Physics Questions and Answers – Energy Consideration: A Quantitative Study

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This set of Physics Online Test for Class 12 focuses on “Energy Consideration: A Quantitative Study”.

1. Identify the significance of energy consideration from the following.
a) Energy consideration provides a link between the force and energy
b) Energy consideration provides a link between the current and energy
c) Energy consideration provides a link between the mechanical and thermal energy
d) Energy consideration provides a link between the voltage and energy
View Answer

Answer: a
Explanation: Energy consideration provides a link the force and energy. Newton’s problems can also be easily solved with the help of energy consideration. This is the significance of energy consideration.
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2. Pick out the expression for power in a rectangular conductor from the following.
a) P=\(\frac {B^2 l^2 v}{R}\)
b) P=\(\frac {B^2 lv^2}{R}\)
c) P=\(\frac {B^2 l^2 v^2}{R}\)
d) P=\(\frac {B^2 l^2 v^2}{R^2}\)
View Answer

Answer: c
Explanation: In a rectangular conductor, the emf (E) is given ➔ Blv
So, current (I) = \(\frac {E}{R} = \frac {Blv}{R}\)
Power (P) = I2R
P=\(\frac {B^2 l^2 v^2}{R}\)
The work done is mechanical and this mechanical energy is dissipated as Joule heat.

3. A circular coil having an area of 3.14 × 10-2 m2 and 30 turns is rotated about its vertical diameter with an angular speed of 70 rad-1 in a uniform horizontal magnetic field of magnitude 5 × 10-2 T. If the coil forms a closed loop of resistance 15 Ω, what is the average power loss due to Joule heating?
a) 0.360
b) 0.362
c) 0.724
d) 0.726
View Answer

Answer: b
Explanation: Given: N = 30; A = 3.14 × 10-2 m2 ; Angular velocity (ω) = 70 rad-1; B = 5 × 10-2 T; R = 15 Ω
Emf (E) = NABω
E = 30 × 3.14 × 10-2 × 5 × 10-2 × 70
E = 3.297 V
Current (I) = \(\frac {E}{R}\)
I = \(\frac {3.297}{15}\)
I = 0.2198 A
Average power loss (P) = \(\frac {I^2 R}{2}\)
P = 0.21982 × \(\frac {15}{2}\)
P = 0.362 W
Therefore, the average power loss due to Joule heating is 0.362 W.
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4. Mechanical energy is initially converted into thermal energy in the case of motional emf.
a) True
b) False
View Answer

Answer: b
Explanation: No, this statement is false. The work done is mechanical and the energy is dissipated as mechanical energy. Then it is initially converted into electric energy and then only finally converted into thermal energy.

5. What is the magnitude of the force on a current carrying conductor of length L in a perpendicular magnetic field B?
a) F=\(\frac {I}{lB}\)
b) F=IlB2
c) F=I2 lB
d) F=IlB
View Answer

Answer: d
Explanation: The current passing through conductor in a perpendicular magnetic field is given as:
I=\(\frac {\varepsilon}{r}\)
I=\(\frac {Blv}{r}\)
Since it is a perpendicular magnetic field, the force ➔ I(l×B) is directed outwards in the direction opposite to the velocity of the rod. The magnitude of the force is given as:
F=IlB
We can also write it as ➔ F=IlB=\(\frac {B^2 l^2 v}{r}\)
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6. Find the true statement.
a) The force on a current carrying conductor in a perpendicular magnetic field is due to the thermal energy dissipated
b) The force on a current carrying conductor in a perpendicular magnetic field is due to the mechanical energy dissipated
c) The force on a current carrying conductor in a perpendicular magnetic field is due to the current passing through it
d) The force on a current carrying conductor in a perpendicular magnetic field is due to the electrical energy initially converted from mechanical energy
View Answer

Answer: d
Explanation: The force on a current carrying conductor in a perpendicular magnetic field is due to drift velocity of charges along the current carrying conductor and the consequent Lorentz force acting on it. So, it is due to the electric energy which was the energy that mechanical energy dissipated in the system was initially converted to.

7. Identify the relation between charge flow and change in magnetic flux.
a) ∆Q = ∆ΦB×r
b) ∆Q = 2∆Φ;B r
c) ∆Q = \(\frac {\triangle \Phi_B}{r}\)
d) ∆Q = \(\frac {\triangle \Phi_B}{2r}\)
View Answer

Answer: c
Explanation: There is a relationship between the charge flow through the circuit and the change in the magnetic flux. According to Faraday’s Law, the induced emf is given by:
|ε|=\( \frac {\triangle \Phi_B}{\triangle t}\)
But, |ε|=Ir=\(( \frac {\triangle Q}{\triangle t} ) \)r
Therefore, the relation is given by:
∆Q = \(\frac {\triangle \Phi_B}{r}\)
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Sanfoundry Global Education & Learning Series – Physics – Class 12.

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