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Rotational Motion NEET Questions — Free MCQ Practice

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A uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle $ \theta $ with the horizontal. The acceleration of the centre of mass of the disc is

A. $ g sin \theta $ B. $ { 2 g sin \theta \over 3 } $ C. $ { g sin \theta \over 3 } $ D. $ {2 g cos \theta \over 3 } $

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Explanation

$ Mg sin \theta - f = Ma $ $ \tau = I . \alpha $ $ As I = { 1 \over 2} MR^2 , \alpha = { a \over R} and \tau = fR$ $ Hence fR = { 1 \over 2} MR^2 . { a \over R } = { 1 \over 2 } MRa $

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A uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle $ \theta $ with the horizontal The frictional force on the disc is

A. $ { Mg sin \theta \over 3 } $ B. $ { 2 Mg sin \theta \over 3 } $ C. $ { Mg sin \theta } $ D. None

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Explanation

$ f = { 1 \over 2} Ma $ $ \therefore f = { Mg sin \theta \over 3 } $

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A uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle $ \theta $ with the horizontal. If the disc is replaced by a ring of the same mass M and the same radius R, the ratio of the frictional force on the ring to that on the disc will be

A. 3/2 B. 2 C. $ \sqrt 2 $ D. 1

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Explanation

For a ring $ I = MR^2 $

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If the angle between the vector $\vec{A} a n d \vec{B}$ is $θ$ , the value of the product $(\vec{B} \times \vec{A}) . \vec{A}$ is equal to:

A.

$B A^{2} \cos θ$

B.

$B A^{2} \sin θ$

C.

$B A^{2} \sin θ \cos θ$

D.

zero

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Explanation

Using vector product:

$Let \vec{R} = (\vec{B} \times \vec{A}) \Rightarrow \vec{R} ⊥ \vec{A} \Rightarrow (\vec{B} \times \vec{A}) . \vec{A} \Rightarrow \vec{R} . \vec{A} = 0$

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

A.

along west

B.

along east

C.

zero

D.

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 -

A.

$\sqrt{29} units$

B.

$31 units$

C.

$\sqrt{37} units$

D.

$\sqrt{41} units$

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

A.

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

B.

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

C.

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

D.

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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Three non zero vectors $\vec{A}, \vec{B} & \vec{C}$ satisfy the relation $\vec{A} \cdot \vec{B} = 0 & \vec{A} \cdot \vec{C} = 0$ . Then $\vec{A}$ can be parallel to:

A.

$\vec{B}$

B.

$\vec{C}$

C.

$\vec{B} \cdot \vec{C}$

D.

$\vec{B} \times \vec{C}$

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Explanation

$(4) As \vec{A} . \vec{B} = 0 and \vec{A} . \vec{C} = 0, so \vec{A} is ⊥ ar to both \vec{B} and \vec{C} . Also, \vec{B} \times \vec{C} is also ⊥ ar to both \vec{B} and \vec{C} . So, \vec{A} can be parallel to \vec{B} \times \vec{C} .$

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The displacement of particle is zero at t=0 and at t=t it is x. It starts moving in the x direction with velocity, which varies as $v = k \sqrt{x}$ , where k is constant. The velocity-

A.

varies with time

B.

is independent to time

C.

inversely proportional to time

D.

inversely proportional to acceleration

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Explanation

$v = k \sqrt{x}$ $\Rightarrow \frac{d x}{d t} = k \sqrt{x}$

$\Rightarrow \int \frac{d x}{\sqrt{x}} = \int k d t \Rightarrow \frac{x^{+ 1 / 2}}{1 / 2} = k t + c$

Given that, at t=0, x=0 $.^{.} .$ c=0

Now, $2 x^{1 / 2} = k t \Rightarrow \sqrt{x} = (1 / 2) k t,$

$∵ x = \frac{k^{2} t^{2}}{4}$

Now, $v = k (1 / 2 k t) = k^{2} t / 2$

Thus velocity varies with time. Hence correct answer is (1)

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The acceleration of a particle is given as $a = 3 x^{2}$ . At t=0, v=0, x=0, the velocity at t =2 sec will be-

A.

0.05 m/s

B.

0.5 m/s

C.

5 m/s

D.

50 m/s

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Explanation

$a = 3 x^{2} \Rightarrow v \frac{d v}{d x} = 3 x^{2}$

$\Rightarrow v d v = 3 x^{2} d x \Rightarrow \frac{v^{2}}{2} = 3 \frac{x^{3}}{3} + c$

At t=0, v=0, x=0;

$∴ c = 0 N o w, \frac{v^{2}}{2} = x^{3}$

$v^{2} = 2 x^{3} \Rightarrow v = \sqrt{2} x^{3 / 2} . . . (1)$

$\Rightarrow \frac{d x}{d t} = \sqrt{2} x^{3 / 2} \Rightarrow d x = \sqrt{2} x^{3 / 2} d t \Rightarrow \frac{d x}{x^{3 / 2}} = \sqrt{2} d t$

Integrating both sides, we get $\frac{- 2}{\sqrt{x}} = \sqrt{2} t + c'$

At t=0, x=0, v=0 $.^{.} .$ c'=0

Now $\frac{- 2}{\sqrt{x}} = \sqrt{2} t$ $\Rightarrow 4 = 2 x t^{2} \Rightarrow x = \frac{2}{t^{2}} . . . (2)$

From (1) and (2) $v = \sqrt{2} {(\frac{2}{t^{2}})}^{3 / 2}$

At t=2 s, v=1/2 m/sec.

Hence correct answer is (2).

Rotational Motion MCQs for NEET — Physics Questions with Answers

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