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Laws of Motion MCQs for NEET — Physics Questions with Answers

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While walking on ice one should take small steps to avoid slipping. This is because smaller steps ensure

Larger friction

Smaller friction

Larger normal force

Smaller normal force

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Forces of 1 N and 2 N act along with the lines x = 0 and y = 0. The equation of the line along which the resultant lies is given by

y - 2x = 0

2y - x = 0

y + x = 0

y - x = 0

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Explanation

The resultant force acts along the line 2y - x = 0. This is because the force along the x-axis is 1 N, and the force along the y-axis is 2 N. The slope of the resultant force is tan(theta) = 2/1 = 2. The equation 2y - x = 0 represents a line with a slope of 2.

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A particle is moving in a vertical circle. The tension in the string when passing through two positions at anele of 30 $°$ and 60 $°$ from vertical (the lowest position) are $T_{1} and T_{2}$ respectively, then:

$T_{1} = T_{2}$

$T_{2} < T_{1}$

$T_{2} > T_{1}$

Tension in the string always remains the same.

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A motorcycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on it:

Increases

decreases

remains the same

fluctuates erratically

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Explanation

As the motorcycle ascends on the overbridge, it experiences a centrifugal force acting outwards. To maintain the circular motion, the normal force must increase to provide the required centripetal force. Hence, the normal force on the motorcycle increases while ascending the overbridge.

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Five persons A, B, C, D & E are pulling a cart of mass 100 kg on a smooth surface and the cart is moving with acceleration 3 $m / s^{2}$ in east direction. When person 'A' stops pulling, it moves with acceleration 1 $m / s^{2}$ in the west direction. When person 'B' stops pulling, it moves with acceleration 24 $m / s^{2}$ in the north direction. The magnitude of the acceleration of the cart when only A & B pull the cart keeping their directions same as the old directions are:

26 $m / s^{2}$

$3 \sqrt{71} m / s^{2}$

25 $m / s^{2}$

30 $m / s^{2}$

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Explanation

$\begin{aligned} {\vec{F}}_{A} + {\vec{F}}_{B} + {\vec{F}}_{C} + {\vec{F}}_{D} + {\vec{F}}_{E} = 100 \times 3 \hat{i} \\ {\vec{F}}_{B} + {\vec{F}}_{C} + {\vec{F}}_{D} + {\vec{F}}_{E} = 100 (- \hat{i}) \\ F_{A} + F_{C} + F_{D} + F_{E} = 100 \times 24 \hat{j} \\ {\vec{F}}_{A} = 400 \hat{i} \\ F_{B} = 300 \hat{i} - 2400 \hat{j} \\ {\vec{a}}_{Net} = \frac{700 \hat{i} - 2400 \hat{j}}{100} \end{aligned}$

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The forces required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If $ϕ$ is angle of friction and $θ$ is the angle which makes with horizontal, then

$\tan θ = \tan ϕ$

$\tan θ = 2 \tan ϕ$

$\tan θ = 3 \tan ϕ$

$\tan ϕ = 3 \tan θ$

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Explanation

$m g (\sin θ + μ \cos θ) = 2 m g (\sin θ - μ \cos θ) \tan θ = 3 μ \tan θ = 3 \tan φ$

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If no external force acts on the system, then the correct statement is:

velocity of centre of mass changes

momentum of centre of mass is conserved and constant

position of the center of mass changes with time

None of these

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Explanation

$F = \frac{dp}{dt}$

If F = 0, then p = constant and conserved

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The engine of a car produces acceleration 4 m/s² in the car. If this car pulls another car of same mass, what will be the acceleration produced

8 m/s²

2 m/s²

4 m/s²

$\frac{1}{2} m / s^{2}$

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Explanation

$F = m \times a$ , If force is constant then $a \propto \frac{1}{m}$ . So If mass is doubled then acceleration becomes half.

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A truck and a car are moving with equal velocity. If equal retarding force is applied on each, then on applying the brakes both will stop after certain distance

Truck will cover less distance before rest

Car will cover less distance before rest

Both will cover equal distance

None

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Explanation

F= ma

$a = \frac{F}{m}$

Stopping distance s = $\frac{v^{2}}{2 a} = \frac{m v^{2}}{2 F}$

If retarding force (F) and velocity (v) are equal then stopping distance ∝ m (mass of vehicle)

As $m_{car} < m_{truck}$ therefore car will cover less distance before coming to rest.

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A constant force acts on a body of mass 0.9 kg at rest for 10s. If the body moves a distance of 250 m, the magnitude of the force is

3 N

3.5 N

4.0 N

4.5 N

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Explanation

$u = 0, S = 250 m, t = 10 \sec$

$S = u t + \frac{1}{2} a t^{2}$ ⇒ $250 = \frac{1}{2} a {[10]}^{2} \Rightarrow a = 5 m / s^{2}$

So, $F = m a = 0 .9 \times 5 = 4 .5 N$

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Yes. You can attempt every Laws of Motion question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.