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Q. A mass-spring system oscillates with a frequency of 2 Hz. What is the angular frequency?

A. 4π rad/s
B. 2π rad/s
C. π rad/s
D. 8π rad/s

Solution

The angular frequency ω is related to the frequency f by the formula ω = 2πf. Therefore, ω = 2π × 2 = 4π rad/s.

Correct Answer: A — 4π rad/s

Q. A mass-spring system oscillates with a frequency of 2 Hz. What is the time period of the oscillation?

A. 0.5 s
B. 1 s
C. 2 s
D. 4 s

Solution

The time period T is the reciprocal of frequency f. T = 1/f = 1/2 Hz = 0.5 s.

Correct Answer: B — 1 s

Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the angular frequency of the system?

A. 3 rad/s
B. 6 rad/s
C. 9 rad/s
D. 12 rad/s

Solution

Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π(3) = 6 rad/s.

Correct Answer: B — 6 rad/s

Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the angular frequency?

A. 3 rad/s
B. 6 rad/s
C. 9 rad/s
D. 12 rad/s

Solution

Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π * 3 ≈ 6 rad/s.

Correct Answer: B — 6 rad/s

Q. A mass-spring system oscillates with a period of 2 seconds. What is the frequency of the oscillation?

A. 0.25 Hz
B. 0.5 Hz
C. 1 Hz
D. 2 Hz

Solution

Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.

Correct Answer: B — 0.5 Hz

Q. A material has a bulk modulus of 200 GPa. If the pressure applied to it is increased by 50 MPa, what is the fractional change in volume?

A. 0.00025
B. 0.0005
C. 0.0025
D. 0.005

Solution

The fractional change in volume is given by ΔV/V = ΔP/B. Here, ΔP = 50 MPa = 0.05 GPa, so ΔV/V = 0.05/200 = 0.00025.

Correct Answer: A — 0.00025

Q. A material has a bulk modulus of 200 GPa. If the pressure on the material is increased by 10 MPa, what is the fractional change in volume?

A. 0.00005
B. 0.0001
C. 0.0002
D. 0.00025

Solution

The fractional change in volume ΔV/V is given by ΔV/V = ΔP/B, where ΔP = 10 MPa and B = 200 GPa, resulting in 0.00005.

Correct Answer: B — 0.0001

Q. A material has a bulk modulus of 200 GPa. What is the change in volume when a pressure of 50 MPa is applied?

A. 0.0125%
B. 0.025%
C. 0.05%
D. 0.1%

Solution

The change in volume ΔV/V = P/B, so ΔV/V = 50 MPa / 200 GPa = 0.025%.

Correct Answer: B — 0.025%

Q. A material is said to be elastic if it:

A. Returns to its original shape after deformation
B. Can be permanently deformed
C. Breaks under stress
D. Has a high tensile strength

Solution

A material is considered elastic if it returns to its original shape after the removal of the applied stress.

Correct Answer: A — Returns to its original shape after deformation

Q. A material is subjected to a tensile stress of 100 MPa and experiences a strain of 0.002. What is its Young's modulus?

A. 50 GPa
B. 100 GPa
C. 200 GPa
D. 500 GPa

Solution

Young's modulus (Y) = Stress / Strain = 100 MPa / 0.002 = 50 GPa.

Correct Answer: B — 100 GPa

Q. A measurement of a physical quantity is reported as 25.0 ± 0.5 units. What is the total uncertainty if this quantity is multiplied by 3?

A. 1.5 units
B. 0.5 units
C. 1.0 units
D. 2.0 units

Solution

Total uncertainty = 3 * 0.5 = 1.5 units.

Correct Answer: A — 1.5 units

Q. A measurement of length is recorded as 5.0 cm with an uncertainty of ±0.1 cm. What is the relative error in this measurement?

A. 0.02
B. 0.01
C. 0.005
D. 0.1

Solution

Relative error = (absolute error / measured value) = 0.1 / 5.0 = 0.02 or 2%.

Correct Answer: B — 0.01

Q. A parallel plate capacitor has a potential difference of V across its plates. What is the electric field between the plates?

A. V/d
B. d/V
C. V²/d
D. d²/V

Solution

The electric field E between the plates of a parallel plate capacitor is given by E = V/d, where d is the separation between the plates.

Correct Answer: A — V/d

Q. A particle is moving in a circular path of radius R with a constant speed v. What is the centripetal acceleration of the particle?

A. v²/R
B. Rv
C. v/R
D. R²/v

Solution

Centripetal acceleration a_c = v²/R.

Correct Answer: A — v²/R

Q. A particle is moving in a circular path with radius r. If the speed of the particle is doubled, how does its angular momentum change?

A. Remains the same
B. Doubles
C. Quadruples
D. Halves

Solution

Angular momentum L = mvr, if v is doubled, L also doubles.

Correct Answer: B — Doubles

Q. A particle is moving in a straight line with a velocity v. If it suddenly starts moving in a circular path of radius r, what will be its angular momentum about the center of the circular path?

A. 0
B. mv
C. mvr
D. mv^2/r

Solution

Angular momentum L = mvr, where v is the linear speed and r is the radius of the circular path.

Correct Answer: C — mvr

Q. A particle is moving in a straight line with a velocity v. What is its angular momentum about a point O located at a distance r from the line of motion?

A. 0
B. mv
C. mvr
D. mv^2

Solution

Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point O.

Correct Answer: B — mv

Q. A particle moves in a circular path of radius 10 m with a constant speed of 5 m/s. What is the period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Circumference = 2πr = 20π m. Period T = distance/speed = 20π m / 5 m/s = 4π s.

Correct Answer: A — 2π s

Q. A particle moves in a circular path of radius 10 m with a speed of 5 m/s. What is the time period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Time period T = 2πr/v = 2π(10 m) / (5 m/s) = 4π s.

Correct Answer: A — 2π s

Q. A particle moves in a circular path of radius r with a constant angular acceleration α. What is the expression for the angular displacement θ after time t?

A. θ = αt²
B. θ = 0.5αt²
C. θ = αt
D. θ = 0.5αt

Solution

Angular displacement θ = 0.5αt² for constant angular acceleration.

Correct Answer: B — θ = 0.5αt²

Q. A particle moves in a circular path of radius r with a constant angular speed ω. What is the time period of the motion?

A. T = 2πr/ω
B. T = ω/2πr
C. T = 2π/ω
D. T = r/ω

Solution

The time period T = 2π/ω for uniform circular motion.

Correct Answer: C — T = 2π/ω

Q. A particle moves in a circular path of radius R with a constant speed v. What is the centripetal acceleration of the particle?

A. v^2/R
B. Rv^2
C. v/R
D. R/v^2

Solution

Centripetal acceleration a_c = v^2/R.

Correct Answer: A — v^2/R

Q. A particle moves in a circular path with a radius of 10 m at a constant speed of 5 m/s. What is the period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Circumference = 2πr = 20π m. Time = distance/speed = 20π m / 5 m/s = 4π s.

Correct Answer: B — 4π s

Q. A particle moves in a straight line under the influence of a constant force of 3 N. If it moves 6 m, what is the work done by the force?

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Work done = Force × Distance = 3 N × 6 m = 18 J.

Correct Answer: D — 18 J

Q. A particle moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 3 m/s, what will be its velocity after 5 seconds?

A. 10 m/s
B. 13 m/s
C. 15 m/s
D. 20 m/s

Solution

Final velocity (v) = u + at = 3 + (2 * 5) = 13 m/s.

Correct Answer: B — 13 m/s

Q. A particle moves in a straight line with an acceleration of 4 m/s². If its initial velocity is 2 m/s, what will be its velocity after 5 seconds?

A. 22 m/s
B. 20 m/s
C. 18 m/s
D. 16 m/s

Solution

Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 2 + 4 * 5 = 22 m/s.

Correct Answer: A — 22 m/s

Q. A particle moves in a straight line with an initial velocity of 10 m/s and accelerates at 2 m/s². What will be its velocity after 5 seconds?

A. 20 m/s
B. 30 m/s
C. 40 m/s
D. 50 m/s

Solution

Final velocity = initial velocity + (acceleration * time) = 10 m/s + (2 m/s² * 5 s) = 20 m/s.

Correct Answer: B — 30 m/s

Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle?

A. mr
B. mr^2
C. m/r
D. m/r^2

Solution

The moment of inertia of a point mass is given by I = mr^2.

Correct Answer: B — mr^2

Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle about the axis?

A. mr
B. mr^2
C. m/r
D. m/r^2

Solution

The moment of inertia of a point mass about an axis is given by I = mr^2.

Correct Answer: B — mr^2

Q. A particle of mass m is moving in a circular path of radius r with a constant speed v. What is the angular momentum of the particle about the center of the circle?

A. mv
B. mvr
C. mr^2
D. mv^2

Solution

Angular momentum L = mvr, where v is the linear speed and r is the radius.

Correct Answer: B — mvr

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