datavorous/entrance-exam-dataset huggingface dataset:
A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D(D \gt d)$ as shown in figure. If it is burning at the rate of $2 \mathrm{~cm} /$ hour then the top of the candle will

Options:
A Remain at the same height
B Fall at the rate of $1 \mathrm{~cm} / \mathrm{hour}$
C Fall at the rate $o f 2 \mathrm{~cm} / \mathrm{how}$
D Go up the rate of $1 \mathrm{~cm} / \mathrm{hour}$

The correct answer is:
Fall at the rate of $1 \mathrm{~cm} / \mathrm{hour}$

Solution:
The candle floats on the water with half its length above and below water level. Let its length be $10 \mathrm{~cm}$. with $5 \mathrm{~cm}$. below the surface and $5 \mathrm{~cm}$, above it. If its length is reduced to $8 \mathrm{~cm}$, It will have $4 \mathrm{~cm}$. above water surface. So we see tip going down by $\mathrm{I} \mathrm{cm}$. So rate of fall of tip $=1 \mathrm{~cm} / \mathrm{hr}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cannon ball is fired with a velocity $200 \mathrm{~m} / \mathrm{sec}$ at an angle of $60^{\circ}$ with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity $100 \mathrm{~m} / \mathrm{sec}$, the second one falling vertically downwards with a velocity $100 \mathrm{~m} / \mathrm{sec}$. The third fragment will be moving with a velocity

Options:
A $100 \mathrm{~m} / \mathrm{s}$ in the horizontal direction
B $300 \mathrm{~m} / \mathrm{s}$ in the horizontal direction
C $300 \mathrm{~m} / \mathrm{s}$ in a direction making an angle of $60^{\circ}$ with the horizontal
D $200 \mathrm{~m} / \mathrm{s}$ in a direction making an angle of $60^{\circ}$ with the horizontal

The correct answer is:
$300 \mathrm{~m} / \mathrm{s}$ in the horizontal direction

Solution:
Momentum of ball (mass $m$) before explosion at the highest point $=m \hat{i}=m u \cos 60^{\circ} \hat{i}$
$=m \times 200 \times \frac{1}{2} \hat{i}=100 \mathrm{mi}_{\mathrm{i}} \mathrm{kms}^{-1}$
Let the velocity of third part after explosion is
After explosion momentum of system =
$\begin{aligned}
& \dot{j}_2+\bar{P}_3 \\
&=\frac{m}{3} \times 100 \hat{j}-\frac{m}{3} \times 100 \hat{j}+\frac{m}{3} \times \hat{w}
\end{aligned}$
By comparing momentum of system before and after the explosion
$\frac{m}{3} \times 100 \hat{j}-\frac{m}{3} \times 100 \hat{j}+\frac{m}{3} \hat{W}=100 m \hat{V}=300 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A cannon shell fired breaks into two equal parts at its highest point. One part retraces the path to the cannon with kinetic energy $E_1$ and kinetic energy of the second part is $E_2$, relation between $E_1$ and $E_2$ is

Options:
A $E_2=15 E_1$
B $E_2=E_1$
C $E_2=4 E_1$
D $E_2=9 E_1$

The correct answer is:
$E_2=9 E_1$

Solution:
At highest point
$m u \cos \theta=-\frac{m}{2} u \cos \theta+\frac{m}{2} v$
$m u \cos \theta+\frac{m}{2} u \cos \theta=\frac{m}{2} v$
$\frac{2 m u \cos \theta+m u \cos \theta}{2}=\frac{m}{2} v$
$$
\begin{aligned}
m v & =3 m u \cos \theta \\
v & =3 u \cos \theta
\end{aligned}
$$
The kinetic energy
$$
\begin{aligned}
& E_1=\frac{1}{2} \times \frac{m}{2} u^2 \cos ^2 \theta \\
& E_1=\frac{1}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
& E_2=\frac{1}{2} \times \frac{m}{2} \times 9 u^2 \cos ^2 \theta \\
& E_2=\frac{9}{4} m u^2 \cos ^2 \theta
\end{aligned}
$$
The relation between the $E_1$ and $E_2$
$$
\begin{aligned}
& E_2=9 \times E_1 \\
& E_2=9 E_1
\end{aligned}
$$
(from the Eq. (ii))

Topic: Physics, Work Power Energy, JEE Main
A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Options:
A 32
B 2
C 16
D 24

The correct answer is:
32

Solution:
Let us assume that we connect n(whole number) capacitors in series such that the potential difference across the combination is 1000 V. This means the potential difference across each capacitor is

Vcapacitor=1000n V≤300 V

n⩾1000300=3.333.

nmin=4

This means we need to connect at least 4 capacitors in series to make sure that the potential across each one is less than 300 V.

Ceq=mnC=2 μF

m4=2⇒m=8

∴ Minimum no. of capacitors =8×4=32

Topic: Physics, Capacitance, JEE Main
A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Options:
A 32
B 2
C 16
D 24

The correct answer is:
32

Solution:
Let us assume that we connect n(whole number) capacitors in series such that the potential difference across the combination is 1000 V. This means the potential difference across each capacitor is

Vcapacitor=1000n V≤300 V

n⩾1000300=3.333.

nmin=4

This means we need to connect at least 4 capacitors in series to make sure that the potential across each one is less than 300 V.

Ceq=mnC=2 μF

m4=2⇒m=8

∴ Minimum no. of capacitors =8×4=32

Topic: Physics, Capacitance, JEE Main, JEE Main 2017 (02 Apr)
A capacitor $\mathrm{C}_1$ is charged up to a voltage $\mathrm{V}=$ $60 \mathrm{~V}$ by connecting it to battery $\mathrm{B}$ through switch (1), Now $\mathrm{C}_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $\mathrm{C}_2=3.0 \mu \mathrm{F}$ and $\mathrm{C}_3=6.0 \mu \mathrm{F}$ through a switch $(2)$ as shown in the figure. The sum of final charges on $\mathrm{C}_2$ and $\mathrm{C}_3$ is:

Options:
A
$36 \mu \mathrm{C}$

B
$20 \mu \mathrm{C}$

C
$54 \mu \mathrm{C}$

D
$40 \mu \mathrm{C}$

The correct answer is:
$36 \mu \mathrm{C}$

Solution:
The sum of final charges on $\mathrm{C}_2$ and $\mathrm{C}_3$ is 36 $\mu \mathrm{C}$.

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A capacitor C1 of capacitance 5 μF is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2 of capacitance 10 μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C2 is _____ μC.

Options:

The correct answer is:
100

Solution:
The initial charge on the 5 μF capacitor will be q=CV=150 μC. If the charge on the 10 μF capacitor is q after the key is closed then,

150-q5=q10⇒q=100 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C2. The energy loss in the process after the charge is distributed between the two capacitors is :

Options:
A 12CV02
B 13CV02
C 14CV02
D 16CV02

The correct answer is:
16CV02

Solution:
Heat Loss is given by,

H=C1C22C1+C2V1-V22

V1 is initial voltage of capacitor C=V0

V2 is the initial voltage of capacitor C2=0

=c×c22c+c2V0-02=C6V02

H=16C0V02

Topic: Physics, Capacitance, JEE Main
A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C2. The energy loss in the process after the charge is distributed between the two capacitors is :

Options:
A 12CV02
B 13CV02
C 14CV02
D 16CV02

The correct answer is:
16CV02

Solution:
Heat Loss is given by,

H=C1C22C1+C2V1-V22

V1 is initial voltage of capacitor C=V0

V2 is the initial voltage of capacitor C2=0

=c×c22c+c2V0-02=C6V02

H=16C0V02

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$ )

Options:
A $1$
B $1.33$
C $0.66$
D $1.50$

The correct answer is:
$1.50$

Solution:
$\begin{aligned} & \frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}\right)} \\ & 0.6=0.2+\frac{0.6}{\mathrm{k}} \\ & \mathrm{k}=\frac{3}{2}\end{aligned}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A capacitor has capacitance 5 µF when its parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness d2 is inserted between the plates. Capacitance of the capacitor in the presence of slab will be ______ µF.

Options:

The correct answer is:
6

Solution:
Capacitance of the parallel plate capacitor when air is present C=ϵ0Ad=5 μF

After insertion of dielectric, the equivalent capacitance of the new capacitor is

Ceq=ϵ0 Ad21.5+d21

=ε0 Ad3+d2=6ε0 A5d

=65×5 μF=6 μF.

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Options:

None

Solution:
The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant $\mathrm{K}$ is
$$
\mathrm{C}=\frac{\mathrm{K} \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
$$
Where ' $\mathrm{K}$ ' is positive and more then one
So, by removing dielectric medium from capacitor keeping A and d constant. So, capacitance of capacitor will decrease, $\mathrm{K}=1$.
When battery and dielectric slab removed from capacitor, then charge stored will remain the same as due to conservation of charge.
The energy stored in charge capacitor $\mathrm{V}=\frac{\mathrm{q}^2}{2 \mathrm{C}}$; where, $\mathrm{q}$ is constant, energy stored $\mathrm{V} \propto 1 / \mathrm{C}$ and $\mathrm{C}$ decreases with the removal of dielectric medium, therefore energy stored in capacitor will increases. Since $q$ is constant and $\mathrm{V}=\mathrm{q}$ / $\mathrm{C}$ and $\mathrm{C}$ decreases which in turn potential $(\mathrm{V})$ will increase and therefore $\mathrm{E}$ increases as $\mathrm{E}=\mathrm{V} / \mathrm{d}$. The distance between plote of capacitor is same and potential is increase. So electric field between the plates of capacitor will increase.

Topic: Physics, Electrostatic Potential and Capacitance
A capacitor is charged and then made to discharge through a resistance. The time constant is \( \tau \). In what time will the potential difference across the capacitor decrease by \( 10 \% ? \)

Options:
A \( \tau \ln 0.1 \)
B \( \tau \ln 0.9 \)
C \( \tau \ln \frac{10}{9} \)
D \( \tau \ln \frac{11}{10} \)

The correct answer is:
\( \tau \ln \frac{10}{9} \)

Solution:
As we know the discharging of capacitor is given by q=q0 e-tτ
Also
V= V o e − t τ
Here
V= 9 10 V o
putting the values
9 10 V o = V o e − t τ
ℓ n ( 9 10 ) = − t τ
ℓ n ( 10 9 ) = t τ
t=τ  ℓ n ( 10 9 )

Topic: Physics, Capacitance, JEE Main
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

Options:
A increases by a factor of 4.
B decreases by a factor of 2.
C remains the same.
D increases by a factor of 2.

The correct answer is:
decreases by a factor of 2.

Solution:
Charge on capacitor,
q=CV
When it is connected with another uncharged capacitor.
Vc=q1+q2C1+C2=q+0C+C
Vc=V2
Initial energy,
Ui=12CV2
Final energy,
Uf=12CV22+12CV22
=CV24
Loss of energy =Ui-Uf
=CV24,
i.e., decreases by a factor of 2.

Topic: Physics, Capacitance, NEET, NEET 2017
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1t2 will be

Options:
A 12
B 13
C 14
D 16

The correct answer is:
16

Solution:
We know that in the discharging circuit the charge is given by,

q=q0e-tRC ⇒tRC=lnq0q

In time t1 the energy stored reduces to half. Hence, q122C=12×q022C⇒q0q1=2. Therefore,

t1RC=ln2=12ln2.

In time t2 the charge stored reduces to 18th of initial value. Hence, q2q0=18. Therefore,

t2RC=ln8=3ln2.

Hence,

t1t2=12ln23ln2=16

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A capacitor is fully charged with a battery and then disconnected. A dielectric is then inserted into the capacitor. How do charges on surface of the dielectric and outer surface of the plates of the capacitor would change respectively?

Options:
A Increases, decreases
B Decreases, increases
C Increases, remain unchanged
D Remains unchanged, increases

The correct answer is:
Increases, remain unchanged

Solution:
As capacitor is disconnected from battery. Its charges at the plates will get stored in the form of electrostatic energy (Case I).
When a dielectric is introduced inside this capacitor, it gets polarised as shown in (Case II). Due to this, charges at surface of dielectric gets increased.
Since, there is no external electric supply of power source.
Hence, charges on capacitor remains unchanged.

Topic: Physics, Capacitance, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
A capacitor is half filled with dielectric \( (\mathrm{K}=2) \) as shown in diagram (1). If the same dielectric is to be filled in the same
capacitor as shown in diagram (2), then what would be the thickness of dielectric so that capacitor still has same
capacitance ?

Options:
A \( \frac{2 \mathrm{~d}}{3} \)
B \( \frac{3 \mathrm{~d}}{2} \)
C \( \frac{3 \mathrm{~d}}{4} \)
D \( \frac{4 \mathrm{~d}}{3} \)

The correct answer is:
\( \frac{2 \mathrm{~d}}{3} \)

Solution:
C′=∈0A2d C"=C0K2
∴C1=C′+C′′
∴C1=C02+C0K2=C02 1+K
(Equivalent capacitance of 1st diagram)
C2=∈0Ad – t+tK
(Equivalent capacitance of 2nd diagram)
Both are same hence :
C02 K+1=∈0Ad – t+tK
∈0A2d 2+1=∈0Ad – t+t2
3d – 3t2=2d
d=3t2 or t=2d3

Topic: Physics, Capacitance, JEE Main
A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is $\frac{A}{3}$ and the height is $d$, the capacitance of the arrangement is :

Options:
A $\frac{13 \epsilon_o A}{17 d}$
B $\frac{11 \epsilon_{\mathrm{o}} \mathrm{A}}{18 \mathrm{~d}}$
C $\frac{18 \epsilon_{\mathrm{o}} \mathrm{A}}{11 \mathrm{~d}}$
D $\frac{11 \epsilon_{\mathrm{o}} \mathrm{A}}{20 \mathrm{~d}}$

The correct answer is:
$\frac{11 \epsilon_{\mathrm{o}} \mathrm{A}}{18 \mathrm{~d}}$

Solution:
All capacitor are in parallel combination.
Also effective area is common area only
$\begin{aligned} & \Rightarrow C_{\text {eq }}=C_1+C_2+C_3 \\ & \Rightarrow C_{\text {eq }}=\frac{A \varepsilon_0}{3 d}+\frac{A \varepsilon_0}{3(2 d)}+\frac{A \varepsilon_0}{3(3 d)} \\ & \Rightarrow C_{\text {eq }}=\frac{A \varepsilon_0}{3}\left(\frac{11}{6 d}\right) \\ & \Rightarrow C_{\text {eq }}=\frac{11 A \varepsilon_0}{18 d}\end{aligned}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A capacitor of $10 \mu \mathrm{F}$ capacitance whose plates are separated by $10 \mathrm{~mm}$ through air and each plate has area $4 \mathrm{~cm}^2$ is now filled equally with two dielectric media of $\mathrm{K}_1=2, \mathrm{~K}_2=3$ respectively as shown in figure. If new force between the plates is $8 \mathrm{~N}$. The supply voltage is _________ $\times 10^{-4} \mathrm{~V}$.
(we modified language of question to make it correct)

Options:

The correct answer is:
93

Solution:
$\begin{aligned}
& \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \\
& \mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F} \\
& \mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F} \\
& \mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F}
\end{aligned}$
Now the charge on
$\begin{aligned}
& \mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c} \\
& \mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C} .
\end{aligned}$
Now force between the plates $\left[F=\frac{Q^2}{2 A \in_0}\right]$
$\begin{aligned}
& \frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8 \\
& 325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85 \\
& \mathrm{~V}^2=\frac{32 \times 8.85 \times 10^{-4}}{325} \\
& \therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}} \\
& \mathrm{~V}=0.93 \times 10^{-2}
\end{aligned}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A capacitor of 50μF is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is _________ μC.

Options:

The correct answer is:
100

Solution:
At steady state current through branch of capacitor is zero. Potential difference across capacitor is 2 V. So, q=50×2=100μC
i=6(2+2+2)×103=1×10-3Amp
VA,B= potential across 50μF
VA,B=i×2×103=1×10-3×2×103=2 volt
Charge on 50μF
q=CV=50×10-6×2⇒100×10-6C=100μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A capacitor of capacitance $5 \mu F$ is connected as shown in the figure. The internal resistance of the cell is
$0.5 \Omega$. The amount of charge on the capacitor plate is

Options:
A $0 \mu C$
B $5 \mu C$
C $10 \mu C$
D $25 \mu C$

The correct answer is:
$10 \mu C$

Solution:
Potential difference a cross line $(2)=$ potential difference a cross capacitor
$=1 \times 2=2 \mathrm{Vth}$
So, charge on capacitor $=5 \times 2=10 \mu \mathrm{C}$
In steady state condition. No current flows through line (1). Hence total current
$i=\frac{2.5}{(1+1+0.5)}=1 \mathrm{~A}$

Topic: Physics, Capacitance, JEE Main
A capacitor of capacitance 10 Î¼F is connected to an AC ammeter. If the source voltage varies as V=502sinâ¡100t, the reading of the ammeter is

Options:
A 50 mA
B 70.7 mA
C 5.0 mA
D 7.07 mA

The correct answer is:
50 mA

Solution:
C=10 μF=10×10-6 F
V=502sin⁡100t
Reading of ammeter I=VrmsXC
XC= impendence of capacitor
XC=1ωC
I=Vrms1ωC
Vrms=V2=5022=50 V
Reading of ammeter I
=50×100×10×10-6A
=50×10-3 A⇒50 mA

Topic: Physics, Alternating Current, JEE Main
A capacitor of capacitance 50pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is____ nJ.

Options:

The correct answer is:
125

Solution:
Initial charge on first capacitor will be, Q=CV

As the second capacitor is identical to the first one, hence potential drop across both capacitor will be equal to V2.

Now, loss of energy, ΔH=Ui-Uf

⇒ΔH=12CV2-122C×V22

⇒ΔH=12CV2-14CV2 ⇒ΔH=14CV2=14×50×10-12×1002=125×10-9 J=125 nJ

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be :

Options:
A 3.2 A
B 1.5 A
C 2.0 A
D 1.2 A

The correct answer is:
1.5 A

Solution:
By energy conservation, it follows that

12CV2=12LImax2 ...1

Equation (1) implies that

Imax=CLV=100×10-66.4×10-3×12=128=1.5 A

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A capacitor of capacitance 150.0 μF is connected to an alternating source of emf given by E=36 sin120πt V. The maximum value of current in the circuit is approximately equal to:

Options:
A 2 A
B 2 A
C 22 A
D 12 A

The correct answer is:
2 A

Solution:
The formula for maximum current is given by

Imax=ωCVm ...(i)

The given data is

ω=120π

Vm=36 VC=150 μF

Substituting the values in equation (i)

Imax=120π×150×10-6×36

=2.036 A

≈2 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A capacitor of capacitance 500 μF is charged completely using a de supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _____ A.

Options:

The correct answer is:
10

Solution:
Energy stored in capacitor

E=12CV2=12500×10-6×104

⇒E=52J

Current will be maximum when whole energy of capacitor becomes energy of inductor.

⇒12LI2=52

⇒I=5L=550×10-3=10 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A capacitor of capacitance 900 µF is charged by a 100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x×10-2 J. The value of x is ______.

Options:

The correct answer is:
225

Solution:
When a charged capacitor is connected to an uncharged capacitor, then there is loss of energy in the form of heat as charges flow from higher potential to lower potential.

So, heat ∆H=12C1C2C1+C2 V1-V22

As the other capacitor is identical therefore charge is equally divided.

=12C22C(100-0)2

=12900×10-62×104=94 Joule

=2.25 Joule.

Hence, the value of x=225.

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

Options:
A $\frac{C_{0}\left(V_{0}-h\right)}{V_{0}}$
B $\frac{C_{0}\left(V-V_{0}\right)}{V_{0}}$
C $\frac{C_{0}\left(V+V_{0}\right)}{V}$
D $\frac{C_{0}\left(V_{0}-V\right)}{V}$

The correct answer is:
$\frac{C_{0}\left(V_{0}-V\right)}{V}$

Solution:
When the switch $S$ is closed, the common potential across the two capacitors becomes $V$. so charge on isolated plates remains same.
$$
\begin{array}{lr}
\therefore \quad C_{0} V_{0}=\left(C_{0}+C\right)V \\
\Rightarrow \quad C V=C_{0} V_{0}-C_{0} V \\
\Rightarrow \quad C=\frac{C_{0}\left(V_{0}-V\right)}{V}
\end{array}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2013
A capacitor of capacitance $C_1=10 \mu \mathrm{F}$ to charged using $9 \mathrm{~V}$ battery. It is then removed from the battery and connected to another capacitor $C_2=20 \mu \mathrm{F}$ as shown in the figure.
The charge on $C_2$ after equilibrium has reached is

Options:
A $6.0 \times 10^{-5} \mathrm{C}$
B $60 \times 10^{-6} \mathrm{C}$
C $3.0 \times 10^{-5} \mathrm{C}$
D $3.0 \times 10^{-6} \mathrm{C}$

The correct answer is:
$6.0 \times 10^{-5} \mathrm{C}$

Solution:
Given,
$$
\begin{aligned}
C_1 & =10 \mu \mathrm{F} \\
& =10^{-5} \mathrm{~F} \\
V_1 & =9 \mathrm{~V}
\end{aligned}
$$
Charge on capacitor $C_1$,
$$
\begin{aligned}
q_1 & =C_1 V_1 \\
& =9 \times 10^{-5} \mathrm{C} \\
C_2 & =20 \mu \mathrm{F}=2 \times 10^{-5} \mathrm{~F}
\end{aligned}
$$
when uncharged capacitor $C_2$ is connected with charged capacitor $C_1$, then charges will flow from capacitor $C_1$ to $C_2$ till both gain the same potential i.e common potential.
$$
\begin{aligned}
& \text { Common potential, } V=\frac{\text { total charge }}{\text { total capacitance }} \\
& =\frac{q_1}{C_1+C_2}=\frac{9 \times 10^{-5}}{10^{-5}+2 \times 10^{-5}} \\
& =\frac{9 \times 10^{-5}}{3 \times 10^{-5}}=3 \mathrm{~V}
\end{aligned}
$$
Charge on capacitor $C_2, q_2=C_2 V$
$$
=2 \times 10^{-5} \times 3=6 \times 10^{-5} \mathrm{C}
$$

Topic: Physics, Capacitance, JEE Main
A capacitor of capacitance $C_1$ charged up to $V$ volt and then connected to an uncharged capacitor $C_2$. Then, the final potential difference across each will be:

Options:
A $\frac{C_2 V}{C_1+C_2}$
B $\frac{C_1 V}{C_1+C_2}$
C $\left(1+\frac{C_2}{C_1}\right) V$
D $\left(1-\frac{C_2}{C_1}\right) V$

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

Solution:
$$
\begin{aligned}
& V_{\text {comma }}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\left(\because V_2=0\right) \\
& \Rightarrow V_{\text {comma }}=\frac{C_1 V_1}{C_1+C_2}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main