datavorous/entrance-exam-dataset huggingface dataset:
Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k^ and -3i^-2j^+k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :

Options:
A i^-2j^+k^
B -3i^-2j^+k^
C -2i^+2k^
D -2i^-j^+2k^

The correct answer is:
i^-2j^+k^

Solution:
As we know, centre of mass is given by, r→com=∑mir→i∑mi

r→com=mı^+2ȷ^+k^+3m-3ı^-2ȷ^+k^m+3m

r→com=-8mı^-4mȷ^+4mk^4m=-2ı^-ȷ^+k^

The magnitude of the position vector of the centre of mass is 22+11+12=6.

It is similar to the magnitude of the option A.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Two bodies of mass 4 g and 25 g are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is :

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

The correct answer is:
2:5

Solution:
The kinetic energy K of an object is given by

K=12mv2=p22m ...1

Following equation (1), for the two given bodies, it can be written that

p122 m1=p222 m2⇒p1p2=m1 m2=425=25

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity g is:

Options:
A g2
B g5
C g10
D g

The correct answer is:
g5

Solution:
a=m2−m1gm1+m2

a=(6−4)g6+4=2g10

a=g5

Topic: Physics, Laws of Motion, JEE Main
Two bodies of mass $10 \mathrm{~kg}$ and $5 \mathrm{~kg}$ moving in concentric orbits of radii $R$ and $r$ such that their periods are the same. Then the ratio between their centripetal acceleration is

Options:
A $R / r$
B $r / R$
C $R^{\wedge} 2 / r^{\wedge} 2$
D $R^{\wedge} 2 / r^{\wedge} 2$

The correct answer is:
$R / r$

Solution:
Angular velocity is given by, $\omega=\frac{2 \pi}{T}$
Centripetal acceleration is given by, $a=\omega^2 r$
$\frac{a_R}{a_r}=\frac{\omega_R^2 \times R}{\omega_r^2 \times r}=\frac{T_r^2}{T_R^2} \times \frac{R}{r}=\frac{R}{r} \quad\left[\text { As } T_r=T_R\right]$

Topic: Physics, Rotational Motion, JEE Main
Two bodies of mass $10 \mathrm{~kg}$ and $5 \mathrm{~kg}$ moving in concentric orbits of radii $R$ and $r$ such that their periods are the same. Then the ratio between their centripetal acceleration is

Options:
A $R / r$
B $r / R$
C $R^2 / r^2$
D $r^2 / R^2$

The correct answer is:
$R / r$

Solution:
$\frac{a_R}{a_r}=\frac{\omega_R^2 \times R}{\omega_r^2 \times r}=\frac{T_r^2}{T_R^2} \times \frac{R}{r}=\frac{R}{r}$
$\left[\right.$ As $\left.T_r=T_R\right]$

Topic: Physics, Motion In Two Dimensions, JEE Main
Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad s-1, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g = 10 m s-2).

Options:
A 20 N

B 60 N

C 40 N

D 10 N

The correct answer is:
60 N

Solution:
Given ω = 2 5 rad/s and A = 1.6 cm
∴ ω=km=25 ⇒k= 25 2
Maximum compression in the spring
=xo+A
=mgk+A
∴ Fmax=4g + k(x0 + A)

Fmax =4g+kmgk+A
= 40 + 20 = 60 N

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (09 Apr Online)
Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad s-1, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g = 10 m s-2).

Options:
A 20 N

B 60 N

C 40 N

D 10 N

The correct answer is:
60 N

Solution:
Given ω = 2 5 rad/s and A = 1.6 cm
∴ ω=km=25 ⇒k= 25 2
Maximum compression in the spring
=xo+A
= mg k + A
=1×10×10025×25+1.6 cm
= 8 5 + 1.6
= 1.6 + 1.6
= 3.2 cm
∴ Fmax=4g + K(x + A)
= 40 + 20 = 60 N

Topic: Physics, Oscillations, JEE Main
Two bodies of masses 2 kg and 4 kg are moving with velocities 20 m s-1 and 10 m s-1 towards each other due to mutual gravitational attraction. The velocity of their centre of mass is

Options:
A 5 m s-1
B 6 m s-1
C 8 m s-1
D Zero

The correct answer is:
Zero

Solution:
Since net force acting on the system is zero, hence position of centre of mass of the system remains unchanged i.e., velocity of the centre of mass is zero

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two bodies of masses $4 \mathrm{~m}$ and $9 \mathrm{~m}$ are separated by a distance ' $r$ '. The gravitational potential at a point on this line joining them where the gravitational field becomes zero is

Options:
A $\frac{-25 G m}{r}$
B $\frac{-4 G m}{r}$
C $\frac{-9 G m}{r}$
D $\frac{-13 G m}{r}$

The correct answer is:
$\frac{-25 G m}{r}$

Solution:
At the position when the gravitational field is zero. (Distance $x$ is measured from $4 \mathrm{~m}$ mass)
$$
\begin{aligned}
& \frac{G(4 \mathrm{~m})}{x^2}=\frac{G(9 \mathrm{~m})}{(r-x)^2} \\
& \frac{4}{9}=\left(\frac{x}{r-x}\right)^2 \Rightarrow \frac{2}{3}=\frac{x}{r-x} \\
& 2 r-2 x=3 x \Rightarrow 2 r=5 x \Rightarrow x=\frac{2 r}{5}
\end{aligned}
$$
The point $P$ is at a distance $\frac{2 r}{5}$ from mass $4 \mathrm{~m}$ and $\left(r-\frac{2}{5} r\right)=\frac{3 r}{5}$ from mass $m$.
$$
\begin{aligned}
& v=-\frac{G(4 \mathrm{~m})}{\left(\frac{2}{5} r\right)}-\frac{G(9 \mathrm{~m})}{\frac{3 r}{5}}=\frac{-5 G m}{r}\left[\frac{4}{2}+\frac{9}{3}\right] \\
& =\frac{-5 G m}{r}[2+3] \quad=-25 \frac{G m}{r}
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main
Two bodies of masses $8 \mathrm{~kg}$ are placed at the vertices $A$ and $B$ of an equilateral triangle $A B C$. A third body of mass $2 \mathrm{~kg}$ is placed at the centroid $G$ of the triangle. If $A G=B G=C G=1 \mathrm{~m}$, where should a
fourth body of mass $4 \mathrm{~kg}$ be placed, so that the resultant force on the $2 \mathrm{~kg}$ body is zero?

Options:
A At $C$
B At a point $P$ on the line $C G$ such that $P G=\frac{1}{\sqrt{2}} \mathrm{~m}$
C At a point $P$ on the line $C G$ such that $P G=0.5 \mathrm{~m}$
D At a point $P$ on the line $C G$ such that $P G=2 \mathrm{~m}$

The correct answer is:
At a point $P$ on the line $C G$ such that $P G=\frac{1}{\sqrt{2}} \mathrm{~m}$

Solution:
According to the question, the arrangement of the masses is as shown below,
Gravitational force between two masses is given as
$F=\frac{G m_{1} m_{2}}{r^{2}}$
where, $G=$ gravitational constant and $r=$ distance between them.
From the given values, we can say that force between masses at $A$ and $G$. $=$ Force between masses at $B$ and $G$. $\begin{array}{ll}\Rightarrow & F_{A}=\frac{G m_{A} m_{G}}{A G^{2}} \text { and } F_{B}=\frac{G m_{B} m_{G}}{B G^{2}} \\ \text { or } & F_{A}=F_{B}=\frac{G \times 8 \times 2}{1^{2}}=16 G...(i)\end{array}$
$\left[\because\right.$ Given, $m_{1}=m_{A}=m_{B}=8 \mathrm{~kg}, m_{2}=m_{G}=2 \mathrm{~kg}$,
$A G=B G=r=1 \mathrm{~m} \text { ] }$
From the figure, resultant of $F_{A}$ and $F_{B}$ is given as
$\begin{aligned} F_{A B} &=\sqrt{F_{A}^{2}+F_{B}^{2}+2 F_{A} F_{B} \cos \theta} \\ &=\sqrt{2 F_{A}^{2}+2 F_{A}^{2} \cos 120^{\circ}} \\ &=\sqrt{F_{A}^{2}}=\sqrt{(16 G)^{2}}=16 G \\ \Rightarrow \quad F_{A B} &=F_{A}=F_{B}=16 G \quad \text { [from Eq. (i)] } \end{aligned}$
For resultant force on $2 \mathrm{~kg}$ body to be zero, $4 \mathrm{~kg}$ body should be placed at a certain distance from $G$ such that
$\mathbf{F}_{C G}=-\mathbf{F}_{A E}$
$\Rightarrow \quad \quad\left|\mathbf{F}_{C G}\right|=\left|\mathbf{F}_{A B}\right|$
or $\frac{G \times m_{C} m_{G}}{x^{2}}=G(\mathbf{1} 6)$
where, $x=$ distance between $4 \mathrm{~kg}$ and $2 \mathrm{~kg}$ body.
Here, $m_{C}=4 \mathrm{~kg}$
$\begin{aligned}
\Rightarrow & & \frac{G \times 4 \times 2}{x^{2}} &=G(16) \\
\Rightarrow & x^{2} &=\frac{1}{2} \\
\Rightarrow & x &=\frac{1}{\sqrt{2}} \mathrm{~m}
\end{aligned}$
This means, $4 \mathrm{~kg}$ body should be placed at point $P$ on line $C G$ such that $P G=\frac{1}{\sqrt{2}} \mathrm{~m}$.

Topic: Physics, Gravitation, JEE Main
Two bodies of masses m1=5 kg and m2=3 kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m1 will be : [Take g=10 m s-2]

âââââââ

Options:
A 30 N
B 40 N
C 50 N
D 60 N

The correct answer is:
40 N

Solution:
The forces acting on two masses are shown below.

Here, N is normal to the incline and tension T along the string up the incline. As the string and the pulley are all light and smooth, the tension in the string is uniform everywhere.

âââââââ

For equilibrium condition, forces should add to zero.

Thus, equating, m2g=m1gsinθ

⇒sinθ=m2m1=35 and cosθ=45

Normal force on m1 is N=m1gcosθ=5gcosθ

=5×10×45=40 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Two bodies of masses m1 and m2 are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity when they are r distance apart is

Options:
A 2G(m1+m2)r
B 2Gm1m2(m1+m2)r
C G(m1+m2)r
D Gm1m2m1+m2r

The correct answer is:
2G(m1+m2)r

Solution:
By conservation of momentum
m1v1-m2v2=0
by conservation of energy
12m1v12+12m2v22-Gm1m2r=0
12m1v12+12m2m1v1m22=Gm1m2r
v122m1+m12m2=Gm1m2r
v12=2Gm22m1+m2r
v1=m22Gm1+m2r
∴ v2=m12Gm1+m2r
Also, vrel=v1+v2
vrel=m2+m12Gm1+m2r
vrel=2Gm1+m2r

Topic: Physics, Gravitation, JEE Main
Two bodies of masses $m_{1}$ and $m_{2}$ are separated by a distance $R$. The distance of the centre of mass of the bodies from the mass $m_{1}$ is

Options:
A $\frac{m_{2} R}{m_{1}+m_{2}}$
B $\frac{m_{1} R}{m_{1}+m_{2}}$
C $\frac{m_{1} m_{2}}{m_{1}+m_{2}} R$
D $\frac{m_{1}+m_{2}}{m_{1}} R$

The correct answer is:
$\frac{m_{2} R}{m_{1}+m_{2}}$

Solution:
The bodies are separated by a distance $R$. so the coordinates of $m_{1}$ and $m_{2}$ will be (0,0) and $(R, 0)$

From the formula of centre of mass,
$$
\begin{aligned}
X_{\mathrm{cm}} &=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} \\
&=\frac{m_{1} \times 0+m_{2} \times R}{m_{1}+m_{2}} \\
X_{\mathrm{cm}} &=\frac{m_{2} \cdot R}{m_{1}+m_{2}}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies of masses $m_1$ and $m_2$ have equal kinetic energies. If $p_1$ and $p_2$ are their respective momentum, then ratio $p_1: p_2$ is equal to

Options:
A $m_1: m_2$
B $m_2: m_1$
C $\sqrt{m_1}: \sqrt{m_2}$
D $m_1^2: m_2^2$

The correct answer is:
$\sqrt{m_1}: \sqrt{m_2}$

Solution:
$P=\sqrt{2 m E} \quad \therefore \propto \sqrt{m}$ (if $E=$ const.). $\frac{P_1}{P_2}=\sqrt{\frac{m_1}{m_2}}$

Topic: Physics, Work Power Energy, JEE Main
Two bodies of masses $m_1$ and $m_2$ have equal kinetic energies. If $p_1$ and $p_2$ are their respective momentum, then ratio $p_1: p_2$ is equal to

Options:
A $m_1: m_2$
B $m_2: m_1$
C $\sqrt{m_1}: \sqrt{m_2}$
D $m_1^2: m_2^2$

The correct answer is:
$\sqrt{m_1}: \sqrt{m_2}$

Solution:
$\begin{aligned}P=\sqrt{2 m E} \quad \therefore \quad P \propto \sqrt{m} \\ \text { (if } E=\text { const.) } \therefore \frac{P_1}{P_2}=\sqrt{\frac{m_1}{m_2}} & \end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $\mathrm{r}$. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

Options:
A
$-\frac{4 \mathrm{Gm}}{\mathrm{r}}$

B
$-\frac{6 \mathrm{Gm}}{\mathrm{r}}$

C
$-\frac{9 \mathrm{Gm}}{r}$

D
zero

The correct answer is:
$-\frac{9 \mathrm{Gm}}{r}$

Solution:
Position of the null point from mass $\mathrm{m}, \mathrm{x}=\frac{\mathrm{r}}{1+\sqrt{\frac{4 \mathrm{~m}}{\mathrm{~m}}}}=\frac{r}{3}$
$$
V=-G m\left(\frac{3}{r}+\frac{12}{2 r}\right)=-9 \frac{G m}{r}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2011
Two bodies of specific heat capacities S1 and S2 have the same heat capacities. They are combined to form a single composite body. The specific heat capacity of the composite body is

Options:
A S1+S2
B S1+S22
C 2S1S2S1+S2
D 1S1+1S2

The correct answer is:
2S1S2S1+S2

Solution:
Given, m1S1=m2S2
m1S1+m2S2=m1+m2S
S=m1S1+m2S2m1+m2=2m1S1m1+m2 ∴m1S1=m2S2
=2m1S1m1+m1S1S2=2S1S2S1+S2

Topic: Physics, Thermodynamics, JEE Main
Two bodies of the same mass are connected with an inelastic string of length l are placed together on a horizontal surface. One of them is projected vertically upwards from the surface with a velocity u=6gl. The maximum height up to which the centre of mass of the system of the two masses rises is

Options:
A 3l4
B l2
C 3l2
D l

The correct answer is:
l

Solution:
The velocity of B just before the string is taut
v1=u2â2gl=6glâ2gl=2gl
After the string is taut, both will move up with same speed, v2=mv1m+m=v12=gl
By this time centre of mass in at h=l2

A further rise in the centre of mass =v222g=gl2g=l2
Hence maximum height =l2+l2=l

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two bodies of the same mass are moving with the same speed, but in different directions in a plane. They have a completely inelastic collision and move together thereafter with a final speed which is half of their initial velocities of the two bodies (in degree) is -

Options:

The correct answer is:
120

Solution:
Momentum conservation along x direction

2mvcosθ=(2m)v2

cosθ=12

θ=60°

then angle between their initial velocities

= 120°

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Two boys are standing at the ends $\mathrm{A}$ and $\mathrm{B}$ of a ground where $\mathrm{AB}=a$. The boy at $\mathrm{B}$ starts running in a direction perpendicular to $\mathrm{AB}$ with velocity $v_1$. The boy at A starts running simultaneously with velocity $v$ and catches the other in time $t$, where $t$ is:

Options:
A $\frac{a}{\sqrt{v^2+v_1^2}}$
B $\frac{a}{v+v_1}$
C $\frac{a}{v-v_1}$
D $\sqrt{\frac{a^2}{v^2-v_1^2}}$

The correct answer is:
$\sqrt{\frac{a^2}{v^2-v_1^2}}$

Solution:
If boy \(A\) catches boy \(B\) in time \(t\),
\(\begin{aligned}
& \Rightarrow(\mathrm{vt})^2=\left(\mathrm{v}_1 \mathrm{t}\right)^2+\mathrm{a}^2 \\
& \Rightarrow \mathrm{t}^2=\frac{\mathrm{a}^2}{\mathrm{v}^2-\mathrm{v}_1^2} \\
& \Rightarrow \mathrm{t}=\frac{\mathrm{a}}{\sqrt{\mathrm{v}^2-\mathrm{v}_1^2}}
\end{aligned}\)

Topic: Physics, Motion In Two Dimensions, JEE Main
Two bulbs consume the same power when operated at 200 V and 300 V respectively. When these bulbs are connected in series across a DC source of 400 V, then the ratio of power consumed across them is

Options:
A 23
B 32
C 49
D 94

The correct answer is:
49

Solution:
R=V2P or RâV2 â´R1R2=20023002=49.
When bulbs are connected in series, the current I is same through each. As P=I2 R or PâR( as I is same in series),
so P1P2=R1R2=49.

Topic: Physics, Current Electricity, JEE Main
Two bulbs of identical volumes connected by a small capillary are initially filled with an ideal gas at temperature T. Bulb 2 is heated to maintain a temperature $2 \mathrm{~T}$ while bulb 1 remains at temperature $\mathrm{T}$. Assume throughout that the heat conduction by the capillary is negligible. Then the ratio of final mass of the gas in bulb 2 to the initial mass of the gas in the same bulb is close to $-$

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

The correct answer is:
$2 / 3$

Solution:
Mole conservation
$\mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$
Initial no. of moles $=\mathrm{n}_{1}=\mathrm{n}_{2}=\frac{\mathrm{n}}{2}$
finally when temp of 1 vessel is $T \&$ another is $2 T$
$\begin{array}{l}
\mathrm{n}_{1}=\frac{\mathrm{PV}}{\mathrm{RT}} \\
\mathrm{n}_{2}=\frac{\mathrm{PV}}{\mathrm{R} 2 \mathrm{~T}} \Rightarrow \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{2}{1} \\
\mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n} \\
\mathrm{n}_{1}=\frac{2 \mathrm{n}}{3} ; \mathrm{n}_{2}=\frac{\mathrm{n}}{3}
\end{array}$
mass of gas $\propto \mathrm{n}_{1}$
$\therefore \frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=\frac{\frac{\mathrm{n}}{3}}{\frac{\mathrm{n}}{2}}=\frac{2}{3}$

Topic: Physics, Kinetic Theory of Gases, KVPY, KVPY 2011 (SB/SX)
Two capacitances of capacity $C_1$ and $C_2$ are connected in series and potential difference $V$ is applied across it. Then the potential difference across ${ }^{C_1}$ will be

Options:
A $v \frac{C_2}{C_1}$
B $V \frac{C_1+C_2}{C_1}$
C $V \frac{C_2}{C_1+C_2}$
D $V \frac{C_1}{C_1+C_2}$

The correct answer is:
$V \frac{C_2}{C_1+C_2}$

Solution:
Charge flowing $=\frac{C_1 C_2}{C_1+C_2} v$. So potential difference across $C_1=\frac{C_1 C_2 V}{C_1+C_2} \times \frac{1}{C_1}=\frac{C_2 V}{C_1+C_2}$

Topic: Physics, Capacitance, JEE Main
Two capacitors $A$ and $B$ are connected in series with a battery as shown in the figure. When the switch $S$ is closed and the two capacitors get charged fully, then

Options:
A The potential difference across the plates of $A$ is $4 \mathrm{~V}$ and across the plates of $B$ is $6 \mathrm{~V}$
B The potential difference across the plates of $A$ is $6 \mathrm{~V}$ and across the plates of $B$ is $4 \mathrm{~V}$
C The ratio of electrical energies stored in $A$ and $B$ is $2: 3$
D The ratio of charges on $A$ and $B$ is $3: 2$

The correct answer is:
The potential difference across the plates of $A$ is $6 \mathrm{~V}$ and across the plates of $B$ is $4 \mathrm{~V}$

Solution:
In series combination of capacitors, voltage distributes on them, in the reverse ratio of their capacitance i.e $\frac{V_A}{V_B}=\frac{3}{2}$ ....[i]
Also $V_A+V_S=10$ ....[ii]
On solving (i) and (ii) $V_A=6 \mathrm{~V}, V_B=4 \mathrm{~V}$

Topic: Physics, Capacitance, JEE Main
Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then:

Options:
A 3C1+5C2=0
B 9C1=4C2
C 5C1=3C2
D 3C1=5C2

The correct answer is:
3C1=5C2

Solution:
For potential to be made zero, after connection
120C1=200C2
⇒3C1=5C2

Topic: Physics, Capacitance, JEE Main, JEE Main 2013 (07 Apr)
Two capacitors C1 and C2 in a circuit are joined as shown in the figure. The potentials of points A and B are V1 and V2 respectively. Then the potential of point D will be

Options:
A V1+V22
B C2V1+C1V2C1+C2
C C1V1+C2V2C1+C2
D C2V1-C1V2C1+C2

The correct answer is:
C1V1+C2V2C1+C2

Solution:
Consider the potential at D be V.
The potential drop across C1 is (V−V1) and C2 is (V2−V)
∴q1=C1(V−V1), q2=C2(V2−V)
As q1=q2 [capacitors are in series]
∴C1(V−V1)=C2(V2−V)
V=C1V1+C2V2C1+C2

Topic: Physics, Electrostatics, JEE Main
Two capacitors each of $1 \mu F$ capacitance are connected in parallel and are then charged by 200 volts $d . c$. supply. The total energy of their charges (in joules) is

Options:
A 0.01
B 0.02
C 0.04
D 0.06

The correct answer is:
0.04

Solution:
$U=\frac{1}{2} C V^2=\frac{1}{2} \times 2 \times(200)^2 \times 10^{-6}=0.04 \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main
Two capacitors each of capacity $2 \mu F$ are connected in parallel. This system is connected in series with a third capacitor of $12 \mu F$ capacity. The equivalent capacity of the system will be

Options:
A $16 \mu F$
B $13 \mu F$
C $4 \mu F$
D $3 \mu F$

The correct answer is:
$3 \mu F$

Solution:
$\Rightarrow C_{A B}=\frac{4 \times 12}{4+12}=3 \mu F$

Topic: Physics, Capacitance, JEE Main
Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is

Options:
A C1C2C1+C2V
B C1+C2C1C2V
C C1+C2V
D C1-C2V

The correct answer is:
C1C2C1+C2V

Solution:
The initial charge on the capacitor C1 will be Q=C1V. If after closing the switch the charge flown through the circuit is q, the charge on C2 will be q and C1 will be Q-q. The potential across the capacitors will be same. Hence,

Q-qC1=qC2⇒q=C2QC1+C2⇒q=C1C2C1+C2V

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

Options:
A 256CV2
B 32CV2
C zero
D 92CV2

The correct answer is:
32CV2

Solution:
Charge on C=CV
Charge on 2C=2C2V

From charge conservation, 2C(2V)-CV=(C+2C)V'
Common potential, V'=V
Final energy, Uf=12CV2+12×2CV2=32CV2

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Two capacitors of capacities $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series and the combination is charged to a potential difference of $120 \mathrm{~V}$. If the charge on the combination is $80 \mu \mathrm{C}$, the energy stored in the capacitor of capacity $C$ in $\mu \mathrm{J}$ is

Options:
A 1800
B 1600
C 14400
D 7200

The correct answer is:
1600

Solution:
Capacitances $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series, then
$C_{\text {eq }}=\frac{C}{1+C}$
Given, $V=120 \mathrm{~V}$ and $q=80 \mu \mathrm{C}$
$\because \quad q=C_{\mathrm{eq}} V$
$80=\frac{C}{C+1} \times 120$
or $\quad C=2 \mu \mathrm{F}$
The energy stored in the capacitor of capacity $C$
$U=\frac{1}{2} \frac{q^2}{C}$
$=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^2}{2 \times 10^{-6}}$
$=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}}$
$U=1600 \mu \mathrm{J}$

Topic: Physics, Capacitance, JEE Main
Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:

Options:
A VK+2
B VK
C 3VK+2
D 3V K

The correct answer is:
3VK+2

Solution:
Now,

VC=2CV+CVKC+2C

=3V K+2

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Two capacitors of capacities $2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ are connected in parallel. A third
capacitor of $6 \mu \mathrm{F}$ capacity is connected in series with this combination. A battery
of $12 \mathrm{~V}$ is connected across this combination. The charge on $2 \mu \mathrm{F}$ capacitor is

Options:
A $12 \mu C$
B $16 \mu C$
C $14 \mu C$
D $11 \mu C$

The correct answer is:
$12 \mu C$

Solution:
Two capacitors of capacities $2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ are connected in parallel. A third capacitor of $6 \mu \mathrm{F}$ capacity is connected in series with this combination. A battery of $12 \mathrm{~V}$ is connected across this combination. The charge on $2 \mu \mathrm{F}$ capacitor is $12 \mu \mathrm{C}$.
Explanation:
The circuit diagram of given situation is as shown below
The equivalent capacitance of $2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ capacitors connected in parallel Is $C_{e q}=2+4=6 \mu \mathrm{F}$
The circuit now becomes
As, both the capacitors are of the same capacitance, so the potential of $12 \mathrm{~V}$ is equally divided in them i.e., $V_{1}=V_{2}=6 V$
In parallel combination, the potential remains the same.
$\therefore$ Charge on $2 \mu \mathrm{F}, \mathrm{Q}=2 \times \mathrm{V}_{1}=2 \times 6=12 \mu \mathrm{C}$
OR
Let $C_{1}=2 \mu F_{1} C_{2}=4 \mu F$ and $C_{3}=6 \mu F$
Equivalent of $C_{1}$ and $C_{2}=C_{4}=6 \mu F$
Equivalent of $C_{3}$ and $C_{4}=C_{5}=3 \mu F$
Charge on $C_{5}=Q=C_{5} V=3 \times 12=36 \mu C$.
$C_{3}$ and $C_{4}$ are in series. Hence charge on them is also $36 \mu C$.
The charge of $36 \mu C$ is divided between $C_{1}$ and $C_{2}$ in proportion to their capacitance.
$\frac{Q_{1}}{Q_{2}}=\frac{C_{1}}{C_{2}}$
$\therefore \frac{Q_{1}}{Q_{1}+Q_{2}}=\frac{C_{1}}{C_{1}+C_{2}}$
$\therefore \frac{Q_{1}}{Q}=\frac{C_{1}}{C_{1}+C_{2}}=\frac{2}{6}=\frac{1}{3}$
$\therefore Q_{1}=\frac{Q}{3}=\frac{36}{3}=12 \mu C$

Topic: Physics, Capacitance, MHT CET, MHT CET 2020 (19 Oct Shift 2)