Work Power And Energy(12th Grade > Physics ) Questions and Answers
Explanation:-
Answer: Option C. -> 24KV:
C
q1+q2=14μC...(1)V=q1c1=q2c2;q1q2=4πϵ0r14πϵ0r2;q1q2=r1r2=34q1=6μC;q2=8μCV2=14πϵ0q2r=9×109×8×10−63V2=24×103V
Explanation:-
Answer: Option A. -> 4.5×105V:
A
Assume the point at which field is zero is at a distance X from 4μC
X=d√q2q1+1=50√94+1=5032+1=1005=20cmV=V1+V2=14πϵ0[q1r1+q2r2]=9×109[4×10−60.2+9×10−60.3]=9×103[20+30]=9×103×50=4.5×105V
Explanation:-
Answer: Option B. -> Zero:
B
Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.
W=→F.→s=Fscosθ=Fscos90∘=0
Explanation:-
Answer: Option A. -> 180 V:
A
12mv2=qV
12×4×10−3×9×10−2=10−6V
V=180V
Explanation:-
Answer: Option A. -> 64 J:
A
Potential at the centre of square
V=4×(9×109×50×10−6√2)=90√2×104V
Work done in bringing a charge (q = 50μC) from ∞ to centre (O) of the square is
W=q(V0−V∞)=qV0
⇒W=50×10−6×90√2×104=64J
Explanation:-
Answer: Option C. -> Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere:
C
After redistribution, charges on them will be different, but they will acquire common potential
i.e. kQ1r1=kQ2r2⇒Q1Q2=r1r2
As σ=Q4πr2⇒σ1σ2=Q1Q2×r22r21⇒σ∝1r
i.e. surface charge density on smaller sphere will be more.
Explanation:-
Answer: Option B. -> WA=WB=WC=0:
B
According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force hence no work is done. WA=WB=WC=0
Explanation:-
Answer: Option B. -> 10 eV:
B
ΔE=2e×5V=10eV⇒ Final kinetic energy =10eV