Q. If the radius of a sphere is halved, how does its volume change?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It reduces to one-eighth
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Solution
Volume of a sphere = (4/3)πr³. If radius is halved, volume becomes (4/3)π(1/2)³ = (4/3)π(1/8) = (1/6)π, which is one-eighth of the original volume.
Correct Answer: D — It reduces to one-eighth
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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
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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.
Correct Answer: B — It halves
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Q. If the radius of a spherical Gaussian surface is doubled, how does the electric field due to a point charge at its center change?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
The electric field due to a point charge is independent of the radius of the Gaussian surface; it remains the same.
Correct Answer: C — It remains the same
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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
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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.
Correct Answer: B — It halves
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Q. If the radius of curvature of a concave mirror is 20 cm, what is its focal length? (2022)
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
The focal length (f) of a concave mirror is given by f = R/2, where R is the radius of curvature. Thus, f = 20 cm / 2 = 10 cm.
Correct Answer: B — 10 cm
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Q. If the radius of curvature of a concave mirror is 40 cm, what is its focal length?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length f of a mirror is given by f = R/2. Thus, f = 40 cm / 2 = 20 cm.
Correct Answer: B — 20 cm
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Q. If the radius of curvature of a convex lens is 20 cm, what is its focal length? (2022)
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length f of a lens is given by f = R/2. Therefore, f = 20 cm / 2 = 10 cm.
Correct Answer: B — 20 cm
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Q. If the radius of curvature of a convex mirror is 20 cm, what is its focal length? (2022)
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length (f) of a mirror is given by f = R/2. For a convex mirror, R is positive, so f = 20 cm / 2 = 10 cm.
Correct Answer: A — 10 cm
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Q. If the radius of curvature of a convex mirror is 30 cm, what is its focal length?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
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Solution
The focal length f of a mirror is given by f = R/2. For a convex mirror, R = 30 cm, so f = 30/2 = 15 cm.
Correct Answer: B — 15 cm
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Q. If the radius of curvature of a lens is 20 cm, what is the focal length of the lens assuming it is made of glass with a refractive index of 1.5?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
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Solution
Using the lens maker's formula, f = R/(n-1) = 20/(1.5-1) = 40 cm.
Correct Answer: A — 10 cm
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Q. If the radius of curvature of a lens is 30 cm and the refractive index is 1.5, what is the focal length of the lens?
A.
10 cm
B.
15 cm
C.
20 cm
D.
25 cm
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Solution
Using the lens maker's formula, f = R/(n-1) = 30/(1.5-1) = 30/0.5 = 60 cm.
Correct Answer: B — 15 cm
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Q. If the radius of curvature of a lens is 30 cm, what is the focal length of the lens assuming it is made of glass with a refractive index of 1.5?
A.
10 cm
B.
15 cm
C.
20 cm
D.
25 cm
Show solution
Solution
Using the lens maker's formula, f = R/(n-1) = 30/(1.5-1) = 30/0.5 = 60 cm.
Correct Answer: B — 15 cm
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Q. If the radius of curvature of a spherical mirror is 30 cm, what is its focal length?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
Show solution
Solution
Focal length f = R/2 = 30 cm / 2 = 15 cm.
Correct Answer: B — 15 cm
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Q. If the radius of curvature of a spherical mirror is 40 cm, what is its focal length?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length (f) of a spherical mirror is given by f = R/2. Here, R = 40 cm, so f = 40/2 = 20 cm.
Correct Answer: B — 20 cm
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Q. If the radius of the Earth is 6400 km, what is the total distance from the center of the Earth to a satellite in a geostationary orbit? (2000)
A.
36000 km
B.
42000 km
C.
32000 km
D.
28000 km
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Solution
The distance from the center of the Earth to a geostationary satellite is approximately 42000 km.
Correct Answer: B — 42000 km
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Q. If the radius of the Earth is doubled, what will be the change in gravitational force experienced by an object on its surface?
A.
It will double
B.
It will remain the same
C.
It will be halved
D.
It will be quartered
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Solution
Gravitational force is inversely proportional to the square of the radius. If the radius is doubled, the force becomes 1/4 of the original.
Correct Answer: D — It will be quartered
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Q. If the radius of the Earth is doubled, what will be the change in gravitational acceleration at its surface?
A.
It will double
B.
It will remain the same
C.
It will be halved
D.
It will increase by 4 times
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Solution
g = G * M / R²; if R is doubled, g becomes g/4.
Correct Answer: C — It will be halved
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Q. If the radius of the Earth is doubled, what will happen to the gravitational force experienced by an object on its surface?
A.
It will double
B.
It will remain the same
C.
It will be halved
D.
It will be quartered
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Solution
Gravitational force is inversely proportional to the square of the radius. If radius is doubled, F becomes F/4.
Correct Answer: D — It will be quartered
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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))
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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.
Correct Answer: A — v = sqrt(GM/(R+h))
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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
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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.
Correct Answer: B — R
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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)
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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.
Correct Answer: A — G * M * m / (R + h)^2
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Q. If the radius of the Earth is R, what is the gravitational acceleration at a height R above the Earth's surface?
A.
g/4
B.
g/2
C.
g
D.
g/8
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Solution
At a height R above the Earth's surface, the gravitational acceleration is g/4, derived from the formula g' = g(R/(R+h))^2.
Correct Answer: A — g/4
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Q. If the radius of the Earth is R, what is the radius of a satellite in a geostationary orbit?
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Solution
The radius of a geostationary orbit is approximately 3R, where R is the radius of the Earth.
Correct Answer: C — 3R
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Q. If the radius of the Earth were to double, what would happen to the gravitational force experienced by a satellite in low Earth orbit?
A.
It would double
B.
It would remain the same
C.
It would decrease to one-fourth
D.
It would increase to four times
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Solution
The gravitational force is inversely proportional to the square of the distance. If the radius doubles, the force decreases to one-fourth.
Correct Answer: C — It would decrease to one-fourth
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Q. If the radius of the Earth were to double, what would happen to the gravitational acceleration at its surface?
A.
It would double
B.
It would remain the same
C.
It would be halved
D.
It would be quartered
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Solution
Gravitational acceleration is inversely proportional to the square of the radius, so it would be quartered.
Correct Answer: D — It would be quartered
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Q. If the radius of the Earth were to double, what would happen to the weight of an object on its surface?
A.
It would double
B.
It would remain the same
C.
It would become four times less
D.
It would become four times more
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Solution
Weight is proportional to 1/r^2; if the radius doubles, the weight becomes four times less.
Correct Answer: C — It would become four times less
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Q. If the radius of the Earth were to increase by 50%, what would happen to the weight of an object on its surface?
A.
It increases by 50%
B.
It increases by 25%
C.
It decreases by 25%
D.
It remains the same
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Solution
Weight is inversely proportional to the square of the radius. If radius increases by 50%, weight decreases by 25%.
Correct Answer: C — It decreases by 25%
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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
Show solution
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.
Correct Answer: C — It would halve
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Q. If the radius of the Earth were to increase while keeping its mass constant, what would happen to the gravitational acceleration at its surface?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
Show solution
Solution
Gravitational acceleration g = GM/R². If R increases, g decreases.
Correct Answer: B — It decreases
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Q. If the radius of the Earth were to increase while keeping its mass constant, what would happen to the value of g? (2023)
A.
g increases
B.
g decreases
C.
g remains the same
D.
g becomes zero
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Solution
If the radius increases while mass remains constant, g decreases because g = G * M / r^2.
Correct Answer: B — g decreases
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