Question 1

The thrust on a rocket ejecting exhaust gases at relative speed v_rel with a burn rate dm/dt is:

Accepted Answer

F_{thrust} = v_{rel}\,\dfrac{dm}{dt}. The reaction thrust equals the rate of momentum carried away by the exhaust: F_thrust = v_rel·(dm/dt).

Question 2

For a rocket in free space (no gravity), the velocity gained when its mass falls from m₀ to m is:

Accepted Answer

v = v_{rel}\,\ln\!\left(\dfrac{m_0}{m}\right). Tsiolkovsky's rocket equation: v = v_rel ln(m₀/m). The velocity depends logarithmically on the mass ratio.

Question 3

A uniform chain falls link by link from rest onto a table. The force it exerts on the table (at any instant) is:

Accepted Answer

The weight of the chain already lying on the table PLUS the rate of momentum delivered by the falling links. The table supports the weight of the accumulated chain and also stops the incoming links; the total force is the sum, giving three times the instantaneous weight on the table.

Question 4

Sand falls vertically at a steady rate dm/dt onto a horizontal conveyor belt moving at speed v. The extra horizontal force needed to keep the belt moving at constant speed is:

Accepted Answer

v\,\dfrac{dm}{dt}. The belt must give each unit mass of sand horizontal momentum: F = v·(dm/dt). (The power delivered is Fv = v²dm/dt, twice the KE gained — the rest is dissipated.)

Question 5

For a time-dependent force F(t) acting on a body, the change in its momentum over a time interval is:

Accepted Answer

\int F\,dt. Δp = ∫F dt (the impulse). Only for constant F does this reduce to F·t.

Question 6

For a position-dependent force F(x), the work done (and hence the change in kinetic energy) as the body moves from x₁ to x₂ is:

Accepted Answer

\int_{x_1}^{x_2} F\,dx. W = ∫F dx = ΔKE (work–energy theorem). Only for constant F does this become F·Δx.

Question 7

The general equation of motion for a variable-mass system (external force F_ext) is:

Accepted Answer

F_{ext} + v_{rel}\dfrac{dm}{dt} = m\dfrac{dv}{dt}. Including the thrust term, F_ext + v_rel(dm/dt) = m(dv/dt); the thrust term accounts for the momentum carried by the added/ejected mass.

Question 8

A chain hanging above a table is released and falls onto it. The reason the peak force on the table exceeds the chain's total weight is that:

Accepted Answer

The table must also destroy the downward momentum of the still-falling links. Besides supporting the weight already on it, the table exerts an additional force to stop the momentum of the arriving links, so the instantaneous force can be up to 3× the weight resting on it.

Question 9

A rocket ejects gas at a relative speed of 1000 m/s and burns fuel at 10 kg/s. The thrust developed is:

Accepted Answer

10000 N. F_thrust = v_rel(dm/dt) = 1000 × 10 = 10000 N.

Question 10

A rocket of instantaneous mass 1000 kg ejects gas at 2000 m/s, burning 20 kg/s (g = 10 m/s²). Its instantaneous upward acceleration is:

Accepted Answer

30 m/s². Thrust = 2000 × 20 = 40000 N; net force = 40000 − mg = 40000 − 10000 = 30000 N; a = 30000/1000 = 30 m/s².

Question 11

Sand falls at 5 kg/s onto a conveyor belt moving horizontally at 2 m/s. The extra force needed to keep the belt moving steadily is:

Accepted Answer

10 N. F = v(dm/dt) = 2 × 5 = 10 N.

Question 12

A uniform chain of linear density 2 kg/m falls onto a table. When a length of 1 m has landed (g = 10 m/s²), the force on the table is:

Accepted Answer

60 N. Force = 3 × (weight of chain already landed) = 3 × (λ·y·g) = 3 × (2 × 1 × 10) = 60 N. Pitfall: it is three times the resting weight, not just the weight.

Question 13

A time-varying force F = 4t (N) acts on a body for 3 s. The impulse delivered is:

Accepted Answer

18 N·s. Impulse = ∫₀³ 4t dt = [2t²]₀³ = 18 N·s.

Question 14

A position-dependent force F = 6x (N) acts on a 3 kg body starting from rest. Its speed after moving 2 m is:

Accepted Answer

≈ 2.83 m/s. W = ∫₀² 6x dx = [3x²]₀² = 12 J = ½mv² ⇒ v = √(2 × 12/3) = √8 ≈ 2.83 m/s.

Question 15

A rocket ejects exhaust at 800 m/s, burning fuel at 5 kg/s. The thrust is:

Accepted Answer

4000 N. F_thrust = v_rel(dm/dt) = 800 × 5 = 4000 N.

Question 16

Gravel drops onto a conveyor belt at 10 kg/s while the belt moves at 3 m/s. The power required to drive the belt against this loading is:

Accepted Answer

90 W. Force = v(dm/dt) = 3 × 10 = 30 N; power = Fv = 30 × 3 = 90 W. (Only half of this, 45 W, becomes KE of the gravel; the rest is dissipated.)

Question 17

A rocket of total mass 1000 kg ejects gas at 2000 m/s. The minimum burn rate needed just to lift it off (g = 10 m/s²) is:

Accepted Answer

5 kg/s. For lift-off thrust = weight: v_rel(dm/dt) = mg ⇒ dm/dt = mg/v_rel = (1000 × 10)/2000 = 5 kg/s.

Question 18

A hose ejects water at 2 kg/s with a speed of 20 m/s. The reaction force (thrust) on the hose is:

Accepted Answer

40 N. F = v(dm/dt) = 20 × 2 = 40 N.

Question 19

A chain of linear density 2 kg/m lying on the floor is lifted straight up at a constant speed of 3 m/s. The extra force (beyond the weight already lifted) needed, due to the momentum of the newly moving links, is:

Accepted Answer

18 N. The momentum term is λv² = 2 × 3² = 18 N (the force to give the just-lifted links their velocity), in addition to the weight of the raised part.

Question 20

A force F = (10 − 2t) N acts on a 2 kg body at rest. The speed of the body at the instant the force becomes zero is:

Accepted Answer

12.5 m/s. F = 0 at t = 5 s. Impulse = ∫₀⁵ (10 − 2t) dt = [10t − t²]₀⁵ = 50 − 25 = 25 N·s; v = 25/2 = 12.5 m/s.

Variable Mass Systems & Advanced Mechanics — JEE Main Physics MCQs with Solutions

Questions with solutions