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

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on
the body in first two seconds of the motion is :

A

12000 J

B

$$-$$ 12000 J

C

$$-$$ 4500 J

D

$$-$$ 9300 J

Here u = 50 m/s , what t = 0

$$\alpha $$ = $${{\Delta v} \over {\Delta t}}$$ = $${{50 - 0} \over {0 - 10}}$$ = $$-$$5 m/s

Speed of the body at t = 2 s

v = u + at

= 50 + ($$-$$ 5) $$ \times $$ 2

= 40 m/s

From work energy theorem,

$$\Delta $$w = $${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$$

= $${1 \over 2}$$ m(v

= $${1 \over 2}$$ $$ \times $$ 10 $$ \times $$ (40

= 5 $$ \times $$ ($$-$$10)(90)

= $$-$$ 4500 J

2

A body of mass m = 10^{–2} kg is moving in a medium and experiences a frictional force F = –kv^{2}. Its initial speed is v_{0} = 10 ms^{–1}. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be:

A

10^{-1} kg m^{-1} s^{-1}

B

10^{-3} kg m^{-1}

C

10^{-3} kg s^{-1}

D

10^{-4} kg m^{-1}

According to the question, final kinetic energy = $${1 \over 8}mv_0^2$$

Let final speed of the body = V_{f}

So final kinetic energy = $${1 \over 2}mv_f^2$$

According to question,

$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$

$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s

Given that, F = –kv^{2}

$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$

$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$

$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$

$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$

$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$

Let final speed of the body = V

So final kinetic energy = $${1 \over 2}mv_f^2$$

According to question,

$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$

$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s

Given that, F = –kv

$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$

$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$

$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$

$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$

$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$

3

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done
by the force during the first 1 sec. will be:

A

18 J

B

4.5 J

C

22 J

D

9 J

Given that, F = 6t

We know, F = ma = $$m{{dv} \over {dt}}$$

$$\therefore$$ $$m{{dv} \over {dt}} = 6t$$

$$ \Rightarrow $$ $$1.{{dv} \over {dt}} = 6t$$ [as m = 1]

$$ \Rightarrow $$ $$\int\limits_0^v {dv} = \int {6t} dt$$

$$ \Rightarrow $$ $$v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1$$

$$ \Rightarrow $$ $$v = {6 \over 2} = 3$$ m/s [ as given t = 1 sec ]

Work done by the body during the first 1 form work-energy theorem,

W = $$\Delta $$K.E = $${1 \over 2}m\left( {{V^2} - {v^2}} \right)$$

= $${1 \over 2}.1.\left( {{3^2} - {0^2}} \right)$$ = 4.5 J

We know, F = ma = $$m{{dv} \over {dt}}$$

$$\therefore$$ $$m{{dv} \over {dt}} = 6t$$

$$ \Rightarrow $$ $$1.{{dv} \over {dt}} = 6t$$ [as m = 1]

$$ \Rightarrow $$ $$\int\limits_0^v {dv} = \int {6t} dt$$

$$ \Rightarrow $$ $$v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1$$

$$ \Rightarrow $$ $$v = {6 \over 2} = 3$$ m/s [ as given t = 1 sec ]

Work done by the body during the first 1 form work-energy theorem,

W = $$\Delta $$K.E = $${1 \over 2}m\left( {{V^2} - {v^2}} \right)$$

= $${1 \over 2}.1.\left( {{3^2} - {0^2}} \right)$$ = 4.5 J

4

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $$ \to $$ $$\infty $$ is :

A

3h

B

$$\infty $$

C

$${5 \over 3}$$h

D

$${8 \over 3}$$h

Let,

Kinetic energy (k) = $${1 \over 2}$$ m $$\upsilon $$^{2} before it hit the ground.

After hitting the ground kinetic energy

(k') = $${1 \over 2}$$ m $$\upsilon $$$$_1^2$$

$$\therefore\,\,\,$$According to the question,

$${1 \over 2}$$ m$$\upsilon $$$$_1^2$$ = $${1 \over 2}$$ $$ \times $$ $${1 \over 2}$$ m$$\upsilon $$^{2}

$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$_{1} = $${v \over {\sqrt 2 }}$$

After hitting the ground the object will bounce

h' = $${{v_1^2} \over {2g}}$$ = $${{{v^2}} \over {4g}}$$ = $${h \over 2}$$ [ as h = $${{{v^2}} \over {2g}}$$ ]

Total distance travelled from the time it first hits the ground to the next time it hits the ground is = $${h \over 2}$$ + $${h \over 2}$$ = h

So, this will create a infinite geometric progression with the common ration $${1 \over 2}$$.

$$\therefore\,\,\,$$ Total distance covered

= h (distance travelled by the obhect when first dropped, before it hits the ground)

+ (h + $${h \over 2}$$ + $${h \over 4}$$ + . . . . . . . .$$ \propto $$)

= h + $${h \over {1 - {1 \over 2}}}$$

= h + 2h

= 3h

Kinetic energy (k) = $${1 \over 2}$$ m $$\upsilon $$

After hitting the ground kinetic energy

(k') = $${1 \over 2}$$ m $$\upsilon $$$$_1^2$$

$$\therefore\,\,\,$$According to the question,

$${1 \over 2}$$ m$$\upsilon $$$$_1^2$$ = $${1 \over 2}$$ $$ \times $$ $${1 \over 2}$$ m$$\upsilon $$

$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$

After hitting the ground the object will bounce

h' = $${{v_1^2} \over {2g}}$$ = $${{{v^2}} \over {4g}}$$ = $${h \over 2}$$ [ as h = $${{{v^2}} \over {2g}}$$ ]

Total distance travelled from the time it first hits the ground to the next time it hits the ground is = $${h \over 2}$$ + $${h \over 2}$$ = h

So, this will create a infinite geometric progression with the common ration $${1 \over 2}$$.

$$\therefore\,\,\,$$ Total distance covered

= h (distance travelled by the obhect when first dropped, before it hits the ground)

+ (h + $${h \over 2}$$ + $${h \over 4}$$ + . . . . . . . .$$ \propto $$)

= h + $${h \over {1 - {1 \over 2}}}$$

= h + 2h

= 3h

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

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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*

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Ray & Wave Optics *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

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