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Physics MCQs for NEET — Practice Questions with Answers

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A lighter body and heavier body both are moving with same momentum and applying same retarding force. Their stoping distance are $s_{1} a n d s_{2}$ respectively. Then the correct relation between $s_{1} a n d s_{2}$ .

$s_{1} > s_{2}$

$s_{1} < s_{2}$

$s_{1} = s_{2}$

None of these

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Explanation

$s = \frac{m u^{2}}{2 F} = \frac{P^{2}}{2 F m} = \frac{K E}{F} s α \frac{1}{m} {P a n d F i s g i v e n} s_{1} > s_{2}$

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A plastic box of mass 5 kg is found to accelerate up at the rate of g / 6, when placed deep inside the water. How much sand should be put inside the box so that it may accelerate down at the rate of g / 6?

2kg

3kg

4kg

5kg

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Explanation

$B - Mg = Ma . . . (1) (M + m) g - B = (M + m) a . . . (2) mg = (2 M + m) a [a = g / 6 \dots \dots given] mg = (10 + m) \frac{g}{6} 6 m - m = 10 m = 2 kg$

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Let $θ$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of bob is m. The tension in the string is mg cos $θ$

always

never

at extreme

at mean position

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A metal ring of mass m and radius R is placed on smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Tension in the ring is:

$\frac{m v^{2}}{R}$

$\frac{m v^{2}}{2 R}$

$\frac{m v^{2}}{2 π R}$

None of these

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Explanation

For a ring of mass m and radius R rotating with a uniform speed v, the tension in the ring is given by (mv^2)/(2Ï€R). This is because the centripetal force required to keep each point of the ring in circular motion is provided by the tension in the ring.

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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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