Q. If the radius of a rotating wheel is halved while keeping the angular velocity constant, what happens to the linear velocity of a point on the edge of the wheel?
A.It doubles
B.It halves
C.It remains the same
D.It becomes zero
Solution
Linear velocity v = rω; if r is halved and ω remains constant, v is halved.
Q. If the radius of a spherical Gaussian surface is doubled while keeping the charge inside constant, how does the electric field change?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
Q. If the radius of a spherical Gaussian surface is doubled, how does the electric field change if the enclosed charge remains constant?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
Q. If the radius of the Earth is R and a satellite is in a circular orbit at a height h above the Earth's surface, what is the expression for the orbital speed v of the satellite?
A.v = sqrt(GM/(R+h))
B.v = sqrt(GM/R)
C.v = sqrt(GM/(R-h))
D.v = sqrt(GM/(R^2 + h^2))
Solution
The orbital speed v of a satellite is given by v = sqrt(GM/(R+h)), where M is the mass of the Earth and G is the gravitational constant.
Q. If the radius of the Earth is R and a satellite is in a geostationary orbit, what is the height of the satellite above the Earth's surface?
A.R/2
B.R
C.R/3
D.R/4
Solution
A geostationary satellite orbits at a height of approximately 36,000 km above the Earth's surface, which is about R (the radius of the Earth) plus the height of the satellite.
Q. If the radius of the Earth is R and a satellite is in a low Earth orbit at a height h, what is the expression for the gravitational force acting on the satellite?
A.G * M * m / (R + h)^2
B.G * M * m / R^2
C.G * M * m / (R - h)^2
D.G * M * m / (R + h)
Solution
The gravitational force acting on the satellite is given by Newton's law of gravitation, which states that F = G * (M * m) / (R + h)^2, where M is the mass of the Earth and m is the mass of the satellite.
Q. If the radius of the Earth were to increase by a factor of 2, what would happen to the gravitational acceleration at its surface?
A.It would double
B.It would remain the same
C.It would halve
D.It would become one-fourth
Solution
Gravitational acceleration is inversely proportional to the square of the radius. If the radius doubles, g becomes 1/(2^2) = 1/4 of the original value.
Q. If the radius of the Earth were to shrink to half its size while keeping its mass constant, what would happen to the gravitational acceleration at the surface?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
Gravitational acceleration is inversely proportional to the square of the radius. If the radius is halved, g becomes 4 times greater.