datavorous/entrance-exam-dataset huggingface dataset:
A stiff spring having spring constant $k=400 \mathrm{~N} / \mathrm{m}$ is attached to the floor vertically. A mass $m=10 \mathrm{~kg}$ is placed on top of the spring. The block oscillates if it is pressed downward and released. Find the extension in the spring at which the block loses contact with spring. (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $25 \mathrm{~cm}$
B $15 \mathrm{~cm}$
C $20 \mathrm{~cm}$
D $22 \mathrm{~cm}$

The correct answer is:
$25 \mathrm{~cm}$

Solution:
The frequency of oscillation
$\omega=\sqrt{\frac{K}{m}}$ ...(i)
The maximum acceleration in SHM,
$\alpha_{\max }=\omega^2 x$
where, $x$ is the maximum extension in the spring.
To maintain contact with spring, the maximum SHM acceleration should be equal to acceleration due to gravity.
i.e.,
$\omega^2 x=g \Rightarrow x=\frac{m g}{K}$[from Eq. (i)]
$=\frac{10 \times 10}{400}=\frac{1}{4} \quad[\therefore K=400 \mathrm{~N} / \mathrm{m}]$
$=0.25 \mathrm{~m}$ or $25 \mathrm{~cm}$

Topic: Physics, Oscillations, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
A stone dropped from the top of the tower touches the ground in $4 \mathrm{sec}$. The height of the tower is about

Options:
A 80 m
B 40 m
C 20 m
D 160 m

The correct answer is:
80 m

Solution:
$h=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times(4)^2=80 \mathrm{~m}$

Topic: Physics, Motion In One Dimension, JEE Main
A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for

Options:
A $6 \mathrm{~s}$
B $5 \mathrm{~s}$
C $7 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$5 \mathrm{~s}$

Solution:
Hints : $u=0$
$$
\begin{aligned}
& \mathrm{S}_3=0+\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 9=45 \\
& \mathrm{~S}_t \text { th }=u+(2 t-1) \frac{g}{2} \\
& \mathrm{~S}_t \text { th }=0+5(2 t-1)=45 \\
& 2 t-1=9 \\
& t=5 \mathrm{sec}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2010
A stone falls from a balloon that is descending at a uniform rate of $12 \mathrm{~m/s}$. The displacement of the stone from the point of release after $10 \mathrm{sec}$ is

Options:
A 490
B 510
C 610
D 725

The correct answer is:
610

Solution:
$\begin{aligned} & u=12 \mathrm{~m} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{sec}^2, t=10 \mathrm{sec} \\ & \text { Displacement }=u t+\frac{1}{2} g t^2 \\ & =12 \times 10+\frac{1}{2} \times 9.8 \times 100=610 \mathrm{~m} \\ & \end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A stone hanging from a massless string of length 15 m is projected horizontally with speed 12 m s-1. The speed of the particle, at the point where the tension in the string is equal to the weight of the particle, is close to

Options:
A 10 m s-1
B 7 m s-1
C 12 m s-1
D 5 m s-1

The correct answer is:
7 m s-1

Solution:
u=12 m s-1

T – mg cos⁡θ=mv2l

As, T=mg we get

mg1-cosθ=mv2l

Also we know that

12mv2-12mu2= – mgll–cos⁡θ

12mv2-12mu2=-mv2

⇒v=u3=123=6.93 m s-1≈7 m s-1

Topic: Physics, Laws of Motion, JEE Main
A stone is dropped from a height $h$. It hits the ground with a certain momentum $p$. If the same stone is dropped from a height $100 \%$ more than the previous height, the momentum when it hits the ground will change by

Options:
A $68 \%$
B $41 \%$
C $200 \%$
D $100 \%$

The correct answer is:
$41 \%$

Solution:
Velocity $v=\sqrt{2 g h}$
and momentum $p=m v$
From Eqs. (i) and (ii), we have
Here
$\frac{p_2}{p_1}=\sqrt{\frac{h_2}{h_1}}$
$$
\begin{aligned}
& \therefore \quad \frac{p_2}{p_1}=\sqrt{\frac{2 h}{h}}=\sqrt{2} \\
& p_2=1.414 p_1 \\
& \% \text { change }=\frac{p_2-p_1}{p_1} \times 100=41 \%
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2012 (Mains)
A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is :

Options:
A 25 m
B 45 m
C 35 m
D 50 m

The correct answer is:
45 m

Solution:
Time for particle to meet =t'=SrelSrel=2010=2 sec

Time taken by Ist particle to reach ground =3 sec

H=12 g32=45 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A stone is executing vertical circular motion with the help of a string as shown below. The string will have least tension at point.

Options:
A A
B B
C C
D D

The correct answer is:
C

Solution:
Te Correct option is C.
The tension at $A$ will be maximum as $\mathrm{mg}$ is acting opposite to tension and it has to exert that much extra cenripetal force to keep the stone in the circle. And tension is least at $\mathrm{C}$ where part of the centripetal force is compensated by $\mathrm{mg}$. This is depicted in the following diagram.

Topic: Physics, Work Power Energy, JEE Main
A stone is projected at angle 30° to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be :

Options:
A 1 : 2
B 1 : 4
C 4 : 1
D 4 : 3

The correct answer is:
4 : 3

Solution:
Let stone is projected with initial velocity v.

Kinetic energy of the stone at point of projection is KE1=12mv2.

Now, velocity of stone at maximum height is =vcos30°, so kinetic energy at that point is KE2=12mvcos30°2

Hence, the ratio KE1KE2=12mv212mvcos30°2=43

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A stone is projected from a point on the ground in such a way so as to hit a bird at height 3h and could attain a maximum height 4h above the ground. If at the instant of projection, the bird flies away horizontally with speed 10 m s-1 and the stone still hits the bird while descending, then the horizontal velocity of stone is

Options:
A 8 m s-1
B 15 m s-1
C 12 m s-1
D 9 m s-1

The correct answer is:
15 m s-1

Solution:
Particle taken t1 time to reach from 0 to 3h and also t1 to reach from 3h to 4h and yet again t1 time from 4h to 3h.
∴ vx3t1-10×3t1=vxt1
⇒ vx = 15 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A stone is projected from the ground with velocity 50 m s-1 at an angle 30Â° with horizontal. It crosses a wall after 3 s. How far beyond the wall the stone will strike the ground? g=10 m s-2

Options:
A 90.2Â m
B 12.5Â m
C 86.5 m
D 216.25 m

The correct answer is:
86.5 m

Solution:
Time, T=2usinâ¡Î¸g
T=2Ã50Ãsinâ¡30o10
=2Ã50Ã1210=5 s
Time spent after crossing the wall, t2=5-3=2 s
â´ Distance, x2=ucosâ¡Î¸Ãt2
=50Ãcosâ¡30oÃ2
=50Ã32Ã2
=503
=50Ã1.73
=86.5 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A stone is projected vertically upwards with speed ' $v$ '. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed ' $v$ '. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right.$, $\left.\cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

Options:
A $4:1$
B $3:2$
C $2:1$
D $1:1$

The correct answer is:
$4:1$

Solution:
P.E. of the stone projected vertically is,
P.E. = mgh
But $\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}$
$\begin{aligned}
\therefore \quad \text { P. } \mathrm{E}_1 & =\mathrm{mg}\left(\frac{\mathrm{v}^2}{2 \mathrm{~g}}\right) \\
& =\frac{\mathrm{mv}^2}{2}... (i)
\end{aligned}$
For the second stone thrown at an angle $\theta$ to the horizontal,
$\begin{array}{ll}
& \mathrm{h}=\frac{\mathrm{v}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{v}^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}=\frac{\mathrm{v}^2}{8 \mathrm{~g}} \\
\therefore \quad & \text { P.E. } 2=\mathrm{mg}\left(\frac{\mathrm{v}^2}{8 \mathrm{~g}}\right)=\frac{\mathrm{mv}^2}{8}... (ii)
\end{array}$
Dividing equation (i) by equation (ii) -
$-\frac{P \cdot E_1}{P \cdot E_2}=\frac{\left(\frac{m v^2}{2}\right)}{\left(\frac{m v^2}{8}\right)}=4: 1$

Topic: Physics, Work Power Energy, JEE Main
A stone is projected with velocity 2gh, so that it just clears two walls of equal height h, at distance of 2h from each other. The time interval of passing between the two walls is

Options:
A hg
B 2hg
C 2hg
D 2hg

The correct answer is:
2hg

Solution:
2h=ut

ux=2hΔt ∵u=ux,t=Δt....i

h=uyt-12gt2

gt2-2uyt+2h=0

t1=2uy+4uy2-8gh2g

t2=2uy-4uy2-8gh2g

Δt=t1-t2=4uy2-8ghg

uy2=g2(Δt)24+2gh

ux2+uy2=u2=(2gh)2

4n2(Δt)2+g2(Δt)24+2gh=4gh

g24Δt4-2ghΔt2+4h2=0

Δt=2gh±4g2h2-4g2h2g2/2=4hg

Δt=2hg

Topic: Physics, Motion In One Dimension, JEE Main
A stone is thrown at 25 m/s at 53o above the horizontal. At what time its velocity is at 45o below the horizontal?

Options:
A 0.5  s
B 4  s
C 3.5  s
D 2.5  s

The correct answer is:
3.5  s

Solution:
Horizontal component of velocity, throughout the motion remain constant
using ucos Î¸=vcos Î±
25cosâ¡53Â°=vcosâ¡45Â°
â¹v=25Ã35.2
=152
â¹vy=vsinâ¡45=15
Now using Vy=uy+ayt in y- direction,
-15=25sinâ¡53Â°-gt
-15=25Ã45-10 t
â¹t=3.5 sec

Topic: Physics, Motion In Two Dimensions, NEET
A stone is thrown at an angle \( \theta \) to the horizontal reaches a maximum heights \( \mathrm{H} \). then the time of flight of stone will be

Options:
A \( \sqrt{\frac{2 H}{g}} \)
B \( 2 \sqrt{\frac{2 H}{g}} \)
C \( \frac{2 \sqrt{2 H \sin \theta}}{g} \)
D \( \frac{\sqrt{2 H \sin \theta}}{g} \)

The correct answer is:
\( 2 \sqrt{\frac{2 H}{g}} \)

Solution:
Maximum height H=u2sin2θ2g
and times of flight T=2usinθg
or T2=4u2sin2θg2
∴ T2H=8g
or T=8Hg=22Hg

Topic: Physics, Motion In Two Dimensions, JEE Main
A stone is thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be

Options:
A $\sqrt{\frac{2 H}{g}}$
B $2 \sqrt{\frac{2 H}{g}}$
C $\frac{2 \sqrt{2 H \sin \theta}}{g}$
D $\frac{\sqrt{2 H \sin \theta}}{g}$

The correct answer is:
$2 \sqrt{\frac{2 H}{g}}$

Solution:
$H=\frac{u^2 \sin ^2 \theta}{2 g}$ and $T=\frac{2 u \sin \theta}{g} \Rightarrow T^2=\frac{4 u^2 \sin ^2 \theta}{g^2}$
$\therefore \frac{T^2}{H}=\frac{8}{g} \Rightarrow T=\sqrt{\frac{8 H}{g}}=2 \sqrt{\frac{2 H}{g}}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A stone is thrown vertically upward with an initial speed \( u \) from the top of a tower, reaches the ground with a speed \( 3 u \). The height of the tower is:

Options:
A \( \frac{3 u^{2}}{g} \)
B \( \frac{4 u^{2}}{g} \)
C \( \frac{6 u^{2}}{g} \)
D \( \frac{9 u^{2}}{g} \)

The correct answer is:
\( \frac{4 u^{2}}{g} \)

Solution:
Let the height of the tower be h.

Taking +ve x-axis along vertically upward direction, we have,

u=u

a=-g

v=-3u

s=AC=-h

As acceleration remains constant.

Now using v2=u2+2as, we get

⇒-3u2=u2+2-g-h

⇒9u2=u2+2gh

⇒2gh=8u2

⇒gh=4u2

⇒h=4u2g.

Topic: Physics, Motion In One Dimension, JEE Main
A stone is thrown vertically upwards from the top of a tower $64 \mathrm{~m}$ high according to the law $s=48 t-16 t^{2}$. The greatest height attained by the stone above ground is

Options:
A $36 \mathrm{~m}$
B $32 \mathrm{~m}$
C $100 \mathrm{~m}$
D $64 \mathrm{~m}$

The correct answer is:
$100 \mathrm{~m}$

Solution:
Given, $s=48 t-16 t^{2}$
$\Rightarrow \quad \frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}$
At the greatest height, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=0$
$\Rightarrow \quad 48-32 t=0$
$\Rightarrow \quad \mathrm{t}=\frac{3}{2}$
$\therefore \quad \mathrm{s}=48 \times \frac{3}{2}-16\left(\frac{3}{2}\right)^{2}$
$=36 \mathrm{~m}$
$\therefore$ Total height $=64+36=100 \mathrm{~m}$

Topic: Mathematics, Parabola, JEE Main
A stone of mass 1.3 kg is being rotated in a horizontal plane as a conical pendulum with the help of a 140 cm long aluminium wire of cross-sectional area 1 mm2. The wire makes an angle θ=75° with the vertical. What is the increment in the length (in mm) of the wire? [Young's modulus of aluminium YAl=7×1010 N m-2, sin75°≈0.97, cos75°≈0.26, g=10 m s-2]

Options:

The correct answer is:
01.00

Solution:
The figure below shows the stone being rotated as a conical pendulum such that the wire makes an angle of 75° with the vertical. Let us assume that the radius OA of the circle is equal to r

For vertical equilibrium of the stone

Tcos75°=Mg

T=Mgcos75°

The elongation in the wire is

∆L=TLAY=MgLAYcos75°

⇒∆L=1.3×10×1.410-6×7×1010×0.26=10-3 m=1 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main
A stone of mass 1kg is tied to end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is:

Options:
A 20 m s-1
B 40 m s-1
C 400 m s-1
D 10 m s-1

The correct answer is:
20 m s-1

Solution:
In horizontal circle, tension towards the centre of the circular path at any moment will provide the required centripetal force.

Therefore,

Tmax=mv2r⇒400=1×v21⇒v=20 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A stone of mass $1 \mathrm{~kg}$ is tied with a string and it is whirled in a vertical circle of radius $1 \mathrm{~m}$. If tension at the highest point is $14 \mathrm{~N}$, then velocity at lowest point will be closest to:

Options:
A $3 \mathrm{~m} / \mathrm{s}$
B $4 \mathrm{~m} / \mathrm{s}$
C $6 \mathrm{~m} / \mathrm{s}$
D $8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$6 \mathrm{~m} / \mathrm{s}$

Solution:
The correct option is: $6 \mathrm{~m} / \mathrm{s}$
Mass of the stone, $\mathrm{m}=1 \mathrm{~kg}$
Radius, $\mathrm{r}=1 \mathrm{~m}$
Tension, $\mathrm{T}=14 \mathrm{~N}$
Let $\mathrm{v}_1$ is velocity at the top point and $\mathrm{v}_2$ is at the bottom point.
The tension at the top point is:
$\begin{aligned}& \mathrm{T}=\frac{\mathrm{mv_{1 } ^ { 2 }}}{\mathrm{r}}+\mathrm{mg} \\& 14=\frac{(1) \mathrm{v}_1^2}{1}+\mathrm{g} \Rightarrow 14=\mathrm{v}_1^2+10 \\& \mathrm{v}_1^2=4 \Rightarrow \mathrm{v}_1=2\end{aligned}$
So, total kinetic energy atthe bottom is:
$\begin{aligned}& \frac{1}{2} \mathrm{mv}_2^2=(\mathrm{mgh})_{\text {bottom }}+\frac{1}{2} \mathrm{mv}_1^2 \\& \frac{1}{2} \mathrm{mv}_2^2=20+2 \\& \mathrm{v}_2^2=44 \Rightarrow \mathrm{v}=\sqrt{44}=6.63 \mathrm{~ms}^{-1}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A stone of mass 1 kg tied to a light inextensible string of length L=103 m is whirling in a circular path of the radius L in the vertical plane. If the ratio of the maximum tension to the minimum tension in the string is 4. What is the speed (in m s-1) of stone at the highest point of the circle? (Take g=10 m s-2 )

Options:

The correct answer is:
10

Solution:
mg + T A = mv A 2 L
T B - mg = mv B 2 L
T B T A = mv B 2 L + mg mv A 2 L - mg = 4
4 v A 2 - v B 2 = 5 gL
By energy conservation
12mvB2=12mvA2+mghA
v B 2 = v A 2 + 4 gL
v A 2 = 3 gL = 3 × 10 × 1 0 3
vA=10 m s-1

Topic: Physics, Laws of Motion, JEE Main
A stone of mass 1 kg tied to a light inextensible string of length L=103 m is whirling in a circular path of the radius L in the vertical plane. If the ratio of the maximum tension to the minimum tension in the string is 2. What is the speed of stone (in m s-1) at the highest point of the circle? (Taking g=10 m s-2)

Options:

The correct answer is:
10

Solution:
TmaxTmin=mv2L+mgmv2L-mg=2
Solving it.
mv2L=3mg
v=3gL
=3101003=10 m s-1

Topic: Physics, Laws of Motion, JEE Main
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10-6 m2 stretched by an amount 0.04 m. The velocity of the projected stone is m s-1. (Young's modulus of rubber =0.5×109 N m-2)

Options:

The correct answer is:
20

Solution:
By energy conservation
12·YAL·x2=12mv2
0.5×109×10-6×(0.04)20.1=201000v2
∴ v2=400
v=20 m s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10-6 m2 stretched by an amount 0.04 m. The velocity of the projected stone is m s-1. (Young's modulus of rubber =0.5×109 N m-2)

Options:

The correct answer is:
20

Solution:
By energy conservation
12·YAL·x2=12mv2
0.5×109×10-6×(0.04)20.1=201000v2
∴ v2=400
v=20 m s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

Options:
A the same throughout the motion.
B minimum when the rope is in the horizontal position.
C minimum at the highest position of the circular path.
D minimum at the lowest position of the circular path.

The correct answer is:
minimum at the highest position of the circular path.

Solution:
As the speed is constant, centripetal force required towards the centre of the circle will be same. At the top, direction of gravitational force and tension acting on the object will be same. Therefore, we can write,

mg+T=mv2r ⇒T=mv2r-mg.

Clearly in this case, tension in the string will be minimum.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A stone of mass $m$, tied to the end of a string, is whirled around in a circle on a horizontal frictionless table. The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by $T=A r^n$, where $A$ is a constant, $r$ is the instantaneous radius of the circle. The value of $n$ is equal to

Options:
A
$-1$

B
$-2$

C
$-4$

D
$-3$

The correct answer is:
$-3$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2012 (26 May Online)
A stone thrown down with a seed $u$ takes a time $t_{1}$ to reach the ground, while another stone, thrown upwards from the same point with the same speed, takes time $t_{2}$. The maximum height the second stone reaches from the ground is

Options:
A $1 / 2 \mathrm{gt}_{1} \mathrm{t}_{2}$
B $\mathrm{g} / 8\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
C $g / 8\left(t_{1}-t_{2}\right)^{2}$
D $1 / 2 \mathrm{gt}_{2}^{2}$

The correct answer is:
$\mathrm{g} / 8\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

Solution:
$-\mathrm{h}=-\mathrm{Ut}_{1}+\frac{-1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2}$
$\mathrm{~h}=\mathrm{Ut}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \ldots .$(1)
$-\mathrm{h}=\mathrm{Ut}_{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2} \ldots \ldots$ (2)
$\mathrm{Ut}_{1}+\mathrm{Ut}_{2}=\frac{1}{2} \mathrm{~g} \mathrm{t}_{2}{ }^{2}-\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}{ }^{2} \quad($ from $1 \& 2)$
$\mathrm{U}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \times\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\mathrm{U}=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)$
$\begin{aligned}
\therefore & \text { Max height }=\mathrm{h}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=(\mathrm{h})+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\mathrm{U} \mathrm{t}_{1}+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} \\
&=\frac{\mathrm{g}}{2} \mathrm{t}_{1}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g} \mathrm{t}_{1}^{2}+\frac{1}{2 \mathrm{~g}} \frac{\mathrm{g}^{2}}{4}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\frac{\mathrm{g}}{2}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \mathrm{t}_{1}+\frac{\mathrm{gt}_{1}{ }^{2}}{2}+\frac{\mathrm{g}}{8}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2} \\
&=\left(\frac{\mathrm{g}}{2}\right)\left\{\left(\mathrm{t}_{2} \mathrm{t}_{1}-\mathrm{t}_{1}{ }^{2}\right)+\mathrm{t}_{1}{ }^{2}+\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)^{2}}{4}\right\} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{4 \mathrm{t}_{2} \mathrm{t}_{1}-4 \mathrm{t}_{1}{ }^{2}+4 \mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1}^{2}-2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left\{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}\right\}}{4} \\
&=\left(\frac{\mathrm{g}}{2}\right) \frac{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}{4} \\
&=\frac{\mathrm{g}}{8}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}
\end{aligned}$
Hence correct Answer is (B).

Topic: Physics, Motion In One Dimension, JEE Main
A stone thrown upward with a speed $u$ from the top of the tower reaches
the ground with a velocity $3 u$. The height of the tower is

Options:
A $3 u^2 / g$
B $4 u^2 / g$
C $6 u^2 / g$
D $9 u^2 / g$

The correct answer is:
$4 u^2 / g$

Solution:
$v^2=u^2+2 g h \Rightarrow(3 u)^2=(-u)^2+2 g h \Rightarrow h=\frac{4 u^2}{g}$

Topic: Physics, Motion In One Dimension, JEE Main
A stone thrown vertically upward satisfies the equation $\mathrm{s}=64 \mathrm{t}-16 \mathrm{t}^{2}$, where s is in meter and $\mathrm{t}$ is in second. What is the time required toreach the maximumheight?

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Given equation is $s=64 t-16 t^{2}$ $\therefore$ On differentiating w.r.t. $t$, we get $\frac{d s}{d t}=64-32 t$
Put $\frac{d s}{d t}=0$ for maximumheight $\Rightarrow 64-32 t=0 \quad \Rightarrow t=2$
Now, $\frac{d^{2} s}{d t^{2}}=-32$
At $t=2, \frac{d^{2} s}{d t^{2}}=-32$
Since, $\left(\frac{d^{2} s}{d t^{2}}\right)_{t=2} < 0$
- Required time $=2$ second

Topic: Mathematics, Application of Derivatives, JEE Main
A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is xu2-gL. The value of x is

Options:
A 2
B 3
C 4
D 1

The correct answer is:
2

Solution:
Applying conservation of mechanical energy at the point A and B, we get12mu2+0=12mv2+mgL

⇒v=u2-2gL

Now in vector form. v→i=u i^ and v→f=v j^

Therefore, change in velocity will be,∆V→=vj^-ui^ and ΔV→=u2+v2

⇒ΔV=u2+u2-2gL

=2u2-gL

Hence, x=2

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (27 Jun Shift 2)
A stone tied to 180 cm long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is 1936x m s-2. The value of x ______.

[Take π=227]

Options:

The correct answer is:
125

Solution:
Angular velocity, ω=28×2π60 rad s-1.

Required centripetal acceleration is given by,

a=ω2R

=5660×2272×1.8

=(44)2225×1.8=1936125

Hence, required x=125.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A stone tied to a string is rotated in a vertical circle. The minimum speed of the stone during a complete vertical circular motion

Options:
A is independent of the mass of the stone
B is independent of the length of the string
C decreases with increasing mass of the stone
D decreases with increasing length of the string

The correct answer is:
is independent of the mass of the stone

Solution:
Minimum speed with which the string is rotating in a vertical circle, (v) = gr

i.e, the minimum speed of stone is independent of the mass of stone.

Topic: Physics, Laws of Motion, JEE Main