Physics Syllabus (JEE Main)
Q. If an object is moved to a height equal to the radius of the Earth, how does the gravitational force acting on it change?
A.
It becomes half
B.
It becomes one-fourth
C.
It remains the same
D.
It becomes zero
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Solution
At a height equal to the radius of the Earth, the gravitational force becomes one-fourth of its value at the surface.
Correct Answer: B — It becomes one-fourth
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Q. If a^3 = 27, then the value of a is?
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Solution
Taking the cube root of both sides, a = 27^(1/3) = 3.
Correct Answer: C — 3
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Q. If a^3 = 27, what is the value of a?
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Solution
Since 27 = 3^3, we have a = 3.
Correct Answer: C — 3
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Q. If a^x = b^y and a = b^k, what is the relationship between x, y, and k?
A.
x = ky
B.
y = kx
C.
x + y = k
D.
x - y = k
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Solution
Substituting a = b^k into a^x = b^y gives (b^k)^x = b^y, leading to kx = y, hence x = ky.
Correct Answer: A — x = ky
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Q. If a^x = b^y, then log_a(b) is equal to?
A.
y/x
B.
x/y
C.
x+y
D.
y-x
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Solution
From a^x = b^y, taking log_a on both sides gives log_a(b^y) = x, hence y * log_a(b) = x, thus log_a(b) = y/x.
Correct Answer: A — y/x
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Q. If light of wavelength 500 nm passes through a diffraction grating with 1000 lines/mm, what is the angle for the first-order maximum?
A.
30 degrees
B.
60 degrees
C.
45 degrees
D.
15 degrees
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Solution
Using the grating equation d sin(θ) = mλ, where d = 1/1000 mm = 1 x 10^-6 m, for m=1, we find θ ≈ 30 degrees.
Correct Answer: A — 30 degrees
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Q. If log_10(1000) = x, what is the value of x?
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Solution
Since 1000 = 10^3, log_10(1000) = 3, thus x = 3.
Correct Answer: B — 3
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Q. If one of the resistors in a Wheatstone bridge is replaced with a variable resistor, what is the effect on the balance condition?
A.
It cannot be balanced
B.
It can be balanced by adjusting the variable resistor
C.
It will always be unbalanced
D.
It will short-circuit the bridge
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Solution
The variable resistor allows for adjustment to achieve balance in the bridge.
Correct Answer: B — It can be balanced by adjusting the variable resistor
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Q. If R1 = 10Ω, R2 = 15Ω, R3 = 5Ω, what should R4 be for the Wheatstone bridge to be balanced?
A.
7.5Ω
B.
10Ω
C.
12.5Ω
D.
15Ω
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Solution
Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (15 * 5) / 10 = 7.5Ω.
Correct Answer: C — 12.5Ω
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Q. If R1 = 10Ω, R2 = 20Ω, and R3 = 30Ω in a Wheatstone bridge, what should R4 be for the bridge to be balanced?
A.
15Ω
B.
20Ω
C.
25Ω
D.
30Ω
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Solution
For balance, R1/R2 = R3/R4, thus R4 = (R2 * R3) / R1 = (20 * 30) / 10 = 60Ω.
Correct Answer: B — 20Ω
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Q. If R1 = 10Ω, R2 = 20Ω, R3 = 30Ω, and R4 = 60Ω in a Wheatstone bridge, is the bridge balanced?
A.
Yes
B.
No
C.
Depends on the voltage
D.
Depends on the current
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Solution
The bridge is not balanced because 10/20 ≠ 30/60.
Correct Answer: B — No
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Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω in a Wheatstone bridge, what is the value of R4 for balance?
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Solution
Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.
Correct Answer: A — 4Ω
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Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω, what is the value of R4 for the Wheatstone bridge to be balanced?
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Solution
Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.
Correct Answer: A — 4Ω
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Q. If R1 = 5Ω, R2 = 10Ω, and R3 = 15Ω in a Wheatstone bridge, what is the value of R4 for balance?
A.
7.5Ω
B.
10Ω
C.
15Ω
D.
20Ω
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Solution
Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (10 * 15) / 5 = 30Ω.
Correct Answer: B — 10Ω
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Q. If R1 = 5Ω, R2 = 15Ω, R3 = 10Ω, what should R4 be for the Wheatstone bridge to be balanced?
A.
30Ω
B.
10Ω
C.
15Ω
D.
5Ω
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Solution
For balance, R4 must be 30Ω since 5/15 = 10/30.
Correct Answer: B — 10Ω
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Q. If the amplitude of a damped oscillator decreases to half its value in 5 seconds, what is the damping ratio?
A.
0.1
B.
0.2
C.
0.3
D.
0.4
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Solution
Using the formula A(t) = A_0 e^(-ζω_nt), we find ζ = 0.2.
Correct Answer: B — 0.2
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Q. If the amplitude of a simple harmonic motion is doubled, how does the maximum velocity change?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
Maximum velocity V_max = Aω. If A is doubled, V_max also doubles.
Correct Answer: A — It doubles
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Q. If the amplitude of a simple harmonic motion is doubled, how does the total energy change?
A.
Remains the same
B.
Doubles
C.
Quadruples
D.
Halves
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Solution
The total energy E in SHM is given by E = (1/2)kA². If A is doubled, E becomes (1/2)k(2A)² = 4(1/2)kA², which is quadrupled.
Correct Answer: C — Quadruples
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Q. If the amplitude of a simple harmonic motion is halved, how does the maximum velocity change?
A.
Halved
B.
Doubled
C.
Remains the same
D.
Quadrupled
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Solution
Maximum velocity V_max = ωA. If A is halved, V_max is also halved.
Correct Answer: A — Halved
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Q. If the amplitude of a simple harmonic oscillator is doubled, how does the total energy change?
A.
Remains the same
B.
Doubles
C.
Quadruples
D.
Halves
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Solution
The total energy in simple harmonic motion is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 2^2 = 4.
Correct Answer: C — Quadruples
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Q. If the amplitude of a simple harmonic oscillator is doubled, what happens to its total energy?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
The total energy of a simple harmonic oscillator is proportional to the square of the amplitude. If the amplitude is doubled, the energy increases by a factor of 2^2 = 4.
Correct Answer: C — It quadruples
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Q. If the amplitude of a simple harmonic oscillator is halved, how does the total energy change?
A.
Remains the same
B.
Halved
C.
Doubled
D.
Quadrupled
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Solution
The total energy in SHM is proportional to the square of the amplitude. If amplitude is halved, energy is reduced to (1/2)^2 = 1/4, which is halved.
Correct Answer: B — Halved
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Q. If the amplitude of a wave is doubled, how does the energy of the wave change?
A.
Remains the same
B.
Doubles
C.
Increases by a factor of four
D.
Increases by a factor of eight
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Solution
The energy of a wave is proportional to the square of the amplitude. If the amplitude is doubled, the energy increases by a factor of 2^2 = 4.
Correct Answer: C — Increases by a factor of four
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Q. If the amplitude of a wave is doubled, what happens to its energy?
A.
Remains the same
B.
Doubles
C.
Increases by a factor of four
D.
Increases by a factor of eight
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Solution
The energy of a wave is proportional to the square of its amplitude. Therefore, if the amplitude is doubled, the energy increases by a factor of four.
Correct Answer: C — Increases by a factor of four
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Q. If the amplitude of a wave is tripled, how does the energy of the wave change?
A.
Increases by a factor of 3
B.
Increases by a factor of 6
C.
Increases by a factor of 9
D.
Remains the same
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Solution
Energy is proportional to the square of the amplitude, so if amplitude is tripled, energy increases by a factor of 3^2 = 9.
Correct Answer: C — Increases by a factor of 9
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Q. If the amplitude of a wave is tripled, what happens to its energy?
A.
Increases by a factor of 3
B.
Increases by a factor of 6
C.
Increases by a factor of 9
D.
Remains the same
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Solution
The energy of a wave is proportional to the square of its amplitude, so if amplitude is tripled, energy increases by a factor of 3^2 = 9.
Correct Answer: C — Increases by a factor of 9
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Q. If the angle between the current element and the line joining the current element to the point of interest is 90 degrees, what is the contribution of that current element to the magnetic field?
A.
Maximum
B.
Minimum
C.
Zero
D.
Undefined
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Solution
If the angle is 90 degrees, the sine of the angle is zero, resulting in zero contribution to the magnetic field from that current element.
Correct Answer: C — Zero
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Q. If the angle between the force and the lever arm is 90 degrees, how does it affect the torque?
A.
Torque is zero
B.
Torque is maximum
C.
Torque is half
D.
Torque is minimum
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Solution
Torque is maximum when the angle is 90 degrees because τ = F × r × sin(θ) and sin(90°) = 1.
Correct Answer: B — Torque is maximum
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Q. If the angle between the force and the lever arm is 90 degrees, what is the torque produced by a 15 N force applied at a distance of 2 m?
A.
0 Nm
B.
15 Nm
C.
30 Nm
D.
45 Nm
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Solution
Torque (τ) = F × d × sin(θ) = 15 N × 2 m × sin(90°) = 30 Nm.
Correct Answer: C — 30 Nm
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Q. If the angle between the transmission axis of two polarizers is 90 degrees, what is the transmitted intensity of light?
A.
Maximum intensity
B.
Half of the original intensity
C.
Zero intensity
D.
One-fourth of the original intensity
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Solution
When the angle between the transmission axes of two polarizers is 90 degrees, no light is transmitted, resulting in zero intensity.
Correct Answer: C — Zero intensity
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