Physics Syllabus (JEE Main)
Q. If the peak voltage of an AC source is 220 V, what is the RMS voltage?
A.
110 V
B.
154 V
C.
220 V
D.
311 V
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Solution
The RMS voltage (V_rms) is given by V_rms = V_peak / √2. Therefore, V_rms = 220 V / √2 = 110 V.
Correct Answer: A — 110 V
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Q. If the period of a satellite in a circular orbit is T, what is the relationship between T and the radius r of the orbit?
A.
T ∝ r
B.
T ∝ r²
C.
T ∝ √r
D.
T ∝ 1/√r
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Solution
T = 2π√(r³/GM), thus T ∝ √r.
Correct Answer: C — T ∝ √r
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Q. If the photoelectric effect is observed, what can be inferred about the incident light?
A.
It has a frequency below the threshold
B.
It has a frequency equal to the threshold
C.
It has a frequency above the threshold
D.
It has any frequency
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Solution
The photoelectric effect occurs only when the frequency of the incident light is above the threshold frequency.
Correct Answer: C — It has a frequency above the threshold
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Q. If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.
Correct Answer: B — It doubles
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Q. If the radius of a circular loop carrying current is doubled, how does the magnetic field at the center change?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The magnetic field at the center of a circular loop is inversely proportional to the radius; thus, doubling the radius halves the magnetic field.
Correct Answer: B — It halves
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Q. If the radius of a circular loop carrying current is doubled, what happens to the magnetic field at the center of the loop?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The magnetic field at the center of a circular loop is given by B = (μ₀I)/(2r). If the radius is doubled, the magnetic field strength is halved.
Correct Answer: B — It halves
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Q. If the radius of a circular loop carrying current is halved, how does the magnetic field at the center change?
A.
Remains the same
B.
Doubles
C.
Halves
D.
Quadruples
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Solution
The magnetic field at the center is inversely proportional to the radius, so it quadruples.
Correct Answer: D — Quadruples
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Q. If the radius of a disc is doubled while keeping its mass constant, how does its moment of inertia change?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
The moment of inertia of a disc is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is quadrupled.
Correct Answer: C — It quadruples
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Q. If the radius of a disk is doubled while keeping its mass constant, how does its moment of inertia change?
A.
Increases by a factor of 2
B.
Increases by a factor of 4
C.
Remains the same
D.
Decreases by a factor of 4
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Solution
The moment of inertia of a disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.
Correct Answer: B — Increases by a factor of 4
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Q. If the radius of a planet is halved while keeping its mass constant, how does the gravitational acceleration at its surface change?
A.
It becomes four times stronger
B.
It becomes twice stronger
C.
It remains the same
D.
It becomes half as strong
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Solution
If the radius is halved, the gravitational acceleration becomes four times stronger, as g is inversely proportional to the square of the radius.
Correct Answer: A — It becomes four times stronger
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Q. If the radius of a planet is halved, what happens to the gravitational acceleration on its surface?
A.
It doubles
B.
It halves
C.
It becomes one-fourth
D.
It remains the same
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Solution
g ∝ 1/R². If R is halved, g becomes 4 times greater.
Correct Answer: A — It doubles
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Q. If the radius of a rotating disc is doubled while keeping the mass constant, how does the angular momentum change if the angular velocity remains the same?
A.
It doubles
B.
It remains the same
C.
It quadruples
D.
It halves
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Solution
Angular momentum L = Iω; if radius is doubled, moment of inertia I increases by a factor of 4, hence L quadruples.
Correct Answer: C — It quadruples
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Q. If the radius of a rotating object is halved while keeping the angular velocity constant, what happens to the linear velocity at the edge?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
Linear velocity v = rω. If r is halved and ω remains constant, v also halves.
Correct Answer: B — It halves
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Q. If the radius of a rotating object is halved while keeping the mass constant, how does its moment of inertia change?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It reduces to one-fourth
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Solution
Moment of inertia I is proportional to the square of the radius, so halving the radius reduces I to one-fourth.
Correct Answer: D — It reduces to one-fourth
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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
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Solution
Linear velocity v = rω; if r is halved and ω remains constant, v is halved.
Correct Answer: B — It halves
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Q. If the radius of a solid disk is doubled while keeping its mass constant, how does its moment of inertia change?
A.
Increases by a factor of 2
B.
Increases by a factor of 4
C.
Remains the same
D.
Decreases by a factor of 2
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Solution
The moment of inertia of a solid disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.
Correct Answer: B — Increases by a factor of 4
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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 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 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 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 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
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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 spherical 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
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 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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