Capacitance
Capacitance

165638 Sixty four water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop as compared to that of smaller drop will be

1 4 times
2 8 times
3 16 times
4 64 times
Capacitance

165639 A parallel plate capacitor has a plate area of 50 $\mathrm{cm}^{2}$ and plate separation of $1.0 \mathrm{~cm}$. A potential difference of 200 volt is applied across the plates with air as the dielectric between plates. The battery is then disconnected and a piece of Bakelite of dielectric constant 4.8 inserted which fills the complete volume between the plates. The capacitance before and after inserting Bakelite are respectively.

1 $44 \mathrm{pF} ; 211.2 \mathrm{pF}$
2 $4.4 \mathrm{pF} ; 211.2 \mathrm{pF}$
3 $4.4 \mathrm{pF} ; 21.12 \mathrm{pF}$
4 $21.12 \mathrm{pF} ; 44 \mathrm{pF}$
Capacitance

165640 In given circuit when switch $\mathbf{S}$ has been closed then charge on capacitor $A$ \& $B$ respectively

1 $3 \mathrm{q}, 6 \mathrm{q}$
2 $6 \mathrm{q}, 3 \mathrm{q}$
3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$
4 $5 \mathrm{q}, 4 \mathrm{q}$
Capacitance

165641 A parallel plate capacitor with air as the dielectric has capacitance C.A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

1 $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$
2 $(\mathrm{K}+2) \frac{\mathrm{C}}{4}$
3 $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$
4 $\frac{\mathrm{KC}}{4}$
Capacitance

165642 Taking the Earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$. Its capacity will be

1 $711 \mu \mathrm{F}$
2 $611 \mu \mathrm{F}$
3 $811 \mu \mathrm{F}$
4 $511 \mu \mathrm{F}$
Capacitance

165638 Sixty four water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop as compared to that of smaller drop will be

1 4 times
2 8 times
3 16 times
4 64 times
Capacitance

165639 A parallel plate capacitor has a plate area of 50 $\mathrm{cm}^{2}$ and plate separation of $1.0 \mathrm{~cm}$. A potential difference of 200 volt is applied across the plates with air as the dielectric between plates. The battery is then disconnected and a piece of Bakelite of dielectric constant 4.8 inserted which fills the complete volume between the plates. The capacitance before and after inserting Bakelite are respectively.

1 $44 \mathrm{pF} ; 211.2 \mathrm{pF}$
2 $4.4 \mathrm{pF} ; 211.2 \mathrm{pF}$
3 $4.4 \mathrm{pF} ; 21.12 \mathrm{pF}$
4 $21.12 \mathrm{pF} ; 44 \mathrm{pF}$
Capacitance

165640 In given circuit when switch $\mathbf{S}$ has been closed then charge on capacitor $A$ \& $B$ respectively

1 $3 \mathrm{q}, 6 \mathrm{q}$
2 $6 \mathrm{q}, 3 \mathrm{q}$
3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$
4 $5 \mathrm{q}, 4 \mathrm{q}$
Capacitance

165641 A parallel plate capacitor with air as the dielectric has capacitance C.A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

1 $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$
2 $(\mathrm{K}+2) \frac{\mathrm{C}}{4}$
3 $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$
4 $\frac{\mathrm{KC}}{4}$
Capacitance

165642 Taking the Earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$. Its capacity will be

1 $711 \mu \mathrm{F}$
2 $611 \mu \mathrm{F}$
3 $811 \mu \mathrm{F}$
4 $511 \mu \mathrm{F}$
Capacitance

165638 Sixty four water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop as compared to that of smaller drop will be

1 4 times
2 8 times
3 16 times
4 64 times
Capacitance

165639 A parallel plate capacitor has a plate area of 50 $\mathrm{cm}^{2}$ and plate separation of $1.0 \mathrm{~cm}$. A potential difference of 200 volt is applied across the plates with air as the dielectric between plates. The battery is then disconnected and a piece of Bakelite of dielectric constant 4.8 inserted which fills the complete volume between the plates. The capacitance before and after inserting Bakelite are respectively.

1 $44 \mathrm{pF} ; 211.2 \mathrm{pF}$
2 $4.4 \mathrm{pF} ; 211.2 \mathrm{pF}$
3 $4.4 \mathrm{pF} ; 21.12 \mathrm{pF}$
4 $21.12 \mathrm{pF} ; 44 \mathrm{pF}$
Capacitance

165640 In given circuit when switch $\mathbf{S}$ has been closed then charge on capacitor $A$ \& $B$ respectively

1 $3 \mathrm{q}, 6 \mathrm{q}$
2 $6 \mathrm{q}, 3 \mathrm{q}$
3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$
4 $5 \mathrm{q}, 4 \mathrm{q}$
Capacitance

165641 A parallel plate capacitor with air as the dielectric has capacitance C.A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

1 $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$
2 $(\mathrm{K}+2) \frac{\mathrm{C}}{4}$
3 $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$
4 $\frac{\mathrm{KC}}{4}$
Capacitance

165642 Taking the Earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$. Its capacity will be

1 $711 \mu \mathrm{F}$
2 $611 \mu \mathrm{F}$
3 $811 \mu \mathrm{F}$
4 $511 \mu \mathrm{F}$
Capacitance

165638 Sixty four water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop as compared to that of smaller drop will be

1 4 times
2 8 times
3 16 times
4 64 times
Capacitance

165639 A parallel plate capacitor has a plate area of 50 $\mathrm{cm}^{2}$ and plate separation of $1.0 \mathrm{~cm}$. A potential difference of 200 volt is applied across the plates with air as the dielectric between plates. The battery is then disconnected and a piece of Bakelite of dielectric constant 4.8 inserted which fills the complete volume between the plates. The capacitance before and after inserting Bakelite are respectively.

1 $44 \mathrm{pF} ; 211.2 \mathrm{pF}$
2 $4.4 \mathrm{pF} ; 211.2 \mathrm{pF}$
3 $4.4 \mathrm{pF} ; 21.12 \mathrm{pF}$
4 $21.12 \mathrm{pF} ; 44 \mathrm{pF}$
Capacitance

165640 In given circuit when switch $\mathbf{S}$ has been closed then charge on capacitor $A$ \& $B$ respectively

1 $3 \mathrm{q}, 6 \mathrm{q}$
2 $6 \mathrm{q}, 3 \mathrm{q}$
3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$
4 $5 \mathrm{q}, 4 \mathrm{q}$
Capacitance

165641 A parallel plate capacitor with air as the dielectric has capacitance C.A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

1 $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$
2 $(\mathrm{K}+2) \frac{\mathrm{C}}{4}$
3 $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$
4 $\frac{\mathrm{KC}}{4}$
Capacitance

165642 Taking the Earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$. Its capacity will be

1 $711 \mu \mathrm{F}$
2 $611 \mu \mathrm{F}$
3 $811 \mu \mathrm{F}$
4 $511 \mu \mathrm{F}$
Capacitance

165638 Sixty four water drops having equal charges combine to form one bigger drop. The capacitance of bigger drop as compared to that of smaller drop will be

1 4 times
2 8 times
3 16 times
4 64 times
Capacitance

165639 A parallel plate capacitor has a plate area of 50 $\mathrm{cm}^{2}$ and plate separation of $1.0 \mathrm{~cm}$. A potential difference of 200 volt is applied across the plates with air as the dielectric between plates. The battery is then disconnected and a piece of Bakelite of dielectric constant 4.8 inserted which fills the complete volume between the plates. The capacitance before and after inserting Bakelite are respectively.

1 $44 \mathrm{pF} ; 211.2 \mathrm{pF}$
2 $4.4 \mathrm{pF} ; 211.2 \mathrm{pF}$
3 $4.4 \mathrm{pF} ; 21.12 \mathrm{pF}$
4 $21.12 \mathrm{pF} ; 44 \mathrm{pF}$
Capacitance

165640 In given circuit when switch $\mathbf{S}$ has been closed then charge on capacitor $A$ \& $B$ respectively

1 $3 \mathrm{q}, 6 \mathrm{q}$
2 $6 \mathrm{q}, 3 \mathrm{q}$
3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$
4 $5 \mathrm{q}, 4 \mathrm{q}$
Capacitance

165641 A parallel plate capacitor with air as the dielectric has capacitance C.A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

1 $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$
2 $(\mathrm{K}+2) \frac{\mathrm{C}}{4}$
3 $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$
4 $\frac{\mathrm{KC}}{4}$
Capacitance

165642 Taking the Earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$. Its capacity will be

1 $711 \mu \mathrm{F}$
2 $611 \mu \mathrm{F}$
3 $811 \mu \mathrm{F}$
4 $511 \mu \mathrm{F}$