Q20. Two equal-density spheres of masses m and 8m fall in the same viscous fluid. If the first terminal speed is v, what is the second?
[RPSC Senior Teacher Grade II Science — 01 Nov 2018]
(A) 0
(B) v
(C) 2v
(D) 4v
Answer: (D) 4v
Explanation: At equal material density, eightfold mass means eightfold volume and twice the radius. Stokes-law terminal speed in a viscous fluid is proportional to radius squared, with other properties fixed. Doubling radius therefore makes terminal speed 4v.
Q19. A dropped stone would take 5 s to hit ground. If stopped after 3 s and released again, how much additional time is needed?
[RPSC Senior Teacher Grade II Mathematics — 29 Oct 2018]
(A) 2 s
(B) 3 s
(C) 4 s
(D) None of these
Answer: (C) 4 s
Explanation: Distance fallen in time t from rest is ½gt². Original total distance corresponds to 25 units of ½g; the first three seconds cover 9 units. Remaining distance is 16 units, so a fresh release takes √16 = 4 s.
Q18. Near Earth, if air friction is negligible, different objects fall with the same what?
[RRB Group D — 09 Oct 2018, Shift 3]
(A) Acceleration
(B) Speed at every starting condition
(C) Velocity under all conditions
(D) Mass
Answer: (A) Acceleration
Explanation: Gravitational force is proportional to an object’s mass, but acceleration equals force divided by that mass. The mass cancels, leaving g. Identically released objects therefore share the same acceleration even if their masses differ greatly.
Q17. Under which condition can a falling object experience gravity alone, without air resistance?
[RRB Group D — 19 Sep 2018, Shift 1]
(A) Underwater
(B) In the atmosphere
(C) In ordinary air
(D) In a vacuum
Answer: (D) In a vacuum
Explanation: A vacuum contains no air to generate drag. A falling body then moves under gravity alone, which is the ideal free-fall condition. In water or air, fluid resistance and possibly buoyancy affect the resulting acceleration.
Q16. Approximately how large is acceleration due to gravity at Earth’s surface?
[RRB ALP CBT-I — 31 Aug 2018, Shift 1]
(A) 9.6 cm/s²
(B) 9.8 m/s²
(C) 10.8 m/s²
(D) 9.8 cm/s²
Answer: (B) 9.8 m/s²
Explanation: The standard near-surface gravitational acceleration is approximately 9.8 metres per second squared. It means a freely falling object’s downward velocity increases by roughly 9.8 metres per second during each second, provided air resistance is negligible.
Q15. The gravitational force acting on a mass m near Earth is equal to which expression?
[SSC CHSL — 17 Mar 2018, Shift 3]
(A) Mass multiplied by gravitational acceleration
(B) Mass divided by gravitational acceleration
(C) Acceleration divided by mass
(D) Reciprocal of mass times acceleration
Answer: (A) Mass multiplied by gravitational acceleration
Explanation: Newton’s second law says F = ma. The force causing free-fall acceleration is the body’s weight, so its magnitude is W = mg. Here m is measured in kilograms and g in m/s², giving newtons.
Q14. A planet has Earth’s mass but twice its radius. How large is its surface gravity relative to Earth’s g?
[SSC CHSL — 15 Mar 2018, Shift 1]
(A) g/4
(B) g/2
(C) 4g
(D) 2g
Answer: (A) g/4
Explanation: Surface gravitational acceleration is GM/R². The mass remains M, but doubling R makes the denominator four times as large. Consequently g′ = GM/(2R)² = g/4. The falling object’s own mass is irrelevant.
Q13. A planet has twice Earth’s mass but the same radius. What is its surface gravity?
[SSC CHSL — 11 Mar 2018, Shift 2]
(A) 2g
(B) g
(C) g/2
(D) 4g
Answer: (A) 2g
Explanation: According to g = GM/R², gravitational acceleration scales directly with planetary mass if the radius is fixed. Replacing M by 2M doubles the acceleration, yielding 2g. This is distinct from doubling the radius, which lowers surface gravity.
Q12. A ball is projected at 40 m/s at 30° to horizontal. With g = 10 m/s², what is its total flight time?
[SSC Scientific Assistant Physics — 25 Nov 2017, Shift 2]
(A) 2 s
(B) 1 s
(C) 4 s
(D) 3 s
Answer: (C) 4 s
Explanation: Time of flight on level ground is T = 2u sinθ/g. Here the initial vertical speed is 40 sin30° = 20 m/s. Thus T = 2 × 20/10 = 4 seconds.
Q11. A cannonball fired at 15° travels 45 m horizontally. For g = 10 m/s², find launch speed.
[SSC Scientific Assistant Physics — 25 Nov 2017, Shift 1]
(A) 20 m/s
(B) 30 m/s
(C) 10 m/s
(D) 40 m/s
Answer: (B) 30 m/s
Explanation: On level ground, range R = u² sin2θ/g. Since sin30° = ½, 45 = u²(½)/10. Therefore u² = 900 and u = 30 m/s, when aerodynamic drag is ignored completely.
