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NEET NEET Questions — Free MCQ Practice

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A vector $\vec{a}$ is turned without a change in its length through a small angle $d θ .$ The value of $| Δ \vec{a} |$ and $Δ a$ are respectively

A.

$0, a d θ$

B.

$a d θ, 0$

C.

0, 0

D.

None of these

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Explanation

When a vector is turned through a small angle dÎ¸ without changing its length, the magnitude of the change in the vector (|Î”a|) is a Ã— dÎ¸, where a is the original vector length. The change in the vector (Î”a) itself is perpendicular to the original vector, with a magnitude of a Ã— dÎ¸.

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Two trains along the same straight rails moving with constant speed 60 km/hr and 30 km/hr respectively towards each other. If at time t = 0, the distance between them is 90 km, the time when they collide is

A.

1 hr

B.

2 hr

C.

3 hr

D.

4 hr

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Explanation

The relative velocity $v_{r e l .} = 60 - (- 30) = 90 k m / h r .$

Distance between the train $s_{r e l .} = 90 k m,$

∴ Time when they collide $= \frac{s_{r e l .}}{v_{r e l .}} = \frac{90}{90} = 1 h r .$

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To a person, going eastward in a car with a velocity of 25 km/hr, a train appears to move towards north with a velocity of $25 \sqrt{3}$ km/hr. The actual velocity of the train will be

A.

25 km/hr

B.

50 km/hr

C.

5 km/hr

D.

$5 \sqrt{3}$ km/hr

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Explanation

$v_{T} = \sqrt{v_{T C}^{2} + v_{C}^{2}}$ = $\sqrt{(25 {\sqrt{3)}}^{2} + {(25)}^{2}}$

= $\sqrt{1875 + 625}$ = $\sqrt{2500}$ = 50 km/hr

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A bus is moving with a velocity 10 m/s on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what velocity should the scooterist chase the bus

A.

50 m/s

B.

40 m/s

C.

30 m/s

D.

20 m/s

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Explanation

Let the velocity of the scooterist =v

Relative velocity of scooterist with respect to bus = (v – 10)

$\Rightarrow S = (v - 10) \times 100 \Rightarrow 1000 = (v - 10) \times 100$

∴ $v = 10 + 10 = 20 m / s$

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The x and y coordinates of the particle at any time are x=5t-2 $t^{2}$ and y=10t respectively, where x and y are in the metres and t is in seconds. The acceleration of the particle at t=2s is :

A.

0

B.

5 $m / s^{2}$

C.

-4 $m / s^{2}$

D.

-8 $m / s^{2}$

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Explanation

(c) Given, x=5t-2 $t^{2}$

The velocity of the particle in the x-direction,

$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (5 t - 2 t^{2}) = 5 - 4 t$

Acceleration, $a_{x} = \frac{d v_{x}}{d t} = - 4 m s^{- 2}$

The velocity of the particle in the y-direction,

$v_{y} = \frac{d y}{d t} = 10 ∴ A c c e l e r a t i o n a_{y} = \frac{d v_{y}}{d t} = 0 ∴ N e t a c c e l e r a t i o n o f t h e p a r t i c l e, {\vec{a}}_{n e t} = a_{x} \hat{i} + a_{y} \hat{j} = (- 4 m s^{2}) \hat{i}$

or $a_{n e t} = - 4 m s^{- 2}$

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Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_{1}$ . On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_{2}$ . The time taken by her to walk up on the moving escalator will be

A.

$\frac{t_{1} - t_{2}}{2}$

B.

$\frac{t_{1} t_{2}}{t_{2} - t_{1}}$

C.

$\frac{t_{1} t_{2}}{t_{2} + t_{1}}$

D.

$t_{1} - t_{2}$

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Explanation

(c)

$S p e e d o f P r e e t i, v_{1} = \frac{h}{t_{1}} S p e e d o f e s c a l a t o r, v_{2} = \frac{h}{t_{2}} S p e e d o f P r e e t i w i t h e s c a l a t o r = v_{1} + v_{2} (V e l o c i t y o f b o t h w i l l c o m b i n e t o g i v e h i g h v e l o c i t y) T i m e = \frac{h}{v_{1} + v_{2}} = \frac{h}{\frac{h}{t_{1}} + \frac{h}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

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A particle moves so that its position vector is given by $r = c o s ω t \hat{x} + s i n ω t \hat{y} w h e r e ω$ is a constant. Which of the following is true?

A.

Velocity and acceleration both are parallel to r.

B.

Velocity is perpendicular to r and acceleration is directed towards to origin.

C.

Velocity is not perpendicular to r and acceleration is directed away from the origin.

D.

Velocity and acceleration both are perpendicular to r.

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A body is moving with velocity 30 m/s towards east. After 10 s its velocity becomes 40 m/s towards north. The average acceleration of the body is

A.

$7 m / s^{2}$

B.

$\sqrt{7} m / s^{2}$

C.

$5 m / s^{2}$

D.

$1 m / s^{2}$

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Explanation

Average acceleration $= \frac{Change in velocity}{Total time}$

$a = \frac{| v_{f} - v_{i} |}{∆ t} = \frac{| 40 j - 30 i |}{∆ t} = \frac{\sqrt{40^{2} + 30^{2}}}{10} = \frac{\sqrt{1600 + 900}}{10} = 5 {ms}^{- 2}$

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A particle moves in the x-y plane according to rule $x = a \sin ωt$ and $y = a \cos ωt$ . The particle follows

A.

an elliptical path

B.

a circular path

C.

a parabolic path

D.

a straight line path inclined equally to x and y-axis

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Explanation

$x = a \sin ωt \Rightarrow \frac{x}{a} = \sin ωt x = a \cos ωt \Rightarrow \frac{y}{a} = \cos ωt \to x^{2} + y^{2} = a^{2}$

This is an equation of a circle, so the particle follows a circular path.

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A particle moves in space such that

$x = 2 t^{3} + 3 t + 4; y = t^{2} + 4 t - 1; z = 2 \sin πt$

Where x, y, z are measured in metre and t in second. The acceleration of the particle at t=3s is

A.

$36 \hat{i} + 2 \hat{j} + \hat{k} {ms}^{- 2}$

B.

$36 \hat{i} + 2 \hat{j} + π \hat{k} {ms}^{- 2}$

C.

$36 \hat{i} + 2 \hat{j} {ms}^{- 2}$

D.

$12 \hat{i} + 2 \hat{j} {ms}^{- 2}$

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Explanation

$x = 2 t^{3} + 3 t + 4 \Rightarrow V_{x} \frac{d x}{d t} = 6 t^{2} + 3 \Rightarrow a_{x} = \frac{d V_{x}}{d t} = 12 t y = t^{2} + 4 t - 1 \Rightarrow V_{y} = 2 t + 4 \Rightarrow a_{y} = \frac{d V_{y}}{d t} = 2 z = 2 sinπ t \Rightarrow V_{z} = \frac{d z}{d t} = 2 πcosπt \Rightarrow a_{z} = \frac{{dV}_{y}}{dt} = - 2 π^{2} \sin πt \Rightarrow \vec{a} = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k} = 12 t \hat{i} + 2 \hat{j} - 2 π^{2} sinπt \hat{k} at t = 3 \sec, \vec{a} = (12 \times 3) \hat{i} + 2 \hat{j} - 2 π^{2} \sin 3 π \hat{k} = 36 \hat{i} + 2 \hat{j}$

NEET Practice Questions (MCQs) with Answers & Solutions

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