1. Putting a dielectric substance between two plates of
condenser, capacity, potential and potential energy
respectively
a) Increase, decrease, decrease
b) Decrease, increase, increase
c) Increase, increase, increase
d) Decrease, decrease, decrease
Explanation: Increase, decrease, decrease
2. A thin metal plate P is inserted half way between the plates
of a parallel plate capacitor of capacitance C in such a way
that it is parallel to the two plates. The capacitance now
becomes
a) C
b) C/2
c) 4C
d) None of these
Explanation: Thin metal plates doesn't affect the capacitance
3.If there are n capacitors in parallel connected to V volt
source, then the energy stored is equal to
a) CV
b) \[\frac{1}{2}nCV^{2}\]
c) \[CV^{2}\]
d) \[\frac{1}{2n}CV^{2}\]
Explanation:
4.If n drops, each of capacitance C, coalesce to form a single
big drop, then the ratio of the energy stored in the big drop
to that in each small drop will be
a) n : 1
b) \[n^{1/3} : 1\]
c) \[n^{5/3} : 1\]
d) \[n^{2} : 1\]
Explanation:
5. A conducting sphere of radius 10cm is charged \[10\mu C\] .
Another uncharged sphere of radius 20 cm is allowed to
touch it for some time. After that if the sphere are
separated, then surface density of charges, on the spheres
will be in the ratio of
a) 1 : 4
b) 1 : 3
c) 2 : 1
d) 1 : 1
Explanation:
6. 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
a) 1 : 64
b) 64 : 1
c) 4 : 1
d) 1 : 4
Explanation:
7.Capacitance (in F) of a spherical conductor with radius 1m
is
a) \[1.1\times 10^{-10}\]
b) \[ 10^{-6}\]
c) \[9\times 10^{-9}\]
d) \[ 10^{-3}\]
Explanation:
8. A condenser has a capacity 2 \[\mu F\] and is charged to a
voltage of 50 V. The energy stored is
a) \[25\times 10^{5}\] Joule
b) 25 Joule
c) \[25\times 10\] erg
d) \[25\times 10^{3}\] erg
Explanation:
9. The energy required to charge a capacitor of 5 \[\mu F\] by
connecting a d.c. source of 20 kV is
a) 10 kJ
b) 5 kJ
c) 2 kJ
d) 1 kJ
Explanation:
10. The capacitance of a parallel plate capacitor is 12 \[\mu F\] . If the
distance between the plates is doubled and area is halved,
then new capacitance will be
a) 8 \[\mu F\]
b) 6 \[\mu F\]
c) 4 \[\mu F\]
d) 3 \[\mu F\]
Explanation:
