166037
$N$ identical drops of mercury are charged simultaneously to $10 \mathrm{~V}$. When combined to form one large drop, the potential is found to be $40 \mathrm{~V}$, the value of $\mathrm{N}$ is
1 4
2 6
3 8
4 10
5 12
Explanation:
: Given, $\mathrm{V}^{\prime}=40 \mathrm{~V}, \mathrm{~V}=10 \mathrm{~V}$ Volume of large drop $=\mathrm{N} \times$ volume of small drop $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3}=\mathrm{Nr}^{3}$ $\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}$ Also potential on big drop of radius, $\mathrm{V}^{\prime}=\left[\frac{\mathrm{K}(\mathrm{NQ})}{\mathrm{R}}\right]=40$ Potential on 1 small drop of radius, $\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{r}}=10$ Hence, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\left[\frac{\mathrm{KNQ}}{\mathrm{R}}\right]}{\mathrm{KQ} / \mathrm{r}}$ $\frac{40}{10}=\frac{\mathrm{Nr}}{\mathrm{R}}$ $4=\frac{\mathrm{N}}{\mathrm{N}^{1 / 3}}$ $\mathrm{~N}^{2 / 3}=4$ $\mathrm{~N}=4^{3 / 2}=8$ $\left[\because \mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}\right]$
[Kerala CEE 2007]
Capacitance
166038
Initially, ' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series. Then,
1 potential difference remains the same and the energy increases ' $n$ ' times.
2 potential difference is ' $\mathrm{nV}$ ' and energy increases ' $n$ ' times.
3 potential difference and the total energy of the combination remain the same.
4 potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Potential difference of each capacitor $=\mathrm{V}$ Connected in series the potential difference $=n V$ In series combination $\left(\mathrm{C}_{1}\right)=\frac{\mathrm{C}}{\mathrm{n}}$ In parallel combination, $\mathrm{C}_{2}=\mathrm{nC}$ Energy $(\mathrm{U})=\frac{1}{2} \mathrm{CV}^{2}$ $\therefore \quad \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \cdot \frac{\mathrm{V}_{2}^{2}}{\mathrm{~V}_{1}^{2}}=\frac{\mathrm{nC}}{\frac{\mathrm{C}}{\mathrm{n}}} \times \frac{\mathrm{V}^{2}}{\mathrm{n}^{2} \mathrm{~V}^{2}}=\frac{\mathrm{n}^{2} \mathrm{C}}{\mathrm{C}} \times \frac{1}{\mathrm{n}^{2}}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\mathrm{n}^{2} \times \frac{1}{\mathrm{n}^{2}}=1$ $\mathrm{U}_{2}=\mathrm{U}_{1}$ Hence, the potential difference becomes $\mathrm{nV}$ and energy remains the same.
[MHT-CET 2020]
Capacitance
166039
Assertion (A): Two capacitor of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cased will be $4: 1$ Reason (R) : In parallel, capacity increases and in series capacity decreases
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $\mathrm{A}$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $\mathrm{A}$
3 A is true, $\mathrm{R}$ is false
4 A is false, $\mathrm{R}$ is true
Explanation:
: Two capacitors of same capacities C. In series, $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2}$ In parallel, $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}=2 \mathrm{C}$ The ratio of $\mathrm{C}_{\mathrm{P}}$ and $\mathrm{C}_{\mathrm{S}}$ is- $\frac{C_{P}}{C_{S}}=\frac{2 C}{C / 2}=\frac{4}{1}$ $C_{P}: C_{S}=4: 1$ In parallel, $\mathrm{C}_{\mathrm{P}}>\mathrm{C} \&$ In series $\mathrm{C}_{\mathrm{s}}>\mathrm{C}$. So, in parallel, capacity is increased but in series, capacity is decreased. Hence, Both $\mathrm{A}$ and $\mathrm{B}$ are true but $\mathrm{R}$ is not a correct explanation for A
[AP EAMCET-23.08.2021
Capacitance
166040
Assertion : If the distance between parallel plate of a capacitor is halved and dielectric constant is three times, then the capacitance become 6 times. Reason : Capacity of the capacitor does not depend upon the nature of the material .
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion
3 If both the Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
Capacitance of a capacitor, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C} \propto \frac{\mathrm{K}}{\mathrm{d}}$ $\therefore \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~d}_{1}} \times \frac{\mathrm{d}_{2}}{\mathrm{~K}_{2}}=\frac{\mathrm{K}}{\mathrm{d}} \times \frac{\mathrm{d} / 2}{3 \mathrm{~K}}=\frac{1}{6}$ $\mathrm{C}_{2}=6 \mathrm{C}_{1}$ Capacity of capacitor $(\mathrm{C})=\frac{\mathrm{Q}}{\mathrm{V}}$ As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, Reason is incorrect.
[AIIMS-1997]
Capacitance
166041
A parallel plate condenser is filled with two dielectric $K_{1}$ and $K_{2}$ as shown in figure. Area of each plate is $A$ and separation is $d$. Its capacitance will be
166037
$N$ identical drops of mercury are charged simultaneously to $10 \mathrm{~V}$. When combined to form one large drop, the potential is found to be $40 \mathrm{~V}$, the value of $\mathrm{N}$ is
1 4
2 6
3 8
4 10
5 12
Explanation:
: Given, $\mathrm{V}^{\prime}=40 \mathrm{~V}, \mathrm{~V}=10 \mathrm{~V}$ Volume of large drop $=\mathrm{N} \times$ volume of small drop $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3}=\mathrm{Nr}^{3}$ $\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}$ Also potential on big drop of radius, $\mathrm{V}^{\prime}=\left[\frac{\mathrm{K}(\mathrm{NQ})}{\mathrm{R}}\right]=40$ Potential on 1 small drop of radius, $\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{r}}=10$ Hence, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\left[\frac{\mathrm{KNQ}}{\mathrm{R}}\right]}{\mathrm{KQ} / \mathrm{r}}$ $\frac{40}{10}=\frac{\mathrm{Nr}}{\mathrm{R}}$ $4=\frac{\mathrm{N}}{\mathrm{N}^{1 / 3}}$ $\mathrm{~N}^{2 / 3}=4$ $\mathrm{~N}=4^{3 / 2}=8$ $\left[\because \mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}\right]$
[Kerala CEE 2007]
Capacitance
166038
Initially, ' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series. Then,
1 potential difference remains the same and the energy increases ' $n$ ' times.
2 potential difference is ' $\mathrm{nV}$ ' and energy increases ' $n$ ' times.
3 potential difference and the total energy of the combination remain the same.
4 potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Potential difference of each capacitor $=\mathrm{V}$ Connected in series the potential difference $=n V$ In series combination $\left(\mathrm{C}_{1}\right)=\frac{\mathrm{C}}{\mathrm{n}}$ In parallel combination, $\mathrm{C}_{2}=\mathrm{nC}$ Energy $(\mathrm{U})=\frac{1}{2} \mathrm{CV}^{2}$ $\therefore \quad \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \cdot \frac{\mathrm{V}_{2}^{2}}{\mathrm{~V}_{1}^{2}}=\frac{\mathrm{nC}}{\frac{\mathrm{C}}{\mathrm{n}}} \times \frac{\mathrm{V}^{2}}{\mathrm{n}^{2} \mathrm{~V}^{2}}=\frac{\mathrm{n}^{2} \mathrm{C}}{\mathrm{C}} \times \frac{1}{\mathrm{n}^{2}}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\mathrm{n}^{2} \times \frac{1}{\mathrm{n}^{2}}=1$ $\mathrm{U}_{2}=\mathrm{U}_{1}$ Hence, the potential difference becomes $\mathrm{nV}$ and energy remains the same.
[MHT-CET 2020]
Capacitance
166039
Assertion (A): Two capacitor of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cased will be $4: 1$ Reason (R) : In parallel, capacity increases and in series capacity decreases
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $\mathrm{A}$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $\mathrm{A}$
3 A is true, $\mathrm{R}$ is false
4 A is false, $\mathrm{R}$ is true
Explanation:
: Two capacitors of same capacities C. In series, $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2}$ In parallel, $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}=2 \mathrm{C}$ The ratio of $\mathrm{C}_{\mathrm{P}}$ and $\mathrm{C}_{\mathrm{S}}$ is- $\frac{C_{P}}{C_{S}}=\frac{2 C}{C / 2}=\frac{4}{1}$ $C_{P}: C_{S}=4: 1$ In parallel, $\mathrm{C}_{\mathrm{P}}>\mathrm{C} \&$ In series $\mathrm{C}_{\mathrm{s}}>\mathrm{C}$. So, in parallel, capacity is increased but in series, capacity is decreased. Hence, Both $\mathrm{A}$ and $\mathrm{B}$ are true but $\mathrm{R}$ is not a correct explanation for A
[AP EAMCET-23.08.2021
Capacitance
166040
Assertion : If the distance between parallel plate of a capacitor is halved and dielectric constant is three times, then the capacitance become 6 times. Reason : Capacity of the capacitor does not depend upon the nature of the material .
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion
3 If both the Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
Capacitance of a capacitor, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C} \propto \frac{\mathrm{K}}{\mathrm{d}}$ $\therefore \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~d}_{1}} \times \frac{\mathrm{d}_{2}}{\mathrm{~K}_{2}}=\frac{\mathrm{K}}{\mathrm{d}} \times \frac{\mathrm{d} / 2}{3 \mathrm{~K}}=\frac{1}{6}$ $\mathrm{C}_{2}=6 \mathrm{C}_{1}$ Capacity of capacitor $(\mathrm{C})=\frac{\mathrm{Q}}{\mathrm{V}}$ As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, Reason is incorrect.
[AIIMS-1997]
Capacitance
166041
A parallel plate condenser is filled with two dielectric $K_{1}$ and $K_{2}$ as shown in figure. Area of each plate is $A$ and separation is $d$. Its capacitance will be
166037
$N$ identical drops of mercury are charged simultaneously to $10 \mathrm{~V}$. When combined to form one large drop, the potential is found to be $40 \mathrm{~V}$, the value of $\mathrm{N}$ is
1 4
2 6
3 8
4 10
5 12
Explanation:
: Given, $\mathrm{V}^{\prime}=40 \mathrm{~V}, \mathrm{~V}=10 \mathrm{~V}$ Volume of large drop $=\mathrm{N} \times$ volume of small drop $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3}=\mathrm{Nr}^{3}$ $\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}$ Also potential on big drop of radius, $\mathrm{V}^{\prime}=\left[\frac{\mathrm{K}(\mathrm{NQ})}{\mathrm{R}}\right]=40$ Potential on 1 small drop of radius, $\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{r}}=10$ Hence, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\left[\frac{\mathrm{KNQ}}{\mathrm{R}}\right]}{\mathrm{KQ} / \mathrm{r}}$ $\frac{40}{10}=\frac{\mathrm{Nr}}{\mathrm{R}}$ $4=\frac{\mathrm{N}}{\mathrm{N}^{1 / 3}}$ $\mathrm{~N}^{2 / 3}=4$ $\mathrm{~N}=4^{3 / 2}=8$ $\left[\because \mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}\right]$
[Kerala CEE 2007]
Capacitance
166038
Initially, ' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series. Then,
1 potential difference remains the same and the energy increases ' $n$ ' times.
2 potential difference is ' $\mathrm{nV}$ ' and energy increases ' $n$ ' times.
3 potential difference and the total energy of the combination remain the same.
4 potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Potential difference of each capacitor $=\mathrm{V}$ Connected in series the potential difference $=n V$ In series combination $\left(\mathrm{C}_{1}\right)=\frac{\mathrm{C}}{\mathrm{n}}$ In parallel combination, $\mathrm{C}_{2}=\mathrm{nC}$ Energy $(\mathrm{U})=\frac{1}{2} \mathrm{CV}^{2}$ $\therefore \quad \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \cdot \frac{\mathrm{V}_{2}^{2}}{\mathrm{~V}_{1}^{2}}=\frac{\mathrm{nC}}{\frac{\mathrm{C}}{\mathrm{n}}} \times \frac{\mathrm{V}^{2}}{\mathrm{n}^{2} \mathrm{~V}^{2}}=\frac{\mathrm{n}^{2} \mathrm{C}}{\mathrm{C}} \times \frac{1}{\mathrm{n}^{2}}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\mathrm{n}^{2} \times \frac{1}{\mathrm{n}^{2}}=1$ $\mathrm{U}_{2}=\mathrm{U}_{1}$ Hence, the potential difference becomes $\mathrm{nV}$ and energy remains the same.
[MHT-CET 2020]
Capacitance
166039
Assertion (A): Two capacitor of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cased will be $4: 1$ Reason (R) : In parallel, capacity increases and in series capacity decreases
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $\mathrm{A}$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $\mathrm{A}$
3 A is true, $\mathrm{R}$ is false
4 A is false, $\mathrm{R}$ is true
Explanation:
: Two capacitors of same capacities C. In series, $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2}$ In parallel, $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}=2 \mathrm{C}$ The ratio of $\mathrm{C}_{\mathrm{P}}$ and $\mathrm{C}_{\mathrm{S}}$ is- $\frac{C_{P}}{C_{S}}=\frac{2 C}{C / 2}=\frac{4}{1}$ $C_{P}: C_{S}=4: 1$ In parallel, $\mathrm{C}_{\mathrm{P}}>\mathrm{C} \&$ In series $\mathrm{C}_{\mathrm{s}}>\mathrm{C}$. So, in parallel, capacity is increased but in series, capacity is decreased. Hence, Both $\mathrm{A}$ and $\mathrm{B}$ are true but $\mathrm{R}$ is not a correct explanation for A
[AP EAMCET-23.08.2021
Capacitance
166040
Assertion : If the distance between parallel plate of a capacitor is halved and dielectric constant is three times, then the capacitance become 6 times. Reason : Capacity of the capacitor does not depend upon the nature of the material .
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion
3 If both the Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
Capacitance of a capacitor, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C} \propto \frac{\mathrm{K}}{\mathrm{d}}$ $\therefore \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~d}_{1}} \times \frac{\mathrm{d}_{2}}{\mathrm{~K}_{2}}=\frac{\mathrm{K}}{\mathrm{d}} \times \frac{\mathrm{d} / 2}{3 \mathrm{~K}}=\frac{1}{6}$ $\mathrm{C}_{2}=6 \mathrm{C}_{1}$ Capacity of capacitor $(\mathrm{C})=\frac{\mathrm{Q}}{\mathrm{V}}$ As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, Reason is incorrect.
[AIIMS-1997]
Capacitance
166041
A parallel plate condenser is filled with two dielectric $K_{1}$ and $K_{2}$ as shown in figure. Area of each plate is $A$ and separation is $d$. Its capacitance will be
166037
$N$ identical drops of mercury are charged simultaneously to $10 \mathrm{~V}$. When combined to form one large drop, the potential is found to be $40 \mathrm{~V}$, the value of $\mathrm{N}$ is
1 4
2 6
3 8
4 10
5 12
Explanation:
: Given, $\mathrm{V}^{\prime}=40 \mathrm{~V}, \mathrm{~V}=10 \mathrm{~V}$ Volume of large drop $=\mathrm{N} \times$ volume of small drop $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3}=\mathrm{Nr}^{3}$ $\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}$ Also potential on big drop of radius, $\mathrm{V}^{\prime}=\left[\frac{\mathrm{K}(\mathrm{NQ})}{\mathrm{R}}\right]=40$ Potential on 1 small drop of radius, $\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{r}}=10$ Hence, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\left[\frac{\mathrm{KNQ}}{\mathrm{R}}\right]}{\mathrm{KQ} / \mathrm{r}}$ $\frac{40}{10}=\frac{\mathrm{Nr}}{\mathrm{R}}$ $4=\frac{\mathrm{N}}{\mathrm{N}^{1 / 3}}$ $\mathrm{~N}^{2 / 3}=4$ $\mathrm{~N}=4^{3 / 2}=8$ $\left[\because \mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}\right]$
[Kerala CEE 2007]
Capacitance
166038
Initially, ' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series. Then,
1 potential difference remains the same and the energy increases ' $n$ ' times.
2 potential difference is ' $\mathrm{nV}$ ' and energy increases ' $n$ ' times.
3 potential difference and the total energy of the combination remain the same.
4 potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Potential difference of each capacitor $=\mathrm{V}$ Connected in series the potential difference $=n V$ In series combination $\left(\mathrm{C}_{1}\right)=\frac{\mathrm{C}}{\mathrm{n}}$ In parallel combination, $\mathrm{C}_{2}=\mathrm{nC}$ Energy $(\mathrm{U})=\frac{1}{2} \mathrm{CV}^{2}$ $\therefore \quad \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \cdot \frac{\mathrm{V}_{2}^{2}}{\mathrm{~V}_{1}^{2}}=\frac{\mathrm{nC}}{\frac{\mathrm{C}}{\mathrm{n}}} \times \frac{\mathrm{V}^{2}}{\mathrm{n}^{2} \mathrm{~V}^{2}}=\frac{\mathrm{n}^{2} \mathrm{C}}{\mathrm{C}} \times \frac{1}{\mathrm{n}^{2}}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\mathrm{n}^{2} \times \frac{1}{\mathrm{n}^{2}}=1$ $\mathrm{U}_{2}=\mathrm{U}_{1}$ Hence, the potential difference becomes $\mathrm{nV}$ and energy remains the same.
[MHT-CET 2020]
Capacitance
166039
Assertion (A): Two capacitor of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cased will be $4: 1$ Reason (R) : In parallel, capacity increases and in series capacity decreases
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $\mathrm{A}$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $\mathrm{A}$
3 A is true, $\mathrm{R}$ is false
4 A is false, $\mathrm{R}$ is true
Explanation:
: Two capacitors of same capacities C. In series, $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2}$ In parallel, $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}=2 \mathrm{C}$ The ratio of $\mathrm{C}_{\mathrm{P}}$ and $\mathrm{C}_{\mathrm{S}}$ is- $\frac{C_{P}}{C_{S}}=\frac{2 C}{C / 2}=\frac{4}{1}$ $C_{P}: C_{S}=4: 1$ In parallel, $\mathrm{C}_{\mathrm{P}}>\mathrm{C} \&$ In series $\mathrm{C}_{\mathrm{s}}>\mathrm{C}$. So, in parallel, capacity is increased but in series, capacity is decreased. Hence, Both $\mathrm{A}$ and $\mathrm{B}$ are true but $\mathrm{R}$ is not a correct explanation for A
[AP EAMCET-23.08.2021
Capacitance
166040
Assertion : If the distance between parallel plate of a capacitor is halved and dielectric constant is three times, then the capacitance become 6 times. Reason : Capacity of the capacitor does not depend upon the nature of the material .
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion
3 If both the Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
Capacitance of a capacitor, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C} \propto \frac{\mathrm{K}}{\mathrm{d}}$ $\therefore \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~d}_{1}} \times \frac{\mathrm{d}_{2}}{\mathrm{~K}_{2}}=\frac{\mathrm{K}}{\mathrm{d}} \times \frac{\mathrm{d} / 2}{3 \mathrm{~K}}=\frac{1}{6}$ $\mathrm{C}_{2}=6 \mathrm{C}_{1}$ Capacity of capacitor $(\mathrm{C})=\frac{\mathrm{Q}}{\mathrm{V}}$ As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, Reason is incorrect.
[AIIMS-1997]
Capacitance
166041
A parallel plate condenser is filled with two dielectric $K_{1}$ and $K_{2}$ as shown in figure. Area of each plate is $A$ and separation is $d$. Its capacitance will be
166037
$N$ identical drops of mercury are charged simultaneously to $10 \mathrm{~V}$. When combined to form one large drop, the potential is found to be $40 \mathrm{~V}$, the value of $\mathrm{N}$ is
1 4
2 6
3 8
4 10
5 12
Explanation:
: Given, $\mathrm{V}^{\prime}=40 \mathrm{~V}, \mathrm{~V}=10 \mathrm{~V}$ Volume of large drop $=\mathrm{N} \times$ volume of small drop $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{3}=\mathrm{Nr}^{3}$ $\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}$ Also potential on big drop of radius, $\mathrm{V}^{\prime}=\left[\frac{\mathrm{K}(\mathrm{NQ})}{\mathrm{R}}\right]=40$ Potential on 1 small drop of radius, $\mathrm{V}=\frac{\mathrm{KQ}}{\mathrm{r}}=10$ Hence, $\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\left[\frac{\mathrm{KNQ}}{\mathrm{R}}\right]}{\mathrm{KQ} / \mathrm{r}}$ $\frac{40}{10}=\frac{\mathrm{Nr}}{\mathrm{R}}$ $4=\frac{\mathrm{N}}{\mathrm{N}^{1 / 3}}$ $\mathrm{~N}^{2 / 3}=4$ $\mathrm{~N}=4^{3 / 2}=8$ $\left[\because \mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r}\right]$
[Kerala CEE 2007]
Capacitance
166038
Initially, ' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series. Then,
1 potential difference remains the same and the energy increases ' $n$ ' times.
2 potential difference is ' $\mathrm{nV}$ ' and energy increases ' $n$ ' times.
3 potential difference and the total energy of the combination remain the same.
4 potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Potential difference of each capacitor $=\mathrm{V}$ Connected in series the potential difference $=n V$ In series combination $\left(\mathrm{C}_{1}\right)=\frac{\mathrm{C}}{\mathrm{n}}$ In parallel combination, $\mathrm{C}_{2}=\mathrm{nC}$ Energy $(\mathrm{U})=\frac{1}{2} \mathrm{CV}^{2}$ $\therefore \quad \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}} \cdot \frac{\mathrm{V}_{2}^{2}}{\mathrm{~V}_{1}^{2}}=\frac{\mathrm{nC}}{\frac{\mathrm{C}}{\mathrm{n}}} \times \frac{\mathrm{V}^{2}}{\mathrm{n}^{2} \mathrm{~V}^{2}}=\frac{\mathrm{n}^{2} \mathrm{C}}{\mathrm{C}} \times \frac{1}{\mathrm{n}^{2}}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\mathrm{n}^{2} \times \frac{1}{\mathrm{n}^{2}}=1$ $\mathrm{U}_{2}=\mathrm{U}_{1}$ Hence, the potential difference becomes $\mathrm{nV}$ and energy remains the same.
[MHT-CET 2020]
Capacitance
166039
Assertion (A): Two capacitor of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cased will be $4: 1$ Reason (R) : In parallel, capacity increases and in series capacity decreases
1 Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is a correct explanation for $\mathrm{A}$
2 Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is not a correct explanation for $\mathrm{A}$
3 A is true, $\mathrm{R}$ is false
4 A is false, $\mathrm{R}$ is true
Explanation:
: Two capacitors of same capacities C. In series, $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C} \times \mathrm{C}}{\mathrm{C}+\mathrm{C}}=\frac{\mathrm{C}}{2}$ In parallel, $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}=2 \mathrm{C}$ The ratio of $\mathrm{C}_{\mathrm{P}}$ and $\mathrm{C}_{\mathrm{S}}$ is- $\frac{C_{P}}{C_{S}}=\frac{2 C}{C / 2}=\frac{4}{1}$ $C_{P}: C_{S}=4: 1$ In parallel, $\mathrm{C}_{\mathrm{P}}>\mathrm{C} \&$ In series $\mathrm{C}_{\mathrm{s}}>\mathrm{C}$. So, in parallel, capacity is increased but in series, capacity is decreased. Hence, Both $\mathrm{A}$ and $\mathrm{B}$ are true but $\mathrm{R}$ is not a correct explanation for A
[AP EAMCET-23.08.2021
Capacitance
166040
Assertion : If the distance between parallel plate of a capacitor is halved and dielectric constant is three times, then the capacitance become 6 times. Reason : Capacity of the capacitor does not depend upon the nature of the material .
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion
2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion
3 If both the Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
Capacitance of a capacitor, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C} \propto \frac{\mathrm{K}}{\mathrm{d}}$ $\therefore \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~d}_{1}} \times \frac{\mathrm{d}_{2}}{\mathrm{~K}_{2}}=\frac{\mathrm{K}}{\mathrm{d}} \times \frac{\mathrm{d} / 2}{3 \mathrm{~K}}=\frac{1}{6}$ $\mathrm{C}_{2}=6 \mathrm{C}_{1}$ Capacity of capacitor $(\mathrm{C})=\frac{\mathrm{Q}}{\mathrm{V}}$ As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, Reason is incorrect.
[AIIMS-1997]
Capacitance
166041
A parallel plate condenser is filled with two dielectric $K_{1}$ and $K_{2}$ as shown in figure. Area of each plate is $A$ and separation is $d$. Its capacitance will be
