165632
A $5 \mu \mathrm{F}$ capacitor is connected in series with a $10 \mu \mathrm{F}$ capacitor. When a $300 \mathrm{~V}$ potential difference is applied across this combination, the total energy stored in the capacitors is
1 $15 \mathrm{~J}$
2 $1.5 \mathrm{~J}$
3 $0.15 \mathrm{~J}$
4 $0.10 \mathrm{~J}$
Explanation:
: Equivalent capacitance of the series combinations, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}$ $\mathrm{C}_{\mathrm{eq}}=\frac{10}{3} \mu \mathrm{F}$ Hence, the energy stored, $\mathrm{E}=\frac{1}{2} \mathrm{C}_{\mathrm{eq}} \mathrm{V}^{2}$ $\mathrm{E}=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times(300)^{2}$ $\mathrm{E}=0.15 \mathrm{~J}$
[WB JEE 2015]
Capacitance
165633
64 identical spheres of charges $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is
1 $64 \mathrm{q}, \mathrm{C}$
2 $16 \mathrm{q}, 4 \mathrm{C}$
3 $64 \mathrm{q}, 4 \mathrm{C}$
4 $16 \mathrm{q}, 64 \mathrm{C}$
Explanation:
: Volume of large sphere $=$ volume of each small sphere $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)$ $\mathrm{R}^{3}=64 \mathrm{r}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\because \quad \mathrm{Q}_{\text {total }}=64 \mathrm{q}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}$ $\text { Now, } \mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\text { Putting the value of } \mathrm{R}=4 \mathrm{r}$ $\mathrm{C}^{\prime}=\left(4 \pi \varepsilon_{\mathrm{o}}\right) 4 \mathrm{r}$ $\mathrm{C}^{\prime}=4 \mathrm{C}$
[WB JEE 2008]
Capacitance
165634
The capacity an isolated conducting sphere of radius $R$ is proportional to
1 $\mathrm{R}^{2}$
2 $\frac{1}{\mathrm{R}^{2}}$
3 $\frac{1}{\mathrm{R}}$
4 $\mathrm{R}$
Explanation:
: Capacitance of the conductor, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ Potential on sphere of radius, $\mathrm{R}$ [Where $\mathrm{q}$ is the charge and $\mathrm{V}$ is the potential] $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}}$ $\therefore \quad \mathrm{C}=\mathrm{Q} \times \frac{4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}}{\mathrm{Q}}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\mathrm{C} \propto \mathrm{R}$
[UP CPMT- 2005]
Capacitance
165636
Consider a spherical drop of mercury of radius $R$ with capacitance $C=4 \pi \epsilon_{0} R$. If two such droplets combine to form a larger one, what would be its capacitance in terms of $C$ ?
1 $3^{1 / 3} \mathrm{C}$
2 $3^{2 / 3} \mathrm{C}$
3 $2^{2 / 3} \mathrm{C}$
4 $2^{1 / 3} \mathrm{C}$
Explanation:
: For spherical drop of mercury, $\text { Radius }=\mathrm{R}$ Capacitance $(C)=4 \pi \varepsilon_{0} R$ When two droplets combine, $\mathrm{r}=\text { smaller radius }$ $\because$ Volume of bigger drop $=$ volume of each small drop $\frac{4}{3} \pi\left(\mathrm{R}^{\prime}\right)^{3}=2 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{\prime}=2^{\frac{1}{3}} \mathrm{r}$ $\therefore$ Capacitance of bigger drop, $\mathrm{C}^{\prime}=4 \pi \varepsilon_{0} \mathrm{R}^{\prime}$ $\mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \times 2^{\frac{1}{3}} \mathrm{r}$ $\mathrm{C}^{\prime}=2^{\frac{1}{3}} \mathrm{C} \quad\left\{\because \mathrm{C}=4 \pi \varepsilon_{0} \mathrm{r}\right\}$
165632
A $5 \mu \mathrm{F}$ capacitor is connected in series with a $10 \mu \mathrm{F}$ capacitor. When a $300 \mathrm{~V}$ potential difference is applied across this combination, the total energy stored in the capacitors is
1 $15 \mathrm{~J}$
2 $1.5 \mathrm{~J}$
3 $0.15 \mathrm{~J}$
4 $0.10 \mathrm{~J}$
Explanation:
: Equivalent capacitance of the series combinations, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}$ $\mathrm{C}_{\mathrm{eq}}=\frac{10}{3} \mu \mathrm{F}$ Hence, the energy stored, $\mathrm{E}=\frac{1}{2} \mathrm{C}_{\mathrm{eq}} \mathrm{V}^{2}$ $\mathrm{E}=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times(300)^{2}$ $\mathrm{E}=0.15 \mathrm{~J}$
[WB JEE 2015]
Capacitance
165633
64 identical spheres of charges $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is
1 $64 \mathrm{q}, \mathrm{C}$
2 $16 \mathrm{q}, 4 \mathrm{C}$
3 $64 \mathrm{q}, 4 \mathrm{C}$
4 $16 \mathrm{q}, 64 \mathrm{C}$
Explanation:
: Volume of large sphere $=$ volume of each small sphere $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)$ $\mathrm{R}^{3}=64 \mathrm{r}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\because \quad \mathrm{Q}_{\text {total }}=64 \mathrm{q}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}$ $\text { Now, } \mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\text { Putting the value of } \mathrm{R}=4 \mathrm{r}$ $\mathrm{C}^{\prime}=\left(4 \pi \varepsilon_{\mathrm{o}}\right) 4 \mathrm{r}$ $\mathrm{C}^{\prime}=4 \mathrm{C}$
[WB JEE 2008]
Capacitance
165634
The capacity an isolated conducting sphere of radius $R$ is proportional to
1 $\mathrm{R}^{2}$
2 $\frac{1}{\mathrm{R}^{2}}$
3 $\frac{1}{\mathrm{R}}$
4 $\mathrm{R}$
Explanation:
: Capacitance of the conductor, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ Potential on sphere of radius, $\mathrm{R}$ [Where $\mathrm{q}$ is the charge and $\mathrm{V}$ is the potential] $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}}$ $\therefore \quad \mathrm{C}=\mathrm{Q} \times \frac{4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}}{\mathrm{Q}}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\mathrm{C} \propto \mathrm{R}$
[UP CPMT- 2005]
Capacitance
165636
Consider a spherical drop of mercury of radius $R$ with capacitance $C=4 \pi \epsilon_{0} R$. If two such droplets combine to form a larger one, what would be its capacitance in terms of $C$ ?
1 $3^{1 / 3} \mathrm{C}$
2 $3^{2 / 3} \mathrm{C}$
3 $2^{2 / 3} \mathrm{C}$
4 $2^{1 / 3} \mathrm{C}$
Explanation:
: For spherical drop of mercury, $\text { Radius }=\mathrm{R}$ Capacitance $(C)=4 \pi \varepsilon_{0} R$ When two droplets combine, $\mathrm{r}=\text { smaller radius }$ $\because$ Volume of bigger drop $=$ volume of each small drop $\frac{4}{3} \pi\left(\mathrm{R}^{\prime}\right)^{3}=2 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{\prime}=2^{\frac{1}{3}} \mathrm{r}$ $\therefore$ Capacitance of bigger drop, $\mathrm{C}^{\prime}=4 \pi \varepsilon_{0} \mathrm{R}^{\prime}$ $\mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \times 2^{\frac{1}{3}} \mathrm{r}$ $\mathrm{C}^{\prime}=2^{\frac{1}{3}} \mathrm{C} \quad\left\{\because \mathrm{C}=4 \pi \varepsilon_{0} \mathrm{r}\right\}$
165632
A $5 \mu \mathrm{F}$ capacitor is connected in series with a $10 \mu \mathrm{F}$ capacitor. When a $300 \mathrm{~V}$ potential difference is applied across this combination, the total energy stored in the capacitors is
1 $15 \mathrm{~J}$
2 $1.5 \mathrm{~J}$
3 $0.15 \mathrm{~J}$
4 $0.10 \mathrm{~J}$
Explanation:
: Equivalent capacitance of the series combinations, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}$ $\mathrm{C}_{\mathrm{eq}}=\frac{10}{3} \mu \mathrm{F}$ Hence, the energy stored, $\mathrm{E}=\frac{1}{2} \mathrm{C}_{\mathrm{eq}} \mathrm{V}^{2}$ $\mathrm{E}=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times(300)^{2}$ $\mathrm{E}=0.15 \mathrm{~J}$
[WB JEE 2015]
Capacitance
165633
64 identical spheres of charges $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is
1 $64 \mathrm{q}, \mathrm{C}$
2 $16 \mathrm{q}, 4 \mathrm{C}$
3 $64 \mathrm{q}, 4 \mathrm{C}$
4 $16 \mathrm{q}, 64 \mathrm{C}$
Explanation:
: Volume of large sphere $=$ volume of each small sphere $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)$ $\mathrm{R}^{3}=64 \mathrm{r}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\because \quad \mathrm{Q}_{\text {total }}=64 \mathrm{q}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}$ $\text { Now, } \mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\text { Putting the value of } \mathrm{R}=4 \mathrm{r}$ $\mathrm{C}^{\prime}=\left(4 \pi \varepsilon_{\mathrm{o}}\right) 4 \mathrm{r}$ $\mathrm{C}^{\prime}=4 \mathrm{C}$
[WB JEE 2008]
Capacitance
165634
The capacity an isolated conducting sphere of radius $R$ is proportional to
1 $\mathrm{R}^{2}$
2 $\frac{1}{\mathrm{R}^{2}}$
3 $\frac{1}{\mathrm{R}}$
4 $\mathrm{R}$
Explanation:
: Capacitance of the conductor, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ Potential on sphere of radius, $\mathrm{R}$ [Where $\mathrm{q}$ is the charge and $\mathrm{V}$ is the potential] $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}}$ $\therefore \quad \mathrm{C}=\mathrm{Q} \times \frac{4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}}{\mathrm{Q}}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\mathrm{C} \propto \mathrm{R}$
[UP CPMT- 2005]
Capacitance
165636
Consider a spherical drop of mercury of radius $R$ with capacitance $C=4 \pi \epsilon_{0} R$. If two such droplets combine to form a larger one, what would be its capacitance in terms of $C$ ?
1 $3^{1 / 3} \mathrm{C}$
2 $3^{2 / 3} \mathrm{C}$
3 $2^{2 / 3} \mathrm{C}$
4 $2^{1 / 3} \mathrm{C}$
Explanation:
: For spherical drop of mercury, $\text { Radius }=\mathrm{R}$ Capacitance $(C)=4 \pi \varepsilon_{0} R$ When two droplets combine, $\mathrm{r}=\text { smaller radius }$ $\because$ Volume of bigger drop $=$ volume of each small drop $\frac{4}{3} \pi\left(\mathrm{R}^{\prime}\right)^{3}=2 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{\prime}=2^{\frac{1}{3}} \mathrm{r}$ $\therefore$ Capacitance of bigger drop, $\mathrm{C}^{\prime}=4 \pi \varepsilon_{0} \mathrm{R}^{\prime}$ $\mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \times 2^{\frac{1}{3}} \mathrm{r}$ $\mathrm{C}^{\prime}=2^{\frac{1}{3}} \mathrm{C} \quad\left\{\because \mathrm{C}=4 \pi \varepsilon_{0} \mathrm{r}\right\}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165632
A $5 \mu \mathrm{F}$ capacitor is connected in series with a $10 \mu \mathrm{F}$ capacitor. When a $300 \mathrm{~V}$ potential difference is applied across this combination, the total energy stored in the capacitors is
1 $15 \mathrm{~J}$
2 $1.5 \mathrm{~J}$
3 $0.15 \mathrm{~J}$
4 $0.10 \mathrm{~J}$
Explanation:
: Equivalent capacitance of the series combinations, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}$ $\mathrm{C}_{\mathrm{eq}}=\frac{10}{3} \mu \mathrm{F}$ Hence, the energy stored, $\mathrm{E}=\frac{1}{2} \mathrm{C}_{\mathrm{eq}} \mathrm{V}^{2}$ $\mathrm{E}=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times(300)^{2}$ $\mathrm{E}=0.15 \mathrm{~J}$
[WB JEE 2015]
Capacitance
165633
64 identical spheres of charges $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is
1 $64 \mathrm{q}, \mathrm{C}$
2 $16 \mathrm{q}, 4 \mathrm{C}$
3 $64 \mathrm{q}, 4 \mathrm{C}$
4 $16 \mathrm{q}, 64 \mathrm{C}$
Explanation:
: Volume of large sphere $=$ volume of each small sphere $\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)$ $\mathrm{R}^{3}=64 \mathrm{r}^{3}$ $\mathrm{R}=4 \mathrm{r}$ $\because \quad \mathrm{Q}_{\text {total }}=64 \mathrm{q}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}$ $\text { Now, } \mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\text { Putting the value of } \mathrm{R}=4 \mathrm{r}$ $\mathrm{C}^{\prime}=\left(4 \pi \varepsilon_{\mathrm{o}}\right) 4 \mathrm{r}$ $\mathrm{C}^{\prime}=4 \mathrm{C}$
[WB JEE 2008]
Capacitance
165634
The capacity an isolated conducting sphere of radius $R$ is proportional to
1 $\mathrm{R}^{2}$
2 $\frac{1}{\mathrm{R}^{2}}$
3 $\frac{1}{\mathrm{R}}$
4 $\mathrm{R}$
Explanation:
: Capacitance of the conductor, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ Potential on sphere of radius, $\mathrm{R}$ [Where $\mathrm{q}$ is the charge and $\mathrm{V}$ is the potential] $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}}$ $\therefore \quad \mathrm{C}=\mathrm{Q} \times \frac{4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}}{\mathrm{Q}}$ $\mathrm{C}=4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}$ $\mathrm{C} \propto \mathrm{R}$
[UP CPMT- 2005]
Capacitance
165636
Consider a spherical drop of mercury of radius $R$ with capacitance $C=4 \pi \epsilon_{0} R$. If two such droplets combine to form a larger one, what would be its capacitance in terms of $C$ ?
1 $3^{1 / 3} \mathrm{C}$
2 $3^{2 / 3} \mathrm{C}$
3 $2^{2 / 3} \mathrm{C}$
4 $2^{1 / 3} \mathrm{C}$
Explanation:
: For spherical drop of mercury, $\text { Radius }=\mathrm{R}$ Capacitance $(C)=4 \pi \varepsilon_{0} R$ When two droplets combine, $\mathrm{r}=\text { smaller radius }$ $\because$ Volume of bigger drop $=$ volume of each small drop $\frac{4}{3} \pi\left(\mathrm{R}^{\prime}\right)^{3}=2 \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{R}^{\prime}=2^{\frac{1}{3}} \mathrm{r}$ $\therefore$ Capacitance of bigger drop, $\mathrm{C}^{\prime}=4 \pi \varepsilon_{0} \mathrm{R}^{\prime}$ $\mathrm{C}^{\prime}=4 \pi \varepsilon_{\mathrm{o}} \times 2^{\frac{1}{3}} \mathrm{r}$ $\mathrm{C}^{\prime}=2^{\frac{1}{3}} \mathrm{C} \quad\left\{\because \mathrm{C}=4 \pi \varepsilon_{0} \mathrm{r}\right\}$
