Physics M.C.Q (1 Marks)

3. At a distance $\frac{\text{l}}2{}$ from O.

Explanation:

It is given that there is no force along x-axis.

COM of rod will remain and will not shift along x-axis (horizontal direction).

Force gravity is acting along y-axis (vertical direction). So, COM will shift along the y-axis by $\frac{\text{l}}2{}$ distance and COM of horizontal rod is at a distance $\frac{\text{l}}2{}$ from one end.

Therefore, lower end of the rod will remain at a distance $\frac{\text{l}}2{}$ from O.

3. 0

Explanation:

The unit vector along the axis of rotation and the unit vector along the resultant force on the particle are perpendicular to each other in a uniform rotation. Therefore, we have:

$\overrightarrow{\text{A}}.\overrightarrow{\text{B}}=0$

4. $\frac{1}{2}\text{m}\omega^2\text{l}$

Explanation:

Consider a small portion of rod at a distance x from the clamped end (as shown in fig.) with width dxand mass dm.

Centripetal force on this portion $=\omega^2\text{xdm}$

$\text{dm}=\Big(\frac{\text{m}}{1}\Big)\text{dx}$

Force on the whole rod $=\text{F}=\int_\limits0^\text{l}\omega^2\text{x}\frac{\text{m}}{\text{l}}\text{dx}$

$\therefore\text{F}=\frac{1}{2}\text{m}\omega^2\text{l}$

3. I_A < I_B

Explanation:

Moment of inertia of circular disc of radius r:

$\text{l}=\frac{1}{2}\text{mr}^2$

Mass = Volume × Density

Volume of disc $=\pi\text{r}^2\text{t}$

Here, tis the thickness of the disc.

As density is same for both the rods, we have:

Moment of inertia, I $\propto$ thickness × (radius)⁴

$\frac{\text{I}_{\text{A}}}{\text{I}_{\text{B}}}=\frac{\text{t}.(\text{r})^4}{\frac{\text{t}}{4}(\text{4r})^4}<1$

$\Rightarrow\frac{\text{I}_{\text{A}}}{\text{I}_{\text{B}}}<1$

$\Rightarrow\text{I}_{\text{A}}<\text{I}_{\text{B}}$

3. It will translate and rotate about the centre.

Explanation:

The given coefficient of friction $\Big(\frac{1}{7}\text{g}\tan\theta\Big)$ is less than the coefficient friction $\Big(\frac{2}{7}\text{g}\tan\theta\Big)$ required for perfect rolling of the sphere on the inclined plane. Therefore, sphere may slip while rolling and it will translate and rotate about the centre.

2. $\frac{\omega\text{M}}{\text{M}+\text{2m}}$

Explanation:

No external torque is applied on the ring; therefore, the angular momentum will be conserved.

$\text{I}\omega=\text{I}'\omega'$

$\Rightarrow\omega'=\frac{\text{I}\omega}{\Gamma}\ \dots(\text{i})$

$\text{I}=\text{Mr}^2$

$\text{I}'=\text{Mr}^2+\text{2mr}^2$

On putting these values in equation (i), we get:

$\omega'=\frac{\omega\text{M}}{\text{M}+\text{2m}}$

3. $\sqrt2\text{v}_0$

Explanation:

For pure rolling, $\omega\text{r}=\text{v}_0$

As shown in the figure, the velocity of the particle will be the resultant of v₀ and $\omega\text{r}.$ Therefore, we have:

$\text{v}_{\text{net}}=\sqrt{\text{v}_0^2+(\omega\text{r})^2}$

$\text{v}_{\text{net}}=\sqrt{\text{v}_0^2+\text{v}_0^2}$

$\text{v}_{\text{net}}=\sqrt2\text{v}_0$

4. $\omega$ is independent of r.

Explanation:

In a pure rotation, angular velocity of all the particles remains same and does not depend on the position of the particle from the axis of rotation.

3. 0

Explanation:

For a purely rotating body, the axis of rotation is always perpendicular to the velocity of the particle. Therefore, we have:

$\overrightarrow{\text{A}}.\overrightarrow{\text{B}}=0$

2. Is zero about a point on the straight line.
3. Is not zero about a point away from the straight line.
4. About any given point remains constant.

Explanation:

2. Angular momentum $=\text{m}\Big(\overrightarrow{\text{r}}\times\overrightarrow{\text{v}}\Big)$

If the point is on the straight line, $\overrightarrow{\text{r}}$ and $\overrightarrow{\text{v}}$ will have the same direction and their cross product will be zero. Hence, angular momentum is zero.

3. If the point is not on the straight line, $\overrightarrow{\text{r}}$ and $\overrightarrow{\text{v}}$ will not have the same direction and their cross product will not be zero. Hence, angular momentum is non-zero.
4. No external torque is applied on the body; therefore, its angular momentum about any given point remains constant.

2. A is true but B is false.

Explanation:

In non-inertial frames, $\frac{\text{dL}}{\text{dt}}=\Gamma_{\text{Total}}$

Here, $\Gamma_{\text{Total}}$ is the total torque on the system due to all the external forces acting on the system. So, equation (B) is not true as in non-inertial frames, pseudo force must be applied to study the motion of the object.

2. The solid sphere reaches the bottom with greater speed.

Explanation:

Acceleration of a sphere on the incline plane is given by:

$\text{a}=\frac{\text{g}\sin\theta}{1+\frac{\text{I}_{\text{COM}}}{\text{mr}^2}}$

I_COM for a solid sphere $=\frac{2}{5}\text{mr}^2$

So, $\text{a}=\frac{\text{g}\sin\theta}{1+\frac{2\text{mr}^2}{5\text{mr}^2}}=\frac{5}{7}\text{g}\sin\theta$

I_COM for a hollow sphere $=\frac{2}{3}\text{mr}^2$

So, $\text{a}'=\frac{\text{g}\sin\theta}{1+\frac{2\text{mr}^2}{3\text{mr}^2}}=\frac{3}{5}\text{g}\sin\theta$

The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

2. 2l

Explanation:

For pure rolling, $\omega\text{r}=\text{v}_0$

As shown in the figure, the velocity of the string will be resultant of v₀ and $\omega\text{r}.$

$\text{v}_{\text{net}}=\text{v}_0+(\omega\text{r})$

$\text{v}_{\text{net}}=2\text{v}_0$

Let:

Linear distance travelled by the cylinder in time $(\text{t}),\ \text{v}_0\text{t}=\text{l}$

$\therefore$ Linear distance travelled by the string in same time, $2\text{v}_0\text{t}=\text{2l}$

1. Increases.

Explanation:

Moment of inertia of a mass is directly proportional to the square of the distance of mass from the axis of rotation.

Therefore, we have:

$\text{I}\propto\text{r}^2$

As the tube is rotated, water is collected at the end of tube because of centrifugal force and distance from the rotation axis increases. Hence, moment of inertia increases.

4. y = 2x

Explanation:

It is given that angular velocity is same for both the wheels

Therefore, we have:

$\text{V}_{\text{A}}=\omega\text{R}$

$\text{V}_{\text{B}}=\omega2\text{R}$

$\text{x}=\text{V}_{\text{A}}\text{t}=\omega\text{Rt}\ \dots(\text{i})$

$\text{y}=\text{V}_{\text{B}}=\omega(\text{2R})\text{t}\ \dots(\text{ii})$

From equations (i) and (ii), we get

$\text{y}=\text{2x}$

1. I₁ < I₂

Explanation:

In the given case, we have:

MOI $\propto$ Density

The density of Iron Is more; therefore, I₂ will be greater.

1. Angular momentum.
2. Linear momentum.

Explanation:

$\overrightarrow{\text{F}}_{\text{ext}}=0$

$\Rightarrow\overrightarrow{\tau}_{\text{ext}}=0$

That is, the change in linear momentum and angular momentum is zero. This is because:

$\frac{\text{d}\overrightarrow{\text{P}}}{\text{dt}}=\overrightarrow{\text{F}}_{\text{ext}}$

And $\frac{\text{d}\overrightarrow{\text{L}}}{\text{dt}}=\overrightarrow{\tau}_{\text{ext}}$

3. If the axes are parallel, I_A < I_B.

Explanation:

If axes A and B are parallel, we get:

I_B = I_A + mr²

Here, r is the distance between two axes and mis the mass of the body.

$\therefore$ I_A < I_B

2. The particles on the diameter mentioned above do not have any linear acceleration.

Explanation:

Linear acceleration of a rotating particle is given as:

$\overrightarrow{\text{a}}=\overrightarrow{\text{r}}\times\overrightarrow{\alpha}$

(b) The sphere is rotating about a diameter; therefore, the position vector of the particles on the diameter is zero. Thus, linear acceleration of the particle is zero. 

(c) and (d): All the particles of the body have the same angular velocity. All the particle on the surface have different linear speeds that depend on the position of the particle from the axis of rotation.

1. It will continue pure rolling.

Explanation:

From the free body diagram of sphere, we have:

Net force on the sphere along the incline,

$\text{F}_{\text{net}}=\text{mg}\sin\theta-\text{ma}\cos\theta\ \dots(\text{i})$

On putting $\text{a}=\text{g}\tan\theta$ in equation (i), we get:

$\text{F}_{\text{net}}=0$

Therefore, if the sphere is set in pure rolling on the incline, it will continue pure rolling.

4. All will take same time.

Explanation:

Let $\theta$ be the inclination angle.

From the free body diagram, we have:

$\text{N}=\text{mg}\cos\theta\ \dots(\text{i})$

$\text{ma}=\text{mg}\sin\theta-\text{f}_{\text{r}}\ \dots(\text{ii})$

Putting $\text{f}_{\text{r}}=\mu\text{N}$ in (ii) we get,

$\text{a}=\text{g}(\sin\theta-\mu\cos\theta)$

The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling; therefore, all the bodies come down with the same acceleration.

3. v_A > v_B

Explanation:

$\tau=\text{I}\alpha$ (Magnitude)

For equal torque, we have:

$\text{I}_{\text{A}\propto\text{A}}=\text{I}_{\text{B}\propto\text{B}}$

$\text{I}_{\text{A}}<\text{I}_{\text{B}}$

$\Rightarrow\alpha_{\text{A}}>\alpha_{\text{B}}\ \dots(\text{i})$

Now, $\omega=\alpha\text{t}$

Or, $\frac{\text{v}}{\text{r}}=\alpha\text{t}$

$\text{v}_{\text{A}}>\text{v}_{\text{B}}$ [Using (i)]

2. The hollow sphere.

Explanation:

Torque is same for all the bodies; therefore, the angular momentum will be conserved. Now, total kinetic energy $=\frac{1}{2}\text{mv}^2+\frac{\text{L}^2}{\text{2I}}$

So, the body with greater value of moment of inertia will have smallest kinetic energy at the bottom of the incline.

Order of the moment of inertia of the bodies: hollow sphere > disc > solid sphere

Hence, the hollow sphere will have the smallest kinetic energy at the bottom.

1. $\vec{\text{r}}.\vec{\text{F}}=0$ and $\overrightarrow{\text{F}}.\overrightarrow{\tau}=0$

Explanation:

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

Thus, $\vec{\tau}$ is perpendicular to $\vec{\text{r}}$ and $\vec{\text{F}}$

Therefore, we have:

$\vec{\text{r}}.\vec{\tau}=0$ and $\vec{\text{F}}.\vec{\tau}=0$

3. The two spheres reach the bottom together.

Explanation:

Acceleration of a sphere on the incline plane is given by:

$\text{a}=\frac{\text{g}\sin\theta}{1+\frac{\text{I}_{\text{COM}}}{\text{mr}^2}}$

I_COM for a solid sphere $=\frac{2}{5}\text{mr}^2$

So, $\text{a}=\frac{\text{g}\sin\theta}{1+\frac{2\text{mr}^2}{5\text{mr}^2}}=\frac{5}{7}\text{g}\sin\theta$

a is independent of mass and radii; therefore, the two spheres reach the bottom together.

3. Remains constant.

Explanation:

For angular momentum. we have:

$\overrightarrow{\text{L}}=\text{m}\Big(\overrightarrow{\text{r}}\times\overrightarrow{\text{v}}\Big)$

$\overrightarrow{\text{v}}=\text{v}\hat{\text{i}}$ and $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$

So, $\overrightarrow{\text{L}}=-\text{mvy}\hat{\text{k}}$

m, v and y are constant; therefore, angular momentum remains constant.

1. $\text{Mr}^2$

Explanation:

Consider an element of length, $\text{dl}=\text{rd}\theta$

$\text{dm}=\frac{\text{M}}{\pi\text{r}}\text{dl}=\frac{\text{M}}{\pi\text{r}}\text{r}\text{d}\theta$

MOI of semicircular wire $=\int_0^{\pi}\text{r}^2\text{dm}$

$\text{I}=\int_{\pi}^0\text{r}^2\frac{\text{m}}{\pi\text{r}}\text{r}\text{d}\theta$

$\Rightarrow\text{I}=\text{mr}^2$

4. $\frac{1}{2}\text{Mga}\sin\theta$

Explanation:

Let N be the normal reaction on the block.

From the free body diagram of the block. it is clear that the forces N and $\text{mg}\cos\theta$ pass through the same line. Therefore, there will be no torque due to N and $\text{mg}\cos\theta.$ The only torque will be produced by $\text{mg}\sin\theta.$

$\therefore​​\vec{\tau}=\vec{\text{F}}\times\vec{\text{r}}$

Since a is the edge of the cube, $\text{r}=\frac{\text{a}}{2}.$

Thus, we have:

$\tau=\text{mg}\sin\theta\times\frac{\text{a}}2{}$

$=\frac{1}{2}\text{mga}\sin\theta$

1. The speed of the particle A is zero.

3. The speed of C is 2v₀.

4. The speed of B is greater than the speed of O.

Explanation:

For pure rolling, $\omega\text{r}=\text{v}_0$

Velocity of the particle at A, B, C and D will be resultant of v₀ and $\omega\text{r}.$

At point B,

$\text{v}_{\text{net}}=\sqrt{\text{v}_0^2+(\omega\text{r})^2}$

$\text{v}_{\text{net}}=\sqrt{\text{v}_0^2+\text{v}_0^2}$

$\text{v}_{\text{net}}=\sqrt{2}\text{v}_0$

At Point C,

$\text{v}_{\text{net}}=\text{v}_0+(\omega\text{r})$

$\text{v}_{\text{net}}=2\text{v}_0$

At Point A,

$\text{v}_{\text{net}}=\text{v}_0-(\omega\text{r})$

$\text{v}_{\text{net}}=0$

At point O,

$\text{r}=0$

$\Rightarrow\text{v}_{\text{net}}=\text{v}_0$