Q. What is the work done by a constant force of 10 N acting over a distance of 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
25 J
Show solution
Solution
Work done W = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer:
A
— 50 J
Learn More →
Q. What is the work done by a constant force of 10 N moving an object 5 m in the direction of the force?
A.
10 J
B.
25 J
C.
50 J
D.
100 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) = 10 N × 5 m = 50 J.
Correct Answer:
C
— 50 J
Learn More →
Q. What is the work done by a force of 10 N moving an object 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
0 J
Show solution
Solution
Work done W = Force x Distance = 10 N x 5 m = 50 J.
Correct Answer:
A
— 50 J
Learn More →
Q. What is the work done by a force of 15 N acting at an angle of 60° while moving an object 4 m? (2021)
A.
30 J
B.
60 J
C.
45 J
D.
90 J
Show solution
Solution
Work = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 × 4 × 0.5 = 30 J.
Correct Answer:
C
— 45 J
Learn More →
Q. What is the work done by a force of 15 N moving an object 4 m in the direction of the force? (2022)
A.
60 J
B.
30 J
C.
45 J
D.
75 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) = 15 N × 4 m = 60 J.
Correct Answer:
A
— 60 J
Learn More →
Q. What is the work done by a force of 25 N moving an object 4 m at an angle of 60° to the direction of the force? (2019)
A.
50 J
B.
100 J
C.
75 J
D.
25 J
Show solution
Solution
Work done = F × d × cos(θ) = 25 N × 4 m × cos(60°) = 25 N × 4 m × 0.5 = 50 J.
Correct Answer:
C
— 75 J
Learn More →
Q. What is the work done by a force of 5 N moving an object 3 m in the direction of the force?
A.
15 J
B.
5 J
C.
3 J
D.
0 J
Show solution
Solution
Work done W = Force × Distance = 5 N × 3 m = 15 J.
Correct Answer:
A
— 15 J
Learn More →
Q. What is the work done by a gas during an isobaric expansion?
A.
Zero
B.
PΔV
C.
ΔU
D.
Q
Show solution
Solution
In an isobaric process, the work done by the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume.
Correct Answer:
B
— PΔV
Learn More →
Q. What is the work done by a gas during an isothermal expansion from volume Vi to Vf?
A.
nRT ln(Vf/Vi)
B.
nR(Tf - Ti)
C.
Zero
D.
nRT
Show solution
Solution
The work done by a gas during an isothermal expansion is W = nRT ln(Vf/Vi).
Correct Answer:
A
— nRT ln(Vf/Vi)
Learn More →
Q. What is the work done by a gas during expansion against a constant external pressure?
Show solution
Solution
The work done by a gas during expansion against a constant external pressure is given by W = PΔV, where P is the pressure and ΔV is the change in volume.
Correct Answer:
A
— PΔV
Learn More →
Q. What is the work done by a gas during expansion at constant pressure?
Show solution
Solution
The work done by a gas during expansion at constant pressure is given by W = PΔV.
Correct Answer:
A
— PΔV
Learn More →
Q. What is the work done by a gas during isothermal expansion?
A.
Zero
B.
Depends on the temperature
C.
Is equal to the heat absorbed
D.
Is equal to the change in internal energy
Show solution
Solution
During isothermal expansion, the work done by the gas is equal to the heat absorbed, as the internal energy remains constant.
Correct Answer:
C
— Is equal to the heat absorbed
Learn More →
Q. What is the work done by an ideal gas during an isobaric expansion if it expands from 2 L to 5 L at a pressure of 3 atm? (2022)
A.
9 J
B.
12 J
C.
15 J
D.
18 J
Show solution
Solution
Work done (W) = P * ΔV = 3 atm * (5 L - 2 L) = 3 atm * 3 L = 9 atm·L = 15 J (1 atm·L = 101.325 J)
Correct Answer:
C
— 15 J
Learn More →
Q. What is the work done by an ideal gas during an isobaric expansion if the pressure is 5 atm and the volume changes from 2 L to 5 L? (2019)
A.
15 L·atm
B.
10 L·atm
C.
5 L·atm
D.
20 L·atm
Show solution
Solution
Work done (W) = PΔV = P(V_final - V_initial) = 5 atm * (5 L - 2 L) = 15 L·atm.
Correct Answer:
A
— 15 L·atm
Learn More →
Q. What is the work done by an ideal gas during an isobaric expansion? (2019)
Show solution
Solution
The work done by an ideal gas during an isobaric (constant pressure) expansion is given by W = PΔV.
Correct Answer:
A
— PΔV
Learn More →
Q. What is the work done by an ideal gas during an isobaric process?
A.
PΔV
B.
nRT
C.
0
D.
nRΔT
Show solution
Solution
The work done by an ideal gas during an isobaric process is given by W = PΔV.
Correct Answer:
A
— PΔV
Learn More →
Q. What is the work done by an ideal gas during an isothermal expansion from volume Vi to Vf?
A.
nRT ln(Vf/Vi)
B.
nR(Tf - Ti)
C.
Zero
D.
nRT
Show solution
Solution
The work done by an ideal gas during an isothermal expansion is W = nRT ln(Vf/Vi).
Correct Answer:
A
— nRT ln(Vf/Vi)
Learn More →
Q. What is the work done by an ideal gas during an isothermal expansion from volume V1 to V2 at temperature T? (2021)
A.
nRT ln(V2/V1)
B.
nRT (V2 - V1)
C.
PV
D.
0
Show solution
Solution
The work done during an isothermal expansion is given by W = nRT ln(V2/V1).
Correct Answer:
A
— nRT ln(V2/V1)
Learn More →
Q. What is the work done by an ideal gas during an isothermal expansion?
A.
Zero
B.
nRT ln(Vf/Vi)
C.
nRT (Vf - Vi)
D.
nR (Tf - Ti)
Show solution
Solution
The work done by an ideal gas during an isothermal expansion is given by W = nRT ln(Vf/Vi).
Correct Answer:
B
— nRT ln(Vf/Vi)
Learn More →
Q. What is the work done by an ideal gas during isothermal expansion? (2021)
A.
0
B.
nRT ln(Vf/Vi)
C.
nRT (Vf - Vi)
D.
nR (Tf - Ti)
Show solution
Solution
For an ideal gas undergoing isothermal expansion, the work done is given by W = nRT ln(Vf/Vi).
Correct Answer:
B
— nRT ln(Vf/Vi)
Learn More →
Q. What is the work done by friction when a 5 kg block slides 2 m on a surface with a coefficient of kinetic friction of 0.4?
A.
-4 N·m
B.
-8 N·m
C.
-10 N·m
D.
-20 N·m
Show solution
Solution
Frictional force F_friction = μk * N = 0.4 * 5 kg * 9.8 m/s² = 19.6 N. Work done by friction = -F_friction * distance = -19.6 N * 2 m = -39.2 N·m, approximately -40 N·m.
Correct Answer:
B
— -8 N·m
Learn More →
Q. What is the work done by the system during an isochoric process?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the temperature
Show solution
Solution
In an isochoric process, the volume remains constant, so no work is done by the system (W = PΔV = 0).
Correct Answer:
C
— Zero
Learn More →
Q. What is the work done by the system in an isochoric process?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the temperature
Show solution
Solution
In an isochoric process, the volume remains constant, so no work is done by the system (W = PΔV = 0).
Correct Answer:
C
— Zero
Learn More →
Q. What is the work done by the system when it expands against a constant external pressure?
Show solution
Solution
The work done by the system during expansion against a constant external pressure is given by W = PΔV.
Correct Answer:
A
— PΔV
Learn More →
Q. What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?
A.
400 J
B.
200 J
C.
100 J
D.
80 J
Show solution
Solution
Work done W = F * d = q * E * d = (2 × 10^-6 C) * (500 N/C) * (0.4 m) = 0.4 J = 80 J.
Correct Answer:
D
— 80 J
Learn More →
Q. What is the work done in moving a charge of +2μC through a potential difference of 10V?
A.
20 μJ
B.
200 μJ
C.
2 μJ
D.
0.2 μJ
Show solution
Solution
W = q * V = 2 × 10^-6 C * 10 V = 20 × 10^-6 J = 20 μJ.
Correct Answer:
B
— 200 μJ
Learn More →
Q. What is the work done in moving a charge of +2μC through a potential difference of 5V?
A.
10 μJ
B.
5 μJ
C.
2 μJ
D.
1 μJ
Show solution
Solution
Work done W = q * V = (2 × 10^-6 C) * (5 V) = 10 μJ.
Correct Answer:
A
— 10 μJ
Learn More →
Q. What is the work done in moving a charge of 2 μC from a point at 50 V to a point at 100 V?
A.
0.1 mJ
B.
0.2 mJ
C.
0.3 mJ
D.
0.4 mJ
Show solution
Solution
Work done W = q * ΔV = 2 × 10^-6 C * (100 V - 50 V) = 2 × 10^-6 * 50 = 0.1 mJ.
Correct Answer:
B
— 0.2 mJ
Learn More →
Q. What is the work done in moving a charge of 2 μC from a point at 50 V to another at 100 V?
A.
100 μJ
B.
200 μJ
C.
150 μJ
D.
50 μJ
Show solution
Solution
Work done W = q * ΔV = (2 × 10^-6 C) * (100 V - 50 V) = 100 μJ.
Correct Answer:
B
— 200 μJ
Learn More →
Q. What is the work done in moving a mass from a point A to point B in a uniform gravitational field?
A.
Depends on the path taken
B.
Zero
C.
Equal to the change in gravitational potential energy
D.
Equal to the gravitational force times distance
Show solution
Solution
The work done is equal to the change in gravitational potential energy, which depends on the initial and final positions.
Correct Answer:
C
— Equal to the change in gravitational potential energy
Learn More →
Showing 22291 to 22320 of 28118 (938 Pages)
