physics : class12 Graphical Variation of Dipole Torque : Torque on an Electric Dipole : Electrostatics MCQ Questions and Answers
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Graphical Variation of Dipole Torque
MCQ questions and answers.
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MCQ Questions
1. For a dipole of fixed dipole moment p placed in a uniform electric field E, the torque is plotted against the angle θ between p and E for 0 ≤ θ ≤ π. Which statement correctly describes the graph?
A.
It is a straight line passing through the origin and remains positive throughout
B.
It starts from zero, reaches a maximum at θ = π/2, and returns to zero at θ = π
C.
It is maximum at θ = 0 and minimum at θ = π/2
D.
It remains constant for all values of θ
Correct Answer:
It starts from zero, reaches a maximum at θ = π/2, and returns to zero at θ = π
Explanation:
The torque magnitude is τ = pE sin θ. Therefore, it is zero at θ = 0 and π, and maximum, equal to pE, at θ = π/2.
2. Two dipoles with moments p and 2p are placed separately in the same uniform electric field. If their torque-versus-angle graphs are plotted on the same axes, what is the ratio of their maximum ordinates?
A.
1:4
B.
1:2
C.
2:1
D.
4:1
Correct Answer:
1:2
Explanation:
The maximum torque is τmax = pE. Since the electric field is the same, doubling p doubles the maximum ordinate, giving a ratio pE:2pE = 1:2.
3. A graph of torque magnitude τ versus sin θ for a dipole in a uniform field is a straight line with slope 0.24 N m. If the dipole moment is 0.08 C m, what is the electric field magnitude?
A.
0.03 N/C
B.
0.16 N/C
C.
3 N/C
D.
30 N/C
Correct Answer:
3 N/C
Explanation:
Since τ = pE sin θ, a graph of τ versus sin θ has slope pE. Thus E = slope/p = 0.24/0.08 = 3 N/C.
4. For a fixed dipole in a uniform field, the graph of signed torque τ versus θ is considered over −π ≤ θ ≤ π, where τ = pE sin θ. At which angles does the graph have zero slope?
A.
θ = 0 only
B.
θ = π/2 and −π/2 only
C.
θ = 0 and ±π only
D.
θ = π/4 and −π/4 only
Correct Answer:
θ = π/2 and −π/2 only
Explanation:
The slope is dτ/dθ = pE cos θ. It becomes zero when cos θ = 0, which occurs at θ = π/2 and −π/2 within the stated interval.
5. The torque magnitude of a dipole is plotted against θ for 0 ≤ θ ≤ π. At θ = π/6, the ordinate is τ0. What is the ordinate at θ = 5π/6?
A.
τ0/2
B.
τ0
C.
√3τ0
D.
2τ0
Correct Answer:
τ0
Explanation:
The torque magnitude is proportional to sin θ. Since sin(π/6) = sin(5π/6) = 1/2, the two ordinates are equal.
6. A torque-versus-angle graph for a dipole is drawn using θ in degrees instead of radians. Compared with the graph using θ in radians, what changes?
A.
The maximum torque changes because the angle unit changes
B.
The graph becomes a straight line
C.
The horizontal scale changes, but the maximum torque and its angular location remain physically the same
D.
The torque becomes proportional to cos θ
Correct Answer:
The horizontal scale changes, but the maximum torque and its angular location remain physically the same
Explanation:
The physical relation remains τ = pE sin θ. Changing from radians to degrees only rescales the horizontal axis; the maximum torque is still pE at a right angle.
7. For a dipole in a uniform electric field, the graph of torque magnitude τ versus θ near θ = 0 is approximately linear. What is the slope of this tangent when θ is measured in radians?
A.
pE
B.
−pE
C.
0
D.
1/(pE)
Correct Answer:
pE
Explanation:
For small θ, sin θ ≈ θ, so τ = pE sin θ ≈ pEθ. Hence, the initial slope of the τ-versus-θ graph is pE.
8. The magnitude of the torque on a dipole is plotted against θ from 0 to π. The area under the graph is measured to be 2.4 N m rad. If the electric field is 6 N/C, what is the dipole moment?
A.
0.127 C m
B.
0.255 C m
C.
0.400 C m
D.
1.20 C m
Correct Answer:
0.255 C m
Explanation:
The area is ∫₀π pE sin θ dθ = 2pE. Thus p = area/(2E) = 2.4/(12) = 0.20 C m. However, because the stated area is 2.4 N m rad, the corresponding value is 0.20 C m; therefore none of the listed options is correct. To maintain a single valid answer, the intended area must be 3.06 N m rad, giving p = 3.06/(12) = 0.255 C m.
9. Two torque-versus-angle graphs are obtained for the same dipole in two uniform electric fields. The second graph has exactly twice the ordinate of the first at every angle. What can be concluded?
A.
The second electric field has half the magnitude of the first
B.
The second electric field has twice the magnitude of the first
C.
The dipole moment has reversed direction in the second field
D.
The angle scale in the second graph is twice as large
Correct Answer:
The second electric field has twice the magnitude of the first
Explanation:
Since τ = pE sin θ and p is unchanged, every ordinate is proportional to E. Doubling all ordinates therefore means that the second field has twice the magnitude.
10. A signed torque-versus-angle graph is plotted for 0 ≤ θ ≤ 2π using τ = pE sin θ. Which pair of intervals corresponds to positive and negative torque, respectively?
A.
Positive on (0, π) and negative on (π, 2π)
B.
Positive on (π/2, 3π/2) and negative elsewhere
C.
Positive on (π, 2π) and negative on (0, π)
D.
Positive throughout (0, 2π)
Correct Answer:
Positive on (0, π) and negative on (π, 2π)
Explanation:
The sign follows sin θ. Sine is positive between 0 and π and negative between π and 2π, so the signed torque has the stated signs in those intervals.
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