HSSC-1 (Class-11) Solved Physics MCQs; Unit-5: Rotational and Circular Motion;  (New Book/KPK Text Board, Peshawar) With Verified Answers 


1. The angular speed in radians/hours for the daily rotation of our earth is?




... Answer is D. 𝝿/12
Angular Speed equation ω = θ/t 👉 Since Earth takes t = 24 hours for to complete 1 revolution, angle θ = 2π rad 👉 ω=(2π rad)/(24 hour) = π/12 rad/hour


2. Linear acceleration a = r𝛂 when '𝞱' is 




... Answer is D. 90°
`\vec a` = `\vec r` x `\vec ∝` = r∝ sin𝞱 👉 when 𝞱 = 90° 👉 a = r∝ sin 90° 👉 sin 90° = 1 👉 a = r∝


3. What is the moment of inertia of a sphere




... Answer is C. `\frac {2}{5}`MR²
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4. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a horizontal plane. It follows




... Answer is C. It moves in a circular path
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5. A body moving in a circular path with constant speed has




... Answer is A. Constant acceleration
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6. Astronauts appear weightless in space because




... Answer is C. satellite is freely falling
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7. Which one is constant for a satellite in orbit? 




... Answer is C. Angular Momentum
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8. If the earth suddenly stops rotating the value of 'g' at the equator would:




... Answer is C. Increase​
Given: Equation for Value of g at equator g' = g - ѠR² 👉 when Ѡ = 0 (angular rotation stop) 👉 g' = g - 0 = g


9. If a solid sphere and solid cylinder of the same mass and density rotate about their own axis, the moment of inertia will be greater for




... Answer is B. Solid cylinder
Moment of inertia of sphere Is = 2/5 mr² 👉 Moment of inertia of cylinder Ic= = 1/2 mr² 👉 1/2 ＞ 2/5 👉 Ic ＞ Is


10. The gravitational force exerted on an astronaut on Earth's surface is 600 N down. When she is in the International Space Station, the gravitational force on her is





... Answer is B. smaller,
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11. A solid cylinder of mass M and radius R rolls down an incline without slipping. Its moment of inertia is about an axis through its center of mass MR²/2. At any instant while in motion, its rotational kinetic energy about its center of mass is what fraction of its total kinetic energy?




... Answer is C. 1/3
👉 K.E (Rotational) = K.E (T)= 1/2 mѠ² 👉 K.E (Translational) K.E (T) = 1/2 mv² = 1/2 mr²Ѡ² = IѠ² 👉 K.E (Total) = K.E (R) + K.E (T) 👉 [K.E (R)]/[K.E (Total)] = [1/2 mѠ²] / [K.E (R) + K.E (T)] = [1/2 mѠ²] / [1/2 mѠ² + IѠ²] 👉 by solving we get [K.E (R)]/[K.E (Total)] = 1/3



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