Newton's Laws of Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Newton's Laws of Motion MCQs with step-by-step solutions (21 questions). Part of Laws of Motion (Std 11). Practise online on Prepizo — no login needed.
▶ Practise Newton's Laws of Motion online (free)
Questions with solutions
Q1 — Newton's Laws of Motion · easy · theory
Newton's first law of motion states that a body continues in its state of rest or uniform motion unless:
A. Its mass changes
B. It is observed from a moving frame
C. It is acted upon by an unbalanced external force ✓ Correct
D. Friction is absent
Solution: This law defines inertia and identifies what a force actually does.
Q2 — Newton's Laws of Motion · easy · theory
The property of a body by which it resists a change in its state of rest or motion is called:
A. Inertia ✓ Correct
B. Force
C. Impulse
D. Momentum
Solution: Mass is the quantitative measure of inertia.
Q3 — Newton's Laws of Motion · easy · theory
Newton's second law of motion states that:
A. A body at rest stays at rest
B. Every action has an equal and opposite reaction
C. Momentum is always conserved
D. The rate of change of momentum equals the applied force ✓ Correct
Solution: For constant mass this reduces to the familiar $F = ma$.
Q4 — Newton's Laws of Motion · medium · theory
Newton's third law states that action and reaction are:
A. Always zero
B. Unequal if the bodies have different masses
C. Equal in magnitude, opposite in direction and act on different bodies ✓ Correct
D. Equal and opposite but act on the same body
Solution: Because they act on different bodies, they never cancel each other out.
Q5 — Newton's Laws of Motion · medium · theory
Action and reaction forces do not cancel each other because they:
A. Act on two different bodies ✓ Correct
B. Act at different times
C. Act in the same direction
D. Are unequal in magnitude
Solution: Only forces acting on the same body can be added to find a net force.
Q6 — Newton's Laws of Motion · medium · theory
A body is in equilibrium when the net external force on it is:
A. Zero, so it is either at rest or moving with uniform velocity ✓ Correct
B. Directed towards the centre
C. Non-zero but constant
D. Zero, so it must be at rest
Solution: Equilibrium does not mean stationary; uniform motion also satisfies it.
Q7 — Newton's Laws of Motion · easy · theory
The mass of a body is a measure of its:
A. Momentum
B. Weight
C. Inertia ✓ Correct
D. Volume
Solution: A larger mass requires a larger force to produce the same acceleration.
Q8 — Newton's Laws of Motion · hard · theory
The most general form of Newton's second law is:
A. $\vec{F} = \dfrac{\vec{p}}{t}$
B. $\vec{F} = \dfrac{d\vec{p}}{dt}$ ✓ Correct
C. $\vec{F} = m\vec{v}$
D. $\vec{F} = m\vec{a}$ always
Solution: It remains valid even when the mass changes, as for a rocket burning fuel.
Q9 — Newton's Laws of Motion · easy · numerical
A force acts on a body of mass $5\text{ kg}$ producing an acceleration of $2\text{ m/s}^2$. The force is:
A. $20\text{ N}$
B. $2.5\text{ N}$
C. $7\text{ N}$
D. $10\text{ N}$ ✓ Correct
Solution: $F = ma = 5 \times 2 = 10\text{ N}$.
Q10 — Newton's Laws of Motion · easy · numerical
A force of $10\text{ N}$ acts on a body of mass $2\text{ kg}$. Its acceleration is:
A. $0.2\text{ m/s}^2$
B. $20\text{ m/s}^2$
C. $12\text{ m/s}^2$
D. $5\text{ m/s}^2$ ✓ Correct
Solution: $a = \dfrac{F}{m} = \dfrac{10}{2} = 5\text{ m/s}^2$.
Q11 — Newton's Laws of Motion · easy · numerical
A force of $20\text{ N}$ acts on a body of mass $5\text{ kg}$. Its acceleration is:
A. $0.25\text{ m/s}^2$
B. $25\text{ m/s}^2$
C. $100\text{ m/s}^2$
D. $4\text{ m/s}^2$ ✓ Correct
Solution: $a = \dfrac{20}{5} = 4\text{ m/s}^2$.
Q12 — Newton's Laws of Motion · hard · numerical
A body of mass $2\text{ kg}$ is acted upon by two perpendicular forces of $6\text{ N}$ and $8\text{ N}$. Its acceleration is:
A. $5\text{ m/s}^2$ at $\tan^{-1}\left(\dfrac{4}{3}\right)$ to the $6\text{ N}$ force ✓ Correct
B. $7\text{ m/s}^2$ along the diagonal
C. $10\text{ m/s}^2$ at $45^\circ$
D. $5\text{ m/s}^2$ at $\tan^{-1}\left(\dfrac{3}{4}\right)$
Solution: Net force $= \sqrt{36 + 64} = 10\text{ N}$, so $a = \dfrac{10}{2} = 5\text{ m/s}^2$ at $\tan^{-1}\left(\dfrac{8}{6}\right)$.
Q13 — Newton's Laws of Motion · easy · numerical
The weight of a body of mass $10\text{ kg}$ is ($g = 10\text{ m/s}^2$):
A. $1000\text{ N}$
B. $100\text{ N}$ ✓ Correct
C. $10\text{ N}$
D. $1\text{ N}$
Solution: $W = mg = 10 \times 10 = 100\text{ N}$.
Q14 — Newton's Laws of Motion · medium · numerical
Forces of $10\text{ N}$ and $6\text{ N}$ act in opposite directions on a body of mass $4\text{ kg}$. Its acceleration is:
A. $2.5\text{ m/s}^2$
B. $1\text{ m/s}^2$ ✓ Correct
C. $4\text{ m/s}^2$
D. $0.25\text{ m/s}^2$
Solution: Net force $= 10 - 6 = 4\text{ N}$, so $a = \dfrac{4}{4} = 1\text{ m/s}^2$.
Q15 — Newton's Laws of Motion · hard · numerical
A man of mass $50\text{ kg}$ stands in a lift accelerating upward at $2\text{ m/s}^2$. The normal reaction on him is ($g = 10\text{ m/s}^2$):
A. $500\text{ N}$
B. $600\text{ N}$ ✓ Correct
C. $400\text{ N}$
D. $100\text{ N}$
Solution: $N = m(g + a) = 50 \times 12 = 600\text{ N}$; he feels heavier.
Q16 — Newton's Laws of Motion · hard · numerical
The same man of mass $50\text{ kg}$ is in a lift accelerating downward at $2\text{ m/s}^2$. The normal reaction is:
A. $600\text{ N}$
B. $500\text{ N}$
C. $400\text{ N}$ ✓ Correct
D. $100\text{ N}$
Solution: $N = m(g - a) = 50 \times 8 = 400\text{ N}$; he feels lighter.
Q17 — Newton's Laws of Motion · hard · numerical
A man stands in a lift whose cable has snapped, so the lift falls freely. The normal reaction on him is:
A. Half his weight
B. Twice his weight
C. Zero ✓ Correct
D. Equal to his weight
Solution: With $a = g$, $N = m(g - g) = 0$ — the condition of apparent weightlessness.
Q18 — Newton's Laws of Motion · easy · numerical
A force of $100\text{ N}$ acts on a body of mass $20\text{ kg}$. Its acceleration is:
A. $2000\text{ m/s}^2$
B. $0.2\text{ m/s}^2$
C. $20\text{ m/s}^2$
D. $5\text{ m/s}^2$ ✓ Correct
Solution: $a = \dfrac{100}{20} = 5\text{ m/s}^2$.
Q19 — Newton's Laws of Motion · hard · numerical
A car of mass $1000\text{ kg}$ travelling at $20\text{ m/s}$ is brought to rest over $50\text{ m}$. The average retarding force is:
A. $6000\text{ N}$
B. $8000\text{ N}$
C. $2000\text{ N}$
D. $4000\text{ N}$ ✓ Correct
Solution: By the work-energy theorem, $F \times 50 = \dfrac{1}{2}(1000)(400)$, giving $F = 4000\text{ N}$.
Q20 — Newton's Laws of Motion · easy · numerical
The net force on a body moving with constant velocity is:
A. Equal to its weight
B. Zero ✓ Correct
C. Equal to $mv$
D. Directed along its motion
Solution: Constant velocity means zero acceleration, and $F = ma = 0$.
Q21 — Newton's Laws of Motion · easy · numerical
A force produces an acceleration of $20\text{ m/s}^2$ in a body of mass $0.5\text{ kg}$. The force is:
A. $20\text{ N}$
B. $40\text{ N}$
C. $0.025\text{ N}$
D. $10\text{ N}$ ✓ Correct
Solution: $F = 0.5 \times 20 = 10\text{ N}$.