12th Grade > Physics
CAPACITANCE MCQs
Total Questions : 26
| Page 2 of 3 pages
Answer: Option D. -> 9 KV
:
D
q1=C1V1=10−3×6×103=6C
q1=C2V2=2×10−3×4×103=8C
Q=q1=6CQ=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=6CQ=CV
6=(C1C2C1+C2)V
6=23×10−3V
V=9×103V
=9KV
Answer: Option A. -> 25 mC
:
A
V=q1+q2C1+C2=q1+q24πϵ0[r1+r2]=150×10−3×9×10915×10−2
V=9×109V
q11=C1V=4πϵ0r1×9×109=19×109×5×10−2×9×109=5×10−2C
q11=50mc
q1−q11=75−50=25mC
:
A
V=q1+q2C1+C2=q1+q24πϵ0[r1+r2]=150×10−3×9×10915×10−2
V=9×109V
q11=C1V=4πϵ0r1×9×109=19×109×5×10−2×9×109=5×10−2C
q11=50mc
q1−q11=75−50=25mC
Answer: Option B. -> 3
:
B
CseriesCparallel=(Cn)nc=1n2
19=1n2
n=3
:
B
CseriesCparallel=(Cn)nc=1n2
19=1n2
n=3
Answer: Option B. -> The total charge on the two spheres is conserved
:
B
The total charge on the spheres will be conserved
:
B
The total charge on the spheres will be conserved
Question 15. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by Δ T, the potential difference V across the capacitance was?
Answer: Option D. -> √2msΔTC
:
D
12CV2=mSΔT
V2=2mSΔTC
V=√2msΔTC
:
D
12CV2=mSΔT
V2=2mSΔTC
V=√2msΔTC
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. -> A decrease in the energy of the system unless Q1R2=Q2R1
:
D
WhenPotentials of spheres are unequal, current will flow in connecting wire so that energy decreases in the form of heat through the connecting wire.
:
D
WhenPotentials 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 C. -> 5
:
C
x=1=[1−1k]
1.6=2[1−1k]
1k=1−0.8
=0.2
k=10.2=5
:
C
x=1=[1−1k]
1.6=2[1−1k]
1k=1−0.8
=0.2
k=10.2=5
Answer: Option B. -> 5V
:
B
Common potential V=C1V1+C2V2C1+C2V=(100×10+0)2C110−3=12×100×10−3=1000200=5V.
:
B
Common potential V=C1V1+C2V2C1+C2V=(100×10+0)2C110−3=12×100×10−3=1000200=5V.
Answer: Option C. -> 2:1
:
C
After redistribution, new charges on sphere are
Q1=(1010+20)×10=103μC
and Q2=(2010+20)×10=203μC
Ratio of charge densities
σ1σ2=Q1Q2×r22r21=(103203)×(2010)2=21
:
C
After redistribution, new charges on sphere are
Q1=(1010+20)×10=103μC
and Q2=(2010+20)×10=203μC
Ratio of charge densities
σ1σ2=Q1Q2×r22r21=(103203)×(2010)2=21