datavorous/entrance-exam-dataset huggingface dataset:
Two blocks of equal masses are connected with a massless spring of spring constant $2500 \mathrm{~N} / \mathrm{m}^2$ and length $10 \mathrm{~cm}$ at rest on the frictionless horizontal plane. If a constant horizontal force $10 \mathrm{~N}$ is applied as shown in the figure, find the maximum distance between the blocks.

Options:
A 10.8 cm
B 10.4 cm
C 10.6 cm
D 10.0 cm

The correct answer is:
10.4 cm

Solution:
In the case of maximum elongation, whole system will be moving with a common acceleration a, so for the system
$F=(m+m) a \Rightarrow a=\frac{F}{2 m}$
The centre of mass of the system will also move with the same acceleration $a$.
Let us suppose that both the masses elongate the spring by a distance $x_1$ and $x_2$, respectively. So, on taking the frame of centre of mass there will act a pseudo force equal to ma on the masses.
On applying the work-energy theorem, taking the centre of mass as a reference point.
$\begin{gathered}
W_{\text {all forces }}=\Delta K \\
(F-m a) x_1+(m a) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
{[\because \text { w.r.t. centre of mass } \Delta K=0]} \\
\Rightarrow\left(F-m \times \frac{F}{2 m}\right) x_1+\left(m \times \frac{F}{2 m}\right) x_2 \\
\Rightarrow\left(F-\frac{F}{2}\right) x_1+\left(\frac{F}{2}\right) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
\left.\Rightarrow \frac{F}{2} \times x_1+\frac{F}{2} \times x_2\right)^2=0 \\
\Rightarrow \frac{F}{2}\left(x_1+x_2\right)=\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\
\Rightarrow \quad x_1+x_2=\frac{F}{K} \Rightarrow x_{\text {total }}=\frac{F}{K}
\end{gathered}$
Here, $F=10 \mathrm{~N}, K=2500 \mathrm{~N} / \mathrm{m}^2$
So, $\quad x_{\text {total }}=\frac{10}{2500} \mathrm{~m}=\frac{1}{250} \mathrm{~m}=\frac{100}{250} \mathrm{~cm}=0.4 \mathrm{~cm}$
So, the maximum distance between the blocks
$\begin{aligned}
\text { will be }= & \text { natural length }+x_{\text {total }} \\
= & 10 \mathrm{~cm}+0.4 \mathrm{~cm}=10.4 \mathrm{~cm} \\
& {[\because \text { Natural length }=10 \mathrm{~cm}(\text { given })] }
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
Two blocks of equal masses are tied to the ends of a light string. The string passes over a mass less pulley fixed on frictionless surface as shown in the figure. The acceleration of the centre of mass of the blocks is ( $g$ - acceleration due to gravity)

Options:
A $\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) \mathrm{g}$
B $\left(\frac{\sqrt{3}+1}{4 \sqrt{2}}\right) \mathrm{g}$
C $\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \mathrm{g}$
D $\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \mathrm{g}$

The correct answer is:
$\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) \mathrm{g}$

Solution:
Since, $\mathrm{mg} \sin 60>\mathrm{mg} \sin 30$. So, block will go left side.
$\mathrm{T}-\mathrm{mg} \sin 30=\mathrm{m}$ a
$\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{m}$ a
$\mathrm{mg} \sin 60-\mathrm{T}=\mathrm{m}$ a
$$
\frac{\sqrt{3} \mathrm{mg}}{2}-\mathrm{T}=\mathrm{ma}
$$
Adding equation (i) and (ii), we have
$$
\begin{aligned}
& \frac{\sqrt{3} \mathrm{mg}}{2}-\frac{\mathrm{mg}}{2}=2 \mathrm{ma} \\
& \mathrm{a}=\frac{(\sqrt{3}-1) \mathrm{g}}{4}
\end{aligned}
$$
The acceleration of the centre of mass of the block is
$$
\begin{aligned}
& \vec{a}_{c m}=\frac{m_1 \vec{a}_1+m_2 \vec{a}_2}{m_1+m_2} \\
& =\frac{m a \hat{l}+m a \hat{j}}{2 m} \\
& =\frac{a}{2}(\hat{i}+\hat{j}) \\
& =\frac{\sqrt{2}}{2} \times \frac{(\sqrt{3}-1)}{4} g=\frac{(\sqrt{3}-1) g}{4 \sqrt{2}}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks of mass 2. 9 kg and 1. 9 kg are suspended from a rigid support S by two inextensible wires each of length 1 m (see figure). The upper wire has negligible mass and the lower wire has a uniform mass of 0. 2 kg/m. The whole system of blocks, wires and support have an upward acceleration of 0. 2 m/s2. The acceleration due to gravity is 9. 8 m/s2.

Find the tension at the midpoint of the upper wire.

Options:

The correct answer is:
50

Solution:
To find tension at mid-point of the upper wire
The equation of motion gives
T2 - 5g = 5a
Therefore T2 = 5(g + a) = 5(9. 8+0. 2) = 50 N

Topic: Physics, Laws of Motion, JEE Main
Two blocks of mass $M_1=20 \mathrm{~kg}$ and $M_2=12 \mathrm{~kg}$ are connected by a metal rod of mass $8 \mathrm{~kg}$. The system is pulled vertically up by applying a force of $480 \mathrm{~N}$ as shown. The tension at the mid-point of the rod is:

Options:
A
$144 \mathrm{~N}$

B
$96 \mathrm{~N}$

C
$240 \mathrm{~N}$

D
$192 \mathrm{~N}$

The correct answer is:
$192 \mathrm{~N}$

Solution:
Acceleration produced in upward direction
$$
\begin{aligned}
& \mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_1+\mathrm{M}_2+\text { Mass of metal rod }} \\
& =\frac{480}{20+12+8}=12 \mathrm{~ms}^{-2}
\end{aligned}
$$
Tension at the mid point
$$
\begin{aligned}
\mathrm{T} & =\left(\mathrm{M}_2+\frac{\text { Mass of rod }}{2}\right) \mathrm{a} \\
& =(12+4) \times 12=192 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2013 (22 Apr Online)
Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is 24π×102 N m-2. What is the minimum radius of the wire ?
( take g=10 m s-2

Options:
A 1250 cm
B 1.25 cm
C 125 cm
D 12.5 cm

The correct answer is:
12.5 cm

Solution:
Tension in the string =2m1m2m1+m2g

=2×3×5×108=752

Stress =TA

24π×102=752πr2

r2=7548×10-2

r2=0.015625

r=0.125 m=12.5 cm

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Two blocks of masses \( 3 \mathrm{~kg} \) and \( 2 \mathrm{~kg} \) are placed side by side on an incline as shown in figure. A force \( F=20 \mathrm{~N} \) is acting on \( 2 \mathrm{~kg} \) block along the incline. The coefficient of friction between the block and the incline is same and equal to \( 0.1 \). Find the normal contact force exerted by \( 2 \mathrm{~kg} \) block on \( 3 \mathrm{~kg} \) block.

Options:
A \( 18 \mathrm{~N} \)
B \( 30 \mathrm{~N} \)
C \( 12 \mathrm{~N} \)
D \( 27.6 \mathrm{~N} \)

The correct answer is:
\( 12 \mathrm{~N} \)

Solution:
According to question a F force act on the block
Now F.B.D. for block 3 kg
Where, N is normal provided by 2 kg block
f is force of friction
μ if coefficient of friction
According to newton's second law
∑F=ma
Applied in the direction of incline
3g sin θ+N-f=3a
f=μN1
Here N1 is normal force applied by the incline on 2 Kg block
3g sin θ+N-3μg cos θ=3a ...(1)

for 2 kg block
∴ F+2g sin θ- N-2μg cos θ=2a ...(2)
adding (1) and (2)
We get
5g sin θ-5μg cos θ+F=5a
given θ = 37°
F=20 N
μ=0.1
⇒ 5 × 10 × 35 - 50(0.1)45 + 20 = 5a ⇒ a=465 m s-2
Now putting the value of acceleration in (2)
We get
20+2035-N-2×45=2×465 20+12-N-85=925
⇒ N=12 N

Topic: Physics, Laws of Motion, JEE Main
Two blocks of masses 5 kg and 10 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal wire is 2Ã109 N m-2. If g=10 m s-2, then what is the minimum radius of the wire which will not break

Options:
A 0.1 mm
B 0.2 mm
C 0.05 mm
D 0.25 mm

The correct answer is:
0.1 mm

Solution:
This is Atwood's machine. Let the tension in the wire be T and acceleration of masses a.
Then, T-5g=5a and 10g-T=10a
This gives a=13g and T⁡=203g=2003N.
Stress =T⁡area=2003×3.14×r⁡2=2×109 (given)
Thus the minimum radius r should be r=2003×3.14×2×109=1.03×10-4 m≈0.1 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m s-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass in m s-1 is 2n, then n=

Options:

The correct answer is:
5

Solution:
vCM=m1v1+m2v2m1+m2
10Ã14+4Ã010+4=10Ã1414=10 m s-1 =2nân=5

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks of masses m1 and m2 are connected by a massless spring and placed at a smooth surface. The spring initially stretched and released. then

Options:
A The momentum of each particle remains constant separately
B The magnitude of momentum of both bodies are same to each other
C The mechanical energy of system remains constant
D Both (b) and (c) are correct.

The correct answer is:
Both (b) and (c) are correct.

Solution:
Since, no external force is present on the system so, conservation principle of momentum is applicable
∴p→i=p→f=p→1+p→2
∴ p→1=-p→2 (∵p→i=0)
∴ p→1=-p→2
∴ p→1=p→2
From this point of view, it is clear that momenta of both particles are equal in magnitude but opposite in direction
Also, friction is absent. So total mechanical energy of system remains conserved

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses m1 and m2 being connected with a light spring of stiffness k are driven with forces F1 and F2 on a smooth horizontal plane. If the blocks move with the same acceleration a0, the value of a0 is:

Options:
A F1-F2m1+m2
B F1m2-m1F2m1+m22
C 12F1m1+F2m2
D decided by the stiffness of the spring

The correct answer is:
F1-F2m1+m2

Solution:
Both the blocks can be chosen as a system, thus,
a0=F1-F2m1+m2

Topic: Physics, Laws of Motion, JEE Main
Two blocks of masses $\mathrm{m}$ and $\mathrm{M}$ are connected by means of a metal wire of cross-sectional area A passing over a frictionless fixed pulley as shown in the figure. The system is then released. If $\mathrm{M}=2 \mathrm{~m}$, then the stress produced in the wire is:

Options:
A
$\frac{2 \mathrm{mg}}{3 \mathrm{~A}}$

B
$\frac{4 \mathrm{mg}}{3 \mathrm{~A}}$

C
$\frac{\mathrm{mg}}{\mathrm{A}}$

D
$\frac{3 \mathrm{mg}}{4 \mathrm{~A}}$

The correct answer is:
$\frac{4 \mathrm{mg}}{3 \mathrm{~A}}$

Solution:
Tension in the wire, $\mathrm{T}=\left(\frac{2 \mathrm{mM}}{\mathrm{m}+\mathrm{M}}\right) \mathrm{g}$
$$
\begin{aligned}
\text { Stress } &=\frac{\text { Force } / \text { Tension }}{\text { Area }}=\frac{2 \mathrm{mM}}{\mathrm{A}(\mathrm{m}+\mathrm{M})} \mathrm{g} \\
&=\frac{2(\mathrm{~m} \times 2 \mathrm{~m}) \mathrm{g}}{\mathrm{A}(\mathrm{m}+2 \mathrm{~m})}(\mathrm{M}=2 \mathrm{~m} \text { given }) \\
&=\frac{4 \mathrm{~m}^2}{3 \mathrm{~mA}} \mathrm{~g}=\frac{4 \mathrm{mg}}{3 \mathrm{~A}}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2013 (25 Apr Online)
Two blocks of masses ' $M$ ' and ' $m$ ' are placed on one another on a smooth horizontal surface as shown in the figure.
The force ' $F$ ' is acting on the mass ' $M$ ' horizontally during time interval ' $t$ '. Assuming no relative sliding between the blocks, the work done by friction on the blocks is .......... .

Options:
A $\frac{F t}{2(M+m)}$
B $\frac{M+m}{m t^2}$
C $\frac{m F^2 t^2}{2(M+m)^2}$
D $\frac{F^2 t^2}{(M+m)}$

The correct answer is:
$\frac{m F^2 t^2}{2(M+m)^2}$

Solution:
From (i) \& (ii), we get $a=\frac{F}{m+M}$ and $f=\frac{m F}{m+M}$ Distances moved in time $t(s)$
$$
=\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{F}{M+m}\right) t^2
$$
Work done by friction $=f \times s$
$$
=\left(\frac{m F}{M+m}\right) \times \frac{1}{2} \frac{F t^2}{(M+m)}=\frac{m F^2 t^2}{2(M+m)^2}
$$

Topic: Physics, Laws of Motion, JEE Main
Two blocks P and Q of masses 0.3 kg and 0.4 kg respectively are stuck to each other by some weak glue as shown in the figure. They hang together at the end of a spring with a spring constant k=200 N m-1. The block Q suddenly falls free due to failure of glue, then find the maximum kinetic energy of the block P during subsequent motion (in mJ).

Options:

The correct answer is:
40

Solution:
Let A be the amplitude, then we have
kA=mg ⇒A=mgk

A=0.4×g200=2 cm
Motion starts from the lower extreme.
KEmax=12mω2A2=12×0.3×km×A2
=12×0.3×2000.3×21002
=0.04 J
=40 mJ

Topic: Physics, Oscillations, JEE Main
Two blocks with masses m1=1 kg and m2=2 kg are connected by a spring constant k=24 N m-1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cm s -1 to the right, the value of amplitude of oscillation (in cm) of the block with respect to other is

Options:

The correct answer is:
2

Solution:
The amplitude of oscillations will be the maximum compression in the spring. At the time of maximum compression velocities of both the blocks are equal say v, then using law of conservation of momentum,
m1v0=m1+m2v
or 1Ã12=1+2v or v=4 cm s-1
Using law of conservation of energy, we have
12m1v02=12kx2+12m1+m2v2
Putting the value and solving we get x=2 cm

Topic: Physics, Oscillations, JEE Main
Two bodies $A$ and $B$ of mass $m$ and $2 m$ respectively are placed on a smooth floor. They are connected by a spring of negligible mass. $A$ third body $C$ of mass $m$ is placed on the floor. The body $C$ moves with a velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain time after the collision it is found that the instantaneous velocities of $A$ and $B$ are same and the compression of the spring is $x_0$. The spring constant $k$ will be

Options:
A
$m \frac{v_0^2}{x_0^2}$

B
$m \frac{v_0}{2 x_0}$

C
$2 m \frac{v_0}{x_0}$

D
$\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

The correct answer is:
$\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

Solution:
Initial momentum of the system block $(C)$ $=m v_0$. After striking with $A$, the block $C$ comes to rest and now both block A and B moves with velocity $\mathrm{v}$ when compression in spring is $x_0$.
By the law of conservation of linear momentum
$$
m v_0=(m+2 m) v \Rightarrow v=\frac{v_0}{3}
$$
By the law of conservation of energy K.E. of block $C=$ K.E. of system + P.E. of system
$$
\begin{aligned}
& \frac{1}{2} m v_0^2=\frac{1}{2}(3 m)\left(\frac{v_0}{3}\right)^2+\frac{1}{2} k x_0^2 \\
& \Rightarrow \quad \frac{1}{2} m v_0^2=\frac{1}{6} m v_0^2+\frac{1}{2} k x_0^2 \\
& \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\
& \therefore \quad k=\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies $A$ and $B$ of mass $m$ and $2 m$ respectively are placed on a smooth floor. They are connected by a spring of negligible mass. $A$ third body $C$ of mass $m$ is placed on the floor. The body $C$ moves with a velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain time after the collision it is found that the instantaneous velocities of $A$ and $B$ are same and the compression of the spring is $x_0$. The spring constant $k$ will be

Options:
A
$m \frac{v_0^2}{x_0^2}$

B
$m \frac{v_0}{2 x_0}$

C
$2 m \frac{v_0}{x_0}$

D
$\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

The correct answer is:
$\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

Solution:
Initial momentum of the system block $(C)$ $=m v_0$. After striking with $A$, the block $C$ comes to rest and now both block A and B moves with velocity $\mathrm{v}$ when compression in spring is $x_0$.
By the law of conservation of linear momentum
$$
m v_0=(m+2 m) v \Rightarrow v=\frac{v_0}{3}
$$
By the law of conservation of energy K.E. of block $C=$ K.E. of system + P.E. of system
$$
\begin{aligned}
& \frac{1}{2} m v_0^2=\frac{1}{2}(3 m)\left(\frac{v_0}{3}\right)^2+\frac{1}{2} k x_0^2 \\
& \Rightarrow \quad \frac{1}{2} m v_0^2=\frac{1}{6} m v_0^2+\frac{1}{2} k x_0^2 \\
& \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\
& \therefore \quad k=\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2012 (12 May Online)
Two bodies $\mathrm{A}$ (of mass $1 \mathrm{~kg}$ ) and $\mathrm{B}$ (of mass $3 \mathrm{~kg}$ ) are dropped from heights of $16 \mathrm{~m}$ and $25 \mathrm{~m}$, respectively. The ratio of the time taken by them to reach the ground is:

Options:
A $4 / 5$
B $5 / 4$
C $12 / 5$
D $5 / 12$

The correct answer is:
$4 / 5$

Solution:
We know that time taken by a body from height $h$ to reach the ground is.
$$
\begin{aligned}
t & =\sqrt{\frac{2 h}{g}} \\
\therefore \quad \frac{t_{\mathrm{A}}}{t_{\mathrm{B}}} & =\frac{\frac{2 h_{\mathrm{A}}}{g}}{\frac{2 h_{\mathrm{B}}}{g}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}} \\
& =\sqrt{\frac{16}{25}}=\frac{4}{5} .
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is x2. Then, the value of x is

Options:

The correct answer is:
3

Solution:
In both, cases is about point of contact ring,

mgh=12Iω2

mgh=122 mR2VR2R2

VR=gh

Solid cylinder

mgh=12Iω2

mgh=1232mR2vC2R2

vC=4gh3

vRvC=32

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36 \mathrm{~g}$ and its density is $9 \mathrm{~g} / \mathrm{cm}^3$. If the mass of the other is $48 \mathrm{~g}$, its density in $\mathrm{g}$ $/ \mathrm{cm}^3$ is

Options:
A $\frac{4}{3}$
B $\frac{3}{2}$
C 3
D 5

The correct answer is:
3

Solution:
Apparent weight $=V(\rho-\sigma) g=\frac{m}{\rho}(\rho-\sigma) g$
where $m=$ mass of the body,
$\rho=$ density of the body
$\sigma=$ density of water
If two bodies are in equilibrium then their apparent weight must be equal.
$\therefore \frac{m_1}{\rho_1}\left(\rho_1-\sigma\right)=\frac{m_2}{\rho_2}\left(\rho_2-\sigma\right)$
$\Rightarrow \quad \frac{36}{9}(9-1)=\frac{48}{\rho_2}\left(\rho_2-1\right)$
By solving we get $\rho_2=3$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is

Options:
A $\frac{4 \pi \rho G m h}{3}$
B $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
C $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
D $\frac{3 \pi \rho \mathrm{Gmh}}{8}$

The correct answer is:
$\frac{8 \pi \rho \mathrm{Gmh}}{3}$

Solution:
Gravitational force on a body of mass $m$ at height $h$ due to earth,
$$
F=\frac{G M_e m}{(R+h)^2}
$$
where, $M_e$ is mass of the earth.
$$
\because \text { Density of earth, } \rho=\frac{\text { mass of earth }\left(M_e\right)}{\text { volume of earth }(V)}
$$
or $M_e=\rho \cdot V=\rho\left(\frac{4}{3} \pi R^3\right)$
$$
\begin{gathered}
\therefore \quad F=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho m}{(R+h)^2} \\
\text { or } \quad F=\frac{G\left(\frac{4}{3} \pi R\right) \rho m}{\left(1+\frac{h}{R}\right)^2} \\
\text { or } \quad F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
\end{gathered}
$$
or
$$
F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
$$
By using Binomial expansion,
$$
F=\frac{4}{3} \pi G R \rho m\left(1-\frac{2 h}{R}\right)
$$
$\left[\because \quad(1+x)^n=1+n x+\frac{n(n-1)}{2 !} \cdot x^2+\ldots\right.$ and neglecting higher terms.]
Difference in weight $=\frac{4}{3} \pi G R \rho m-\frac{4}{3} \pi G \rho m R\left(\frac{2 h}{R}\right)$
Hence, error in weight $=\frac{8}{3} \pi \rho G m h$.

Topic: Physics, Gravitation, JEE Main
Two bodies have their moments of inertia $l$ and $2 l$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio:

Options:
A $2: 1$
B $1: 2$
C $\sqrt{2}: 1$
D $1: \sqrt{2}$

The correct answer is:
$1: \sqrt{2}$

Solution:
K.E. rotation $K=\frac{1}{2} I \omega^2$
Angular momenta $L=100$
$\therefore \quad K=\frac{L^2}{2}$
Here $K_1+K_2, I_1=I_1, I_2=21$
$$
\therefore \quad \frac{L_1}{L_2}=\sqrt{\frac{2 I_1 K_1}{2 I_2 K_2}}
$$
$$
\begin{aligned}
& =\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{I}{2 I}} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
Two bodies of different masses ${ }^{m_a}$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio of the time taken by the two to cover these distances are

Options:
A a:b
B b:a
C $\sqrt{a}: \sqrt{b}$
D $a^2: b^2$

The correct answer is:
$\sqrt{a}: \sqrt{b}$

Solution:
$\begin{aligned} & h=\frac{1}{2} g t^2 \Rightarrow t=\sqrt{2 h / g} \\ & t_a=\sqrt{\frac{2 a}{g}} \text { and } t_b=\sqrt{\frac{2 b}{g}} \Rightarrow \frac{t_a}{t_b}=\sqrt{\frac{a}{b}}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
Two bodies of equal masses are heated at a uniform rate under identical conditions. The change in temperature in the two cases is shown graphically. The ratio of their specific heats in the solid state is

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

The correct answer is:
1:3

Solution:
P(Δt)=m s(Δθ)
(Δθ)(Δt)=Pms

ΔθΔt1=602=30

ΔθΔt2=404=10

s1s2=ΔθΔt2ΔθΔt1=13

Topic: Physics, Thermodynamics, JEE Main
Two bodies of equal masses are some distance apart. If $20 \%$ of mass is transferred from the first body to the second body, then the gravitational force between them

Options:
A increase by $4 \%$
B increase by $14 \%$
C decreases by $4 \%$
D decreases by $14 \%$

The correct answer is:
decreases by $4 \%$

Solution:
(c) Initially force,
$$
F_1=\frac{G m^2}{r^2}
$$
When $20 \%$ mass is transferred from one body to another, then their masses are
$$
m_1=\frac{80}{100} m \text { and } m_2=\frac{120}{100} m
$$
So, force is finally
$$
F_2=\frac{G m^2}{r^2} \times\left(\frac{24}{25}\right)
$$
Hence, reduction in $F_2$ is $\frac{F_1-F_2}{F_1} \times 100 \%=-4 \%$

Topic: Physics, Gravitation, JEE Main
Two bodies of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ have position vectors $\hat{i}+2 \hat{j}+\hat{k}$ and $-3 \hat{i}-2 \hat{j}+\hat{k}$, respectively. The centre of mass of this system has a position vector

Options:
A $-2 \hat{i}+2 \hat{k}$
B $-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
C $2 \hat{i}-\hat{j}-2 \hat{k}$
D $-\hat{i}+\hat{j}+\hat{k}$

The correct answer is:
$-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Solution:
The position vector of centre of mass
$$
\begin{aligned}
\vec{r} & =\frac{m_1 \overrightarrow{r_1}+m_2 \vec{r}_2}{m_1+m_2} \\
& =\frac{1(\hat{i}+2 \hat{j}+\hat{k})+3(-3 \hat{i}-2 \hat{j}+\hat{k})}{1+3} \\
& =\frac{1}{4}(-8 \hat{i}-4 \hat{j}+4 \hat{k}) \\
& =-2 \hat{i}-\hat{j}+\hat{k}
\end{aligned}
$$
The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ have position vectors $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$, respectively. The centre of mass of this system has a position vector :

Options:
A $-\hat{i}+\hat{j}+\hat{k}$
B $-2 \hat{i}+2 \hat{k}$
C $-2 \hat{i}-\hat{j}+\hat{k}$
D $2 \hat{i}-\hat{j}-2 \hat{k}$

The correct answer is:
$-2 \hat{i}-\hat{j}+\hat{k}$

Solution:
$\quad \overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2}{\mathrm{~m}_1+\mathrm{m}_2}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ have position vectors $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$, respectively. The centre of mass of this system has a position vector :

Options:
A $-\hat{i}+\hat{j}+\hat{k}$
B $-2 \hat{i}+2 \hat{k}$
C $-2 \hat{i}-\hat{j}+\hat{k}$
D $2 \hat{i}-\hat{j}-2 \hat{k}$

The correct answer is:
$-2 \hat{i}-\hat{j}+\hat{k}$

Solution:
$\quad \overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2}{\mathrm{~m}_1+\mathrm{m}_2}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2009 (Mains)