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NEET Practice Questions (MCQs) with Answers & Solutions

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A vector A points, vertically upward and, B points towards north. The vector product $A \times B$ is -

along west

along east

zero

vertically downward

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The linear velocity of a rotating body is given by $v = ω \times r$ , where $ω$ is the angular velocity and r is the radius vector. The angular velocity of a body $ω = \hat{i} - 2 \hat{j} + 2 \hat{k}$ and their radius vector $r = 4 \hat{j} - 3 \hat{k}, | v |$ is -

$\sqrt{29} units$

$31 units$

$\sqrt{37} units$

$\sqrt{41} units$

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Explanation

(A)

$\vec{V} = \vec{ω} \times \vec{r} = (\hat{i} - 2 \hat{j} + 2 \hat{k}) \times (4 \hat{j} - 3 \hat{k}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 0 & 4 & - 3 \end{matrix}| = (- 2 \times - 3 - 2 \times 4) \hat{i} + (1 \times - 3 - 2 \times 0) (- \hat{j}) + (1 \times 4 - (- 2) \times 0) \hat{k} \vec{V} = - 2 \hat{i} + 3 \hat{j} + 4 \hat{k} Magnitude of \vec{v}, |\vec{V}| = \sqrt{{(- 2)}^{2} + {(3)}^{2} + {(4)}^{2}} = \sqrt{29} units$

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The displacement x of a particle along a straight line at time t is given by $x = a_{0} + a_{1} t + a_{2} t^{2}$ . The acceleration of the particle is

$a_{0}$

$a_{1}$

$2 a_{2}$

$a_{2}$

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Explanation

(C)

$v = \frac{d x}{d t} = = a_{1} t^{1 - 1} + 2 a_{2} t^{2 - 1} = a_{1} + 2 a_{2} t a = \frac{d v}{d t} = 2 a_{2} t^{1 - 1} = 2 a_{2}$

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The displacement of a particle is given by $y = a + bt + {ct}^{2} - {dt}^{4}$ . The initial velocity and acceleration are respectively

b, -4d

-d, 2c

b, 2c

2c, -4d

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Explanation

$G i v e n, y = a + b t + c t^{2} - d t^{4} \Rightarrow v = \frac{d v}{d t} = b + 2 c t - 4 d t^{3} \Rightarrow a = \frac{d v}{d t} = 2 c - 12 d t^{2} a t t = 0, v = b a n d a = 2 c$

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The momentum is given by p=4t+1, the force at t=2s is

4 N

8 N

10 N

15 N

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Explanation

(A)

$F = \frac{d p}{d t} = 4 N S o f o r c e i s c o n s \tan t a l w a y s$

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A particle moves along a straight line such that its displacement at any time t is given by $s = t^{3} - 3 t^{2} + 6 t - 12$ metres. The velocity when the acceleration is zero is:

$3 {ms}^{- 1}$

$- 12 {ms}^{- 1}$

$42 {ms}^{- 1}$

$- 9 {ms}^{- 1}$

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Explanation

Given,

$x = t^{3} - 3 t^{2} + 6 t - 12 v = \frac{d x}{d t} = 3 t^{2} - 6 t + 6 a = \frac{d v}{d t} = 6 t - 6 w h e n, a = 0 \Rightarrow t = 1 s e c \Rightarrow v = 3 m / s$

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The position x of particle varies with time t as $x = {at}^{2} - {bt}^{3}$ . The acceleration of the particle will be zero at time equal to

$\frac{a}{b}$

$\frac{2 a}{b}$

$\frac{a}{3 b}$

Zero

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Explanation

Diffrentiating x wrt t ,

$v = \frac{d x}{d t}$

$v = 2 a t - 3 b t^{2}$

$a = 2 a - 6 b t$

At $t = \frac{a}{3 b} a = 0$

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A body is moving according to the equation $x = at + {bt}^{2} - {ct}^{3}$ where x = displacement and a, b and c are constants. The acceleration of the body is

$a + 2 bt$

$2 b + 6 ct$

$2 b - 6 ct$

$3 b - 6 {ct}^{2}$

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Explanation

Diffrentiating x wrt t,

$v = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}$

$a = 2 b - 6 c t$

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A particle moves along X-axis in such a way that its X coordinate varies with time t according to the equation $x = (2 - 5 t + 6 t^{2}) m$ . The initial velocity of the particle is

-5 m/s

6 m/s

3 m/s

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Explanation

$v = \frac{d x}{d t} = - 5 + 12 t A t t = 0 v = - 5 m / s$

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If the momentum of a particle is given by P=(180-8t) kg m/s, then its force will be

Zero

8 N

-8 N

4 N

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Explanation

$F = \frac{d P}{d t}$

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