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Physics MCQs for NEET — Practice Questions with Answers

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The vector having a magnitude of 10 and perpendicular to the vector $3 \hat{i} - 4 \hat{j}$ is

$4 \hat{i} + 3 \hat{j}$

$5 \sqrt{2} \hat{i} - 5 \sqrt{2} \hat{j}$

$8 \hat{i} + 6 \hat{j}$

$8 \hat{i} - 6 \hat{j}$

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Explanation

$(3 i + 4 j) . (8 i + 6 j) = 0$

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A force $\vec{F} = 3 \hat{i} + c \hat{j} + 2 \hat{k}$ acting on a particle causes a displacement $\vec{d} = - 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . If the work done is 6J then the value of 'c' is

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Explanation

$W = \vec{F} . \vec{d} \Rightarrow 6 = \vec{F} . \vec{d} \Rightarrow 6 = - 12 + 2 C + 6 \Rightarrow C = 6$

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A particle moving along a straight line according to the law $x = At + {Bt}^{2} + {Ct}^{3}$ , where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.

Its velocity at t=0 is A

Its acceleration at t=0 is B

Its velocity at t=0 is B

Its acceleration at t=0 is C

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Explanation

$(A) v = \frac{d x}{d t} = A + 2 B t + 3 C t^{2} a = \frac{d v}{d t} = 2 B + 6 C t A t t = 0 v = A a = 2 B$

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If the velocity of a particle moving on x-axis is given by $v = 3 t^{2} - 12 t + 6$ . At which time is the acceleration of particle zero?

$2$ sec

$2 + \sqrt{2}$ sec

$2 - \sqrt{2}$ sec

zero

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Explanation

$(A) a = \frac{d v}{d t} = 6 t - 12 F o r a = 0, 6 t - 12 = 0 t = 2 s e c$

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A particle moves along straight line such that at time t its position from a fixed point O on the line is $x = 3 t^{2} - 2$ . The velocity of the particle when t=2 is:

$8 {ms}^{- 1}$

$4 {ms}^{- 1}$

$12 {ms}^{- 1}$

$0$

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Explanation

(C)

$v = \frac{d x}{d t} = 6 t A t t = 2 s e c v = 12 m / s$

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Coordinates of a moving particle are given by $x = {ct}^{2}$ and $y = {bt}^{2}$ . The speed of the particle is given by

$2 t (c + b)$

$2 t \sqrt{c^{2} - b^{2}}$

$t \sqrt{c^{2} + b^{2}}$

$2 t \sqrt{c^{2} + b^{2}}$

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Explanation

$\vec{v} = \overset{}{v_{x} \hat{i} + v_{y} \hat{j}} S p e e d |\vec{v}| = \sqrt{{v^{2}}_{x} + {v_{y}}^{2}} \vec{v} = \frac{d x}{d t} \hat{i} + \frac{d y}{d t} \hat{j} = 2 c t \hat{i} + 2 b t \hat{j} ∴ |\vec{v}| = \sqrt{{(2 c t)}^{2} + {(2 b t)}^{2}} = 2 t \sqrt{c^{2} + b^{2}}$

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The x and y components of vector $\vec{A}$ are 4m and 6m respectively. The x, y components of vector $\vec{A} + \vec{B}$ are 10m and 9m respectively. The length of $\vec{B}$ is ______ and angle that $\vec{B}$ makes with the x axis is given by _______.

$\sqrt{0.45} m, \tan^{- 1} (1 / 2)$

$\sqrt{4.5} m, \tan^{- 1} (1 / 2)$

$\sqrt{45} m, \tan^{- 1} (1 / 2)$

$4 \sqrt{5} m, \tan^{- 1} (1 / 2)$

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A particle travels with speed 50 m/s from the point (3, 7) in a direction $7 \hat{i} - 24 \hat{j}$ . Find its position vector after 3 seconds.

$151 \hat{i} + 45 \hat{j}$

$45 \hat{i} - 137 \hat{j}$

$151 \hat{i} + 45 \hat{j}$

$4.5 \hat{i} - 151 \hat{j}$

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If $\vec{A} = 2 \hat{i} + \hat{j} & \vec{B} = \hat{i} - \hat{j}$ . Find component of $\vec{A}$ along $\vec{B}$ & perpendicular to $\vec{B}$ .

$\frac{\hat{i} - \hat{j}}{2}, \frac{3}{2} (\hat{i} + \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, - \frac{2}{3} (\hat{i} + \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, - \frac{3}{2} (\hat{i} - \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, \frac{2}{3} (\hat{i} - \hat{j})$

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If $\vec{a}$ is a vector and x is a non-zero scalar, then

xa is a vector in the direction of $\vec{a}$

xa is a vector collinear to $\vec{a}$

xa and $\vec{a}$ have independent directions

xa is a vector perpendicular to $\vec{a}$

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Explanation

(B)

$I f x i s p o s i t i v e, x \vec{a} i s p a r a l l e l t o \vec{a} I f x i s n e g a t i v e, x \vec{a} i s a n t i - p a r a l l e l t o \vec{a} S o, x \vec{a} i s a l w a y s c o l l i n e a r t o \vec{a}$

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