datavorous/entrance-exam-dataset huggingface dataset:
A body of mass \(\mathrm{I} \mathrm{kg}\) is suspended from a spring of negligible mass. Another body of mass \(500 \mathrm{~g}\) moving vertically upwards hits the suspended body with a velocity of \(3 \mathrm{~ms}^{-1}\) and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is \(\frac{10}{\pi} \mathrm{Hz}\), the amplitude of the motion and the spring constant are respectively,

Options:
A \(5 \mathrm{~cm}, 300 \mathrm{Nm}^{-1}\)
B \(10 \mathrm{~cm}, 300 \mathrm{Nm}^{-1}\)
C \(10 \mathrm{~cm}, 600 \mathrm{Nm}^{-1}\)
D \(5 \mathrm{~cm}, 600 \mathrm{Nm}^{-1}\)

The correct answer is:
\(5 \mathrm{~cm}, 600 \mathrm{Nm}^{-1}\)

Solution:
Given, mass of body, \(M=1 \mathrm{~kg}\), frequency of oscillation, \(f=\frac{10}{\pi} \mathrm{Hz}\),
mass of hitting body, \(m=0.5 \mathrm{~kg}\) and speed of body, \(v=3 \mathrm{~ms}^{-1}\)
As frequency of oscillation of spring mass system,
\(f=\frac{1}{2 \pi} \sqrt{\frac{k}{M_{\text {body }}}}\)
Since, hitting body get embedded to the initial mass, so mass of the body changed to
\(M_{\text {body }}=M+m=1+0.5=1.5 \mathrm{~kg}\)
Spring constant, \(\quad k=(2 \pi /)^2 M_{\text {body }}\)
\(=\left(2 \pi \times \frac{10}{\pi}\right)^2(1.5)=600 \mathrm{Nm}^{-1}\)
Now, velocity of body after collision,
\(\begin{aligned}
& (M+m) v^{\prime}=m v \Rightarrow(1+1 / 2) v^{\prime}=1 / 2 \times 3 \\
& v^{\prime}=1 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
The maximum amplitude of the sping-mass system is \(v^{\prime}=A \omega \Rightarrow A=\frac{v^{\prime}}{2 \pi f}=\frac{1}{20}=0.05 \mathrm{~m}=5 \mathrm{~cm}\)
Hence, the correct answer is (d).

Topic: Physics, Oscillations, JEE Main
A body of mass $x \mathrm{~kg}$ is moving with a velocity of $100 \mathrm{~ms}^{-1}$. Its de-Broglie wavelength is $6.62 \times 10^{-35} \mathrm{~m}$. Hence, $\mathrm{x}$ is $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}\right.$ )

Options:
A $0.1 \mathrm{~kg}$
B $0.25 \mathrm{~kg}$
C $0.15 \mathrm{~kg}$
D $0.2 \mathrm{~kg}$

The correct answer is:
$0.1 \mathrm{~kg}$

Solution:
$\begin{aligned} \lambda &=\frac{\mathrm{h}}{\mathrm{mv}} \\ &=\frac{6.62 \times 10^{-34}}{6.62 \times 10^{-35} \times 100} \\ &=0.1 \mathrm{~kg} \end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main
A body of weight $2 \mathrm{~kg}$ is suspended as shown in the figure. The tension $T_1$ in the horizontal string (in $\mathrm{kg} w t$ ) is

Options:
A $2 / \sqrt{3}$
B $\sqrt{3} / 2$
C $2 \sqrt{3}$
D 2

The correct answer is:
$2 \sqrt{3}$

Solution:
$\begin{aligned} & T \sin 30=2 \mathrm{~kg} w t \\ & \Rightarrow T=4 \mathrm{~kg} w t \\ & T_1=T \cos 30^{\circ} \\ & =4 \cos 30^{\circ} \\ & =2 \sqrt{3}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are 0.6 and 0.4 respectively, the acceleration of the body will be (Acceleration due to gravity = g=10 m s-2)

Options:

The correct answer is:
2

Solution:
Weight of the body = 64 N
so the mass of the body m=6.4 kg, μ k =0.4

Net acceleration = Applied force -Kinetic friction Mass of the body = μ s mg− μ k mg m =( μ s − μ k )g=(0.6−0.4)g=0.2 g =2 m s-2

Topic: Physics, Laws of Motion, JEE Main
A body of weight $200 \mathrm{~N}$ is suspended from a tree branch through a chain of mass $10 \mathrm{~kg}$. The branch pulls the chain by a force equal to (if $g=10 \mathrm{~m} / \mathrm{s}^2$ ) :

Options:
A $100 \mathrm{~N}$
B $200 \mathrm{~N}$
C $300 \mathrm{~N}$
D $150 \mathrm{~N}$

The correct answer is:
$300 \mathrm{~N}$

Solution:
Chain block system is in equilibrium so
$\mathrm{T}=200+100=300 \mathrm{~N} \text {. }$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A body P floats in water with half its volume immersed. Another body Q floats in a liquid of density 34th of the density of water with two-third of the volume immersed. The ratio of density of P to that of Q is

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

The correct answer is:
1 :1

Solution:
Let, for body P , volume =VP
Given, Immersed volume =VP2
For body Q volume =VQ
Immersed volume =23VQ
According to Archimedes principle,
âµ Weight of body = weight of fluid displaced
For body P,
VpÏpg=12VpÏÏgSo, ÏPÏw=12 .....(i)
For body Q ,
VQÏQg=23VQ34Ïwg
âµ density of liquid=34Ïw
So, ÏQÏw=12 .....(ii)
From Equations (i) and (ii),
ÏPÏQ=1 or ÏP :ÏQ=1 :1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

Options:
A at the instant just before the body hits the earth
B it remains constant all through
C at the instant just after the body is projected
D at the highest position of the body

The correct answer is:
at the instant just before the body hits the earth

Solution:
We know
$$
P=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{v}}=F y \cos \theta
$$
So just before hitting $\theta$ is zero and both $F$ and $v$ are maximum.

Topic: Physics, Gravitation, NEET, NEET 2011 (Screening)
A body projected vertically upwards with a velocity $u$ returns to the starting point in 4 seconds. If $g=10 \mathrm{~m} / \mathrm{sec}^2$, the value of $u$ is

Options:
A 5 m/sec
B 10 m/sec
C 15 m/sec
D 20 m/sec

The correct answer is:
20 m/sec

Solution:
Time of flight $T=\frac{2 u}{g}=4 \mathrm{sec};u=20 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In One Dimension, JEE Main
A body released from a great height falls freely towards the earth. Another body is released from the same height exactly one second later. The separation (in m) between the two bodies two seconds after the release of the second body is (g=9.8 m s-2)

Options:

The correct answer is:
24.5

Solution:
Separation of the body after t second after the release of 1st body is
∆x=12gt2-12gt-12
=12gt2-t-12=12g2t-1t=2+1=3
∆x=12×9.8×5m=24.5 m

Topic: Physics, Motion In One Dimension, JEE Main
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is:

Options:
A solid sphere
B solid cylinder
C hollow cylinder
D ring

The correct answer is:
solid cylinder

Solution:
12Iω2=12×12mv2

I=12 mR2

Body is solid cylinder.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 sec. The number of rotations made in the process is

Options:

The correct answer is:
200

Solution:
ωf=ω0+αt

α=1200×6

θ=ω0t+12αt2

=600×1060+12×1200×6×136

θ=200

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A body starting at t=0 from origin and oscillates simple harmonically with a period of 4 s. After what time will its kinetic energy be 75% of its total energy

Options:
A 12 s
B 13 s
C 14 s
D 1 s

The correct answer is:
13 s

Solution:
As we know, KE=12mv2=12mω2A2-x2

and TE=12mω2A2

As per question,

KE=75100TE⇒12mω2A2-x2=12mω2A2⇒x=±A2

From mean position, time taken to reach half of the amplitude will be, T12=412=13 s

Topic: Physics, Oscillations, JEE Main
A body, starting from rest, is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

Options:
A t1/2
B t3/4
C t3/2
D t1/4

The correct answer is:
t3/2

Solution:
Power, P=Fv=constant
Or maat=constant

Or a2t=constant
⇒a∝1t ....... (1)
Now, S=2 at2 ∵ u=0
⇒S∝at2
∴S∝t2t [Using (1)]
⇒S∝t3/2

Topic: Physics, Work Power Energy, JEE Main
A body starts falling freely from height H hits an inclined plane in its path at height h. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of Hh for which the body will take the maximum time to reach the ground is _____.

Options:

The correct answer is:
2

Solution:
Total time of flight =T can be written as,

T=2hg+2H-hg

For maximum time =dTdh=0

⇒2g12h-12H-h=0

⇒H-h=h

⇒h=H2⇒Hh=2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10 s is S1, next 10 s is S2 and the last 10 s is S3. Then S1 : S2 : S3 is the same as

Options:
A 1 : 2 : 4

B 1 : 2 : 5

C 1 : 3 : 5

D 1 : 3 : 9

The correct answer is:
1 : 3 : 5

Solution:
a = constant
S1=v0t+12at2
v1=v0+at
S2=v1t+12at2 ⇒v0t+32at2
v2=v1+at
=v0+2at
S3=v2t+12at2
=v0t+52at2
v0=0
S1=12at2; S2=32at2; S3=52at2
S1 : S2 : S3=1 : 3 : 5

Topic: Physics, Motion In One Dimension, JEE Main
A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the nth second to the distance covered in n second is

Options:
A 2n-1n2
B 1n2-1n
C 2n2-1n
D 2n+1n2

The correct answer is:
2n-1n2

Solution:
Here, sn=1nan2
snth = distance travelled in n second - distance travelled in (n-1) second
=2n-12a
∴ snthsn=2n-1n2=2n-1n2

Topic: Physics, Motion In One Dimension, JEE Main
A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the nth second to the distance covered in \( \mathrm{n} \) seconds is

Options:
A \( \frac{2}{n}-\frac{1}{n^{2}} \)
B \( \frac{1}{n^{2}}-\frac{1}{n} \)
C \( \frac{2}{n^{2}}-\frac{1}{n} \)
D \( \frac{2}{n}+\frac{1}{n^{2}} \)

The correct answer is:
\( \frac{2}{n}-\frac{1}{n^{2}} \)

Solution:
Given initial velocity u=0 and constant acceleration.
-Distance travelled in nth second Snth=u+2n-12a=0+2n-12a=2n-12a

Distance travelled in n seconds is Sn=ut+12at2=0+12an2=12an2

∴ SnthSn=2n-1a2×2an2
= 2 n - 1 n 2

Topic: Physics, Motion In One Dimension, JEE Main
A body starts from rest on a long inclined plane of slope $45^{\circ}$. The coefficient of friction between the body and the plane varies as $\mu=0.3 x$, where $x$ is distance travelled down the plane. The body will have maximum speed (for $g=10 \mathrm{~m} / \mathrm{s}^2$ ) when $x=$

Options:
A
$9.8 \mathrm{~m}$

B
$27 \mathrm{~m}$

C
$12 \mathrm{~m}$

D
$3.33 \mathrm{~m}$

The correct answer is:
$3.33 \mathrm{~m}$

Solution:
When the body has maximum speed then
$$
\begin{aligned}
& \mu=0.3 \mathrm{x}=\tan 45^{\circ} \\
& \therefore \mathrm{x}=3.33 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2013 (22 Apr Online)
A body starts moving from rest with constant acceleration covers displacement S1 in first (p-1) seconds and S2 in first p seconds. The displacement S1+S2 will be made in time :

Options:
A (2p+1) s
B 2p2-2p+1 s
C (2p-1) s
D 2p2-2p+1 s

The correct answer is:
2p2-2p+1 s

Solution:
For the displacement S1 of the body, it can be written that

S1=12a(p-1)2 ...1

And, for the displacement S2 of the body, it follows that

S2=12a(p)2 ...2

If t be the required time, it follows that

S1+S2=12at2 ...3

Equations (1), (2) and (3) imply that

(p-1)2+p2=t2⇒t=2p2-2p+1 s

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A body starts to fall freely under gravity. The distances covered by it in first,second and third second are in ratio

Options:
A 1:3:5
B 1:2:3
C 1:4:9
D 1:5:6

The correct answer is:
1:3:5

Solution:
$S_n=u+\frac{g}{2}(2 n-1) ;$ when $u=0, S_1: S_2: S_3=1: 3: 5$

Topic: Physics, Motion In One Dimension, JEE Main
A body takes 4 min to cool from 61°C to 59°C. If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49°C is:

Options:
A 4 min.
B 3 min.
C 8 min.
D 6 min.

The correct answer is:
6 min.

Solution:
ΔTΔt=KTt-TS Tt= average temperature

TS= surrounding temperature

61-594=K61+592-30 ...(i)

51-49t=K51+492-30 ...ii

Divide (1) and (2)

t4=60-3050-30=3020

So, t=6 min

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A body takes 10 min to cool from 60 â to 50 â. If the temperature of the surroundings is 25 â and 527 â respectively. The final temperature of the body is

Options:
A 48 Â°C
B 46 °C
C 49 °C
D 42.85 °C

The correct answer is:
42.85 °C

Solution:
According to Newton's law
Î¸1-Î¸2t=KÎ¸1+Î¸22-Î¸0
â´ 60-5010=K60+502-25 ....(i)
Let Î¸ be the temperature after another 10 min
â´ 50-Î¸10=KÎ¸+502-25 ....(ii)
Dividing Eq.(i) by Eq. (ii), we get
1050-Î¸ = 30Ã2Î¸ â´ Î¸=42.85 â

Topic: Physics, Thermodynamics, JEE Main
A body takes 10 minutes to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$. The temperature of surroundings is constant at $25^{\circ} \mathrm{C}$. Then, the temperature of the body after next 10 minutes will be approximately

Options:
A
$43^{\circ} \mathrm{C}$

B
$47^{\circ} \mathrm{C}$

C
$41^{\circ} \mathrm{C}$

D
$45^{\circ} \mathrm{C}$

The correct answer is:
$43^{\circ} \mathrm{C}$

Solution:
According to Newton's law of cooling,
$$
\begin{aligned}
&\left(\frac{\theta_1-\theta_2}{t}\right)=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) \\
&\left(\frac{60-50}{10}\right)=K\left(\frac{60+50}{2}-25\right) \\
&\text { and, }\left(\frac{50-\theta}{10}\right)=K\left(\frac{50+\theta}{2}-25\right)
\end{aligned}
$$
Dividing eq. (i) by (ii),
$$
\frac{10}{(50-\theta)}=\frac{60}{\theta} \Rightarrow \theta=42.85^{\circ} \mathrm{C} \cong 43^{\circ} \mathrm{C}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A body takes time t to reach the bottom of an inclined plane of angle Î¸ with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of friction of the rough surface is

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

The correct answer is:
3 4 tanθ

Solution:
μ=tanθ( 1− 1 n 2 ) =tanθ( 1− 1 2 2 )= 3 4 tanθ

Topic: Physics, Laws of Motion, JEE Main
A body travels $102.5 \mathrm{~m}$ in $\mathrm{n}^{\text {th }}$ second and $115.0 \mathrm{~m}$ in $(\mathrm{n}+2)^{\text {th }}$ second. The acceleration is :

Options:
A $6.25 \mathrm{~m} / \mathrm{s}^2$
B $12.5 \mathrm{~m} / \mathrm{s}^2$
C $9 \mathrm{~m} / \mathrm{s}^2$
D $5 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$6.25 \mathrm{~m} / \mathrm{s}^2$

Solution:
$\begin{aligned}
& \text { Given, } 102.5=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) \& \\
& 115=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}+3) \\
& \Rightarrow 102.5=\mathrm{u}+\mathrm{an}-\frac{\mathrm{a}}{2} \& \\
& 115=\mathrm{u}+\mathrm{an}+\frac{3 \mathrm{a}}{2} \\
& 12.5=2 \mathrm{a} \Rightarrow \mathrm{a}=6.25 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (04 Apr Shift 1)
A body travels a distance $s$ in $t$ seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration $\mathrm{f}$ and in the second part with constant retardation $\mathrm{r}$. The value of $\mathrm{t}$ is given by

Options:
A
$\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}$

B
$2 s\left(\frac{1}{f}+\frac{1}{r}\right)$

C
$\frac{2 \mathrm{~s}}{\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{r}}}$

D
$\sqrt{2 s(f+r)}$

The correct answer is:
$\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}$

Solution:
Portion OA, OB corresponds to motion with acceleration ' $\mathrm{f}$ ' and retardation ' $r$ ' respectively.
Area of $\triangle O A B=S$ and $O B=t$. Let $O L=t_1$,
$\mathrm{LB}=\mathrm{t}_2$ and $\mathrm{AL}=\mathrm{v}, \mathrm{S}=\frac{1}{2} \mathrm{OB} \cdot \mathrm{AL}=\frac{1}{2} \mathrm{t} \cdot \mathrm{v} ; \mathrm{v}=\frac{2 \mathrm{~S}}{\mathrm{t}}$
Also, $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{t}_1}, \mathrm{t}_1=\frac{\mathrm{v}}{\mathrm{f}}=\frac{2 \mathrm{~s}}{\mathrm{tf}}$ and $\mathrm{r}=\frac{\mathrm{v}}{\mathrm{t}_2}, \mathrm{t}_2=\frac{\mathrm{v}}{\mathrm{r}}=\frac{2 \mathrm{~s}}{\mathrm{tr}} ; \mathrm{t}=\mathrm{t}_1+\mathrm{t}_2=\frac{2 \mathrm{~s}}{\mathrm{tf}}+\frac{2 \mathrm{~s}}{\mathrm{tr}}$
$t=\left(\frac{1}{f}+\frac{1}{r}\right) \frac{2 s}{t} \Rightarrow t=\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2003
A body travels for $15 \mathrm{sec}$ starting from rest with constant acceleration. If it travels distances $s_1, s_2$ and $s_3$ in the first five seconds, second five seconds and next five seconds respectively the relation between $s_1, s_2$ and $S_3$ is

Options:
A $S_1=S_2=S_3$
B $5 S_1=3 S_2=S_3$
C $S_1=\frac{1}{3} S_2=\frac{1}{5} S_3$
D $s_1=\frac{1}{5} s_2=\frac{1}{3} s_3$

The correct answer is:
$S_1=\frac{1}{3} S_2=\frac{1}{5} S_3$

Solution:
If the body starts from rest and moves with constant acceleration then the ratio of distances in consecutive equal time interval $S_1: S_2: S_3=1: 3: 5$

Topic: Physics, Motion In One Dimension, JEE Main
A body, under the action of a force $\overrightarrow{\mathrm{F}}=6 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$, acquires an acceleration of $1 \mathrm{~m} / \mathrm{s}^2$. The mass of this body must be :

Options:
A $10 \sqrt{2} \mathrm{~kg}$
B $2 \sqrt{10} \mathrm{~kg}$
C $10 \mathrm{~kg}$
D $20 \mathrm{~kg}$

The correct answer is:
$10 \sqrt{2} \mathrm{~kg}$

Solution:
$$
\begin{aligned}
\overrightarrow{\mathrm{F}} & =6 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+10 \hat{\mathrm{k}} \\
|\overrightarrow{\mathrm{F}}| & =\sqrt{6^2+8^2+10^2}=10 \sqrt{2} \\
\mathrm{~m} & =\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{a}}=\frac{10 \sqrt{2}}{1}=10 \sqrt{2} \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2009 (Mains)
A body weighing $13 \mathrm{~kg}$ is suspended by two strings $5 \mathrm{~m}$ and $12 \mathrm{~m}$ long, their other ends being fastened to the extremities of a rod $13 \mathrm{~m}$ long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are

Options:
A
$12 \mathrm{~kg}$ and $13 \mathrm{~kg}$

B
$5 \mathrm{~kg}$ and $5 \mathrm{~kg}$

C
$5 \mathrm{~kg}$ and $12 \mathrm{~kg}$

D
$5 \mathrm{~kg}$ and $13 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$ and $12 \mathrm{~kg}$

Solution:
$\mathrm{T}_2 \cos \left(\frac{\pi}{2}-\theta\right)=\mathrm{T}_1 \cos \theta \Rightarrow \mathrm{T}_1 \cos \theta=\mathrm{T}_2 \sin \theta$
$\mathrm{T}_1 \sin \theta+\mathrm{T}_2 \cos \theta=13$.
$\because \mathrm{OC}=\mathrm{CA}=\mathrm{CB}$
$\Rightarrow \angle \mathrm{AOC}=\angle \mathrm{OAC}$ and $\angle \mathrm{COB}=\angle \mathrm{OBC}$
$\therefore \sin \theta=\sin \mathrm{A}=\frac{5}{13}$ and $\cos \theta=\frac{12}{13}$
$\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{5}{12} \Rightarrow \mathrm{T}_1=\frac{5}{12} \mathrm{~T}_2$
$\mathrm{~T}_2\left(\frac{5}{12} \cdot \frac{5}{13}+\frac{12}{13}\right)=13$
$\mathrm{~T}_2\left(\frac{169}{12 \cdot 13}\right)=13$
$\mathrm{~T}_2=12 \mathrm{~kgs} . \Rightarrow \mathrm{T}_1=5 \mathrm{~kgs}$.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2007
A body weighing $20 \mathrm{~kg}$ just slides down a rough inclined plane that rises 5 in 12 . The coefficient of friction is

Options:
A $0.46$
B $4.6$
C $0.52$
D $0.12$

The correct answer is:
$0.46$

Solution:
The given situation can be shown as
As the plane rises 5 in 12
$$
\therefore \quad \sin \theta=\frac{5}{12}
$$
and $\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1+\left(\frac{5}{12}\right)^{2}}$
$$
=\frac{\sqrt{119}}{12}
$$
So, the coefficient of friction,
$$
\mu=\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{5}{12} \times \frac{12}{\sqrt{119}}=\frac{5}{\sqrt{119}}=0.46
$$

Topic: Physics, Laws of Motion, JEE Main
A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use g=GMR2=9.8 m s-2 and radius of earth, R=6400 km.]

Options:
A 49.17 N
B 48.83 N
C 49.83 N
D 49 N

The correct answer is:
48.83 N

Solution:
Weight of pole =mg=49 N

At equator due to rotation =ge=g-Rω2

so W=mge=mg-Rω2

∴ WP>We Wp=49 N

So, We=48.83 N. We<49 N

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?

Options:
A 150 N
B 200 N
C 250 N
D 100 N

The correct answer is:
100 N

Solution:
Weight =mg=mGMR2=W
At half way down to the centre, W'=mGMR3R23R22
=mGM2R2
=100 N

Topic: Physics, Gravitation, JEE Main
A body weighs W at the surface of the earth. Its weight at a height equal to half the radius of the earth will be

Options:
A W 2
B 2W 3
C 4W 9
D 8W 27

The correct answer is:
4W 9

Solution:
The weight of an object at the surface of the earth is

W=mg=mGMR2

At a height h=R2, the weight would be

W'=mGMR+h2=49×mGMR2

∴W'=49W

Topic: Physics, Gravitation, JEE Main
A body weight W, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:

Options:
A W91
B W100
C W9
D W3

The correct answer is:
W100

Solution:
Final distance from the centre of earth wil be R+9R=10R.

Now acceleration due to gravity at any height(distance from centre r) is given by,

g=GMr2.

Therefore,

g'=GM10R2

Now, value of g at surface will be, GMR2.

Hence,

g'=g100.

Then, W'=mg'=mg100=W100

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A body whose momentum is constant must have constant

Options:
A acceleration
B force
C velocity
D mass

The correct answer is:
velocity

Solution:
Momentum of a moving body with velocity $v$ is given as
$$
p=m v
$$
Since mass $m$ of a body remains always constant. Hence, if momentum of body is constant, then its velocity must be constant.

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
A Bohr hydrogen atom undergoes a transition n=5ân=4 and emits a photon of frequency Î½. Frequency of circular motion of an electron in n=4 orbit is Î½4. The ratio Î½Î½4 is

Options:
A 1825
B 1625
C 925
D 825

The correct answer is:
1825

Solution:
En=-mz2e48ε02n2h2
So, hν=+mz2e48ε02h2116-125
∴ ν=mz2e48ε02h3916×25 ...(1)
And frequency ν4=1T=v2πr
=Ze22ε0nh12π πmze2ε0h2n2=z2e4m4ε02n3h3 ...(2)
∴ νν4=1825=0.72

Topic: Physics, Atomic Physics, JEE Main
A bomb is dropped by a fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :

Options:
A straight line vertically down the plane
B parabola in a direction opposite to the motion of plane
C parabola in the direction of motion of plane
D hyperbola

The correct answer is:
straight line vertically down the plane

Solution:
The bomb after being dropped from the plane will have only horizontal velocity but zero vertical velocity. This scenario is as same as if a body is horizontally projected from the top of a tower, and it hits the ground tracing a parabolic path. Hence, the trajectory of the bomb is also a parabola, in the direction of motion of the plane. But if we look the trajectory with respect to observer sitting in the plane then this looks like straight line vertically down the plane due to same horizontal velocity.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A bomb is dropped by an aero plane flying horizontally with a velocity $200 \mathrm{~km} / \mathrm{hr}$ and at a height of $980 \mathrm{~m}$. At the time of dropping a bomb, the distance of the aero plane from the target on the ground to hit directly is $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ )

Options:
A $\frac{\sqrt{2} \times 10^4}{9} \mathrm{~m}$
B $\frac{10^4}{9} \mathrm{~m}$
C $\frac{10^4}{9 \sqrt{2}} \mathrm{~m}$
D $\frac{10^4}{18} \mathrm{~m}$

The correct answer is:
$\frac{10^4}{9 \sqrt{2}} \mathrm{~m}$

Solution:
The plane is flying horizontally. Hence initial vertical component of the velocity is zero.
If it reaches the ground in time $t$, then
$$
\begin{aligned}
& \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \\
& \mathrm{t}^2=\frac{2 \mathrm{~h}}{\mathrm{~g}}=\frac{2 \times 980}{9.8}=200 \\
& \mathrm{t}=10 \sqrt{2} \mathrm{~s}
\end{aligned}
$$
The horizontal component of the velocity is
$$
\mathrm{V}=200 \mathrm{~km} / \mathrm{hr}=200 \times \frac{5}{18}=\frac{1000}{18} \mathrm{~m} / \mathrm{s}
$$
The horizontal distance to be covered is
$$
\mathrm{d}=\mathrm{Vt}=\frac{1000}{18} \times 10 \sqrt{2}=\frac{10^4}{9 \sqrt{2}} \mathrm{~m}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main