Class 11 Physics, Chapter 4 (NCERT). 15 questions by concept, 7 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
In an inertial frame, a car travelling along a straight rough road has a constant velocity of 30 km/h. Which statement gives the net force?
(b) Zero, because its velocity is constant.
Constant velocity means zero acceleration. The external forces balance; friction or drag and other forces can still act individually.
A massless string junction is in equilibrium. A vertical lower string pulls it downward with 60 N, an applied force pulls right with 50 N, and the upper string runs upward-left at angle θ from the vertical. What is tan θ?
(c) 5/6
Resolve the upper-string tension at angle θ to the vertical.
Equilibrium requires Tcosθ = 60 N and Tsinθ = 50 N.
Divide horizontal by vertical: tanθ = 50/60.
Therefore, the requested answer is 5/6.
An upper string makes angle θ with the vertical and supports a junction pulled downward by 60 N. A horizontal force balances its horizontal component. Given cos θ = 4/5 and sin θ = 3/5, what is the upper tension?
(c) 75 N
The angle θ is measured from the vertical, so the upward component is Tcosθ.
Vertical equilibrium gives Tcosθ = 60 N.
Thus T = 60/(4/5) N = 75 N. Use the cosine, not the sine, for this vertical component.
Therefore, the requested answer is 75 N.
A body of mass 13 kg experiences only two horizontal forces: 5 N east and 12 N north. What is the magnitude of its acceleration?
(b) 1 m/s²
The two force directions are perpendicular. The resultant magnitude is √(Fx² + Fy²) = 13 N; divide by 13 kg.
The position-time graph describes a particle of mass 4 kg. What is the net force during the open straight rising interval, excluding its endpoints?
(c) 0 N
A straight position-time segment has constant slope and therefore constant velocity. Acceleration and net force vanish throughout this interval, though the velocity is nonzero.
The position-time graph describes a particle of mass 4 kg. At the corner where the rising segment becomes horizontal, what signed impulse acts on the particle? Positive impulse points along positive x.
(a) -3 N s
Immediately before this corner velocity is the rising-segment slope h/t; afterward it is zero. The impulse is m(0 − h/t), negative because the particle is stopped.
Identical balls arrive with equal momentum and are brought to rest. Catch A takes 0.02 s; catch B takes 4 times as long. What is average force magnitude(B)/average force magnitude(A)?
(c) 1/4
Both catches produce the same momentum change. Average force is impulse divided by duration, so its ratio is the reciprocal of the duration ratio.
Blocks A, B and C lie in that order from left to right on a smooth horizontal surface. Their masses are 10, 20 and 30 kg. They remain in contact as a force of 60 N pushes A to the right. What is the force A exerts on B?
(c) 50 N
All three blocks accelerate together on the smooth surface: a_system = 60/(10 + 20 + 30).
Consider blocks B and C as one subsystem. A’s contact force must accelerate their combined mass 20 + 30.
Therefore F_A→B = (20 + 30) × 60/(10 + 20 + 30) N.
Therefore, the requested answer is 50 N.
Masses A = 10 kg and B = 20 kg are connected by a light string on a smooth horizontal surface. A pull of 600 N is applied first to A away from B, then in a separate trial to B away from A. What are the string tensions in these trials, respectively?
(b) 400 N, 200 N
The common acceleration magnitude in either trial is F/(mA + mB). The string accelerates the trailing mass: B in the first trial, A in the second.
An initially stationary isolated system separates into a 1/50 kg part moving forward at 80 m/s and a 100 kg part. Speeds are measured in the original laboratory frame. What velocity does the heavier part have?
(b) 0.016 m/s backward
The initial total momentum is zero. M V = −m v, so the heavier part recoils with 0.016 m/s backward.
On a smooth surface, cart A pushes cart B with 12 N eastward; cart B simultaneously pushes cart A with 12 N westward. A student adds these two forces when finding the force on cart A alone. What is the error?
(c) The two forces act on different bodies; only the westward force acts on A.
Identify each force by its receiver and agent. Only the force exerted by B on A belongs in A’s free-body equation.
A 70 kg person stands on a force-reading scale in a lift that accelerates upward at 5 m/s². Taking g = 10 m/s², what support force does the scale report?
(c) 1050 N
Choose upward positive: N − mg = ma_y. Hence N = m(g + a_y), including the negative sign for downward acceleration.
A block is just about to slide down a plane at angle θ above horizontal. Given sin θ = 3/5 and cos θ = 4/5, what is the coefficient of static friction?
(a) 3/4
At the threshold, mg sin θ = μ_s N and N = mg cos θ. Dividing gives μ_s = tan θ = sin θ/cos θ.
A frictionless road is banked at angle θ to the horizontal, with tan θ = 1/4. A car follows a horizontal circle of radius 40 m. Taking g = 10 m/s², what is the square of its design speed?
(c) 100 m²/s²
For the frictionless design speed, Ncosθ = mg and Nsinθ = mv²/R.
Divide the equations to obtain tanθ = v²/(Rg), so v² = Rg tanθ.
Substitute R = 40 m, g = 10 m/s² and tanθ = 1/4.
Therefore, the requested answer is 100 m²/s².
A 400 kg car travels at the frictionless design speed on a bank with cos θ = 4/5, where θ is the angle above horizontal. Taking g = 10 m/s², what is the normal reaction magnitude?
(a) 5000 N
At the frictionless design speed the vertical forces balance: Ncosθ = mg.
Hence N = mg/cosθ.
Substitute m = 400 kg, g = 10 m/s² and cosθ = 4/5. The normal force exceeds mg because it also supplies the inward horizontal force.
Therefore, the requested answer is 5000 N.
Wrong
Equal action and reaction forces cancel, so nothing accelerates.
Why it feels right. I add both arrows as if they act on one body.
Right
Third-law force pairs act on different bodies.
Worked example
