Non-Inertial Frames & Pseudo Forces — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Non-Inertial Frames & Pseudo Forces MCQs with step-by-step solutions (20 questions). Part of Newton's Laws of Motion & Friction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Non-Inertial Frames & Pseudo Forces · medium · theory
The pseudo (fictitious) force that must be added to apply Newton's laws in a frame accelerating with a⃗_f is:
A. $\vec{F}_p = -m\vec{g}$
B. $\vec{F}_p = -m\vec{a}_f$ ✓ Correct
C. $\vec{F}_p = +m\vec{a}_f$
D. $\vec{F}_p = \dfrac{m}{\vec{a}_f}$
Solution: In a non-inertial frame, a pseudo force −m a⃗_f (opposite to the frame's acceleration) is added to every body of mass m so that F = ma still works.
Q2 — Non-Inertial Frames & Pseudo Forces · medium · theory
Pseudo forces are introduced because:
A. Newton's laws never hold anywhere
B. Gravity disappears in accelerating frames
C. Newton's second law is not valid as-is in a non-inertial (accelerating) frame ✓ Correct
D. Real forces vanish in accelerating frames
Solution: In an accelerating frame the observed acceleration does not equal (real force)/m; adding the pseudo force restores the form F_net = ma in that frame.
Q3 — Non-Inertial Frames & Pseudo Forces · hard · theory
A person of mass m stands on a weighing scale in a lift accelerating upward with acceleration a. The scale reads:
A. ma
B. mg
C. m(g + a) ✓ Correct
D. m(g − a)
Solution: Normal reaction N = m(g + a) when accelerating upward. For downward acceleration N = m(g − a); in free fall (a = g) N = 0 (weightlessness).
Q4 — Non-Inertial Frames & Pseudo Forces · medium · theory
In a uniformly accelerating frame, the "effective gravity" g_eff experienced by bodies is:
A. Always equal to g
B. The vector combination of real g and the pseudo-force acceleration ✓ Correct
C. Independent of the frame's acceleration
D. Always zero
Solution: g_eff = g − a_f (vector). For a horizontally accelerating frame, g_eff = √(g² + a²), tilted from the vertical.
Q5 — Non-Inertial Frames & Pseudo Forces · hard · theory
A simple pendulum hangs from the roof of a car moving with a horizontal acceleration a. In equilibrium the string makes an angle θ with the vertical given by:
A. $\cos\theta = \dfrac{a}{g}$
B. $\tan\theta = \dfrac{g}{a}$
C. $\tan\theta = \dfrac{a}{g}$ ✓ Correct
D. $\sin\theta = \dfrac{a}{g}$
Solution: In the car's frame, the pseudo force ma (backward) and weight mg give tanθ = a/g, the bob tilting opposite to the acceleration.
Q6 — Non-Inertial Frames & Pseudo Forces · medium · theory
The centrifugal force is:
A. The reaction to the normal force
B. A real force directed towards the centre
C. Always equal to the weight
D. The pseudo force appearing in a rotating (non-inertial) frame, directed radially outward ✓ Correct
Solution: In a frame rotating with the body, the pseudo (centrifugal) force mω²r acts outward, balancing the inward real forces in that frame.
Q7 — Non-Inertial Frames & Pseudo Forces · hard · theory
A person standing inside a lift in free fall (a = g downward) experiences an apparent weight of:
A. 2mg
B. mg
C. Zero (weightlessness) ✓ Correct
D. mg/2
Solution: N = m(g − a) = m(g − g) = 0. The pseudo force exactly cancels gravity in the free-falling frame.
Q8 — Non-Inertial Frames & Pseudo Forces · medium · theory
In a non-inertial frame accelerating with a_f, the pseudo force acts:
A. On every body, with magnitude ma_f, opposite to the frame's acceleration ✓ Correct
B. Along the frame's acceleration
C. Only on the heaviest body
D. Only on bodies in contact with the ground
Solution: Every mass m in the frame feels a pseudo force ma_f directed opposite to the frame's acceleration.
Q9 — Non-Inertial Frames & Pseudo Forces · medium · numerical
A 50 kg person stands on a scale in a lift accelerating upward at 2 m/s² (g = 10 m/s²). The scale reading is:
A. 600 N ✓ Correct
B. 100 N
C. 400 N
D. 500 N
Solution: N = m(g + a) = 50(10 + 2) = 600 N.
Q10 — Non-Inertial Frames & Pseudo Forces · medium · numerical
The same 50 kg person now stands in a lift accelerating downward at 2 m/s² (g = 10 m/s²). The scale reads:
A. 400 N ✓ Correct
B. 500 N
C. 600 N
D. 100 N
Solution: N = m(g − a) = 50(10 − 2) = 400 N.
Q11 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A pendulum in a car accelerating horizontally at 10 m/s² (g = 10 m/s²) settles at an angle to the vertical of:
A. 45° ✓ Correct
B. 90°
C. 30°
D. 60°
Solution: tanθ = a/g = 10/10 = 1 ⇒ θ = 45°.
Q12 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A pendulum in a car accelerating at 5 m/s² (g = 10 m/s²) makes an angle with the vertical of:
A. 30°
B. tan⁻¹(2) ≈ 63.4°
C. tan⁻¹(0.5) ≈ 26.6° ✓ Correct
D. 45°
Solution: tanθ = a/g = 5/10 = 0.5 ⇒ θ = tan⁻¹(0.5) ≈ 26.6°.
Q13 — Non-Inertial Frames & Pseudo Forces · medium · numerical
The apparent weight of a 5 kg mass in a lift that is in free fall is:
A. 25 N
B. 0 N ✓ Correct
C. 50 N
D. 100 N
Solution: In free fall a = g, so N = m(g − g) = 0 (weightlessness).
Q14 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A frame accelerates horizontally at 5 m/s² (g = 10 m/s²). The effective gravity in this frame is:
A. ≈ 11.2 m/s² ✓ Correct
B. 15 m/s²
C. 5 m/s²
D. ≈ 8.7 m/s²
Solution: g_eff = √(g² + a²) = √(10² + 5²) = √125 ≈ 11.2 m/s².
Q15 — Non-Inertial Frames & Pseudo Forces · medium · numerical
A 2 kg block rests on the smooth floor of a truck that accelerates at 3 m/s². The pseudo force on the block (in the truck's frame) is:
A. 6 N ✓ Correct
B. 3 N
C. 1.5 N
D. 20 N
Solution: F_p = ma_f = 2 × 3 = 6 N, directed opposite to the truck's acceleration.
Q16 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A spring balance in a lift accelerating upward at 2 m/s² carries a 5 kg mass (g = 10 m/s²). Its reading is:
A. 60 N ✓ Correct
B. 50 N
C. 40 N
D. 10 N
Solution: Reading = m(g + a) = 5(10 + 2) = 60 N.
Q17 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A block rests on a smooth 37° wedge fixed in a truck. The horizontal acceleration of the truck for which the block does not slide on the wedge is (tan37° = 0.75, g = 10 m/s²):
A. 13.3 m/s²
B. 5 m/s²
C. 10 m/s²
D. 7.5 m/s² ✓ Correct
Solution: For no sliding, the pseudo force must balance the tendency to slide: a = g tanθ = 10 × 0.75 = 7.5 m/s².
Q18 — Non-Inertial Frames & Pseudo Forces · medium · numerical
The pseudo force on a 3 kg body in a frame accelerating at 4 m/s² has magnitude:
A. 7 N
B. 12 N ✓ Correct
C. 0.75 N
D. 30 N
Solution: F_p = ma_f = 3 × 4 = 12 N.
Q19 — Non-Inertial Frames & Pseudo Forces · hard · numerical
For the apparent weight of a person in a lift to become double the true weight, the lift must accelerate upward at (g = 10 m/s²):
A. 2 m/s²
B. 10 m/s² ✓ Correct
C. 5 m/s²
D. 20 m/s²
Solution: N = 2mg = m(g + a) ⇒ a = g = 10 m/s².
Q20 — Non-Inertial Frames & Pseudo Forces · hard · numerical
A simple pendulum has a period of 2 s at rest. In a lift accelerating upward at 2 m/s² (g = 10 m/s²), its period becomes:
A. ≈ 1.83 s ✓ Correct
B. ≈ 2.19 s
C. 2 s
D. ≈ 1.0 s
Solution: T ∝ 1/√g_eff, with g_eff = g + a = 12. T = T₀√(g/g_eff) = 2√(10/12) ≈ 1.83 s. Pitfall: upward acceleration increases g_eff, so the period decreases.