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NEET NEET Questions — Free MCQ Practice

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The displacement x of a particle moving in one dimension under the action of the constant force is related to time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. Find the displacement of the particle from t = 0 sec to t = 6 s.

A.

0

B.

12 m

C.

6 m

D.

18 m

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Explanation

$t = \sqrt{x} + 3 x = {(t - 3)}^{2} A t t = 0 s e c, x_{1} = 9 m A t t = 6 s e c, x_{2} = 9 m D i s p l a c e m e n t = x_{2} - x_{1} = 0$

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The acceleration a (in ${ms}^{- 2}$ ) of a body, starting from rest varies with time t(in s) following the equation a =3t+4. The velocity of the body at time t = 2s will be :

A.

10 ${ms}^{- 1}$

B.

18 ${ms}^{- 1}$

C.

14 ${ms}^{- 1}$

D.

26 ${ms}^{- 1}$

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Explanation

$a = 3 t + 4 \int_{0}^{v} d v = \int_{0}^{2} (3 t + 4) d t V = 3 {[\frac{t^{2}}{2}]}_{0}^{2} + 4 {[t]}_{0}^{2} = \frac{3}{2} \times 4 + 4 \times 2 = 14 m / s$

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A point moves in a straight line under the retardation a $v^{2}$ . If the initial velocity is u, the distance covered in 't' seconds is-

A.

$aut$

B.

$\frac{1}{a} \ln (aut)$

C.

$\frac{1}{a} \ln (1 + aut)$

D.

$a \ln (aut)$

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Explanation

$- \frac{dv}{dt} = {av}^{2} \int_{u}^{v} \frac{dv}{v^{2}} = - \int_{0}^{t} adt {[\frac{- 1}{v}]}_{v}^{u} = - at \frac{1}{v} - \frac{1}{u} = at v = \frac{u}{1 + uat} \int ds = \int (\frac{u}{1 + uat}) dt s = \frac{1}{a} \log_{e} (1 + uat)$

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A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation of 2 $m / s^{2}$ . The ratio of time of ascent to the time of descent is: $[g = 10 m / s^{2}]$

A.

1:1

B.

$\sqrt{\frac{2}{3}}$

C.

$\frac{2}{3}$

D.

$\sqrt{\frac{3}{2}}$

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A bullet loses $\frac{1}{20}$ of its velocity passing through a plank. The least number of planks required to stop the bullet is (All planks offers same retardation)

A.

10

B.

11

C.

12

D.

23

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Explanation

For a bullet to be stopped, it must lose all its velocity. If it loses 1/20th of its velocity after passing through one plank, then to lose its entire velocity, it needs to pass through 20 planks. Therefore, the least number of planks required to stop the bullet is 11.

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A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4 t^{3} - 2 t)$ , where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

A.

28 m/s²

B.

22 m/s²

C.

12 m/s²

D.

10 m/s²

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Explanation

$v = 4 t^{3} - 2 t$ (given) ∴ $a = \frac{d v}{d t} = 12 t^{2} - 2$

and $x = \int_{0}^{t} v d t = \int_{0}^{t} (4 t^{3} - 2 t) d t = t^{4} - t^{2}$

When particle is at 2m from the origin $t^{4} - t^{2} = 2$

⇒ $t^{4} - t^{2} - 2 = 0$ ⇒ $(t^{2} - 2) (t^{2} + 1) = 0$ ⇒ $t = \sqrt{2} \sec$

Acceleration at $t = \sqrt{2} \sec$ given by,

$a = 12 t^{2} - 2 = 12 \times 2 - 2$ = $22 m / s^{2}$

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The relation between time and distance is $t = α x^{2} + β x$ , where α and β are constants. The retardation is

A.

$2 α v^{3}$

B.

$2 β v^{3}$

C.

$2 α β v^{3}$

D.

$2 β^{2} v^{3}$

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Explanation

$\frac{d t}{d x} = 2 α x + β \Rightarrow v = \frac{1}{2 α x + β} L e t, 2 α x + β = p \Rightarrow \frac{d p}{d x} = 2 α \Rightarrow v = \frac{1}{p} \Rightarrow \frac{d v}{d p} = - \frac{1}{p^{2}} = \frac{- 1}{{(2 α x + β)}^{2}} N o w, \frac{d v}{d x} = \frac{d v}{d p} \times \frac{d p}{d x} = \frac{- 2 α}{{(2 α x + β)}^{2}}$

$∵ a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t}$

$a = v \frac{d v}{d x} = \frac{- v . 2 α}{{(2 α x + β)}^{2}} = - v . 2 α \times \frac{1}{{(2 α x + β)}^{2}} = - 2 α . v . v^{2} = - 2 α v^{3}$

∴ Retardation $= 2 α v^{3}$

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A point moves with uniform acceleration and v₁, v₂ and v₃ denote the average velocities in the three successive intervals of time t₁, t₂ and t₃. Which of the following relations is correct ?

A.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} + t_{3})$

B.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} + t_{2}) : (t_{2} + t_{3})$

C.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{1} - t_{3})$

D.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} - t_{3})$

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Explanation

Let $u_{1}, u_{2}, u_{3}$ and u₄ be velocities at time $t = 0, t_{1}, (t_{1} + t_{2})$ and $(t_{1} + t_{2} + t_{3})$ respectively and acceleration is a then $v_{1} = \frac{u_{1} + u_{2}}{2}, v_{2} = \frac{u_{2} + u_{3}}{2} and v_{3} = \frac{u_{3} + u_{4}}{2}$

Also $u_{2} = u_{1} + a t_{1}, u_{3} = u_{1} + a (t_{1} + t_{2})$

and $u_{4} = u_{1} + a (t_{1} + t_{2} + t_{3})$

By solving, we get $\frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{(t_{1} + t_{2})}{(t_{2} + t_{3})}$

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The acceleration of a moving body can be found from

A.

Area under velocity-time graph

B.

Area under distance-time graph

C.

Slope of the velocity-time graph

D.

Slope of distance-time graph

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Explanation

Acceleration $a = \tan θ$ , where θ is the angle of tangent drawn on the graph with the time axis.

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The initial velocity of a particle is u (at t = 0) and the acceleration f is given by at. Which of the following relation is valid

A.

$v = u + a t^{2}$

B.

$v = u + a \frac{t^{2}}{2}$

C.

$v = u + a t$

D.

v = u

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Explanation

If acceleration is variable (depends on time) then

$f = \frac{d v}{d t} d v = f d t \int_{u}^{v} d v = \int_{0}^{t} (a t) d t$

$v = u + \frac{a t^{2}}{2}$

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