Q. A mass m is attached to a string and is whirled in a vertical circle. At the top of the circle, the tension in the string is T. What is the expression for the tension at the bottom of the circle?
A.
T + mg
B.
T - mg
C.
T
D.
T + 2mg
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Solution
At the bottom, T + mg = mv²/r; T = mv²/r - mg.
Correct Answer: A — T + mg
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Q. A mass m is attached to a string of length L and is swung in a vertical circle. At the highest point of the circle, what is the minimum speed required to keep the mass in circular motion?
A.
√(gL)
B.
√(2gL)
C.
gL
D.
2gL
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Solution
At the highest point, the centripetal force must equal the weight: mv²/L = mg, thus v = √(gL).
Correct Answer: A — √(gL)
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Q. A mass m is lifted to a height h in a uniform gravitational field. What is the work done against gravity?
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Solution
The work done against gravity when lifting a mass m to a height h is given by W = mgh.
Correct Answer: A — mgh
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Q. A mass on a spring 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
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Solution
The angular frequency ω is given by ω = 2πf. For f = 2 Hz, ω = 2π(2) = 4π rad/s.
Correct Answer: A — 4π rad/s
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Q. A mass-spring system is subjected to a periodic force. If the amplitude of oscillation is 0.1 m and the frequency is 2 Hz, what is the maximum velocity of the mass?
A.
0.4 m/s
B.
0.2 m/s
C.
0.1 m/s
D.
0.8 m/s
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Solution
Maximum velocity (v_max) = Aω = A(2πf) = 0.1 * (2π * 2) = 0.4 m/s.
Correct Answer: A — 0.4 m/s
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Q. A mass-spring system is subjected to a periodic force. If the amplitude of the forced oscillation is 0.1 m and the damping coefficient is 0.2 kg/s, what is the maximum velocity of the oscillation?
A.
0.1 m/s
B.
0.2 m/s
C.
0.3 m/s
D.
0.4 m/s
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Solution
Maximum velocity (v_max) = Aω, where ω = 2πf. Assuming f = 1 Hz, v_max = 0.1 * 2π * 1 = 0.2 m/s.
Correct Answer: B — 0.2 m/s
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Q. A mass-spring system oscillates with a frequency of 2 Hz. If the system is damped, what is the relationship between the damped frequency and the natural frequency?
A.
Damped frequency is greater
B.
Damped frequency is equal
C.
Damped frequency is less
D.
Damped frequency is unpredictable
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Solution
In a damped system, the damped frequency is always less than the natural frequency.
Correct Answer: C — Damped frequency is less
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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
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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
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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
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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
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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
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Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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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
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Solution
Angular frequency (ω) is given by ω = 2πf. Thus, ω = 2π * 3 ≈ 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass-spring system oscillates with a frequency of 3 Hz. What is the period of the oscillation?
A.
0.33 s
B.
0.5 s
C.
1 s
D.
2 s
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Solution
Period (T) is the reciprocal of frequency (f). T = 1/f = 1/3 = 0.33 s.
Correct Answer: A — 0.33 s
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Q. A mass-spring system oscillates with a frequency of 5 Hz. What is the period of the motion?
A.
0.2 s
B.
0.5 s
C.
1 s
D.
2 s
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Solution
Period T = 1/f = 1/5 = 0.2 s.
Correct Answer: A — 0.2 s
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Q. A mass-spring system oscillates with a natural frequency of 3 Hz. If a damping force is applied, what is the new frequency of oscillation if the damping ratio is 0.1?
A.
2.8 Hz
B.
2.9 Hz
C.
3.0 Hz
D.
3.1 Hz
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Solution
New frequency (ω_d) = ω_n√(1-ζ²) = 3√(1-0.1²) ≈ 2.9 Hz.
Correct Answer: B — 2.9 Hz
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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
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Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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Q. A mass-spring system oscillates with a period of 4 seconds. What is the frequency of the oscillation?
A.
0.25 Hz
B.
0.5 Hz
C.
1 Hz
D.
2 Hz
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Solution
Frequency f is the reciprocal of the period T. Thus, f = 1/T = 1/4 = 0.25 Hz.
Correct Answer: A — 0.25 Hz
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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
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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
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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
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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
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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%
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Solution
The change in volume ΔV/V = P/B, so ΔV/V = 50 MPa / 200 GPa = 0.025%.
Correct Answer: B — 0.025%
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Q. A material has a Poisson's ratio of 0.3. What is the ratio of lateral strain to longitudinal strain?
A.
0.3
B.
0.7
C.
1.0
D.
0.5
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Solution
Poisson's ratio is defined as the negative ratio of lateral strain to longitudinal strain, so a Poisson's ratio of 0.3 means the ratio is 0.3.
Correct Answer: A — 0.3
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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
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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
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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
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Solution
Young's modulus (Y) = Stress / Strain = 100 MPa / 0.002 = 50 GPa.
Correct Answer: B — 100 GPa
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Q. A material with a Poisson's ratio of 0.5 is considered to be:
A.
Perfectly elastic
B.
Perfectly plastic
C.
Incompressible
D.
Brittle
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Solution
A Poisson's ratio of 0.5 indicates that the material is incompressible.
Correct Answer: C — Incompressible
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Q. A measurement of 15.5 m has a relative error of 0.03. What is the absolute error?
A.
0.465 m
B.
0.5 m
C.
0.3 m
D.
0.45 m
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Solution
Absolute error = Relative error * Measured value = 0.03 * 15.5 m = 0.465 m.
Correct Answer: A — 0.465 m
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Q. A measurement of 30 m has an error of ±0.1 m. What is the true value if the measurement is taken as the average?
A.
30.1 m
B.
30 m
C.
29.9 m
D.
30.05 m
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Solution
True value is taken as the average of the measured value and the error, which is (30 + 29.9) / 2 = 30.05 m.
Correct Answer: D — 30.05 m
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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
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Solution
Total uncertainty = 3 * 0.5 = 1.5 units.
Correct Answer: A — 1.5 units
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Q. A measurement of length is recorded as 12.3 cm with an uncertainty of ±0.1 cm. What is the relative error in the measurement?
A.
0.0081
B.
0.008
C.
0.01
D.
0.1
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Solution
Relative error = (absolute error / measured value) = 0.1 / 12.3 = 0.0081.
Correct Answer: B — 0.008
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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
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Solution
Relative error = (absolute error / measured value) = 0.1 / 5.0 = 0.02 or 2%.
Correct Answer: B — 0.01
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Q. A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, Specific heat of metal = 0.9 J/g°C)
A.
30°C
B.
40°C
C.
50°C
D.
60°C
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Solution
Using conservation of energy: m1*c1*(T1-Tf) = m2*c2*(Tf-T2). Solving gives Tf = 50°C.
Correct Answer: C — 50°C
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Q. A motor has a power rating of 1500 Watts. How much work can it do in 10 minutes?
A.
90000 Joules
B.
15000 Joules
C.
25000 Joules
D.
30000 Joules
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Solution
Work done = Power × Time = 1500 W × (10 × 60 s) = 90000 J.
Correct Answer: A — 90000 Joules
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