Rotational Motion — Page 9 | JEE Physics

Q81

Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :

A

{{M{l^2}} \over 6}

B

{M{l^2}}

C

{{M{l^2}} \over {12}}

D

{2 \over 3}M{l^2}

Correct Answer

Option D

Solution

According to perpendicular Axis theorem. Ix + Iy = Iz Iz $\Rightarrow$

{{m{l^2}} \over 3} + {{m{l^2}} \over 3}

= {{2m{l^2}} \over 3}

Q82

The torque of a force

5 \hat{i}+3 \hat{j}-7 \hat{k}

about the origin is

\tau

. If the force acts on a particle whose position vector is

2 i+2 j+k

, then the value of

\tau

will be

A

11 \hat{i}+19 \hat{j}-4 \hat{k}

B

-11 \hat{i}+9 \hat{j}-16 \hat{k}

C

-17 \hat{i}+19 \hat{j}-4 \hat{k}

D

17 \hat{i}+9 \hat{j}+16 \hat{k}

Correct Answer

Option C

Solution

$\vec{\tau}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{array}\right|$

\begin{aligned} &=\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10) \\\\ &=-17 \hat{i}+19 \hat{j}-4 \hat{k} \end{aligned}

Q83

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : An electric fan continues to rotate for some time after the current is switched off. Reason R : Fan continues to rotate due to inertia of motion. In the light of above statements, choose the most appropriate answer from the options given below.

A A is not correct but R is correct

B A is correct but R is not correct

C Both A and R are correct and R is the correct explanation of A

D Both A and R are correct but R is NOT the correct explanation of A

Correct Answer

Option C

Solution

The correct answer is Both A and R are correct and R is the correct explanation of A.

Explanation: Assertion A: An electric fan continues to rotate for some time after the current is switched off.

This is a correct statement.

When you switch off the fan, it doesn't stop immediately but continues to rotate for some time.

Reason R: Fan continues to rotate due to inertia of motion.

This is also a correct statement.

Inertia is the resistance of any physical object to any change in its state of motion.

This includes changes to the object's speed or direction of motion.

An object will stay in its state of motion unless a force acts on it.

In the case of the fan, after the current is switched off, the fan blades have inertia and continue to move due to this inertia until the frictional forces (like air resistance and friction in the fan's bearings) cause it to stop.

Therefore, inertia of motion is the correct reason for the fan's continued motion after the current is switched off.

Q84

A particle of mass

\mathrm{m}

is projected with a velocity '

\mathrm{u}

' making an angle of

30^{\circ}

with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height

\mathrm{h}

is :

A

\dfrac{\mathrm{\mu}^3}{\sqrt{2} \mathrm{~g}}

B zero

C

\dfrac{\sqrt{3}}{2} \dfrac{\mathrm{\mu}^2}{\mathrm{~g}}

D

\dfrac{\sqrt{3}}{16} \dfrac{\mathrm{\mu}^3}{\mathrm{~g}}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{L}=m u \cos \theta H \\ & =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g} \end{aligned}

Q85

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be

t_1

and

t_2

, respectively, then

A

t_1< t_2

B

t_1=2 t_2

C

t_1 >t_2

D

t_1=t_2

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{a}_{\mathrm{cm}}}} \\ & \mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^2}} \\ & \mathrm{a}_1=\mathrm{a}_{\mathrm{cm}_1}=\frac{5 \mathrm{~g} \sin \theta}{7} \ldots . . \text{ Solid } \\ & \mathrm{a}_2=\mathrm{a}_{\mathrm{cm}_2}=\frac{3 \mathrm{~g} \sin \theta}{5} \ldots . \text{ Hollow } \\ & \mathrm{a}_1>\mathrm{a}_2 \\ & \mathrm{t}_1<\mathrm{t}_2 \end{aligned}

Q86

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

A

\dfrac{3}{4}

B

\dfrac{4}{3}

C

\dfrac{5}{2}

D

\dfrac{2}{5}

Correct Answer

Option C

Solution

\begin{aligned} & \frac{\text{ Linear KE }}{\text{ Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^2}{\frac{1}{2} \mathrm{I} \omega^2} \\ & \frac{\mathrm{mv}_{\mathrm{cm}}^2}{\frac{2}{5} \mathrm{mR}^2 \omega^2}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R}) \end{aligned}

Q87

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

A

\dfrac{\lambda L^3}{8 \pi^2}

B

\dfrac{\lambda L^3}{16 \pi^2}

C

\dfrac{\lambda L^3}{4 \pi^2}

D

\dfrac{\lambda L^3}{12}

Correct Answer

Option A

Solution

First, relate the length of the rod to the circumference of the ring: $L = 2\pi R$ The mass of the ring, denoted by $M$ , is the product of the linear mass density and the length of the rod: $M = \lambda \times L$ The moment of inertia of a ring about a diameter is given by the formula: $I = \dfrac{MR^2}{2}$ Substituting the expression for mass $M$ and using the relation for $R$ derived from the circumference: $I = \dfrac{\lambda \times L}{2} \times \left(\dfrac{L}{2\pi}\right)^2$ Simplifying the expression, we arrive at: $I = \dfrac{\lambda L^3}{8\pi^2}$ This is the moment of inertia of the ring about any of its diameters.

Q88

Which of the following are correct expression for torque acting on a body? A.

\vec{\tau}=\vec{r} \times \vec{L}

B.

\vec{\tau}=\dfrac{d}{d t}(\vec{r} \times \vec{p})

C.

\vec{\tau}=\vec{r} \times \dfrac{d \vec{p}}{d t}

D.

\vec{\tau}=I \vec{\alpha}

E.

\vec{\tau}=\vec{r} \times \vec{F}

(

\vec{r}=

position vector;

\vec{p}=

linear momentum;

\vec{L}=

angular momentum;

\vec{\alpha}=

angular acceleration;

I=

moment of inertia;

\vec{F}=

force;

t=

time) Choose the correct answer from the options given below:

A A, B, D and E Only

B C and D Only

C B, C, D and E Only

D B, D and E Only

Correct Answer

Option C

Solution

Let's examine each expression step by step:

\vec{\tau} = \vec{r} \times \vec{L}

Here,

\vec{L}

is the angular momentum. However, torque is defined as the time derivative of angular momentum:

\vec{\tau} = \frac{d\vec{L}}{dt},

not as the cross product of the position vector with the angular momentum. In fact, if you write

\vec{r} \times \vec{L} = \vec{r} \times (\vec{r} \times \vec{p}),

you don't obtain the standard expression for torque. Thus, this expression is not correct.

\vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p})

For a particle, the angular momentum is defined as

\vec{L} = \vec{r} \times \vec{p}.

Taking the time derivative gives

\frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}.

Since

\frac{d\vec{r}}{dt} = \vec{v}

and

\vec{p} = m\vec{v},

the term

\vec{v} \times m\vec{v}

is zero. This simplifies to

\vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt},

which is a standard expression for torque. So, this expression is correct.

\vec{\tau} = \vec{r} \times \frac{d \vec{p}}{d t}

This is the standard definition of torque, as

\frac{d \vec{p}}{d t}

is the net force

\vec{F}.

Hence, we can also write

\vec{\tau} = \vec{r} \times \vec{F}.

This expression is correct.

\vec{\tau} = I \vec{\alpha}

This relation applies to rigid bodies rotating about a fixed axis (where the moment of inertia

I

is constant and can be treated as a scalar).

It is a common form used in rotational dynamics, although one must be cautious since it is a special case.

In the context of this problem, it is acceptable as a correct expression.

\vec{\tau} = \vec{r} \times \vec{F}

This is the fundamental definition of torque in physics.

It directly relates the force applied to a particle and its lever arm.

This expression is clearly correct.

To summarize: Expression A is not a standard or generally valid expression for torque.

Expressions B, C, D, and E are acceptable under the usual assumptions in mechanics.

Looking at the provided options, the correct answer is: Option C: B, C, D and E Only.