Physics Syllabus (JEE Main)
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 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 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 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 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
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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 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
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Solution
Gravitational acceleration is inversely proportional to the square of the radius. If the radius is halved, g becomes 4 times greater.
Correct Answer: D — It quadruples
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Q. If the radius of the orbit of a satellite is doubled, what happens to its orbital speed?
A.
It remains the same
B.
It doubles
C.
It increases by a factor of √2
D.
It decreases by a factor of √2
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Solution
The orbital speed v is given by v = √(GM/r). If r is doubled, v decreases by a factor of √2.
Correct Answer: D — It decreases by a factor of √2
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Q. If the rate of change of current in an inductor is 2 A/s, what is the induced EMF if the inductance is 3 H?
A.
6 V
B.
3 V
C.
2 V
D.
1 V
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Solution
Induced EMF (ε) = -L(dI/dt) = -3 H * 2 A/s = -6 V.
Correct Answer: A — 6 V
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Q. If the refractive index of a medium is 1.33, what is the critical angle for total internal reflection?
A.
48.6°
B.
60.0°
C.
30.0°
D.
45.0°
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Solution
Critical angle θc = sin⁻¹(1/1.33) ≈ 48.6°.
Correct Answer: A — 48.6°
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Q. If the refractive index of a medium is 1.33, what is the maximum angle of incidence for total internal reflection when light travels to air?
A.
41.8°
B.
48.6°
C.
53.1°
D.
60.0°
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Solution
The critical angle θc can be calculated as θc = sin^(-1)(n2/n1) = sin^(-1)(1.00/1.33) ≈ 48.6°.
Correct Answer: A — 41.8°
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Q. If the refractive index of a medium is 1.33, what is the speed of light in that medium if the speed of light in vacuum is 3 x 10^8 m/s?
A.
2.25 x 10^8 m/s
B.
2.5 x 10^8 m/s
C.
2.75 x 10^8 m/s
D.
3 x 10^8 m/s
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Solution
Speed of light in medium = c/n = (3 x 10^8 m/s) / 1.33 ≈ 2.25 x 10^8 m/s.
Correct Answer: A — 2.25 x 10^8 m/s
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Q. If the refractive index of a medium is 1.5, what is the maximum angle of incidence for total internal reflection when light travels to air?
A.
41.8°
B.
48.6°
C.
60.0°
D.
90.0°
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Solution
The critical angle θc is given by sin(θc) = n2/n1 = 1/1.5, which gives θc ≈ 41.8°.
Correct Answer: A — 41.8°
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Q. If the refractive index of a medium is 1.5, what is the speed of light in that medium? (Speed of light in vacuum = 3 x 10^8 m/s)
A.
2 x 10^8 m/s
B.
1.5 x 10^8 m/s
C.
3 x 10^8 m/s
D.
4.5 x 10^8 m/s
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Solution
Speed of light in medium = c/n = (3 x 10^8 m/s) / 1.5 = 2 x 10^8 m/s.
Correct Answer: A — 2 x 10^8 m/s
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Q. If the refractive index of a medium is 1.5, what is the speed of light in that medium?
A.
2 x 10^8 m/s
B.
3 x 10^8 m/s
C.
1.5 x 10^8 m/s
D.
4.5 x 10^8 m/s
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Solution
The speed of light in a medium is given by v = c/n. Here, c = 3 x 10^8 m/s and n = 1.5, so v = 3 x 10^8 / 1.5 = 2 x 10^8 m/s.
Correct Answer: A — 2 x 10^8 m/s
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Q. If the refractive index of a medium is 1.5, what is the wavelength of light in that medium if the wavelength in vacuum is 600 nm?
A.
400 nm
B.
600 nm
C.
800 nm
D.
900 nm
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Solution
Wavelength in medium = λ/v = λ0/n = 600 nm / 1.5 = 400 nm.
Correct Answer: A — 400 nm
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Q. If the refractive index of a medium is 2.0, what is the critical angle for total internal reflection at the interface with air?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
Using the formula sin(θc) = n2/n1, we have sin(θc) = 1.00/2.00, leading to θc ≈ 30°.
Correct Answer: C — 60°
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Q. If the refractive index of a medium is 2.0, what is the critical angle for total internal reflection when light travels from this medium to air?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
Using the formula sin(θc) = n2/n1, where n1 = 2.0 (medium) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.0 = 0.5, leading to θc = 60°.
Correct Answer: C — 60°
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Q. If the refractive index of a medium is 2.0, what is the critical angle for total internal reflection?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
Using the formula θc = sin^(-1)(1/n), where n = 2.0, we find θc = sin^(-1)(1/2) = 30°.
Correct Answer: C — 60°
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Q. If the refractive index of a medium is 2.0, what is the maximum angle of incidence for total internal reflection when light travels to air?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.00/2.00) ≈ 30°; thus, maximum angle of incidence for total internal reflection is 60°.
Correct Answer: C — 60°
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Q. If the refractive index of a medium is greater than 1, how does it affect the speed of light in that medium?
A.
Increases speed
B.
Decreases speed
C.
No effect
D.
Depends on wavelength
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Solution
The speed of light in a medium is given by v = c/n, where n is the refractive index. If n > 1, the speed decreases.
Correct Answer: B — Decreases speed
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Q. If the refractive index of a medium is greater than 1, what happens to the speed of light in that medium?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes infinite
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Solution
The speed of light in a medium is given by v = c/n, where n is the refractive index. If n > 1, then v < c, meaning the speed decreases.
Correct Answer: B — It decreases
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Q. If the refractive index of a medium is greater than 1, what happens to the speed of light in that medium compared to vacuum?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes infinite
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Solution
The speed of light in a medium is less than that in vacuum, given by v = c/n, where n > 1.
Correct Answer: B — It decreases
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Q. If the refractive index of a medium is greater than 1, what happens to the wavelength of light in that medium?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
The wavelength of light decreases in a medium with a refractive index greater than 1, as λ' = λ/n.
Correct Answer: B — It decreases
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Q. If the refractive index of a medium is greater than 1, what happens to the wavelength of light in that medium compared to its wavelength in vacuum?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
The wavelength of light decreases in a medium with a refractive index greater than 1, as it is given by λ' = λ/n.
Correct Answer: B — It decreases
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Q. If the refractive index of a thin film is greater than that of the surrounding medium, what happens to the phase of the reflected wave?
A.
No phase change
B.
Phase change of π
C.
Phase change of 2π
D.
Phase change of λ/2
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Solution
When light reflects off a medium with a higher refractive index, it undergoes a phase change of π (180 degrees).
Correct Answer: B — Phase change of π
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Q. If the refractive index of diamond is 2.42, what is the critical angle for total internal reflection when light travels from diamond to air?
A.
24.4°
B.
30.0°
C.
36.9°
D.
41.8°
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Solution
Using the formula sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.00 (air), we find θc ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. If the resistance in a circuit is doubled while keeping the voltage constant, what happens to the current?
A.
It doubles.
B.
It halves.
C.
It remains the same.
D.
It quadruples.
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Solution
According to Ohm's Law (I = V/R), if resistance (R) is doubled and voltage (V) remains constant, the current (I) will be halved.
Correct Answer: B — It halves.
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