Further Mechanics and Thermal Physics Practice Questions
A-Levels · A-Level Physics · 144 free MCQs with instant results and detailed explanations.
144
Total
47
Easy
77
Medium
20
Hard
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Sample Questions from Further Mechanics and Thermal Physics
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Q1
Easy
A solid sphere of radius R rolls down a frictionless inclined plane of height h. Which of the following equations correctly relates the potential energy lost to the kinetic energy gained?
A.
mgh = (1/2)mvยฒ + (2/5)mvยฒ
B.
mgh = (1/2)mvยฒ + (1/2)mvยฒ
C.
mgh = (1/3)mvยฒ + (1/2)mvยฒ
D.
mgh = (1/2)mvยฒ
Show Answer & Explanation
Correct Answer: A
As the sphere rolls down, its potential energy (mgh) is converted into translational and rotational kinetic energy. The translational kinetic energy is (1/2)mvยฒ and the rotational kinetic energy for a sphere is (2/5)mvยฒ, resulting in the equation mgh = (1/2)mvยฒ + (2/5)mvยฒ.
Q2
Easy
If the temperature of a gas is increased while keeping its volume constant, what happens to the pressure of the gas according to Gay-Lussac's Law?
A.
Pressure increases
B.
Pressure decreases
C.
Pressure remains constant
D.
Pressure fluctuates randomly
Show Answer & Explanation
Correct Answer: A
According to Gay-Lussac's Law, at constant volume, the pressure of a gas is directly proportional to its temperature (in Kelvin). Thus, if the temperature increases, the pressure also increases.
Q3
Easy
A gas occupies a volume of 5.0 mยณ at a pressure of 100,000 Pa. If the volume is compressed to 2.5 mยณ and the temperature remains constant, what is the new pressure of the gas?
A.
200,000 Pa
B.
150,000 Pa
C.
100,000 Pa
D.
50,000 Pa
Show Answer & Explanation
Correct Answer: A
According to Boyle's Law, P1V1 = P2V2. Rearranging gives P2 = P1 * (V1/V2). Substituting the values: P2 = 100,000 Pa * (5.0 mยณ / 2.5 mยณ) = 200,000 Pa.
Q4
Medium
A block of ice melts at its melting point of 0ยฐC. If the latent heat of fusion of ice is 334,000 J/kg, how much energy is required to melt 2 kg of ice?
A.
334,000 J
B.
668,000 J
C.
1,000,000 J
D.
667,000 J
Show Answer & Explanation
Correct Answer: B
The energy required to melt ice can be calculated using the formula Q = mL, where m is the mass and L is the latent heat. Hence, Q = 2 kg ร 334,000 J/kg = 668,000 J.
Q5
Medium
A particle moves in a circular path with a radius of 5 m at a constant speed of 10 m/s. What is the centripetal acceleration of the particle?
A.
5 m/sยฒ
B.
10 m/sยฒ
C.
20 m/sยฒ
D.
25 m/sยฒ
Show Answer & Explanation
Correct Answer: C
Centripetal acceleration (a_c) is calculated using a_c = vยฒ/r. Here, v = 10 m/s and r = 5 m, so a_c = (10 m/s)ยฒ / 5 m = 100/5 = 20 m/sยฒ.
Q6
Medium
A gas expands isothermally at a temperature of 300 K and does 500 J of work. What is the change in internal energy of the gas during this process?
A.
0 J
B.
250 J
C.
500 J
D.
1000 J
Show Answer & Explanation
Correct Answer: A
In an isothermal process, the change in internal energy (ฮU) of an ideal gas is zero because internal energy depends only on temperature. Thus, ฮU = 0 J.
Q7
Medium
Two objects of masses 3 kg and 5 kg collide elastically. If the 3 kg object is moving at 6 m/s and the 5 kg object is stationary, what is the final velocity of the 3 kg object after the collision?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show Answer & Explanation
Correct Answer: C
Using the conservation of momentum and kinetic energy for elastic collisions, the final velocity of the 3 kg object can be calculated as v_f = (m1 - m2)/(m1 + m2) * u1, which gives v_f = (3 - 5)/(3 + 5) * 6 = 4 m/s.
Q8
Hard
A block of mass m is resting on a frictionless inclined plane at an angle ฮธ. What is the acceleration of the block down the plane?
A.
g sin(ฮธ)
B.
g cos(ฮธ)
C.
g tan(ฮธ)
D.
0
Show Answer & Explanation
Correct Answer: A
The acceleration of the block down the plane is determined by the component of gravitational force acting parallel to the incline, which is given by g sin(ฮธ). This is because the gravitational force is split into two components: one perpendicular to the incline and one parallel to it.
Q9
Hard
A 5 kg object is moving in a circular path of radius 10 m at a constant speed of 15 m/s. What is the net centripetal force acting on the object?
A.
11.25 N
B.
75 N
C.
37.5 N
D.
22.5 N
Show Answer & Explanation
Correct Answer: B
Centripetal force is given by the formula F_c = mvยฒ/r. Here, m = 5 kg, v = 15 m/s, and r = 10 m. Substituting these values, F_c = 5 kg * (15 m/s)ยฒ / 10 m = 75 N. Thus, the net centripetal force acting on the object is 75 N.
Q10
Hard
A block of mass 5 kg is sliding down a frictionless incline of angle 30 degrees. Calculate the acceleration of the block down the incline. (Take g = 9.81 m/sยฒ)
A.
4.9 m/sยฒ
B.
9.81 m/sยฒ
C.
5.0 m/sยฒ
D.
2.45 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration down the incline can be calculated using the formula: a = g * sin(ฮธ). Here, ฮธ = 30 degrees, so a = 9.81 * sin(30) = 9.81 * 0.5 = 4.9 m/sยฒ.
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Practice All 144 Questions →Further Mechanics and Thermal Physics โ A-Levels A-Level Physics Practice Questions Online
This page contains 144 practice MCQs for the chapter Further Mechanics and Thermal Physics in A-Levels A-Level Physics. The questions are organized by difficulty โ 47 easy, 77 medium, 20 hard โ so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.