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Physics MCQs for NEET — Practice Questions with Answers

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At the top of the trajectory of a projectile, the magnitude of the acceleration is

Maximum

Minimum

Zero

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Explanation

Acceleration through out the projectile motion remains constant and equal to g.

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A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

$25^{o} 8^{'}$

$33^{o} 7^{'}$

$42^{o} 8^{'}$

$53^{o} 8^{'}$

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Explanation

$R = 4 H \cot θ,$ if R = 3H then $\cot θ = \frac{3}{4}$ ⇒ $θ = 53 ° 8'$

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Two bodies are projected with the same velocity. If one is projected at an angle of 30° and the other at an angle of 60° to the horizontal, the ratio of the maximum heights reached is

3 : 1

1 : 3

1 : 2

2 : 1

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Explanation

As $H = \frac{u^{2} \sin^{2} θ}{2 g}$ ∴ $\frac{H_{1}}{H_{2}} = \frac{\sin^{2} θ_{1}}{\sin θ_{2}} = \frac{\sin^{2} 30 °}{\sin^{2} 60}$ = $\frac{1 / 4}{3 / 4} = \frac{1}{3}$

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If a body A of mass M is thrown with velocity V at an angle of 30° to the horizontal and another body B of the same mass is thrown with the same speed at an angle of 60° to the horizontal. The ratio of horizontal range of A to B will be

1 : 3

1 : 1

$1 : \sqrt{3}$

$\sqrt{3} : 1$

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Explanation

For complementary angles range will be equal.

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Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15^o, 30^o, 45^o and 60^o with the horizontal respectively. The body having shortest range is

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Explanation

When the angle of projection is very far from 45° then range will be minimum.

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Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event

Speed before he jumps and his weight

The direction in which he leaps and the initial speed

The force with which he pushes the ground and his speed

None of these

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Explanation

Range $= \frac{u^{2} \sin 2 θ}{g} .$

It is clear that range is proportional to the direction (angle) and the initial speed.

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In a projectile motion, velocity at maximum height is

$\frac{u \cos θ}{2}$

$u \cos θ$

$\frac{u \sin θ}{2}$

None of these

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Explanation

Only horizontal component of velocity $(u \cos θ)$ .

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The equation of motion of a projectile are given by x = 36 t metre and 2y = 96 t – 9.8 t² metre. The angle of projection is

$\sin^{- 1} (\frac{4}{5})$

$\sin^{- 1} (\frac{3}{5})$

$\sin^{- 1} (\frac{4}{3})$

$\sin^{- 1} (\frac{3}{4})$

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Explanation

x = 36 t ∴ $v_{x} = \frac{d x}{d t} = 36 m / s$

$y = 48 t - 4 .9 t^{2}$ ∴ $v_{y} = 48 - 9 .8 t$

at t = 0, v_x = 36 and v_y = 48 m/s

So, angle of projection $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) = \tan^{- 1} (\frac{4}{3})$

Or $θ = \sin^{- 1} (4 / 5)$

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For a given velocity, a projectile has the same range R for two angles of projection. If t₁ and t₂ are the times of flight in the two cases then :

$t_{1} t_{2} \propto R^{2}$

$t_{1} t_{2} \propto R$

$t_{1} t_{2} \propto \frac{1}{R}$

$t_{1} t_{2} \propto \frac{1}{R^{2}}$

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Explanation

For the same range, the angle of projection should be θ and 90–θ.

So, the time of flights $t_{1} = \frac{2 u \sin θ}{g}$ and

$t_{2} = \frac{2 u \sin (90 - θ)}{g} = \frac{2 u \cos θ}{g}$

By multiplying $= t_{1} t_{2} = \frac{4 u^{2} \sin θ \cos θ}{g^{2}}$

$t_{1} t_{2} = \frac{2}{g} \frac{(u^{2} \sin 2 θ)}{g} = \frac{2 R}{g}$ ⇒ $t_{1} t_{2} \propto R$

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A body of mass m is thrown upwards at an angle θ with the horizontal with velocity v. While rising up the velocity of the mass after t seconds will be

$\sqrt{{(v \cos θ)}^{2} + {(v \sin θ)}^{2}}$

$\sqrt{{(v \cos θ - v \sin θ)}^{2} - g t}$

$\sqrt{v^{2} + g^{2} t^{2} - (2 v \sin θ) g t}$

$\sqrt{v^{2} + g^{2} t^{2} - (2 v \cos θ) g t}$

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Explanation

Instantaneous velocity of rising mass after t sec will be $v_{t} = \sqrt{v_{x}^{2} + v_{y}^{2}}$

where $v_{x} = v \cos θ =$ Horizontal component of velocity

$v_{y} = v \sin θ - g t =$ Vertical component of velocity

$v_{t} = \sqrt{{(v \cos θ)}^{2} + {(v \sin θ - g t)}^{2}}$

$v_{t} = \sqrt{v^{2} + g^{2} t^{2} - 2 v \sin θ g t}$

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