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 4 of 11 pages

Question 31. Two point charges each of +1

μ

C are at a distance of 3 m apart. The work done in bringing the two charges closer to a distance 1 m apart is

0.006 J
0.008 J
0.0006 J
0.06 J

Discuss Question

Answer: Option A. -> 0.006 J
:
A

W = U_{2} - U_{1} = \frac{1}{4 π ϵ_{0}} q_{1} q_{2} [\frac{1}{r_{2}} - \frac{1}{r_{1}}] = 9 \times 10^{9} \times 10^{- 12} [\frac{1}{1} - \frac{1}{3}] = 9 \times \frac{2}{3} \times 10^{- 3} W = 0.006 J

Question 32. Two Charges

Q_{1}

and

Q_{2}

are as shown in the figure. A third charge

Q_{3}

is moved from points A to B along a circular path. Change in potential energy of the system is

Two Charges Q1 And Q2 Are As Shown In The Figure. A Third ...

kQ1Q2Q3
4kQ1Q2
4kQ2Q3
23kQ2Q3

Discuss Question

Answer: Option C. -> 4kQ2Q3
:
C

Δ U = U_{f} - U_{i} U_{f} = k [\frac{Q_{1} Q_{2}}{0.6} + \frac{Q_{2} Q_{3}}{0.2} + \frac{Q_{3} Q_{1}}{0.8}] Δ U = U_{f} - U_{i} = k Q_{2} Q_{3} [\frac{1}{0.2} - 1] = 4 k Q_{2} Q_{3}

Question 33. If an electron starts from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be

3.2×10−10J
3.2×10−18J
1J
1 dyne

Discuss Question

Answer: Option B. -> 3.2×10−18J
:
B

K . E . = q_{0} (V_{A} - V_{B}) = 1.6 \times 10^{- 19} (70 - 50) = 3.2 \times 10^{- 18} J

Question 34. Three particles, each of charge

10 μ C

are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is
Given

(\frac{1}{4 π ϵ} = 9 \times 10^{9} N - m^{2} / C^{2})

Zero
Infinite
27 J
100 J

Discuss Question

Answer: Option C. -> 27 J
:
C
For pair of charge

U = \frac{1}{4 π ϵ_{0}} [\frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}}]

= 3 \times 9 \times 10^{9} \times \frac{100 \times 10^{- 12} \times 100}{10} = 27 J

Question 35. A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. The electric potential at the centre of shell will be

\frac{1}{4 π ϵ_{0}}

times:

q2R
4q3R
5q6R
2q3R

Discuss Question

Answer: Option C. -> 5q6R
:
C
Charge of -q will be induced on the the inner surface and +q on the outer surface

V = V_{1} + V_{2} + V_{3} = \frac{1}{4 π ϵ_{0}} [\frac{q}{R} - \frac{q}{2 R} + \frac{q}{3 R}]

V = \frac{1}{4 π ϵ_{0}} [\frac{6 q - 3 q + 2 q}{6 R}]

V = \frac{1}{4 π ϵ_{0}} [\frac{5 q}{6 R}]

Question 36. An electron (charge =

1.6 \times 10^{- 19}

coulomb) is accelerated through a potential of 1,00,000 volts. The energy required by the electron is

1.6×10−24 joule
1.6×10−14 joule
0.53×10−14 joule
1.6×10−14 joule

Discuss Question

Answer: Option D. -> 1.6×10−14 joule
:
D
Energy = qV=

1.6 \times 10^{- 19} \times 100000 = 1.6 \times 10^{- 14} J

Question 37. Three charges each of charge ‘q’ are placed at the corners of an equilateral triangle. The fourth charge to be placed at the centre of the triangle so that electrostatic potential energy of the system is zero is

q√3
−q√3
q
−q√2

Discuss Question

Answer: Option B. -> −q√3
:
B
PE=0

3 \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{a} + 3 \frac{1}{4 π ϵ_{0}} \times \frac{q x}{(\frac{a}{\sqrt{3}})} = 0 \sqrt{3} x = - q x = - \frac{q}{\sqrt{3}}

Question 38. Two charges

q_{1}

and

q_{2}

are placed 30 cm apart as shown in the figure. A third charge

q_{3}

is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is

\frac{q_{3}}{4 π ϵ_{0}} k

, where k is

Two Charges Q1 And Q2 Are Placed 30 Cm Apart As Shown In The...

Discuss Question

Answer: Option A. -> 8q2
:
A

Change in potential energy

(Δ U) = U_{f} - U_{i}

\Rightarrow Δ U = \frac{1}{4 π ϵ_{0}} [(\frac{q_{1} q_{3}}{0.4} + \frac{q_{2} q_{3}}{0.1}) - (\frac{q_{1} q_{3}}{0.4} + \frac{q_{2} q_{3}}{0.5})]

\Rightarrow Δ U = \frac{1}{4 π ϵ_{0}} [8 q_{2} q_{3}] = \frac{q_{3}}{4 π ϵ_{0}} (8 q_{2})

∴ k = 8 q_{2}

Question 39. Charges of

+ \frac{10}{3} \times 10^{- 9} C

are placed at each of the four corners of a square of side 8 cm. The potential at the intersection of the diagonals is

150√2 volt
1500√2 volt
900√2 volt
900 volt

Discuss Question

Answer: Option B. -> 1500√2 volt
:
B
Potential at the centre O,

V = 4 \times \frac{1}{4 π ϵ_{0}} . \frac{Q}{\frac{a}{\sqrt{2}}}

where

Q = \frac{10}{3} \times 10^{- 9} C

and

a = 8 c m = 8 \times 10^{- 2} m

Charges Of +103×10−9C Are Placed At Each Of The Four Cor...

V = 4 \times 9 \times 10^{9} \times \frac{\frac{10}{3} \times 10^{- 9}}{\frac{8 \times 10^{- 2}}{\sqrt{2}}} = 1500 \sqrt{2}

volt

Question 40. A positively charged conductor

Is always at +ve potential
Is always at zero potential
Is always at negative potential
May be at +ve, zero or -ve potential

Discuss Question

Answer: Option D. -> May be at +ve, zero or -ve potential
:
D
May be at positive, zero or negative potential, it is according to the way one defines the zero potential.