Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 200 N/m. What is the angular frequency of the oscillator?
A.
5 rad/s
B.
10 rad/s
C.
20 rad/s
D.
15 rad/s
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Solution
The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.
Correct Answer: B — 10 rad/s
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Q. A simple harmonic oscillator has a mass of 2 kg and a spring constant of 50 N/m. What is the angular frequency of the oscillator?
A.
5 rad/s
B.
10 rad/s
C.
15 rad/s
D.
20 rad/s
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Solution
The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.
Correct Answer: B — 10 rad/s
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Q. A simple harmonic oscillator has a maximum displacement of 0.1 m and a maximum speed of 2 m/s. What is the angular frequency?
A.
10 rad/s
B.
20 rad/s
C.
5 rad/s
D.
15 rad/s
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Solution
ω = V_max/A = 2/0.1 = 20 rad/s.
Correct Answer: A — 10 rad/s
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Q. A simple harmonic oscillator has a maximum displacement of 0.1 m. What is the maximum potential energy if the spring constant is 200 N/m?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
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Solution
Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.
Correct Answer: B — 2 J
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Q. A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
T = 2π√(m/k) = 2π√(2/200) = 1 s.
Correct Answer: B — 1 s
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Q. A simple harmonic oscillator has a total energy E. If the amplitude is halved, what will be the new total energy?
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Solution
The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.
Correct Answer: A — E/4
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Q. A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?
A.
200 N/m
B.
500 N/m
C.
1000 N/m
D.
2000 N/m
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Solution
Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.
Correct Answer: B — 500 N/m
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Q. A simple harmonic oscillator has a total energy of 50 J. If the amplitude is doubled, what will be the new total energy?
A.
50 J
B.
100 J
C.
200 J
D.
400 J
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Solution
Total energy in SHM is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 4.
Correct Answer: C — 200 J
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Q. A simple harmonic oscillator has an amplitude A and a maximum speed v_max. What is the relationship between v_max and A?
A.
v_max = Aω
B.
v_max = A/ω
C.
v_max = A²ω
D.
v_max = A/2ω
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Solution
The maximum speed v_max of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.
Correct Answer: A — v_max = Aω
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Q. A simple harmonic oscillator has an amplitude A. What is the maximum speed of the oscillator?
A.
Aω
B.
A/ω
C.
A²ω
D.
A/2ω
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Solution
The maximum speed (v_max) of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.
Correct Answer: A — Aω
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Q. A simple harmonic oscillator has an amplitude of 5 cm. What is the maximum displacement from the mean position?
A.
0 cm
B.
2.5 cm
C.
5 cm
D.
10 cm
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Solution
The maximum displacement in simple harmonic motion is equal to the amplitude, which is 5 cm.
Correct Answer: C — 5 cm
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Q. A small insect can walk on the surface of water due to which of the following phenomena?
A.
Viscosity
B.
Surface tension
C.
Capillarity
D.
Buoyancy
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Solution
The insect can walk on water due to surface tension, which creates a 'skin' on the surface of the water.
Correct Answer: B — Surface tension
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Q. A small insect can walk on the surface of water due to which property?
A.
Viscosity
B.
Surface tension
C.
Capillarity
D.
Density
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Solution
The insect can walk on water because of surface tension, which creates a 'skin' on the surface that can support the weight of the insect.
Correct Answer: B — Surface tension
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Q. A small insect can walk on the surface of water without sinking. This is primarily due to:
A.
Buoyancy
B.
Surface tension
C.
Viscosity
D.
Density
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Solution
The insect can walk on water due to surface tension, which creates a 'skin' on the surface that can support the weight of the insect.
Correct Answer: B — Surface tension
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Q. A soap bubble has a radius of 0.1 m. What is the surface area of the bubble?
A.
0.04 m²
B.
0.12 m²
C.
0.25 m²
D.
0.31 m²
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Solution
Surface area of a sphere = 4πr² = 4π(0.1)² = 0.1256 m², approximately 0.31 m².
Correct Answer: D — 0.31 m²
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Q. A soap bubble has a radius of 5 cm. What is the surface area of the bubble?
A.
100π cm²
B.
50π cm²
C.
25π cm²
D.
20π cm²
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Solution
Surface area of a sphere = 4πr² = 4π(5)² = 100π cm².
Correct Answer: A — 100π cm²
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Q. A soap solution is added to water. What happens to the surface tension?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Adding soap to water decreases the surface tension because soap molecules disrupt the cohesive forces between water molecules.
Correct Answer: B — Decreases
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Q. A soap solution is added to water. What is the effect on the surface tension?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes negative
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Solution
Adding soap decreases the surface tension of water.
Correct Answer: B — Decreases
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Q. A solenoid has a length of 1 m and a cross-sectional area of 0.01 m². If the magnetic field inside it is 0.4 T, what is the magnetic flux?
A.
0.004 Wb
B.
0.04 Wb
C.
0.4 Wb
D.
0.1 Wb
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Solution
Magnetic flux (Φ) = B * A = 0.4 T * 0.01 m² = 0.004 Wb.
Correct Answer: B — 0.04 Wb
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Q. A solenoid has a length of 1 m and a cross-sectional area of 0.01 m². If the magnetic field inside it is 0.2 T, what is the magnetic flux?
A.
0.002 Wb
B.
0.01 Wb
C.
0.02 Wb
D.
0.1 Wb
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Solution
Magnetic flux (Φ) = B * A. Here, Φ = 0.2 T * 0.01 m² = 0.002 Wb.
Correct Answer: C — 0.02 Wb
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Q. A solenoid of length L and cross-sectional area A carries a current I. What is the magnetic field inside the solenoid?
A.
μ₀nI
B.
μ₀I/n
C.
μ₀I/(nA)
D.
μ₀I/(2n)
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Solution
The magnetic field inside a solenoid is given by B = μ₀nI, where n is the number of turns per unit length.
Correct Answer: A — μ₀nI
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Q. A solenoid produces a magnetic field when an electric current passes through it. What happens to the magnetic field if the current is reversed?
A.
The magnetic field disappears
B.
The magnetic field direction reverses
C.
The magnetic field strength increases
D.
The magnetic field strength decreases
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Solution
Reversing the current in a solenoid reverses the direction of the magnetic field according to the right-hand rule.
Correct Answer: B — The magnetic field direction reverses
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Q. A solenoid produces a uniform magnetic field inside it. What factors affect the strength of this magnetic field?
A.
Length of the solenoid
B.
Number of turns per unit length
C.
Current through the solenoid
D.
All of the above
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Solution
The strength of the magnetic field inside a solenoid is affected by the number of turns per unit length and the current flowing through it, as well as the length of the solenoid.
Correct Answer: D — All of the above
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Q. A solenoid produces a uniform magnetic field inside it. What happens to the magnetic field strength if the current through the solenoid is doubled?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
The magnetic field strength inside a solenoid is directly proportional to the current flowing through it.
Correct Answer: B — It doubles
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Q. A solenoid with a length of 1 m and a cross-sectional area of 0.01 m² carries a current of 5 A. If the magnetic field inside the solenoid is uniform, what is the magnetic field strength?
A.
0.1 T
B.
0.2 T
C.
0.5 T
D.
1 T
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Solution
The magnetic field inside a solenoid is given by B = μ₀ * (N/L) * I. Assuming N/L = 1 for simplicity, B = μ₀ * I = 4π × 10^-7 T*m/A * 5 A = 0.5 T.
Correct Answer: C — 0.5 T
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Q. A solid cone rolls down a slope. If its height is h, what is the speed of the cone at the bottom?
A.
√(gh)
B.
√(2gh)
C.
√(3gh)
D.
√(4gh)
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Solution
Using conservation of energy, the potential energy mgh converts to kinetic energy. For a solid cone, the final speed at the bottom is v = √(2gh).
Correct Answer: B — √(2gh)
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Q. A solid cone rolls down an incline. If its height is h, what is the relationship between its potential energy and kinetic energy at the bottom?
A.
PE = KE
B.
PE = 2KE
C.
PE = 3KE
D.
PE = 4KE
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Solution
For a solid cone rolling down an incline, the potential energy at height h is converted into translational and rotational kinetic energy, leading to PE = 2KE.
Correct Answer: B — PE = 2KE
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Q. A solid cone rolls down an incline. What is the moment of inertia about its axis?
A.
(3/10)mR^2
B.
(1/10)mR^2
C.
(1/3)mR^2
D.
(2/5)mR^2
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Solution
The moment of inertia of a solid cone about its axis is I = (1/10)mR^2.
Correct Answer: C — (1/3)mR^2
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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 will have a greater speed at the bottom?
A.
Solid cylinder
B.
Hollow cylinder
C.
Both have the same speed
D.
Depends on the mass
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Solution
The solid cylinder has a smaller moment of inertia compared to the hollow cylinder, thus it will have a greater speed at the bottom.
Correct Answer: A — Solid cylinder
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Q. A solid cylinder and a hollow cylinder of the same mass and radius roll down the same incline. Which one reaches the bottom first?
A.
Solid cylinder
B.
Hollow cylinder
C.
Both reach at the same time
D.
Depends on the angle of incline
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Solution
The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.
Correct Answer: A — Solid cylinder
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