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

A conical pendulum of length 1 m makes an angle $$\theta $$ = 45^{o} w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms^{−2} )

A

0.4 m/s

B

4 m/s

C

0.2 m/s

D

2 m/s

FBD of pendulum is :

$$\therefore\,\,\,$$ T sin $$\theta $$ = $${{m{v^2}} \over r}$$

T cos $$\theta $$ = mg

$$\therefore\,\,\,$$ tan $$\theta $$ = $${{{v^2}} \over {rg}}$$

$$ \Rightarrow $$$$\,\,\,$$ tan45^{o}^{} = $${{{v^2}} \over {rg}}$$

$$ \Rightarrow $$$$\,\,\,$$ v^{2} = rg

$$ \Rightarrow $$$$\,\,\,$$ v = $$\sqrt {0.4 \times 10} $$ = 2 m/s

$$\therefore\,\,\,$$ T sin $$\theta $$ = $${{m{v^2}} \over r}$$

T cos $$\theta $$ = mg

$$\therefore\,\,\,$$ tan $$\theta $$ = $${{{v^2}} \over {rg}}$$

$$ \Rightarrow $$$$\,\,\,$$ tan45

$$ \Rightarrow $$$$\,\,\,$$ v

$$ \Rightarrow $$$$\,\,\,$$ v = $$\sqrt {0.4 \times 10} $$ = 2 m/s

2

Two masses m_{1} = 5 kg and m_{2} = 10 kg, connected by an inextensible
string over a frictionless pulley, are moving as shown in the figure. The
coefficient of friction of horizontal surface is 0.15. The minimum
weight m that should be put on top of m_{2} to stop the motion is :

A

10.3 kg

B

18.3 kg

C

27.3 kg

D

43.3 kg

Moving block will stop when the friction force between m_{2} and surface is $$ \ge $$ tension force.

So condition for stopping the moving block,

$$f \ge T$$

$$ \Rightarrow \mu N \ge T$$

$$ \Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g$$

When m is minimum then,

$$\mu \left( {m + {m_2}} \right)g = {m_1}g$$

$$ \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$$

$$ \Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}}$$ = 23.33 kg

So if m $$ \ge $$ 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg.

So condition for stopping the moving block,

$$f \ge T$$

$$ \Rightarrow \mu N \ge T$$

$$ \Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g$$

When m is minimum then,

$$\mu \left( {m + {m_2}} \right)g = {m_1}g$$

$$ \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$$

$$ \Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}}$$ = 23.33 kg

So if m $$ \ge $$ 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg.

3

A given object takes n times more time to slide down a $${45^ \circ }$$ rough inclined plane as it takes to slide down a perfectly smooth $${45^ \circ }$$ incline. The coefficient of kinetic friction between the object and the incline is :

A

$${1 \over {2 - {n^2}}}$$

B

$$1 - {1 \over {{n^2}}}$$

C

$$\sqrt {1 - {1 \over {{n^2}}}} $$

D

$$\sqrt {{1 \over {1 - {n^2}}}} $$

Let, t_{1} and t_{2} are time taken to move on the smooth and rough surface for smooth surface,

S = $${1 \over 2}$$ g sin45^{o} $$t_1^2$$

$$ \Rightarrow $$$$\,\,\,\,$$ t_{1} = $$\sqrt {{{2\sqrt 2 S} \over g}} $$

For rough surface,

S = $${1 \over 2}$$ g (sin45^{o} $$-$$ $$\mu $$_{k} cos45^{o}) $$t_2^2$$

$$ \Rightarrow $$$$\,\,\,\,$$ t_{2} = $$\sqrt {{{2\sqrt 2 S} \over {g\left( {1 - {\mu _k}} \right)}}} $$

Here $${\mu _k}$$ = Kinetic friction.

According to question,

t_{2} = n t_{1}

$$ \Rightarrow $$$$\,\,\,\,$$ $${{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}$$ = n^{2} $$ \times $$ $${{2\sqrt 2 \,S} \over g}$$

$$ \Rightarrow $$$$\,\,\,\,$$ 1 $$-$$ $$\mu $$_{k} = $${1 \over {{n^2}}}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $${\mu _k}$$ = 1 $$-$$ $${1 \over {{n^2}}}$$

S = $${1 \over 2}$$ g sin45

$$ \Rightarrow $$$$\,\,\,\,$$ t

For rough surface,

S = $${1 \over 2}$$ g (sin45

$$ \Rightarrow $$$$\,\,\,\,$$ t

Here $${\mu _k}$$ = Kinetic friction.

According to question,

t

$$ \Rightarrow $$$$\,\,\,\,$$ $${{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}$$ = n

$$ \Rightarrow $$$$\,\,\,\,$$ 1 $$-$$ $$\mu $$

$$ \Rightarrow $$$$\,\,\,\,$$ $${\mu _k}$$ = 1 $$-$$ $${1 \over {{n^2}}}$$

4

A body of mass 2 kg slides down with an acceleration of 3 m/s^{2} on a rough inclined plane having a slope of $${30^o}$$. The external force required to take the same body up the plane with the same acceleration will be : (g = 10 m/s^{2})

A

14 N

B

20 N

C

6 N

D

4 N

When mass slide down then,

Mgsin$$\theta $$ $$-$$ $$\mu $$Mg cos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ a = g(sin30^{o} $$-$$ $$\mu $$ cos30^{o})

When mass pushed upward with force F,

F $$-$$ Mgsin$$\theta $$ $$-$$ $$\mu $$ Mgcos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ F = Mg(sin30^{o} + $$\mu $$ cos 30^{o}) + Mg(sin30^{o} $$-$$ $$\mu $$ cos30^{o})

$$ \Rightarrow $$ $$\,\,\,$$ F = 2Mg sin30^{o}

= 2 $$ \times $$ 2 $$ \times $$ 10 $$ \times $$ $${1 \over 2}$$

= 20 N

Mgsin$$\theta $$ $$-$$ $$\mu $$Mg cos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ a = g(sin30

When mass pushed upward with force F,

F $$-$$ Mgsin$$\theta $$ $$-$$ $$\mu $$ Mgcos$$\theta $$ = Ma

$$ \Rightarrow $$ $$\,\,\,$$ F = Mg(sin30

$$ \Rightarrow $$ $$\,\,\,$$ F = 2Mg sin30

= 2 $$ \times $$ 2 $$ \times $$ 10 $$ \times $$ $${1 \over 2}$$

= 20 N

Number in Brackets after Paper Name Indicates No of Questions

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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*

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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*

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Ray & Wave Optics *keyboard_arrow_right*

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