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 force acts on a 2 kg object so that its position is given as a function of time as x = 3t^{2} + 5. What is the work done by this force in first 5 seconds ?

A

850 J

B

950 J

C

875 J

D

900 J

Displacement,

x = 3t^{2} + 5

$$ \therefore $$ v = $${{dx} \over {dt}} = 6t$$

At t = 0, velocity = 6 $$ \times $$ 0 = 0

at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($$\Delta $$K)

= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$

= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)^{2} $$-$$ 0

= 900 J

x = 3t

$$ \therefore $$ v = $${{dx} \over {dt}} = 6t$$

At t = 0, velocity = 6 $$ \times $$ 0 = 0

at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($$\Delta $$K)

= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$

= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)

= 900 J

2

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -

A

$${{m{g^2}{t^2}} \over 8}$$

B

$${{3m{g^2}{t^2}} \over 8}$$

C

$$-$$ $${{m{g^2}{t^2}} \over 8}$$

D

0

N $$-$$ mg = $${{mg} \over 2}$$ $$ \Rightarrow $$ N = $${{3mg} \over 2}$$

The distance travelled by the system in time t is

S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$

Now, work done

W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$

$$ \Rightarrow $$ W = $${{3m{g^2}{t^2}} \over 8}$$

The distance travelled by the system in time t is

S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$

Now, work done

W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$

$$ \Rightarrow $$ W = $${{3m{g^2}{t^2}} \over 8}$$

3

A particle which is experiencing a force, given by $$\overrightarrow F = 3\widehat i - 12\widehat j,$$ undergoes a displacement of $$\overrightarrow d = 4\overrightarrow i $$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

A

9 J

B

10 J

C

12 J

D

15 J

Work done = $$\overrightarrow F \cdot \overrightarrow d $$

$$=$$ 12 J

work energy theorem

w_{net} $$=$$ $$\Delta $$K.E.

12 $$=$$ K_{f} $$-$$ 3

K_{f} = 15 J

$$=$$ 12 J

work energy theorem

w

12 $$=$$ K

K

4

A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 10^{6} N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms^{–2}
, the value of x will be close to :

A

8 cm

B

4 cm

C

40 cm

D

80 cm

velocity of 1 kg block just before it collides with 3kg block

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s

initial compression of spring

1.25 $$ \times $$ 10^{6} x_{0} = 30 $$ \Rightarrow $$ x_{0} $$ \approx $$ 0

applying work energy theorem,

W_{g} + W_{sp} = $$\Delta $$KE

$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 10^{6} (0^{2} $$-$$ x^{2})

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v^{2}

solving x $$ \approx $$ 4 cm

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s

initial compression of spring

1.25 $$ \times $$ 10

applying work energy theorem,

W

$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 10

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v

solving x $$ \approx $$ 4 cm

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (2) *keyboard_arrow_right*

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

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Simple Harmonic Motion *keyboard_arrow_right*

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Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

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Magnetics *keyboard_arrow_right*

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Ray & Wave Optics *keyboard_arrow_right*

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