Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the spring constant if the mass is 2 kg? (2023)
A.
18 N/m
B.
12 N/m
C.
6 N/m
D.
9 N/m
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Solution
Frequency (f) = (1/2π)√(k/m) => k = (2πf)² * m = (2π*3)² * 2 ≈ 18 N/m.
Correct Answer: A — 18 N/m
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Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the spring constant if the mass is 0.5 kg? (2022)
A.
18 N/m
B.
12 N/m
C.
6 N/m
D.
24 N/m
Show solution
Solution
Frequency (f) = 1/(2π)√(k/m) => k = (2πf)²m = (2π × 3)² × 0.5 ≈ 18 N/m.
Correct Answer: A — 18 N/m
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Q. A mass-spring system oscillates with a frequency of 2 Hz. What is the period of oscillation? (2019)
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
Period (T) = 1 / frequency = 1 / 2 Hz = 0.5 s.
Correct Answer: B — 1 s
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Q. A parallel plate capacitor has a capacitance of 5μF. If the distance between the plates is halved, what will be the new capacitance? (2023)
A.
5μF
B.
10μF
C.
2.5μF
D.
20μF
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Solution
Capacitance C is inversely proportional to the distance d between the plates. If d is halved, C doubles, so the new capacitance is 10μF.
Correct Answer: B — 10μF
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Q. A pendulum has a length of 1 m. What is its time period? (2023)
A.
2 s
B.
1 s
C.
0.5 s
D.
0.25 s
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Solution
Time period (T) = 2π√(L/g) = 2π√(1/9.8) ≈ 2 s
Correct Answer: A — 2 s
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Q. A plane mirror produces an image that is: (2023)
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
A plane mirror always produces a virtual and erect image, which is the same size as the object.
Correct Answer: B — Virtual and erect
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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 20 m/s. What is the maximum height it reaches? (2022)
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using the formula: H = (u² * sin²θ) / (2g). H = (20² * (1/2)) / (2 * 9.8) = 10.2 m, approximately 10 m.
Correct Answer: B — 10 m
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Q. A ray of light passes through the center of curvature of a concave mirror. What will be the nature of the image formed? (2020)
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
When a ray passes through the center of curvature, it reflects back on itself, forming a real and inverted image at the same distance as the object.
Correct Answer: A — Real and inverted
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Q. A ray of light strikes a glass surface at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction? (2022)
A.
20 degrees
B.
30 degrees
C.
40 degrees
D.
50 degrees
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Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Therefore, sin(r) = (1 * sin(30))/1.5 = 0.333. Thus, r = sin^(-1)(0.333) ≈ 19.5 degrees, which is approximately 20 degrees.
Correct Answer: A — 20 degrees
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Q. A ray of light strikes a plane surface at an angle of incidence of 30 degrees. What is the angle of reflection? (2023)
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
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Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 30 degrees.
Correct Answer: A — 30 degrees
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Q. A rotating body has an angular momentum L. If its moment of inertia is doubled and angular velocity is halved, what will be the new angular momentum? (2021)
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Solution
New angular momentum L' = I'ω' = (2I)(ω/2) = L.
Correct Answer: A — L/2
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Q. A rotating object has a moment of inertia of 3 kg·m² and is spinning with an angular velocity of 4 rad/s. What is its kinetic energy? (2023)
A.
12 J
B.
24 J
C.
48 J
D.
6 J
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Solution
The rotational kinetic energy is given by KE = 0.5 I ω² = 0.5 * 3 * (4)² = 24 J.
Correct Answer: B — 24 J
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Q. A rotating object has a moment of inertia of 5 kg·m² and is rotating with an angular velocity of 4 rad/s. What is its kinetic energy? (2022)
A.
40 J
B.
20 J
C.
10 J
D.
80 J
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Solution
The rotational kinetic energy is given by KE = (1/2)Iω² = (1/2)(5 kg·m²)(4 rad/s)² = 40 J.
Correct Answer: A — 40 J
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Q. A solid cylinder and a hollow cylinder of the same mass and radius are released from rest at the same height. Which one reaches the ground first? (2022)
A.
Solid cylinder
B.
Hollow cylinder
C.
Both reach at the same time
D.
Depends on the height
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Solution
The solid cylinder has a lower moment of inertia, thus it accelerates faster and reaches the ground first.
Correct Answer: A — Solid cylinder
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Q. A solid sphere and a hollow sphere of the same mass and radius are rolling down an incline. Which one will reach the bottom first? (2021)
A.
Solid sphere
B.
Hollow sphere
C.
Both will reach at the same time
D.
It depends on the angle of incline
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Solution
The solid sphere has a smaller moment of inertia compared to the hollow sphere, allowing it to accelerate faster down the incline.
Correct Answer: A — Solid sphere
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Q. A solid sphere of mass M and radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass of the sphere when it reaches the bottom? (2021)
A.
√(2gh)
B.
√(5gh/7)
C.
√(3gh/5)
D.
√(gh)
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Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, v = √(5gh/7).
Correct Answer: B — √(5gh/7)
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Q. A sound wave travels through air at a speed of 340 m/s. What is the wavelength of a sound wave with a frequency of 1700 Hz? (2022)
A.
0.2 m
B.
0.5 m
C.
1.0 m
D.
2.0 m
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Solution
Using the formula λ = v/f, where v is the speed and f is the frequency, we have λ = 340 m/s / 1700 Hz = 0.2 m.
Correct Answer: A — 0.2 m
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Q. A spring is compressed by 0.1 m and has a spring constant of 200 N/m. What is the potential energy stored in the spring? (2019)
A.
1 J
B.
2 J
C.
0.5 J
D.
4 J
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Solution
Potential Energy (PE) = 1/2 * k * x^2 = 1/2 * 200 N/m * (0.1 m)^2 = 1 J
Correct Answer: A — 1 J
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Q. A spring is compressed by 0.2 m and has a spring constant of 500 N/m. What is the potential energy stored in the spring? (2023)
A.
10 J
B.
5 J
C.
20 J
D.
2 J
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Solution
Potential Energy in Spring (PE) = 1/2 kx^2 = 1/2 * 500 N/m * (0.2 m)^2 = 10 J
Correct Answer: A — 10 J
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Q. A stone is thrown downward with an initial velocity of 5 m/s from a height of 45 m. How long will it take to hit the ground? (2020)
A.
3 s
B.
4 s
C.
5 s
D.
6 s
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Solution
Using the equation of motion: h = ut + 0.5gt². Solving for t gives t ≈ 4 seconds.
Correct Answer: B — 4 s
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Q. A torque of 15 N·m is applied to a wheel with a moment of inertia of 3 kg·m². What is the angular acceleration? (2023)
A.
3 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
15 rad/s²
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Solution
Using τ = Iα, we have α = τ/I = 15/3 = 5 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A torque τ is applied to a rigid body with a moment of inertia I. If the torque is doubled, what happens to the angular acceleration? (2019)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Newton's second law for rotation, τ = Iα. If τ is doubled, α also doubles.
Correct Answer: A — It doubles
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Q. A train starts from rest and accelerates uniformly at 2 m/s². What is its speed after 10 seconds? (2021)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
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Solution
Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 0 + 2 * 10 = 20 m/s.
Correct Answer: B — 20 m/s
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Q. A tuning fork vibrates at a frequency of 440 Hz. What is the time period of its vibration? (2020)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
Show solution
Solution
Time period (T) = 1 / frequency = 1 / 440 Hz ≈ 0.00227 s
Correct Answer: A — 0.00227 s
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Q. A tuning fork vibrates at a frequency of 440 Hz. What is the time period of the tuning fork? (2020)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
Show solution
Solution
Time period (T) = 1 / frequency = 1 / 440 Hz ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular speed of the rod just before it hits the ground? (2019)
A.
√(3g/L)
B.
√(2g/L)
C.
√(g/L)
D.
√(4g/L)
Show solution
Solution
Using conservation of energy, potential energy at the top converts to rotational kinetic energy at the bottom. The angular speed ω = √(3g/L).
Correct Answer: B — √(2g/L)
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Q. A wave has a frequency of 50 Hz and a wavelength of 2 m. What is the speed of the wave? (2021)
A.
25 m/s
B.
50 m/s
C.
100 m/s
D.
200 m/s
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Solution
The speed of a wave is given by the formula v = fλ. Here, v = 50 Hz * 2 m = 100 m/s.
Correct Answer: B — 50 m/s
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Q. A wave has a frequency of 500 Hz and a wavelength of 0.5 m. What is the speed of the wave? (2023)
A.
250 m/s
B.
500 m/s
C.
1000 m/s
D.
2000 m/s
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Solution
The speed of the wave can be calculated using the formula v = fλ. Here, v = 500 Hz * 0.5 m = 250 m/s.
Correct Answer: B — 500 m/s
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Q. A wave has a frequency of 500 Hz and a wavelength of 0.68 m. What is the speed of the wave? (2021)
A.
340 m/s
B.
500 m/s
C.
680 m/s
D.
250 m/s
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Solution
The speed of the wave can be calculated using the formula v = fλ. Thus, v = 500 Hz * 0.68 m = 340 m/s.
Correct Answer: A — 340 m/s
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Q. A wheel is rotating with an angular acceleration of 2 rad/s². If it starts from rest, what will be its angular velocity after 5 seconds? (2022)
A.
5 rad/s
B.
10 rad/s
C.
15 rad/s
D.
20 rad/s
Show solution
Solution
Using the formula ω = ω₀ + αt, where ω₀ = 0, α = 2 rad/s², and t = 5 s, we get ω = 0 + 2*5 = 10 rad/s.
Correct Answer: B — 10 rad/s
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