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NEET Practice Questions (MCQs) with Answers & Solutions

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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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A vector that is perpendicular to both the vectors $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is

$- \hat{i} + \hat{k}$

$- \hat{i} - 2 \hat{j} + \hat{k}$

$\hat{i} - 2 \hat{j} + \hat{k}$

$\hat{i} + \hat{k}$

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Explanation

(A)

$\hat{i} + \hat{k} = - \hat{i} + 0 \hat{j} + \hat{k} \vec{a} . (- \hat{i} + \hat{k}) = - 1 + 0 + 1 = 0 \vec{b} . (- \hat{i} + \hat{k}) = - 1 + 0 + 1 = 0$

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If $θ$ is the angle between vectors $\vec{a}$ and $\vec{b}$ , and $| \vec{a} \times \vec{b} | = \vec{a} . \vec{b}$ , then $θ$ is equal to

$0 °$

$180 °$

$135 °$

$45 °$

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Explanation

$G i v e n, | \vec{a} \times \vec{b} | = \vec{a} . \vec{b} \Rightarrow a b \sin θ = a b \cos θ \Rightarrow \tan θ = 1 \Rightarrow θ = 45 °$

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The vector $\vec{b}$ , which is collinear with the vector $\vec{a}$ =(2, 1, -1) and satisfies the condition $\vec{a}$ . $\vec{b}$ =3, is

(1, 1/2, -1/2)

(2/3, 1/3, -1/3)

(1/2, 1/4, -1/4)

(1, 1, 0)

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Explanation

$Going through the options; Only (A) satifies the condition \vec{a} . \vec{b} = 3 .$

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If $3 \hat{i} + 2 \hat{j} + 8 \hat{k}$ and $2 \hat{i} + x \hat{j} + \hat{k}$ are at right angle then x=

-7

-4

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Explanation

If two vectors are at right angle, then their dot product = 0

$\Rightarrow (3 \hat{i} + 2 \hat{j} + 8 \hat{k}) . (2 \hat{i} + x \hat{j} + \hat{k}) = 0 \Rightarrow 6 + 2 x + 8 = 0 \Rightarrow x = - 7$

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If a, b, c are three non-zero vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ the value of $\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ is

Less than zero

equal to zero

greater than zero

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