Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Concrete mixture is made by mixing cement, stone and sand in a rotating
cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute(rpm) to ensure proper mixing is close to :

(Take the radius of the drum to be 1.25 m and its axle to be horizontal) :

(Take the radius of the drum to be 1.25 m and its axle to be horizontal) :

A

0.4

B

1.3

C

8.0

D

27.0

2

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

3

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 $${V = \pi {R^2}l}$$ ]

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

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

4

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, $$I$$ = $${{M{l^2}} \over 3}$$

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

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

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (5) *keyboard_arrow_right*

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Simple Harmonic Motion *keyboard_arrow_right*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*