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Physics NEET Questions — Free MCQ Practice

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The coordinate of an object is given as a function of time by $x = 7 t - 3 t^{2}$ , where x is in meters and t is in seconds. Its average velocity over the interval from t=0 to t=4 is:

A.

5 m/s

B.

-5 m/s

C.

11 m/s

D.

-11 m/s

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Explanation

$Average velocity = \frac{∆ x}{∆ t} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}} = \frac{28 - 48}{4} = - 5 m / s$

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A particle moves along a straight line and its position as function of time is given by $x = t^{3} - 3 t^{2} + 3 t + 3$ then particle

A.

stops at t=1s and reverses its direction of motion

B.

stops at t=1s and continues further without change of direction

C.

stops at t=2s and reverses its direction of motion

D.

stops at t=2s and continues further without change of direction

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Explanation

$x = t^{3} - 3 t^{2} + 3 t + 3 \Rightarrow v = \frac{d x}{d t} = 3 t^{2} - 6 t + 3 W h e n i t s t o p s, v = 0 \Rightarrow 3 t^{2} - 6 t + 3 = 0 \Rightarrow t^{2} - 2 t + 1 = 0 \Rightarrow {(t - 1)}^{2} = 0 \Rightarrow t = 1 s e c S o, t h e p a r t i c l e s t o p s o n l y o n c e a t t = 1 s e c$

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A particle moves in a straight line, according to the law $x = 4 a [t + a \sin (\frac{t}{a})]$ , where x is its position in meters, t is in sec & a is some constant, then the velocity is zero at :

A.

$x = 4 a^{2} π meters$

B.

$t = π \sec$

C.

$t = 0 \sec$

D.

none

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Explanation

(A)

$x = 4 a [t + a \sin (\frac{t}{a})] v = \frac{dx}{dt} v = 4 a [1 + \cos (\frac{t}{a})] For velocity to be zero, 1 + \cos (\frac{t}{a}) = 0 \Rightarrow t = πa Putting in x, x = 4 {πa}^{2}$

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A point moves in a straight line so that its displacement is x m at time t sec, given by $x^{2} = t^{2} + 1$ . Its acceleration in $m / s^{2}$ at time 1 sec is:

A.

$\frac{1}{x}$

B.

$\frac{1}{x} - \frac{1}{x^{3}}$

C.

$\frac{1}{x^{3}}$

D.

$- \frac{t^{2}}{x^{3}}$

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Explanation

$x^{2} = t^{2} + 1 2 x (\frac{dx}{dt}) = 2 t v = \frac{t}{x} a = \frac{x - vt}{x^{2}} = \frac{1}{x} - \frac{t^{2}}{x^{3}} At t = 1 \sec a = \frac{1}{x} - \frac{1}{x^{3}}$

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The motion of a body is given by the equation, $\frac{d v}{d t} = 4 - 2 v$ where v is the speed in m/s and t in second. If the body was at rest at t=0, then find speed of body as a function of time.

A.

$2 (1 - e^{- 2 t})$

B.

$1 - e^{- 2 t}$

C.

$2 (1 + e^{- 2 t})$

D.

$1 - e^{- t}$

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Explanation

Given that $\frac{d v}{d t} = 4 - 2 v$

dv=(4-2v)dt or $\frac{d v}{4 - 2 v} = d t$

or $d v = {}_{0}^{v}\int \frac{d v}{4 - 2 v} = {}_{0}^{t}\int d t$ or ${[\frac{\log_{e} (4 - 2 v)}{- 2}]}_{0}^{v} = {[t]}_{0}^{t}$

or $\log_{e} \{4 - 2 v\} - \log_{e} (4) = - 2 t$ or $\log_{e} \{\frac{4}{4} - \frac{2 v}{4}\} = - 2 t$

$\Rightarrow \frac{4 - 2 v}{4} = e^{- 2 t} \Rightarrow \frac{4}{4} - \frac{2 v}{4} = e^{- 2 t} \Rightarrow 1 - \frac{2 v}{4} = e^{- 2 t}$

$\Rightarrow \frac{2 v}{4} = 1 - e^{- 2 t}$ or $= 2 (1 - e^{- 2 t})$

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A particle is projected at an angle $θ$ with horizontal with an initital speed u. When it makes and angle $α$ with horizontal, its speed is

A.

$u \cos θ$

B.

$u \cos θ u \cos α$

C.

$\frac{u \sin θ}{\sin α}$

D.

$\frac{u \cos θ}{\cos α}$

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Explanation

Horizontal component of velocity remains constant during projectile motion

so,

$v \cos α = u \cos θ \Rightarrow v = \frac{u \cos θ}{\cos α}$

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A body is projected with velocity $20 \sqrt{3}$ m/s with an angle of projection 60 $°$ with horizontal. Calculate velocity on that point where body makes an angle 30 $°$ with the horizontal.

A.

20 m/s

B.

$\frac{20}{\sqrt{3}} m / s$

C.

$10 \sqrt{3} m / s$

D.

10 m/s

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Explanation

Horizontal velocity is always constant during projectile motion, when only gravitational force act on the body.

So, $v \cos α = u \cos θ$

$v \times \frac{\sqrt{3}}{2} = 20 \sqrt{3} \times \frac{1}{2} \Rightarrow v = 20 m / s$

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A body thrown vertically so as to reach its maximum height in t second. The toal time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

A.

$\sqrt{2} t$

B.

$(1 + \frac{1}{\sqrt{2}}) t$

C.

$\frac{3 t}{2}$

D.

$\frac{t}{\sqrt{2}}$

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Explanation

$Let time to reach P from A be t H = \frac{1}{2} {gt}^{2} t = \sqrt{\frac{2 H}{g}} Time to reach C from P be t' \frac{H}{2} = \frac{1}{2} g {(t')}^{2} t' = \sqrt{\frac{H}{g}} = \sqrt{\frac{2 H}{2 g}} = \frac{t}{\sqrt{2}} So total time to go from A to C t + t' = t + \frac{t}{\sqrt{2}} = t [1 + \frac{1}{\sqrt{2}}]$

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A particle is projected with a velocity u making an angle $θ$ with the horizontal. At any instant, its velocity V is at right angles to its initial velocity u; then V is:

A.

u cos $θ$

B.

u tan $θ$

C.

u cot $θ$

D.

u sec $θ$

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A projectile is given an initial velocity of $\hat{i} + 2 \hat{j}$ . The cartesian equation of its path is (g = 10 ${ms}^{- 2}$ )

A.

$y = 2 x - 5 x^{2}$

B.

$y = x - 5 x^{2}$

C.

$4 y = 2 x - 5 x^{2}$

D.

$y = 2 x - 25 x^{2}$

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Explanation

$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ} \{\begin{cases} u \cos θ = 1 \\ u \sin θ = 2 \end{cases} y = 2 x - \frac{10 x^{2}}{2 \times 1} = 2 x - 5 x^{2}$

Physics MCQs for NEET — Practice Questions with Answers

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