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Physics M.C.Q (1 Marks)

Electrostatics · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 69 questions.

Questions · Page 2 of 2

M.C.Q (1 Marks)

MCQ 511 Mark
A parallel plate air filled capacitor of capacitance ' $C$ ' has plate area ' $A$ ' and the distance between the plates ' $d$. When a metal sheet of thickness $\left(\frac{d}{2}\right)$ and of the same area ' $A$ ' is introduced between the plates, its capacitance becomes ' $C_2$ ' The ratio $C_2: C_1$ is
Answer
Correct option: A.
$2: 1$
(a) : As $C_1=\frac{\varepsilon_0 A}{d}$. . . . . .(i)
Here $A$ area of plate $d$ is distance between the plates. For dielectric medium and thickness $t$, capacitance is, $C_2=\frac{\varepsilon_0 \cdot A}{d-t+\frac{t}{k}}$
$\begin{array}{ll}C_2=\frac{\varepsilon_0 A}{d-\frac{d}{2}+0}=\frac{2 \varepsilon_0 A}{d} & (k=\infty \text { for metal) } \\ C_2=2 C_1  .  . . . . .(From equation (i))& \\ \frac{C_2}{C_1}=\frac{2}{1} & \end{array}$
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MCQ 521 Mark
The network of 4 capacitors is connected to a battery as shown. The ratio of charges on capacitors ' $C_2$ ' and ' $C_4$ ' is
Image
Answer
Correct option: A.
$\frac{3}{22}$
(a):
Image
As capacitor $C_1, C_2$ and $C_3$ are in series.
So equivalent capacitance is given by
$
\begin{aligned}
& \frac{1}{C_{\text {eqf }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{C_2 C_3+C_3 C_1+C_2 C_1}{C_1 C_2 C_3} \\
& \Rightarrow C_{\text {el }}=\frac{C_1 C_2 C_3}{C_1 C_2+C_2 C_3+C_1 C_3}
\end{aligned}
$
Give, $\left(C_1=1 C_1 C_2=2 C_1 C_3=3 C\right)$
$
\Rightarrow C_{\text {eq }}=\frac{C(2 C)(3 C)}{C(2 C)+(2 C)(3 C)+C(3 C)} \Rightarrow C_{\text {eq }}=\frac{6}{11}
$
Charge on capacitors $C_1, C_2$ and $C_3=C_{\text {eq }} V=\frac{6 C}{11} V$
Charge on capacitor $C _4=4 CV$
Now, $\frac{\text { Charge on Capacitor } C_2}{\text { Charge on } C_4}=\frac{6 C}{\frac{11}{4 C V}} V=\frac{3}{22}$
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MCQ 531 Mark
Two charges of equal magnitude $q$ are placed in air at a distance $2 a$ apart and third charge $-2 q$ is placed at midpoint. The potential energy of the system is ( $\epsilon_0=$ permittivity of free sapce)
Answer
Correct option: D.
$-\frac{7 q^2}{8 \pi \epsilon_0 a}$
(d) : The potential energy of the system is
$
U=\frac{1}{4 \pi \epsilon_0} \frac{(q)(-2 q)}{a}+\frac{1}{4 \pi \epsilon_0} \frac{(q)(q)}{2 a}+\frac{1}{4 \pi \epsilon_0} \frac{(q)(-2 q)}{a}
$
Image
$\begin{aligned} & =\frac{1}{4 \pi \epsilon_0}\left[\frac{-2 q^2}{a}+\frac{q^2}{2 a}+\frac{-2 q^2}{a}\right] \\ & =\frac{1}{4 \pi \epsilon_0}\left[\frac{-4 q^2+q^2-4 q^2}{2 a}\right]=-\frac{7 q^2}{8 \pi \epsilon_0 a}\end{aligned}$
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MCQ 541 Mark
A charged cylinder of radius $3 mm$ has surface charge density $4 \mu C / m ^2$. It is placed in a medium of dielectric constant 628 . The electric intensity at a point at a distance of $1.5 m$ from its axis is
Answer
Correct option: A.
$1.44 V / m$
(a) : The electric field intensity at a point outside a charged cylinder is
$
E=\frac{\sigma R}{\varepsilon_0 K r}
$
where, $\sigma=$ surface charge density,
$R=$ radius of the cylinder,
$K=$ dielectric constant,
$r=$ distance of point from axis of the cylinder
Here,
$\sigma=4 \mu C / m ^2=4 \times 10^{-6} C / m ^2$
$
\begin{aligned}
& R=3 mm =3 \times 10^{-3} m \\
& K=628, r=1.5 m
\end{aligned}
$
$
\therefore E=\frac{\left(4 \times 10^{-6} C ^{-} m ^2\right)\left(3 \times 10^{-3} m \right)}{\left(8.85 \times 10^{-12} C ^2 / N m ^2\right)(628)(1.5 m )}=1.44 V / m
$.
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MCQ 551 Mark
In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is
Answer
Correct option: A.
$32 \mu F$
(a) : When a dielectric medium is filled between plates of a capacitor, then new capacitance
$
\begin{aligned}
& C^{\prime}=\frac{K \varepsilon_0 A}{d^{\prime}}=\frac{4 K \varepsilon_0 A}{d} \\
& {\left[\because d^{\prime}=\frac{d}{4}\right]} \\
& =4 \times K \times C \\
& =(4 \times 2 \times 4) \mu F =32 \mu F \\
& {\left[\because C=\frac{\varepsilon_0 A}{d}=4 \mu F \right]} \\
&
\end{aligned}
$
$
\left[\because C=\frac{\varepsilon_0 A}{d}=4 \mu F \right]
$
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MCQ 561 Mark
A series combination of $N_1$ capacitors (each of capacity $C_1$ ) is charged to potential difference ' $3 V$. Another parallel combination of $N _2$ capacitors (each of capacity $C_2$ ) is charged to potential difference ' $V$ '. The total energy stored in both combinations is same. The value of $C_1$ in terms of $C_2$ is
Answer
Correct option: A.
$\frac{C_2 N_1 N_2}{9}$
(a) : For series combination of $N_1$ capacitors,
$
C_{\text {eq }}=\frac{C_1}{N_1} ; V_1=3 V
$
Energy stored is, $E_1=\frac{1}{2} C_{e q} V_1^2=\frac{1}{2} \frac{C_1}{N_1} 9 V^2=\frac{9}{2} \frac{C_1}{N_1} V^2$
For parallel combination of $N_2$ capacitors,
$
C_{\text {eq }}=N_2 C_2 ; V_2=V ; E_2=\frac{1}{2} C_{e q} V^2=\frac{1}{2} C_2 N_2 V^2
$
Since energy stored in both the combination is equal $E_1=E_2$
$
\frac{9}{2} \frac{C_1}{N_1} V^2=\frac{C_2 N_2 V^2}{2} \Rightarrow C_1=C_2 \frac{N_2 N_1}{9}
$
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MCQ 581 Mark
Two parallel plate air capacitors of same capacity $C$ are connected in series to a battery of emf $E$. Then one of the capacitors is completely filled with dielectric material of constant $K$. The change in the effective capacity of the series combination is
Answer
Correct option: A.
$\frac{C}{2}\left[\frac{K-1}{K+1}\right]$
(a): When the two parallel capacitors are connected in series then effective capacitance
$
\frac{1}{C_1}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C} \text { or } C_1=\frac{C}{2}
$
When one of the capacitor is filled with dielectric material of constant $K$, then effective capacitance
$
\frac{1}{C_2}=\frac{1}{C}+\frac{1}{K C}=\frac{1}{C}\left(1+\frac{1}{K}\right) ; \frac{1}{C_2}=\frac{K+1}{K C} \text { or } C_2=\frac{K C}{K+1}
$
Then change in effective capacitance is
$
\Delta C=C_2-C_1=\frac{K C}{K+1}-\frac{C}{2}=\frac{C}{2}\left(\frac{K-1}{K+1}\right)
$
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MCQ 591 Mark
A parallel plate air capacitor has capacity $C$ farad, potential $V$ volt and energy $E$ joule. When the gap between the plates is completely filled with dielectric
Answer
Correct option: B.
Both $V$ and $E$ decrease
(b): When the gap between the plates is completely filled with dielectric of dielectric constant $K$, then potential is
$
V=\frac{Q d}{A \varepsilon_0 K}...(i)
$
and electric field is
$
E=\frac{Q}{A \varepsilon_0 K}....(ii)
$
From equations (i) and (ii), both electric field and potential decrease.
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MCQ 601 Mark
When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is $3.75 \mu F$. The capacity of each capacitor is
Answer
Correct option: A.
$5 \mu F$
(a): The resultant capacitance
Image
$
C_{\text {resultant }}=\frac{3 C \times C}{3 C+C}=\frac{3 C^2}{4 C}=\frac{3}{4} C
$
Given : $C_{\text {resultann }}=3.75 \mu F$
$
C=\frac{3.75 \times 4}{3}=5 \mu F
$

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MCQ 611 Mark
The amount of work done in increasing the voltage across the plates of capacitor from $5 V$ to $10 V$ is $W$. The work done in increasing it from $10 V$ to $15 V$ will be
Answer
Correct option: D.
$1.67 W$
(d) : The amount of work done in increasing the voltage across the plates of capacitor is stored in the form of electrostatic energy.
$
\begin{aligned}
\therefore \quad W & =U_2-U_1 \\
& =\frac{1}{2} C V_2^2-\frac{1}{2} C V_1^2=\frac{1}{2} C\left(10^2-5^2\right)=\frac{75}{2} C
\end{aligned}
$. . . . . .(i)
Similarly, work done in increasing the voltage from $10 V$ to $15 V$ is
$
W^{\prime}=\frac{1}{2} C\left(15^2-10^2\right)=\frac{125}{2} C
$. . . . . .(ii)
From eqns. (i) and (ii)
$
\frac{W^{\prime}}{W}=\frac{125}{75} \text { or } W^{\prime}=1.67 W
$
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MCQ 621 Mark
Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitor is completely filled with dielectric material of constant $K$, then potential difference of the other capacitor will become
Answer
(b) : Let the capacitance of air filled parallel plate capacitor be $C$. Then capacitance of dielectric filled capacitor is $KC$.

$\therefore$ Equivalent capacitance, $C _{\text {eq }}=\frac{C \times K C}{C+K C}=\frac{K C}{(K+1)}$
Charge in the circuit, $q=C_{\text {eq }} V=\frac{K C V}{(K+1)}$
$\therefore$ Potential difference of air filled capacitor is
$
V^{\prime}=\frac{q}{C}=\frac{K V}{(K+1)}
$
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MCQ 631 Mark
Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{A}{3}$ and the separation between the plates is $d, 2 d$ and $3 d$ respectively. The equivalent capacity of combination is ( $\epsilon_0=$ absolute permittivity of free space)
Answer
Correct option: B.
$\frac{11 \epsilon_0 A}{18 d}$
(b) :$
\begin{gathered}
C_p=C_1+C_2+C_3=\frac{\epsilon_0 A}{3 d}+\frac{\epsilon_0 A}{3(2 d)}+\frac{\epsilon_0 A}{3(3 d)} \\
=\frac{\epsilon_0 A}{3 d}+\frac{\epsilon_0 A}{6 d}+\frac{\epsilon_0 A}{9 d}=\frac{11 \epsilon_0 A}{18 d}
\end{gathered}
$
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MCQ 641 Mark
A capacitor $C_1=4 \mu F$ is connected in series with another capacitor $C_2=1 \mu F$. The combination is connected across $d.c.$ source of $200 V$. The ratio of potential across $C_2$ to $C_1$ is
  • A
    $2: 1$
  • ✓
    $4: 1$
  • C
    $8: 1$
  • D
    $16: 1$
Answer
Correct option: B.
$4: 1$
Given situation is shown in the figure.Here, $C_1=4 \mu F$$C_2=1 \mu F$
$V=200 V , \frac{V_2}{V_1}=?$Net capacitance of the circuit,$C=\frac{C_1 C_2}{C_1+C_2} ; $
$C=\frac{4 \times 1}{4+1}=\frac{4}{5} \mu F$
Charge flows through circuit,$q=C V=\left(\frac{4}{5} \mu F \right) \times(200 V )$
$=160 \mu C =160 \times 10^{-6} C
$Potential across $C_1, V_1=\frac{q}{C_1}=\frac{160 \times 10^{-6}}{4 \times 10^{-6}}=40 V$
Potential across $C_2, V_2=V-V_1=200-40=160 V$
$\therefore \frac{V_2}{V_1}=\frac{160}{40}=\frac{4}{1}$
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MCQ 651 Mark
The electric field intensity at a point near and outside the surface of a charged conductor of any shape is ' $E_1$. The electric field intensity due to uniformly charged infinite thin plane sheet is ' $E_2$ ' The relation between ' $E_1$ ' and ' $E_2$ ' is
Answer
Correct option: C.
$E_1=2 E_2$
(c) : Electric field near surface of charged conductor
$
E_1=\frac{\sigma}{\varepsilon_0}
$. . . . . .(i)
Electric field due to uniformly charged infinite thin plane sheet,
$
\begin{aligned}
& E_2=\frac{\sigma}{2 \varepsilon_0}=\frac{E_1}{2} \quad \text { [Using eqn. (i)] } \\
& \therefore \quad E_1=2 E_2
\end{aligned}
$
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MCQ 661 Mark
The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is $6 \mu F$. The capacity of each capacitor is
Answer
Correct option: B.
$4 \mu F$
(b) : Suppose capacity of each capacitor is $C$.
When two capacitors are connected in parallel their equivalent capacity is given by,
$
C_1=C+C=2 C
$
When two capacitors are connected in series their equivalent capacity is given by
$
C_2=\frac{C \times C}{C+C}=\frac{C}{2}
$
According to question, $C_1-C_2=6 \mu F$
$
\begin{aligned}
& 2 C-\frac{C}{2}=6 \mu F ; \quad \frac{3 C}{2}=6 \mu F \\
& \therefore \quad C=4 \mu F
\end{aligned}
$
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MCQ 671 Mark
An electron in potentiometer wire experiences a force $2.4 \times 10^{-19} N$. The length of potentiometer wire is $6 m$. The e.m.f. of the battery connected across the wire is (electronic charge $=1.6 \times 10^{-19} C$ )
Answer
Correct option: B.
$9 V$
(b) : Force on an electron, $F=e E$
$\therefore \quad$ Electric field,
$
E=\frac{F}{e}=\frac{2.4 \times 10^{-19} N }{1.6 \times 10^{-19} C }=\frac{3}{2} N C ^{-1}
$
Potential difference across the wire is
$
\begin{aligned}
& V=E l=\frac{3}{2} N C ^{-1} \times 6 m =9 V \\
& \therefore \quad \text { Emf of the battery }=V=9 V
\end{aligned}
$
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MCQ 681 Mark
In air, a charged soap bubble of radius $r$ is in equilibrium having outside and inside pressures being equal. The charge on the drop is ( $\epsilon_0=$ permittivity of free space, $T$ = surface tension of soap solution)
Answer
Correct option: D.
$4 \pi r^2 \sqrt{\frac{8 T \epsilon_0}{r}}$
(d) : Excess pressure inside the soap bubble is
$
P=\frac{4 T}{r}
$
As the pressure inside and outside is the same, so pressure of the charged bubble is
$
P_{\text {electro }}=\frac{4 T}{r} ; \frac{\sigma^2}{2 \epsilon_0}=\frac{4 T}{r}
$
where $\sigma$ is the surface charge density of the bubble
$
\sigma=\sqrt{\frac{8 T \epsilon_0}{r}}
$
$\therefore \quad$ Charge on the bubble is
$
Q=4 \pi r^2 \sigma=4 \pi r^2 \sqrt{\frac{8 T \epsilon_0}{r}}
$
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MCQ 691 Mark
Two concentric spheres kept in air have radii $R$ and $r$. They have similar charge and equal surface charge density $\sigma$. The electric potential at their common centre is ( $\epsilon_0=$ permittivity of free space)
Answer
(a) :
Image
As both the spheres have equal surface charge density $\sigma$, therefore charges on the spheres are $Q_1=\sigma 4 \pi r^2$ and $\quad Q_2=\sigma 4 \pi R^2$
Electric potential at their common centre $O$ is
$
\begin{aligned}
V & =\frac{1}{4 \pi \epsilon_0} \frac{Q_1}{r}+\frac{1}{4 \pi \epsilon_0} \frac{Q_2}{R} \\
& =\frac{1}{4 \pi \epsilon_0}\left[\frac{\sigma 4 \pi r^2}{r}+\frac{\sigma 4 \pi R^2}{R}\right]=\frac{4 \pi \sigma}{4 \pi \epsilon_0}[r+R]=\frac{\sigma(R+r)}{\epsilon_0}
\end{aligned}
$
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