Work Power And Energy(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

WORK POWER AND ENERGY MCQs

Work And Energy, Sources Of Energy, Electric Potential Energy Potential And Dipoles

Total Questions : 109 | Page 5 of 11 pages

Question 41. An infinite number of charges of equal magnitude q, but of opposite sign are placed alternately starting with positive charge along x- axis at x = 1 m; x = 2 m,x = 4 m, x= 8 m and so on. The electric potential at the point x = 0 due to these charges will be (in S.I units and k =

\frac{1}{4 π ϵ_{0}}

kq2
kq3
k2q3
k3q2

Discuss Question

Answer: Option C. -> k2q3
:
C

V = V_{1} + V_{2} + V_{3} + . . . . . . . = \frac{1}{4 π ϵ_{0}} [\frac{q_{1}}{r_{1}} - \frac{q_{2}}{r_{2}} + \frac{q_{3}}{r_{3}} - \frac{q_{4}}{r_{4}} + . . . . . . .] = k [\frac{q}{1} - \frac{q}{2} + \frac{q}{4} - \frac{q}{8} + . . . . . .] = k q [1 - \frac{1}{2} + \frac{1}{2^{2}} - \frac{1}{2^{3}} + . . . . . .] = k q [\frac{1}{1 + \frac{1}{2}}] = \frac{2 k q}{3}

Question 42. A uniform electric field of magnitude

E_{0}

is directed along the positive X - axis. If the potential V is zero at x = 0, then its value at X = + x will be

Vx=+xE0
Vx=−xE0
Vx=+x2E0
Vx=−x2E0

Discuss Question

Answer: Option B. -> Vx=−xE0
:
B

∵ E = - \frac{d V}{d X} \Rightarrow V_{x} = - x E_{0}

Question 43. An electric dipole of moment p is placed in the position of stable equilibrium in uniform electric field of intensity E. If it is rotated through an angle

θ

from the initial position, the potential energy of electric dipole in the final position is

pE cosθ
pE sinθ
pE(1−cosθ)
−pE cosθ

Discuss Question

Answer: Option D. -> −pE cosθ
:
D
Potential energy of dipole in electric field is

U = - p E c o s θ

; where

θ

is the angle between electric fieldand dipole moment vector.

Question 44. Three charges 2q, -q, -q are located at the vertices of an equilateral triangle. At the centre of the triangle

The field is zero but potential is non-zero
The field is non-zero but potential is zero
Both field and potential are zero
Both field and potential are non-zero

Discuss Question

Answer: Option B. -> The field is non-zero but potential is zero
:
B
Obviously, from charge configuration, at the centre electric field is non-zero. Potential at the centre due to 2q charge is

V_{2 q} = \frac{2 q}{r}

Three Charges 2q, -q, -q Are Located At The Vertices Of An ...

and the potential due to – q charge is

V_{- q} = - \frac{q}{r}

(r = distance of centre)

∴

Total potential

V = V_{2 q} + V_{- q} + V_{- q} = 0

Question 45. Two point charges 20mC and - 60 mC are separated by a distance of 16cm. The net electric potential is zero on the line joining them

At a distance of 4cm from 20mC in between the charges
At a distance of 8cm from 20mC and 24cm from -60 mC
Both of the above
None of the above

Discuss Question

Answer: Option C. -> Both of the above
:
C

x = \frac{d}{\frac{q_{2}}{q_{1}} \pm 1}; x_{1} = \frac{d}{\frac{q_{2}}{q_{1}} + 1} = \frac{6}{\frac{60}{20} + 1} = 4 c m x_{2} = \frac{16}{\frac{60}{20} - 1} = \frac{16}{2} = 8 c m

point outside

Question 46. The work done in carrying a charge of

5 μ C

from a point A to a point B in an electric field is 10mJ. The potential difference

(V_{B} - V_{A})

+ 2kV
- 2 kV
+ 200 V
- 200 V

Discuss Question

Answer: Option A. -> + 2kV
:
A
Work done

W = Q (V_{B} - V_{A}) \Rightarrow (V_{B} - V_{A}) = \frac{W}{Q} = \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}} J / C = 2 k V

Question 47. The blocks A and B shown in fig. Have masses

M_{A} = 5 k g

and

M_{B} = 4 k g

. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is

The Blocks A And B Shown In Fig. Have Masses MA=5kg And MB=4...

√32√g
√34√g
√g2√3
√g2

Discuss Question

Answer: Option C. -> √g2√3
:
C
If A moves down the incline by 1 m, B shal move up by

\frac{1}{2} m

.
If the speed of B is v, then the speed of A will be 2v.
From conservation of energy,
Gain in KE = loss in PE

\frac{1}{2} m_{A} (2 v)^{2} + \frac{1}{2} m_{B} v^{2} = m_{A} g \times \frac{3}{5} - m_{B} g \times \frac{1}{2}

Solving, we get

v = \frac{\sqrt{g}}{2 \sqrt{3}}

Question 48. A man pushes a wall and fails to displace it. He does

Negative work
Positive but not maximum work
No work at all
Maximum work

Discuss Question

Answer: Option C. -> No work at all
:
C
No displacement is there.Hence work is zero.

Question 49. The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a force F such that the force F is directly proportional to t is