M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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19 questions · 6 auto-graded MCQ + 13 self-marked written.

MCQ 11 Mark

Two metal spheres of capacitances $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy:

A
$\frac{\text{Q}_1}{\text{Q}_2}<\frac{\text{C}_1}{\text{C}_2}$
✓
$\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_1}{\text{C}_2}$
C
$\frac{\text{Q}_1}{\text{Q}_2}>\frac{\text{C}_1}{\text{C}_2}$
D
$\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_2}{\text{C}_1}$

Answer

Correct option: B.

$\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_1}{\text{C}_2}$

When the spheres are connected, charges flow between them until they both acquire the same common potential $V.$
The final charges on the spheres are given by:
$Q_1= C_{1}V$ and $Q_2 = C_2V$
$\therefore\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_1\text{V}}{\text{C}_2\text{V}}=\frac{\text{C}_1}{\text{C}_2}$

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MCQ 21 Mark

A parallel$-$plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let $Q_+$ and $Q_-$ be the charges appearing on the positive and negative plates respectively:

A
$Q_+ > Q_-$
✓
$Q_{+ }= Q_-$
C
$Q_+ < Q_-$
D
The information is not sufficient to decide the relation between $Q_+$ and $Q_-.$

Answer

Correct option: B.

$Q_{+ }= Q_-$

The charge induced on the plates of a capacitor is independent of the area of the plates.
$\therefore Q_{+ }= Q_-$

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MCQ 31 Mark

The equivalent capacitance of the combination shown in figure is:

A
$C$
✓
$2C$
C
$\frac{\text{C}}{2}$
D
None of these.

Answer

Correct option: B.

$2C$

Since the potential at point $A$ is equal to the potential at point $B$, no current will flow along arm $AB.$
Hence, the capacitor on arm $AB$ will not contribute to the circuit.
Also, because the remaining two capacitors are connected in parallel.
the net capacitance of the circuit is given by
$C_{eq} = C + C = 2C$

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MCQ 41 Mark

Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors:

A
$C_1 > C_2$
B
$C_1 = C_2$
✓
$C_1 < C_2$
D
The information is not sufficient to decide the relation between $C_1$ and $C_2.$

Answer

Correct option: C.

$C_1 < C_2$

Region $AB$ shows the potential difference across capacitor $C_{1}$ and region $CD$ shows the potential difference across capacitor $C_2.$
Now, we can see from the graph that region $AB$ is greater than region $CD.$
Therefore, the potential difference across capacitor $C_1$ is greater than that across capacitor $C_2.$
$\because$ Capacitance, $\text{C}=\frac{\text{Q}}{\text{V}}$
$\therefore\text{C}_1<\text{C}_2(Q$ remains the same in series connection$).$

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Question 51 Mark

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

The electric field in the capacitor.
The charge on the capacitor.
The potential difference between the plates.
The stored energy in the capacitor.

Answer

The charge on the capacitor.

Explanation:
When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.
Thus, the net effect is a reduced electric field.
Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by
$\text{U}=\frac{\text{qV}}{2}$
Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.

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MCQ 61 Mark

If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be:

A
$2C$ and $2V$
B
$C$ and $2V$
✓
$2C$ and $V$
D
$C$ and $V$

Answer

Correct option: C.

$2C$ and $V$

In a parallel combination of capacitors, the potential difference across the capacitors remain the same,
as the right$-$hand$-$side plates and the left$-$hand$-$side plates of both the capacitors are connected to the same terminals of the battery.
Therefore, the potential remains the same, that is, $V.$
For the parallel combination of capacitors, the capacitance is given by
$C_{eq} = C_1 + C_2$
Here,
$C_1 = C_2 = C$
$\therefore C_{eq} = 2C$

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Question 71 Mark

Following operations can be performed on a capacitor:
$X -$ connect the capacitor to a battery of emf $\epsilon.$
$Y -$ disconnect the battery.
$Z -$ reconnect the battery with polarity reversed.
$W -$ insert a dielectric slab in the capacitor.

In $\text{XYZ} ($perform $X,$ then $Y,$ then $Z)$ the stored electric energy remains unchanged and no thermal energy is developed.
The charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XYW}$.
The electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}.$
The electric field in the capacitor after the action $\text{XW}$ is the same as that after $W$.

Answer

The charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XYW}.$
The electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}.$
The electric field in the capacitor after the action $\text{XW}$ is the same as that after $\text{WX}$.

Justification of option $(b)$:
If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of $K$ on inserting a dielectric of a dielectric constant $K$ between the plates of the capacitor.
Mathematically,
$q = Kq_{0 }$
Here $, q_0$ and q are the charges without dielectric and with dielectric, respectively.
The amount of charge stored does not depend upon the polarity of the plates.
Thus, the charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XYZ}$.
Justification of option $(c)$:
Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, $q = q_{0},$ because after the battery is disconnected, the capacitor gets no source to store charge from.
In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy $U$ as $\frac{1}{2}\text{q}\epsilon.$
Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor.
So, the energy stored in the capacitor will also not change after the action $\text{XYW}$.
However, during the action $\text{WXY},$ the amount of charge that will get stored in the capacitor will get increased by a factor of $K,$ as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of $K$.
Thus, the electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}.$
Justification of option $(d)$:
The electric field between the plates $E$ depends on the potential across the capacitor and the distanced between the plates of the capacitor.
Mathematically,
$\text{E}=\frac{\epsilon}{\text{d}}$
In either case, that is, during actions $\text{XW}$ and $\text{WX},$ the potential remains the same, that is, $\epsilon.$
Thus, the electric field $E$ remains the same.
Denial of option $(a)$ :
During the action $\text{XYZ},$ the battery has to do extra work equivalent to $\frac{1}{2}\text{CV}^2$ to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{CV}^2.$
This extra work done will be dissipated as heat energy.
Thus, thermal energy is developed.
However, the stored electric energy remains unchanged, that is, $\frac{1}{2}\text{CV}^2.$

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Question 81 Mark

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will:

Increase.
Decrease.
Remain unchanged.
Become zero.

Answer

Remain unchanged.

Explanation:
The force between the plates is given by
$\text{F}=\frac{\text{q}^2}{2\epsilon_0\text{A}}$
Since the capacitor is isolated, the charge on the plates remains constant.
We know that the charge is conserved in an isolated system.
Thus, the force acting between the plates remains unchanged.

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Question 91 Mark

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a batter. Now:

The facing surfaces of the capacitor have equal and opposite charges.
The two plates of the capacitor have equal and opposite charges.
The battery supplies equal and opposite charges to the two plates.
The outer surfaces of the plates have equal charges.

Answer

The facing surfaces of the capacitor have equal and opposite charges.
The two plates of the capacitor have equal and opposite charges.

The outer surfaces of the plates have equal charges.

Explanation:
In H.C Verma the answer is (a), (c), (d). But according to us the answer should be (a), (b), (d) all these options are the properties of a capacitor and the option (c) is incorrect because the battery is a source of energy not charge. Moreover if a capacitor plates have equal charge on outside and equal charge on inside then one can think that the charge on the plates must be also equal so option (b) cant be incorrect.

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MCQ 101 Mark

Two capacitors each having capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and the breakdown voltage of the combination will be:

A
$\text{2C}\ \text{and}\ \text{2V}$
B
$\frac{\text{C}}{2}\ \text{and}\ \frac{\text{V}}{2}$
C
$\text{2C}\ \text{and}\ \frac{\text{V}}{2}$
✓
$\frac{\text{C}}{2}\ \text{and}\ \text{2V}$

Answer

Correct option: D.

$\frac{\text{C}}{2}\ \text{and}\ \text{2V}$

Since the voltage gets added up when the capacitors are connected in series, the voltage of the combination is $2V.$
Also, the capacitance of a series combination is given by
$\frac{1}{\text{C}_{\text{net}}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}$
Here,
$C_{net} =$ Net capacitance of the combination
$\text{C}_1=\text{C}_2=\text{C}$
$\therefore\text{C}_{\text{net}}=\frac{\text{C}}{2}$

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Question 111 Mark

Three capacitors of capacitances $6\mu\text{F}$ each are available. The minimum and maximum capacitances, which may be obtained are:

$6\mu\text{F},\ 18\mu\text{F}$
$3\mu\text{F},\ 12\mu\text{F}$
$2\mu\text{F},\ 12\mu\text{F}$
$2\mu\text{F},\ 18\mu\text{F}$

Answer

$2\mu\text{F},\ 18\mu\text{F}$

Explanation:
The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow\text{C}=2\mu\text{F}$
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
$\text{C}=6+6+6=18\mu\text{F}$

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Question 121 Mark

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor:

The battery will supply more charge.
The capacitance will increase.
The potential difference between the plates will increase.
Equal and opposite charges will appear on the two faces of the metal plate.

Answer

Equal and opposite charges will appear on the two faces of the metal plate.

Explanation:
The capacitance of the capacitor in which a dielectric slab of dielectric constant K, area A and thickness t is inserted between the plates of the capacitor of area A and separated by a distance d is given by:
$\text{C}=\frac{\epsilon_0\text{A}}{(\text{d}-\text{t})+\big(\frac{\text{t}}{\text{K}}\big)}$
Since it is given that the thickness of the sheet is negligible, the above formula reduces to $\text{C}=\frac{\epsilon_0\text{A}}{\text{d}}.$ In other words, there will not be any change in the electric field, potential or charge.
Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor.

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Question 131 Mark

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates (figure). The capacitance now becomes:

$\frac{\text{C}}{2}$
2C
0
Indeterminate.

Answer

Indeterminate.

Explanation:
The thin metal plate inserted between the plates of a parallel-plate capacitor of capacitance C connects the two plates of the capacitor; hence, the distance d between the plates of the capacitor reduces to zero. It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously.
Mathematically,
$\text{C}=\frac{\epsilon_0\text{A}}{\text{d}}$
In this case, d = 0.
$\therefore\text{C}=\infty$

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Question 141 Mark

The energy density in the electric field created by a point charge falls off with the distance from the point charge as:

$\frac{1}{\text{r}}$
$\frac{1}{\text{r}^2}$
$\frac{1}{\text{r}^3}$
$\frac{1}{\text{r}^4}$

Answer

$\frac{1}{\text{r}^4}$

Explanation:
Energy density U is given by
$\text{U}=\frac{1}{2}\epsilon_0\text{E}^2\ \dots(1)$
The electric field created by a point charge at a distance r is given by
$\text{E}=\frac{\text{q}}{4\pi\epsilon_0\text{r}^2}$
On putting the above form of E in eq. 1, we get
$\text{U}=\frac{1}{2}\epsilon_0\Big(\frac{\text{q}}{4\pi\epsilon_0\text{r}^2}\Big)^2$
Thus, U is directly proportional to $\frac{1}{\text{r}^4}.$

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Question 151 Mark

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

Charge on the capacitor.
Potential difference across the capacitor.
Energy of the capacitor.
Energy density between the plates.

Answer

Potential difference across the capacitor.
Energy of the capacitor.

Explanation:
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$\text{V}=\frac{\text{Qd}}{\epsilon_0\text{A}}$
Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by:
$\text{E}=\frac{\text{qV}}{2}$
So, it will also increase.
Energy density u, that is, energy stored per unit volume in the electric field is given by:
$\text{u}=\frac{1}{2}\epsilon_0\text{E}^2$
So, u will remain constant with increase in distance between the plates.

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Question 161 Mark

Two metal plates having charges Q, -Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will:

Increase.
Decrease.
Remain the same.
Become zero.

Answer

Increase.

Explanation:
Oil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of $\frac{1}{\text{K}}$ of the original field when we insert a dielectric between the plates of a capacitor (K is the dielectric constant of the dielectric). So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed.

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Question 171 Mark

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is:

$\frac{\text{CV}}{\epsilon_0}$
$\frac{\text{2CV}}{\epsilon_0}$
$\frac{\text{CV}}{2\epsilon_0}$
$\text{Zero}.$

Answer

$\text{Zero}.$

Explanation:

Since the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.
$\phi=\oint\text{E.ds}=\frac{\text{q}}{\epsilon_0}=0$
Here,
$\phi=$ Electric flux
q = Total charge enclosed by the Gaussian surface.

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Question 181 Mark

The capacitance of a capacitor does not depend on:

The shape of the plates.
The size of the plates.
The charges on the plates.
The separation between the plates.

Answer

The charges on the plates.

Explanation:
The capacitance of a capacitor is given by:
$\text{C}=\frac{\epsilon_0\text{A}}{\text{d}}$
Here, A is the area of the plates of the capacitor and d is the distance between the plates.
So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates.

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Question 191 Mark

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q':

Q' may be larger than Q.
Q' must be larger than Q.
Q' must be equal to Q.
Q' must be smaller than Q.

Answer

Q' must be smaller than Q.

Explanation:
The relation between the induced charge Q' and the charge on the capacitor Q is given by:
$\text{Q}'=\text{Q}\Big(1-\frac{1}{\text{K}}\Big)$
Here, K is the dielectric constant that is always greater than or equal to 1.
So, we can see that for K > 1, Q' will always be less than Q.

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M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip