🩺 NEET🧲 Physics

Motion in a Plane

5 NEET Physics previous year questions on Motion in a Plane — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.

Q1 FREE PREVIEW

The trajectory of projectile, projected from the ground is given by \(y=x-\frac{{x}^{2}}{20}\). Where x and y are measured in meter. The maximum height attained by the projectile will be:

a

\(5\mathrm{m}\)

b

\(10\sqrt{2}\mathrm{m}\)

c

\(200\mathrm{m}\)

d

\(10\mathrm{m}\)

✓ Correct answer: a)

\(5\mathrm{m}\)

Explanation

\(y=x-\frac{{x}^{2}}{20}\)

For maximum height,
\(\frac{dy}{dx}=0\Rightarrow \frac{d}{dx}\left(x-\frac{{x}^{2}}{20}\right)=0\\ 1-\frac{2x}{20}=0\\ x=10\)

So, \({y}_{\max }=10-\frac{100}{20}=5m\)

Q2

A bullet is fired from a gun at the speed of \(280{\mathrm{ms}}^{-1}\) in the direction \(30^\circ\) above the horizontal. The maximum height attained by the bullet is \(\left(\mathrm{g}=9.8{\mathrm{ms}}^{-2},\sin 30^\circ =0.5\right)\):-

a

2000 m

b

1000 m

c

3000 m

d

2800 m

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Q3

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

a

\( \theta=\tan ^{-1}\left(\frac{1}{4}\right) \)

b

\( \theta=\tan ^{-1}(4) \)

c

\( \theta=\tan ^{-1}(2) \)

d

\( \theta=45^{\circ} \)

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Q4

A car starts from rest and accelerates at \( 5 \mathrm{~m} / \mathrm{s}^{2} \). At \( \mathrm{t}=4 \mathrm{~s} \), a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at \( \mathrm{t}=6 \mathrm{~s} \) ?

[NEET 2021]

a

\( 20 \mathrm{~m} / \mathrm{s}, 0 \)

b

\( 20 \sqrt{2} \mathrm{~m} / \mathrm{s}, 0 \)

c

\( 20 \sqrt{2} \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}^{2} \)

d

\( 20 \mathrm{~m} / \mathrm{s}, 5 \mathrm{~m} / \mathrm{s}^{2} \)

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Q5

A particle moving in a circle of radius \( \mathrm{R} \) with a uniform speed takes a time \( \mathrm{T} \) to complete one revolution. If this particle were projected with the same speed at an angle \( \theta \) to the horizontal, the maximum height attained by it equals \( 4 \mathrm{R} \). The angle of projection, \( \theta \), is then given by:

[NEET 2021]

a

\( \theta=\cos ^{-1}\left(\frac{\pi^{2} \mathrm{R}}{\mathrm{gT}^{2}}\right)^{1 / 2} \)

b

\(\theta ={\sin }^{-1}{\left(\frac{{\pi }^{2}\mathrm{R}}{{\mathrm{gT}}^{2}}\right)}^{1/2}\)

c

\( \theta=\sin ^{-1}\left(\frac{2 g \mathrm{~T}^{2}}{\pi^{2} \mathrm{R}}\right)^{1 / 2} \)

d

\( \theta=\cos ^{-1}\left(\frac{\mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}\right)^{1 / 2} \)

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