Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point, assuming it is 2 meters from the mass creating the field?
A.-20 J/kg
B.-10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.
Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point, assuming it is at a distance of 2 meters from the mass creating the field?
A.-20 J/kg
B.-10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.
Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point assuming the reference point is at infinity?
A.-10 J/kg
B.10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
The gravitational potential V is related to the gravitational field strength g by V = -g * r. If we consider r to be 1 meter, V = -10 * 1 = -10 J/kg.
Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point, assuming it is 2 meters above the reference point?
A.-20 J/kg
B.-10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
Gravitational potential V = -g * h = -10 N/kg * 2 m = -20 J/kg.
Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point assuming it is at a distance of 2 m from the mass?
A.-20 J/kg
B.-10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
Gravitational potential V = -g * r = -10 * 2 = -20 J/kg.
Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point, assuming it is at a distance of 2 m from the mass creating the field?
A.-20 J/kg
B.-10 J/kg
C.0 J/kg
D.-5 J/kg
Solution
Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.
Q. If the gravitational potential at a point is -15 J/kg and the gravitational field strength is constant at 3 N/kg, what is the distance from the mass?
A.5 m
B.10 m
C.15 m
D.20 m
Solution
Using V = -g * r, we have -15 = -3 * r, thus r = 15/3 = 5 m.
Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons in the photoelectric effect?
A.It doubles
B.It remains the same
C.It is halved
D.It becomes zero
Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons?
A.It doubles
B.It remains the same
C.It is halved
D.It becomes zero
Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Q. If the internal resistance of a cell is negligible, how does it affect the accuracy of a potentiometer measurement?
A.It increases accuracy
B.It decreases accuracy
C.It has no effect
D.It makes the measurement impossible
Solution
If the internal resistance of a cell is negligible, it increases the accuracy of the potentiometer measurement as it does not affect the voltage being measured.
Q. If the length of a side of a cube is measured as 2.0 ± 0.1 m, what is the maximum possible error in the volume of the cube?
A.0.8 m³
B.0.4 m³
C.0.2 m³
D.0.1 m³
Solution
Volume V = L³. The maximum error in volume can be calculated using the formula: ΔV = 3L²ΔL. Here, ΔL = 0.1 m, L = 2.0 m, so ΔV = 3(2.0)²(0.1) = 1.2 m³.
