11th And 12th > Physics
CAPACITANCE MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option C. ->
:
C
:
C
Q1=Q2+Q3 because in series combination charge is same on both the
condenser and V=V1+V2 because in parallel combination V2=V3
Hence V=V1+V2
Answer: Option B. ->
The total charge on the two spheres is conserved
:
B
:
B
The total charge on the spheres will be conserved
Answer: Option C. ->
The field between the plates
:
C
In a charged capacitor energy is stored in the form electric field
:
C
In a charged capacitor energy is stored in the form electric field
Answer: Option D. ->
All of the above
:
D
:
D
As the capacitor is isolated, the potential decreases as the capcitance increases
Answer: Option D. ->
A decrease in the energy of the system unless Q1R2=Q2R1
:
D
:
D
When Potentials of spheres are unequal, current will flow in connecting wire so that energy decreases in the form of heat through the connecting wire.
Answer: Option B. ->
3
:
B
CseriesCparallel=(Cn)nc=1n2
19=1n2
n=3
:
B
CseriesCparallel=(Cn)nc=1n2
19=1n2
n=3
Answer: Option D. ->
4q3
:
D
Initial charge on sphere of radius R = q
Charge on this sphere after joining =q′=Rq+(−2q)×RR+2R=−q×R3R=−q3
So charge flown between them =q−(−q3)=4q3
:
D
Initial charge on sphere of radius R = q
Charge on this sphere after joining =q′=Rq+(−2q)×RR+2R=−q×R3R=−q3
So charge flown between them =q−(−q3)=4q3
Answer: Option C. ->
6
:
C
C=ϵ0Ad;C1=kϵ0A(d2)
C1C=2k;
12=2k; k=122=6
:
C
C=ϵ0Ad;C1=kϵ0A(d2)
C1C=2k;
12=2k; k=122=6
Answer: Option D. ->
9 KV
:
D
q1=C1V1=10−3×6×103=6C
q1=C2V2=2×10−3×4×103=8C
Q=q1=6C Q=CV
6=(C1C2C1+C2)V
6=23×10−3V
V=9×103V
=9KV
:
D
q1=C1V1=10−3×6×103=6C
q1=C2V2=2×10−3×4×103=8C
Q=q1=6C Q=CV
6=(C1C2C1+C2)V
6=23×10−3V
V=9×103V
=9KV
Answer: Option B. ->
4 V
:
B
q=C1V1=C1V2
V1V2=C1C1=94.5=21
V1+V2V2=2+11
V2=123=4V
:
B
q=C1V1=C1V2
V1V2=C1C1=94.5=21
V1+V2V2=2+11
V2=123=4V