Physics Syllabus (JEE Main)
Q. If a metal has a work function of 4 eV, what is the minimum wavelength of light required to cause the photoelectric effect?
A.
310 nm
B.
620 nm
C.
1240 nm
D.
2480 nm
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Solution
The minimum wavelength can be calculated using the equation λ = hc/W. Substituting h = 4.14 x 10^-15 eV·s, c = 3 x 10^8 m/s, and W = 4 eV gives λ = 310 nm.
Correct Answer: A — 310 nm
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Q. If a metal has a work function of 4 eV, what is the threshold frequency for the photoelectric effect?
A.
1.5 x 10^14 Hz
B.
3.2 x 10^14 Hz
C.
5.0 x 10^14 Hz
D.
6.3 x 10^14 Hz
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Solution
The threshold frequency can be calculated using the formula f = φ/h, where h is Planck's constant (4.14 x 10^-15 eV·s).
Correct Answer: B — 3.2 x 10^14 Hz
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Q. If a metal rod is heated at one end, how does the heat travel to the other end?
A.
By convection
B.
By radiation
C.
By conduction
D.
By insulation
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Solution
Heat travels through the metal rod by conduction, as the kinetic energy is transferred from one atom to another.
Correct Answer: C — By conduction
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Q. If a needle is carefully placed on the surface of water, it floats due to:
A.
Cohesion
B.
Adhesion
C.
Surface tension
D.
Density
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Solution
The needle floats because the surface tension of water creates a force that counteracts the weight of the needle.
Correct Answer: C — Surface tension
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Q. If a person lifts a 10 kg object to a height of 5 m in 2 seconds, what is the power exerted?
A.
25 W
B.
50 W
C.
75 W
D.
100 W
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Solution
First, calculate the work done: W = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J. Then, P = W/t = 490 J / 2 s = 245 W.
Correct Answer: B — 50 W
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Q. If a person lifts a 10 kg object to a height of 5 m in 5 seconds, what is the power exerted?
A.
100 W
B.
200 W
C.
300 W
D.
400 W
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Solution
First, calculate the work done: W = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J. Then, power is P = W/t = 490 J / 5 s = 98 W.
Correct Answer: B — 200 W
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Q. If a person lifts a 10 kg weight to a height of 2 meters in 4 seconds, what is the power exerted?
A.
50 W
B.
100 W
C.
200 W
D.
250 W
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Solution
First, calculate the work done: W = mgh = 10 kg * 9.8 m/s² * 2 m = 196 J. Then, power is P = W/t = 196 J / 4 s = 49 W, approximately 50 W.
Correct Answer: B — 100 W
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Q. If a person lifts a 50 kg box to a height of 2 meters in 5 seconds, what is the power exerted?
A.
200 W
B.
400 W
C.
500 W
D.
600 W
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Solution
The work done against gravity is W = mgh = (50 kg)(9.8 m/s²)(2 m) = 980 J. Power is P = W/t = 980 J / 5 s = 196 W, approximately 200 W.
Correct Answer: B — 400 W
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Q. If a person lifts a 50 kg object to a height of 2 meters in 5 seconds, what is the power exerted?
A.
200 W
B.
400 W
C.
500 W
D.
600 W
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Solution
The work done is W = mgh = 50 kg * 9.8 m/s^2 * 2 m = 980 J. Power is W/t = 980 J / 5 s = 196 W, approximately 200 W.
Correct Answer: B — 400 W
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Q. If a person lifts a weight of 100 N to a height of 2 m in 4 seconds, what is the power exerted?
A.
50 Watts
B.
100 Watts
C.
200 Watts
D.
25 Watts
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Solution
Work done = Force × Distance = 100 N × 2 m = 200 J. Power = Work / Time = 200 J / 4 s = 50 W.
Correct Answer: A — 50 Watts
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Q. If a photon has a wavelength of 300 nm, what is its energy?
A.
4.14 eV
B.
6.2 eV
C.
8.2 eV
D.
10.0 eV
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Solution
Energy (E) = hc / λ = (1240 nm·eV) / 300 nm = 4.13 eV.
Correct Answer: B — 6.2 eV
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Q. If a photon has energy of 5 eV, what is its corresponding frequency?
A.
6.2 x 10^14 Hz
B.
7.5 x 10^14 Hz
C.
8.0 x 10^14 Hz
D.
9.0 x 10^14 Hz
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Solution
Frequency (ν) = E/h = 5 eV / (4.14 x 10^-15 eV·s) = 7.5 x 10^14 Hz.
Correct Answer: B — 7.5 x 10^14 Hz
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Q. If a photon with energy 5 eV strikes a metal with work function 3 eV, what is the kinetic energy of the emitted electron?
A.
2 eV
B.
3 eV
C.
5 eV
D.
8 eV
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Solution
Kinetic energy (K.E.) = Photon energy - Work function = 5 eV - 3 eV = 2 eV.
Correct Answer: A — 2 eV
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Q. If a planet rotates about its axis, which of the following statements is true regarding its angular momentum?
A.
It is zero
B.
It is constant
C.
It changes with time
D.
It depends on the distance from the sun
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Solution
The angular momentum of a rotating planet is constant if no external torques act on it.
Correct Answer: B — It is constant
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface of radius R, what is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀R²
D.
Q/4πε₀
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Solution
The total electric flux through the Gaussian surface is given by Φ = Q/ε₀, where Q is the charge enclosed.
Correct Answer: B — Q/ε₀
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface of radius R, what is the electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/2ε₀
D.
Q/4ε₀
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Solution
The electric flux through a closed surface is given by Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.
Correct Answer: B — Q/ε₀
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface, what is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀
D.
4πQ/ε₀
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Solution
According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed divided by ε₀, thus Φ = Q/ε₀.
Correct Answer: B — Q/ε₀
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Q. If a potentiometer is used to compare two cells, what is the effect of increasing the length of the wire?
A.
It increases the accuracy of the measurement
B.
It decreases the accuracy of the measurement
C.
It has no effect on the measurement
D.
It makes the cells equal
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Solution
Increasing the length of the wire increases the potential gradient, which can improve the accuracy of the measurement by allowing for finer adjustments.
Correct Answer: A — It increases the accuracy of the measurement
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Q. If a potentiometer is used to compare two emfs, what is the key requirement for the two circuits?
A.
They must have the same resistance
B.
They must have the same length of wire
C.
They must be connected in series
D.
They must have the same potential gradient
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Solution
For accurate comparison of two emfs using a potentiometer, both circuits must have the same potential gradient to ensure that the readings are directly comparable.
Correct Answer: D — They must have the same potential gradient
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Q. If a potentiometer is used to measure a voltage of 12 V and the length of the wire used is 6 m, what is the potential gradient?
A.
2 V/m
B.
4 V/m
C.
6 V/m
D.
8 V/m
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Solution
The potential gradient is calculated as the potential difference divided by the length of the wire: 12 V / 6 m = 2 V/m.
Correct Answer: A — 2 V/m
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Q. If a potentiometer is used to measure the EMF of a cell and the balance point is found at 3 m, what can be inferred about the cell's EMF if the potential gradient is 1.5 V/m?
A.
1.5 V
B.
3 V
C.
4.5 V
D.
6 V
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Solution
The EMF of the cell can be calculated as EMF = potential gradient × length = 1.5 V/m × 3 m = 4.5 V.
Correct Answer: C — 4.5 V
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Q. If a potentiometer is used to measure the potential difference across a resistor, what must be true about the circuit?
A.
The resistor must be in series with the potentiometer.
B.
The potentiometer must be connected in parallel with the resistor.
C.
The potentiometer must be connected in series with the power supply.
D.
The resistor must be short-circuited.
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Solution
To measure the potential difference across a resistor, the potentiometer must be connected in parallel with the resistor.
Correct Answer: B — The potentiometer must be connected in parallel with the resistor.
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Q. If a potentiometer wire has a potential difference of 12 V and a length of 6 m, what is the potential gradient?
A.
2 V/m
B.
1 V/m
C.
0.5 V/m
D.
3 V/m
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Solution
The potential gradient is calculated as the potential difference divided by the length of the wire: 12 V / 6 m = 2 V/m.
Correct Answer: A — 2 V/m
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Q. If a potentiometer wire has a potential difference of 12V across it and is 4m long, what is the potential gradient?
A.
3 V/m
B.
4 V/m
C.
2 V/m
D.
1 V/m
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Solution
The potential gradient is calculated as V/L = 12V/4m = 3 V/m.
Correct Answer: A — 3 V/m
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Q. If a potentiometer wire has a resistance of 10 ohms and a current of 0.5 A flows through it, what is the potential difference across the wire?
A.
2.5 V
B.
5 V
C.
10 V
D.
15 V
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Solution
Using Ohm's law, V = IR = 0.5 A * 10 ohms = 5 V.
Correct Answer: B — 5 V
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Q. If a ray of light passes from air into a medium with a refractive index of 2, what is the angle of refraction if the angle of incidence is 30°?
A.
15°
B.
30°
C.
45°
D.
60°
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Solution
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). 1 * sin(30°) = 2 * sin(θ2) => sin(θ2) = 0.25 => θ2 = 15°.
Correct Answer: C — 45°
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Q. If a ray of light passes through the optical center of a lens, what happens to the ray?
A.
It bends towards the normal.
B.
It bends away from the normal.
C.
It continues in a straight line.
D.
It converges to a point.
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Solution
A ray of light passing through the optical center of a lens continues in a straight line without bending.
Correct Answer: C — It continues in a straight line.
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Q. If a resistor has a resistance of 5 ohms and a current of 2 amperes flows through it, what is the voltage across the resistor?
A.
10 V
B.
5 V
C.
2.5 V
D.
1 V
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Solution
Using Ohm's Law, V = I * R = 2 A * 5 Ω = 10 V.
Correct Answer: A — 10 V
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Q. If a resistor is connected to a 9V battery and the current flowing through it is 3A, what is the resistance of the resistor?
A.
1 Ω
B.
3 Ω
C.
9 Ω
D.
27 Ω
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Solution
Using Ohm's Law, R = V / I = 9 V / 3 A = 3 Ω.
Correct Answer: C — 9 Ω
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Q. If a rolling object has a mass m and radius R, what is the expression for its rotational kinetic energy?
A.
(1/2)Iω^2
B.
(1/2)mv^2
C.
(1/2)mv^2/R^2
D.
(1/2)mv^2 + (1/2)Iω^2
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Solution
The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia.
Correct Answer: A — (1/2)Iω^2
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