22n-1

Solution:
Let t be the thickness of the plank

υ01-1n2=υ02-2at

∴ 2at=υ02-υ021-1n2

If m number of planks are required to stop a bullet, then

0=υ02-2atm

∴ m=υ022at

=υ02υ021-1-1n2

=n2n2-n-12=n22n-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2014 (19 Apr Online)
A bullet of 4 g mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms1, the impulse imparted to the gun and velocity of recoil of gun are

Options:
A 0.4 kg m s-1, 0.1 m s-1
B 0.2 kg m s-1, 0.05 m s-1
C 0.2 kg m s-1, 0.1 m s-1
D 0.4 kg m s-1, 0.05 m s-1

The correct answer is:
0.2 kg m s-1, 0.05 m s-1

Solution:
By momentum conservation,
4×10-3(50-v)-4v=0
v=4×10-3×504+4×10-3≈0.05 m s-1
Impulse J=mv=4×.05=0.2 kg m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A bullet of 10 g leaves the barrel of gun with a velocity of 600 m s-1. If the barrel of gun is 50 cm long and mass of gun is 3 kg, then value of impulse supplied to the gun will be:

Options:
A 6 N s
B 3 N s
C 36 N s
D 12 N s

The correct answer is:
6 N s

Solution:
The formula to calculate the impulse J on the gun is given by

J=mv-u ...1

where, m is the mass, u is the initial velocity and v is the final velocity of the bullet.

Substitute the values of the known parameters into equation (1) to calculate the required impulse.

J=10 g×1 kg1000 g×600-0 m s-1= 6 N s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m s-1. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v in m s-1, so that the pendulum describes a circle. (Assume the string to be inextensible and g=10 m s-2 )

Options:

The correct answer is:
400

Solution:
Final velocity of system (bullet + mass) should be =5gr

Now M5gr=m(v+100)

thus v=Mm5gr-100=500-100=400 m s-1

v=400 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of the bullet before hitting the wood is 10 m s-1 and, it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is x N. The value of x to the nearest integer is,

Options:

The correct answer is:
10

Solution:
v2=u2+2as⇒0=(10)2+2(-a)12
⇒a=100 m s-2
⇒F=ma=(0.1)(100)=10 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A bullet of mass 0.1 kg moving horizontally with speed 400 m s-1 hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is _______.

Options:
A 0.90
B 0.50
C 0.65
D 0.25

The correct answer is:
0.25

Solution:
Let the mass of the bullet be m and mass of the block be M. It is given that V=400 m s-1. By conservation of momentum,

mV=(m+M)v

⇒0.1×400=(0.1+3.9)v
⇒v=10 m s-1

The force acting on the block

f=μm'g⇒a=μg

Using v'2-v2=-2aS

⇒ 02-102=2(-μg)(S)
⇒μ=1002×10×20=0.25

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A bullet of mass $0.02 \mathrm{~kg}$ travelling horizontally with velocity $250 \mathrm{~ms}^{-1}$ strikes a block of wood of mass $0.23 \mathrm{~kg}$ which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of $40 \mathrm{~m}$. The coefficient of sliding friction of the rough surface is $\left(g=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A 0.75
B 0.61
C 0.51
D 0.30

The correct answer is:
0.51

Solution:
After impact the bullet and block move together and comes to rest after covering a distance of $40 \mathrm{~m}$.
By conservation of momentum
$\begin{gathered}
m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \\
\text { or } \quad 0.02 \times 250+0.23 \times 0=0.02 v+0.23 v
\end{gathered}$
$\begin{array}{r}
5+0=v(0.25) \\
\frac{500}{25}=v=20 \mathrm{~ms}^{-1}
\end{array}$
Now, by conservation of energy
$\frac{1}{2} M v^2=\mu R \cdot d$
$\begin{array}{ll}\text { or } & \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow & \mu=\frac{200}{9.8 \times 40}=0.51\end{array}$

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