166015
A capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2 \mathrm{C}$. The final charges on the capacitors are :
1 $\frac{\mathrm{Q}}{2}, \frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{4}, \frac{3 \mathrm{Q}}{4}$
3 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$
4 $\frac{\mathrm{Q}}{5}, \frac{4 \mathrm{Q}}{5}$
Explanation:
: A capacitor is charged by battery and then battery was replaced with another uncharged capacitor of equal capacitance. The charge stored in the first capacitor $=\mathrm{Q}$ Charge stored in second capacitor $=0$ The two capacitors attain common potential $\left(\mathrm{V}_{\mathrm{C}}\right)$ The final charges on two capacitors are- $\mathrm{Q}_{1}=\mathrm{CV}_{\mathrm{C}}=\frac{\mathrm{CQ}}{3 \mathrm{C}}=\frac{\mathrm{Q}}{3}$ $\mathrm{Q}_{2}=2 \mathrm{CV}_{\mathrm{C}}=\frac{2 \mathrm{Q}}{3}$
[Karnataka CET-2019]
Capacitance
166017
Capacitance of a parallel plate capacitor becomes $4 / 3$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates : [d is the separation between the plates]. The dielectric constant of the slab is :
1 4
2 8
3 2
4 6
Explanation:
: Given that, Capacitance, $\mathrm{C}^{\prime}=\frac{4}{3} \mathrm{C}$ $\operatorname{Thickness}(\mathrm{t})=\frac{\mathrm{d}}{2}$ We know that, $\text { Capacitance is air }(C)=\frac{\varepsilon_{0} A}{d}$ Now Let ' $t$ ' thickness dielectric slab introduced $C^{\prime}=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$ Now, $\frac{4}{3} \mathrm{C}=\mathrm{C}^{\prime}$ $\frac{4}{3} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}-\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2 \mathrm{~K}}}$ $4\left(1-\frac{1}{2}+\frac{1}{2 \mathrm{~K}}\right)=3$ $\left(1-\frac{1}{2}\right)+\frac{1}{2 \mathrm{~K}}=\frac{3}{4}$ $\frac{\mathrm{K}+1}{2 \mathrm{~K}}=\frac{3}{4}$ $\mathrm{~K}=2$
[Karnataka CET-2003]
Capacitance
166018
64 small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is :
1 $64: 1$
2 $1: 64$
3 $1: 4$
4 $4: 1$
Explanation:
: We know that, $\quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}}$ $64 \mathrm{r}^{3}=\mathrm{R}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\sigma_{\mathrm{s}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \quad \text { (for small surface) }$ $\sigma_{l}=\frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}} \quad \text { (for large surface) }$ Ratio of small drop to large drop is- $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \div \frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}}$ $=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \times \frac{4 \pi(4 \mathrm{r})^{2}}{64 \mathrm{q}}$ $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{1}{4}$ $\sigma_{\mathrm{s}}: \sigma_{l}=1: 4$
[Karnataka CET-2002]
Capacitance
166020
A dielectric of dielectric constant $K$ is introduced such that half of area of capacitor of capacity $C$ is occupied by it. The new capacity is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 $\frac{\mathrm{C}}{2}(1+\mathrm{K})$
4 $2 \mathrm{C}(1+\mathrm{K})$
Explanation:
: The dielectric is introduced such that, half of its area is occupied by it. In the given case the two capacitors are in parallel. $\because \mathrm{C}^{\prime}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $\mathrm{C}_{1} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}} \text { and } \mathrm{C}_{2}=\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}}+\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{C}}{2}(1+\mathrm{K})$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
166015
A capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2 \mathrm{C}$. The final charges on the capacitors are :
1 $\frac{\mathrm{Q}}{2}, \frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{4}, \frac{3 \mathrm{Q}}{4}$
3 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$
4 $\frac{\mathrm{Q}}{5}, \frac{4 \mathrm{Q}}{5}$
Explanation:
: A capacitor is charged by battery and then battery was replaced with another uncharged capacitor of equal capacitance. The charge stored in the first capacitor $=\mathrm{Q}$ Charge stored in second capacitor $=0$ The two capacitors attain common potential $\left(\mathrm{V}_{\mathrm{C}}\right)$ The final charges on two capacitors are- $\mathrm{Q}_{1}=\mathrm{CV}_{\mathrm{C}}=\frac{\mathrm{CQ}}{3 \mathrm{C}}=\frac{\mathrm{Q}}{3}$ $\mathrm{Q}_{2}=2 \mathrm{CV}_{\mathrm{C}}=\frac{2 \mathrm{Q}}{3}$
[Karnataka CET-2019]
Capacitance
166017
Capacitance of a parallel plate capacitor becomes $4 / 3$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates : [d is the separation between the plates]. The dielectric constant of the slab is :
1 4
2 8
3 2
4 6
Explanation:
: Given that, Capacitance, $\mathrm{C}^{\prime}=\frac{4}{3} \mathrm{C}$ $\operatorname{Thickness}(\mathrm{t})=\frac{\mathrm{d}}{2}$ We know that, $\text { Capacitance is air }(C)=\frac{\varepsilon_{0} A}{d}$ Now Let ' $t$ ' thickness dielectric slab introduced $C^{\prime}=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$ Now, $\frac{4}{3} \mathrm{C}=\mathrm{C}^{\prime}$ $\frac{4}{3} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}-\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2 \mathrm{~K}}}$ $4\left(1-\frac{1}{2}+\frac{1}{2 \mathrm{~K}}\right)=3$ $\left(1-\frac{1}{2}\right)+\frac{1}{2 \mathrm{~K}}=\frac{3}{4}$ $\frac{\mathrm{K}+1}{2 \mathrm{~K}}=\frac{3}{4}$ $\mathrm{~K}=2$
[Karnataka CET-2003]
Capacitance
166018
64 small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is :
1 $64: 1$
2 $1: 64$
3 $1: 4$
4 $4: 1$
Explanation:
: We know that, $\quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}}$ $64 \mathrm{r}^{3}=\mathrm{R}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\sigma_{\mathrm{s}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \quad \text { (for small surface) }$ $\sigma_{l}=\frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}} \quad \text { (for large surface) }$ Ratio of small drop to large drop is- $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \div \frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}}$ $=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \times \frac{4 \pi(4 \mathrm{r})^{2}}{64 \mathrm{q}}$ $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{1}{4}$ $\sigma_{\mathrm{s}}: \sigma_{l}=1: 4$
[Karnataka CET-2002]
Capacitance
166020
A dielectric of dielectric constant $K$ is introduced such that half of area of capacitor of capacity $C$ is occupied by it. The new capacity is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 $\frac{\mathrm{C}}{2}(1+\mathrm{K})$
4 $2 \mathrm{C}(1+\mathrm{K})$
Explanation:
: The dielectric is introduced such that, half of its area is occupied by it. In the given case the two capacitors are in parallel. $\because \mathrm{C}^{\prime}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $\mathrm{C}_{1} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}} \text { and } \mathrm{C}_{2}=\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}}+\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{C}}{2}(1+\mathrm{K})$
166015
A capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2 \mathrm{C}$. The final charges on the capacitors are :
1 $\frac{\mathrm{Q}}{2}, \frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{4}, \frac{3 \mathrm{Q}}{4}$
3 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$
4 $\frac{\mathrm{Q}}{5}, \frac{4 \mathrm{Q}}{5}$
Explanation:
: A capacitor is charged by battery and then battery was replaced with another uncharged capacitor of equal capacitance. The charge stored in the first capacitor $=\mathrm{Q}$ Charge stored in second capacitor $=0$ The two capacitors attain common potential $\left(\mathrm{V}_{\mathrm{C}}\right)$ The final charges on two capacitors are- $\mathrm{Q}_{1}=\mathrm{CV}_{\mathrm{C}}=\frac{\mathrm{CQ}}{3 \mathrm{C}}=\frac{\mathrm{Q}}{3}$ $\mathrm{Q}_{2}=2 \mathrm{CV}_{\mathrm{C}}=\frac{2 \mathrm{Q}}{3}$
[Karnataka CET-2019]
Capacitance
166017
Capacitance of a parallel plate capacitor becomes $4 / 3$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates : [d is the separation between the plates]. The dielectric constant of the slab is :
1 4
2 8
3 2
4 6
Explanation:
: Given that, Capacitance, $\mathrm{C}^{\prime}=\frac{4}{3} \mathrm{C}$ $\operatorname{Thickness}(\mathrm{t})=\frac{\mathrm{d}}{2}$ We know that, $\text { Capacitance is air }(C)=\frac{\varepsilon_{0} A}{d}$ Now Let ' $t$ ' thickness dielectric slab introduced $C^{\prime}=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$ Now, $\frac{4}{3} \mathrm{C}=\mathrm{C}^{\prime}$ $\frac{4}{3} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}-\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2 \mathrm{~K}}}$ $4\left(1-\frac{1}{2}+\frac{1}{2 \mathrm{~K}}\right)=3$ $\left(1-\frac{1}{2}\right)+\frac{1}{2 \mathrm{~K}}=\frac{3}{4}$ $\frac{\mathrm{K}+1}{2 \mathrm{~K}}=\frac{3}{4}$ $\mathrm{~K}=2$
[Karnataka CET-2003]
Capacitance
166018
64 small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is :
1 $64: 1$
2 $1: 64$
3 $1: 4$
4 $4: 1$
Explanation:
: We know that, $\quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}}$ $64 \mathrm{r}^{3}=\mathrm{R}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\sigma_{\mathrm{s}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \quad \text { (for small surface) }$ $\sigma_{l}=\frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}} \quad \text { (for large surface) }$ Ratio of small drop to large drop is- $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \div \frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}}$ $=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \times \frac{4 \pi(4 \mathrm{r})^{2}}{64 \mathrm{q}}$ $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{1}{4}$ $\sigma_{\mathrm{s}}: \sigma_{l}=1: 4$
[Karnataka CET-2002]
Capacitance
166020
A dielectric of dielectric constant $K$ is introduced such that half of area of capacitor of capacity $C$ is occupied by it. The new capacity is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 $\frac{\mathrm{C}}{2}(1+\mathrm{K})$
4 $2 \mathrm{C}(1+\mathrm{K})$
Explanation:
: The dielectric is introduced such that, half of its area is occupied by it. In the given case the two capacitors are in parallel. $\because \mathrm{C}^{\prime}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $\mathrm{C}_{1} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}} \text { and } \mathrm{C}_{2}=\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}}+\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{C}}{2}(1+\mathrm{K})$
166015
A capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2 \mathrm{C}$. The final charges on the capacitors are :
1 $\frac{\mathrm{Q}}{2}, \frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{4}, \frac{3 \mathrm{Q}}{4}$
3 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$
4 $\frac{\mathrm{Q}}{5}, \frac{4 \mathrm{Q}}{5}$
Explanation:
: A capacitor is charged by battery and then battery was replaced with another uncharged capacitor of equal capacitance. The charge stored in the first capacitor $=\mathrm{Q}$ Charge stored in second capacitor $=0$ The two capacitors attain common potential $\left(\mathrm{V}_{\mathrm{C}}\right)$ The final charges on two capacitors are- $\mathrm{Q}_{1}=\mathrm{CV}_{\mathrm{C}}=\frac{\mathrm{CQ}}{3 \mathrm{C}}=\frac{\mathrm{Q}}{3}$ $\mathrm{Q}_{2}=2 \mathrm{CV}_{\mathrm{C}}=\frac{2 \mathrm{Q}}{3}$
[Karnataka CET-2019]
Capacitance
166017
Capacitance of a parallel plate capacitor becomes $4 / 3$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates : [d is the separation between the plates]. The dielectric constant of the slab is :
1 4
2 8
3 2
4 6
Explanation:
: Given that, Capacitance, $\mathrm{C}^{\prime}=\frac{4}{3} \mathrm{C}$ $\operatorname{Thickness}(\mathrm{t})=\frac{\mathrm{d}}{2}$ We know that, $\text { Capacitance is air }(C)=\frac{\varepsilon_{0} A}{d}$ Now Let ' $t$ ' thickness dielectric slab introduced $C^{\prime}=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$ Now, $\frac{4}{3} \mathrm{C}=\mathrm{C}^{\prime}$ $\frac{4}{3} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}-\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2 \mathrm{~K}}}$ $4\left(1-\frac{1}{2}+\frac{1}{2 \mathrm{~K}}\right)=3$ $\left(1-\frac{1}{2}\right)+\frac{1}{2 \mathrm{~K}}=\frac{3}{4}$ $\frac{\mathrm{K}+1}{2 \mathrm{~K}}=\frac{3}{4}$ $\mathrm{~K}=2$
[Karnataka CET-2003]
Capacitance
166018
64 small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is :
1 $64: 1$
2 $1: 64$
3 $1: 4$
4 $4: 1$
Explanation:
: We know that, $\quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}}$ $64 \mathrm{r}^{3}=\mathrm{R}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\sigma_{\mathrm{s}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \quad \text { (for small surface) }$ $\sigma_{l}=\frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}} \quad \text { (for large surface) }$ Ratio of small drop to large drop is- $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \div \frac{64 \mathrm{q}}{4 \pi \mathrm{R}^{2}}$ $=\frac{\mathrm{q}}{4 \pi \mathrm{r}^{2}} \times \frac{4 \pi(4 \mathrm{r})^{2}}{64 \mathrm{q}}$ $\frac{\sigma_{\mathrm{s}}}{\sigma_{l}}=\frac{1}{4}$ $\sigma_{\mathrm{s}}: \sigma_{l}=1: 4$
[Karnataka CET-2002]
Capacitance
166020
A dielectric of dielectric constant $K$ is introduced such that half of area of capacitor of capacity $C$ is occupied by it. The new capacity is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 $\frac{\mathrm{C}}{2}(1+\mathrm{K})$
4 $2 \mathrm{C}(1+\mathrm{K})$
Explanation:
: The dielectric is introduced such that, half of its area is occupied by it. In the given case the two capacitors are in parallel. $\because \mathrm{C}^{\prime}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $\mathrm{C}_{1} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}} \text { and } \mathrm{C}_{2}=\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{A} \varepsilon_{0}}{2 \mathrm{~d}}+\frac{\mathrm{KA} \varepsilon_{0}}{2 \mathrm{~d}}$ $\mathrm{C}^{\prime} =\frac{\mathrm{C}}{2}(1+\mathrm{K})$
