Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :

A

16 cm

B

12 cm

C

14 cm

D

18 cm

Consider an element of radius x and thickness dx

Mass of element, dm = $$\sigma 2\pi x\left( {dx} \right)$$

Here, $$\sigma $$ = mass per unit area = $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$$

Moment of inertia of element, dI = (dm)x^{2}

$$ \Rightarrow $$ I = $$\sigma 2\pi \int\limits_r^R {{x^3}dx} $$

= $$\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$$

= $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$$

= $${m \over 2}\left( {{R^2} + {r^2}} \right)$$ .....(i)

Moment of inertia of thin cylinder of same mass,

I = m$$r_0^2$$ ......(ii)

$$ \Rightarrow $$ m$$r_0^2$$ = $${m \over 2}\left( {{R^2} + {r^2}} \right)$$

$$ \Rightarrow $$ $$r_0^2$$ = 250

$$ \Rightarrow $$ r_{0} $$ \simeq $$ 16 cm

Mass of element, dm = $$\sigma 2\pi x\left( {dx} \right)$$

Here, $$\sigma $$ = mass per unit area = $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$$

Moment of inertia of element, dI = (dm)x

$$ \Rightarrow $$ I = $$\sigma 2\pi \int\limits_r^R {{x^3}dx} $$

= $$\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$$

= $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$$

= $${m \over 2}\left( {{R^2} + {r^2}} \right)$$ .....(i)

Moment of inertia of thin cylinder of same mass,

I = m$$r_0^2$$ ......(ii)

$$ \Rightarrow $$ m$$r_0^2$$ = $${m \over 2}\left( {{R^2} + {r^2}} \right)$$

$$ \Rightarrow $$ $$r_0^2$$ = 250

$$ \Rightarrow $$ r

2

MCQ (Single Correct Answer)

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :

A

16 rad/s^{2}

B

20 rad/s^{2}

C

10 rad/s^{2}

D

12 rad/s^{2}

40 + f = m(R$$\alpha $$) . . . .(i)

40 $$ \times $$ R $$-$$ f $$ \times $$ R = mR

40 $$-$$ f = mR$$\alpha $$ . . . .(ii)

From (i) and (ii)

$$\alpha $$ = $${{40} \over {mR}}$$ = 16

3

MCQ (Single Correct Answer)

A circular disc D_{1} of mass M and radius R has two identical discs D_{2} and D_{3} of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO' ,passing through the centre of D_{1} as shown in the figure, will be :

A

3MR^{2}

B

MR^{2}

C

$${2 \over 3}$$ MR^{2}

D

$${4 \over 5}$$ MR^{2}

I $$=$$ $${{M{R^2}} \over 2} + 2\left( {{{M{R^2}} \over 4} + M{R^2}} \right)$$

$$ = {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$$

$$=$$ $$3M{R^2}$$

$$ = {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$$

$$=$$ $$3M{R^2}$$

4

MCQ (Single Correct Answer)

The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :

A

$${\pi \over 8}$$

B

$${\pi \over 6}$$

C

$${\pi \over 4}$$

D

$${\pi \over 3}$$

2.5 = 1 $$ \times $$ 5 sin $$\theta $$

sin$$\theta $$ = 0.5 = $${1 \over 2}$$

$$\theta $$ = $${\pi \over 6}$$

sin$$\theta $$ = 0.5 = $${1 \over 2}$$

$$\theta $$ = $${\pi \over 6}$$

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