Capacitance
Capacitance

165618 If the distance between the plates of a parallel plate capacitor of capacity $10 \mu \mathrm{F}$ is doubled, then new capacity will be

1 $5 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $10 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165619 A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, then capacitance of capacitor becomes

1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Capacitance

165620 If the distance between the plates of parallel plate capacitor is halved and the dielectric constant is doubled, then its capacity will

1 increase by 16 times
2 increase by 4 times
3 increase by 2 times
4 remain the same
Capacitance

165621 A parallel plate capacitor of capacitance 100 $\mathrm{pF}$ is to be constructed by using paper sheets of $1 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is 4 , the number of circular metal foils of diameter $2 \mathrm{~cm}$ each required for the purpose is

1 40
2 20
3 30
4 10
Capacitance

165618 If the distance between the plates of a parallel plate capacitor of capacity $10 \mu \mathrm{F}$ is doubled, then new capacity will be

1 $5 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $10 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165619 A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, then capacitance of capacitor becomes

1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Capacitance

165620 If the distance between the plates of parallel plate capacitor is halved and the dielectric constant is doubled, then its capacity will

1 increase by 16 times
2 increase by 4 times
3 increase by 2 times
4 remain the same
Capacitance

165621 A parallel plate capacitor of capacitance 100 $\mathrm{pF}$ is to be constructed by using paper sheets of $1 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is 4 , the number of circular metal foils of diameter $2 \mathrm{~cm}$ each required for the purpose is

1 40
2 20
3 30
4 10
Capacitance

165618 If the distance between the plates of a parallel plate capacitor of capacity $10 \mu \mathrm{F}$ is doubled, then new capacity will be

1 $5 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $10 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165619 A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, then capacitance of capacitor becomes

1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Capacitance

165620 If the distance between the plates of parallel plate capacitor is halved and the dielectric constant is doubled, then its capacity will

1 increase by 16 times
2 increase by 4 times
3 increase by 2 times
4 remain the same
Capacitance

165621 A parallel plate capacitor of capacitance 100 $\mathrm{pF}$ is to be constructed by using paper sheets of $1 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is 4 , the number of circular metal foils of diameter $2 \mathrm{~cm}$ each required for the purpose is

1 40
2 20
3 30
4 10
Capacitance

165618 If the distance between the plates of a parallel plate capacitor of capacity $10 \mu \mathrm{F}$ is doubled, then new capacity will be

1 $5 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $10 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165619 A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, then capacitance of capacitor becomes

1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Capacitance

165620 If the distance between the plates of parallel plate capacitor is halved and the dielectric constant is doubled, then its capacity will

1 increase by 16 times
2 increase by 4 times
3 increase by 2 times
4 remain the same
Capacitance

165621 A parallel plate capacitor of capacitance 100 $\mathrm{pF}$ is to be constructed by using paper sheets of $1 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is 4 , the number of circular metal foils of diameter $2 \mathrm{~cm}$ each required for the purpose is

1 40
2 20
3 30
4 10
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