Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $${{BC} \over {AB}}$$ is close to :

A

1.85

B

1.37

C

1.5

D

3

Here AB = x

and BC = y

and $$\lambda $$ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60

$$ \therefore $$ X

$$ \Rightarrow $$ $${x \over 2}$$ = $${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$$

$$ \Rightarrow $$ $${x \over 2}$$ = $${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$$

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x

$$ \therefore $$ x = $${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$$

= $${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$$

= $$-$$ y $$ \pm $$ $$\sqrt 3 $$y

$${x \over y} \ne - \sqrt 3 - 1$$ as $${x \over y}$$ = positive.

$$ \therefore $$ $${x \over y}$$ = $$\sqrt 3 - 1$$

$$ \Rightarrow $$ $${y \over x}$$ = $${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$$

= $${{\sqrt 3 + 1} \over 2}$$

= $${{2.732} \over 2}$$

= 1.366 $$ \simeq $$ 1.37

2

The moment of inertia of a uniform cylinder of length $$l$$ and radius R about its perpendicular bisector is $$I$$.
What is the ratio $${l \over R}$$ such that the moment of inertia is minimum?

A

$${3 \over {\sqrt 2 }}$$

B

$$\sqrt {{3 \over 2}} $$

C

$${{\sqrt 3 } \over 2}$$

D

1

The volume of the cylinder V = $$\pi {R^2}l$$

$$\therefore$$ $${R^2} = {V \over {\pi l}}$$

We know, moment of inertia of a uniform cylinder of length $$l$$ and radius R about its perpendicular bisector is,

$$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$$

[ Putting $${R^2} = {V \over {\pi l}}$$ in this equation]

$$ \Rightarrow $$ $$I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}$$

Here $$I$$ is a function of $$l$$ as M and V are constant.

$$I$$ will be maximum or minimum when $${{{dI} \over {dl}}}$$ = 0.

$$ \Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0$$

$$ \Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}$$

$$ \Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}$$ [ as $${V = \pi {R^2}l}$$ ]

$$ \Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}$$

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

$$\therefore$$ $${R^2} = {V \over {\pi l}}$$

We know, moment of inertia of a uniform cylinder of length $$l$$ and radius R about its perpendicular bisector is,

$$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$$

[ Putting $${R^2} = {V \over {\pi l}}$$ in this equation]

$$ \Rightarrow $$ $$I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}$$

Here $$I$$ is a function of $$l$$ as M and V are constant.

$$I$$ will be maximum or minimum when $${{{dI} \over {dl}}}$$ = 0.

$$ \Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0$$

$$ \Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}$$

$$ \Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}$$ [ as $${V = \pi {R^2}l}$$ ]

$$ \Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}$$

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

3

A slender uniform rod of mass M and length $$l$$ is pivoted at one end so that it
can rotate in a vertical plane (see figure). There is negligible friction at the
pivot. The free end is held vertically above the pivot and then released. The
angular acceleration of the rod when it makes an angle $$\theta$$ with the vertical is

A

$${{2g} \over {3l}}\cos \theta $$

B

$${{3g} \over {2l}}\sin \theta $$

C

$${{2g} \over {3l}}\sin \theta $$

D

$${{3g} \over {3l}}\sin \theta $$

Forces acting on the rod are shown below.
Torque about pivot point O due to force N_{x} and N_{y} are zero.

Mgcos$$\theta $$ is passing through O, so torque will be zero due to this force.

So torque about the point O is

$$\tau = Mg\sin \theta \times {l \over 2}$$

We know, $$\tau = I\alpha $$

$$\therefore$$ $$I\alpha = Mg\sin \theta \times {l \over 2}$$

Moment of inertia of rod about point O, $$I$$ = $${{M{l^2}} \over 3}$$

$$\therefore$$ $${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$$

$$ \Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta $$

Mgcos$$\theta $$ is passing through O, so torque will be zero due to this force.

So torque about the point O is

$$\tau = Mg\sin \theta \times {l \over 2}$$

We know, $$\tau = I\alpha $$

$$\therefore$$ $$I\alpha = Mg\sin \theta \times {l \over 2}$$

Moment of inertia of rod about point O, $$I$$ = $${{M{l^2}} \over 3}$$

$$\therefore$$ $${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$$

$$ \Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta $$

4

Moment of inertia of an equilateral triangular lamina ABC, about the axis
passing through its centre O and perpendicular to its plane is I_{o} as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

A

$${7 \over 8}$$ I_{o}

B

$${15 \over 16}$$ I_{o}

C

$${{3\,{{\rm I}_o}} \over 4}$$

D

$${{31\,{{\rm I}_o}} \over 32}$$

Let, side of triangle ABC = $$\ell $$

According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,

I_{O} = $${1 \over {12}}$$ m$$\ell $$^{2}

In, triangle DEF,

DE = DF = EF = $${1 \over 2}$$ AB = $${1 \over 2}$$ $$\ell $$

$$\therefore\,\,\,$$ moment of inertia of triangle DEF,

I_{DEF} = $${1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}$$

= $${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$$

= $${{{I_O}} \over {16}}$$

$$\therefore\,\,\,$$ Moment of inertia of the remaining part,

I_{remain} = I_{O} $$-$$ $${{{{\rm I}_O}} \over {16}}$$ = $${{15\,{{\rm I}_O}} \over {16}}$$

According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,

I

In, triangle DEF,

DE = DF = EF = $${1 \over 2}$$ AB = $${1 \over 2}$$ $$\ell $$

$$\therefore\,\,\,$$ moment of inertia of triangle DEF,

I

= $${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$$

= $${{{I_O}} \over {16}}$$

$$\therefore\,\,\,$$ Moment of inertia of the remaining part,

I

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