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 neutron moving with a speed ‘v’ makes a head on collision with a stationary
hydrogen atom in ground state. The minimum kinetic energy of the neutron for
which inelastic collision will take place is :

A

10.2 eV

B

16.8 eV

C

12.1 eV

D

20.4 eV

Let, velocity offer collision = v_{1}

$$ \therefore $$ From conservation of momentum,

mv = (m + m) v_{1}

$$ \Rightarrow $$ v_{1} = $${v \over 2}$$

$$ \therefore $$ Loss in kinetic energy

= $${1 \over 2}$$ mv^{2} $$-$$ $${1 \over 2}$$ (2m) $$ \times $$ $${\left( {{v \over 2}} \right)^2}$$

= $${1 \over 4}\,$$mv^{2}

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$$ \therefore $$ $${1 \over 4}$$mv^{2} = (13.6 $$-$$ 3.4) eV = 10.2 eV

$$ \Rightarrow $$ $${1 \over 2}$$mv^{2} = 20.4 eV

$$ \therefore $$ From conservation of momentum,

mv = (m + m) v

$$ \Rightarrow $$ v

$$ \therefore $$ Loss in kinetic energy

= $${1 \over 2}$$ mv

= $${1 \over 4}\,$$mv

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$$ \therefore $$ $${1 \over 4}$$mv

$$ \Rightarrow $$ $${1 \over 2}$$mv

2

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

3

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

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

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

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

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

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

4

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

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*

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*