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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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A cyclist turns around a curve at 15 miles/hour. If he turns at double the speed, the tendency to overturn is

Doubled

Quadrupled

Halved

Unchanged

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Explanation

$F = \frac{m v^{2}}{r}$ ⇒ $F \propto v^{2}$ . If v becomes double then F (tendency to overturn) will become four times.

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If the overbridge is concave instead of being convex, the thrust on the road at the lowest position will be

$m g + \frac{m v^{2}}{r}$

$m g - \frac{m v^{2}}{r}$

$\frac{m^{2} v^{2} g}{r}$

$\frac{v^{2} g}{r}$

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A car sometimes overturns while taking a turn. When it overturns, it is

The inner wheel which leaves the ground first

The outer wheel which leaves the ground first

Both the wheels leave the ground simultaneously

Either wheel leaves the ground first

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Explanation

Because the reaction on inner wheel decreases and becomes zero. So it leaves the ground first.

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Yes. You can attempt every NEET 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.