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NEET Practice Questions (MCQs) with Answers & Solutions

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The coordinates of a moving particle at a time t, are given by, x=5 sin 10t, y=5 cos 10t. The speed of the particle is:

$50 \sqrt{2}$

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Explanation

(B)

$x = 5 \sin 10 t y = 5 \cos 10 t v_{x} = \frac{dx}{dt} v_{y} = \frac{dy}{dt} = 5 \times 10 \cos l 0 t = - 50 \sin l 0 t = 50 \cos l 0 t Velocity \vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = 50 \cos l 0 t \hat{i} - 50 \sin l 0 t \hat{j} Speed = Magnitude of velocity = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(50 \cos l 0 t)}^{2} + {(- 50 \sin l 0 t)}^{2}} = 50 \sqrt{\cos^{2} 10 t + \sin^{2} 10 t} units = 50 units$

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For a rocket moving in free space, the fraction of mass to be disposed of Off to attain a speed equal to two times the exhaust speed is given by (given $e^{2}$ = 7.4)

0.40

0.37

0.50

0.86

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Explanation

$\begin{aligned} v & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} o r \begin{aligned} 2 v_{r} & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} 2 & = \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} \frac{m_{0}}{m} & = e^{2} = 7.4 \end{aligned} o r \begin{aligned} \frac{m}{m_{0}} & = 0.14 \end{aligned}$

This is a fraction of mass remaining. Hence fraction of mass disposed of = 0.86.

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A simple pendulum hangs from the roof of a train moving on horizontal rails. If the string is inclined towards the front of the train, then train is

Moving with constant velocity

In accelerated motion

In retarded motion

At rest

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Explanation

Using concept of inertia

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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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