Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

From a solid sphere of mass $$M$$ and radius $$R$$ a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:

A

$${{4M{R^2}} \over {9\sqrt {3\pi } }}$$

B

$${{4M{R^2}} \over {3\sqrt {3\pi } }}$$

C

$${{M{R^2}} \over {32\sqrt {2\pi } }}$$

D

$${{M{R^2}} \over {16\sqrt {2\pi } }}$$

From the diagram you can see diagonal of the cube is equal to the diameter of the sphere.

$$\therefore$$ $$\sqrt 3 a = 2R$$

$$ \Rightarrow $$ $$a = {2 \over {\sqrt 3 }}R$$

Let M is the mass of sphere and M' is the mass of the cube.

So $${M \over {M'}} = {{{4 \over 3}\pi {R^3}} \over {{a^3}}}$$

$$ = {{{4 \over 3}\pi {R^3}} \over {{{\left( {{2 \over {\sqrt 3 }}R} \right)}^3}}} = {{\sqrt 3 } \over 2}\pi $$

$$\therefore$$ $$M' = {{2M} \over {\sqrt 3 \pi }}$$

Moment of inertia of the cube about the axis passing through its center and perpendicular to one of its face,

$$I = {{M'{a^2}} \over 6} = {{{{2M} \over {\sqrt 3 \pi }} \times \left( {{2 \over {\sqrt 3 }}R} \right)} \over 6}$$

$$ = {{4M{R^2}} \over {9\sqrt 3 \pi }}$$

$$\therefore$$ $$\sqrt 3 a = 2R$$

$$ \Rightarrow $$ $$a = {2 \over {\sqrt 3 }}R$$

Let M is the mass of sphere and M' is the mass of the cube.

So $${M \over {M'}} = {{{4 \over 3}\pi {R^3}} \over {{a^3}}}$$

$$ = {{{4 \over 3}\pi {R^3}} \over {{{\left( {{2 \over {\sqrt 3 }}R} \right)}^3}}} = {{\sqrt 3 } \over 2}\pi $$

$$\therefore$$ $$M' = {{2M} \over {\sqrt 3 \pi }}$$

Moment of inertia of the cube about the axis passing through its center and perpendicular to one of its face,

$$I = {{M'{a^2}} \over 6} = {{{{2M} \over {\sqrt 3 \pi }} \times \left( {{2 \over {\sqrt 3 }}R} \right)} \over 6}$$

$$ = {{4M{R^2}} \over {9\sqrt 3 \pi }}$$

2

MCQ (Single Correct Answer)

Distance of the center of mass of a solid uniform cone from its vertex is $$z{}_0$$. If the radius of its base is $$R$$ and its height is $$h$$ then $$z{}_0$$ is equal to :

A

$${{5h} \over 8}$$

B

$${{3{h^2}} \over {8R}}$$

C

$${{{h^2}} \over {4R}}$$

D

$${{3h} \over 4}$$

Let the density of solid cone $$\rho $$.

$$dm = \rho \pi {r^2}dy$$

$${y_{cm}} = {{\int {ydm} } \over {\int {dm} }}$$

$$ = {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}$$

$$ = {{3h} \over 4}$$

$$dm = \rho \pi {r^2}dy$$

$${y_{cm}} = {{\int {ydm} } \over {\int {dm} }}$$

$$ = {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}$$

$$ = {{3h} \over 4}$$

3

MCQ (Single Correct Answer)

A mass $$'m'$$ is supported by a massless string wound around a uniform hollow cylinder of mass $$m$$ and radius $$R.$$ If the string does not slip on the cylinder, with what acceleration will the mass fall or release?

A

$${{2g} \over 3}$$

B

$${{g} \over 2}$$

C

$${{5g} \over 6}$$

D

$$g$$

Here string is not slipping over pulley.

From figure,

Acceleration $$a = R\alpha \,\,\,\,...(i)$$

Applying Newton's second law on the block we get,

$$mg-T=ma$$ $$...(ii)$$

Torque on cylinder due to tension in string about the fixed point,

$$T \times R = I\alpha$$

$$T \times R = m{R^2}\alpha = m{R^2}\left( {{a \over R}} \right)$$

[ as $$I$$ = MR

or $$T=ma$$

$$ \Rightarrow \,\,mg - ma = ma$$

$$ \Rightarrow \,\,\,a = {g \over 2}$$

4

MCQ (Single Correct Answer)

A bob of mass $$m$$ attached to an inextensible string of length $$l$$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $$\omega \,rad/s$$ about the vertical. About the point of suspension:

A

angular momentum is conserved

B

angular momentum changes in magnitude but not in direction.

C

angular momentum changes in direction but not in magnitude.

D

angular momentum changes both in direction and magnitude.

Torque working on the bob of mass $$m$$ is, $$\tau = mg \times \ell \,\sin \,\theta .$$ (Direction of torque by the weight is parallel to plane of rotation of the particle)

As $$\tau $$ is perpendicular to the angular momentum $$\overrightarrow L$$ of the bob, so the direction of $$L$$ changes but magnitude remains same.

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