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 car of weight W is on an inclined road that rises by 100 m over a distance of 1 km
and applies a constant frictional force $${W \over 20}$$ on the car. While moving uphill on the road at a speed of 10 ms^{ā1}, the car needs power P. If it needs power $${p \over 2}$$ while moving downhill at speed v then value of $$\upsilon $$ is :

A

20 ms^{$$-$$1}

B

15 ms^{$$-$$1}

C

10 ms^{$$-$$1}

D

5 ms^{$$-$$1}

Here, tan$$\theta $$ = $${{100} \over {1000}} = {1 \over {10}}$$

$$ \therefore $$ sin$$\theta $$ = $${1 \over {10}}$$ (as $$\theta $$ is very small),

when car is moving uphill :

P = f $$ \times $$ u

= (wsin$$\theta $$ + f) $$ \times $$ u

= $$\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$$

P = $${{3w} \over {20}} \times 10$$ = $${{3w} \over 2}$$

When car is moving down hill :

$$ \therefore $$ $${P \over 2}$$ = (wsin$$\theta $$ $$-$$ f) $$ \times $$ v

$$ \Rightarrow $$ $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} \right)$$ $$ \times $$ v

$$ \Rightarrow $$ $${{w \over {20}} \times }$$ v = $${{3w} \over 4}$$

$$ \Rightarrow $$ v = 15 m/s

$$ \therefore $$ sin$$\theta $$ = $${1 \over {10}}$$ (as $$\theta $$ is very small),

when car is moving uphill :

P = f $$ \times $$ u

= (wsin$$\theta $$ + f) $$ \times $$ u

= $$\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$$

P = $${{3w} \over {20}} \times 10$$ = $${{3w} \over 2}$$

When car is moving down hill :

$$ \therefore $$ $${P \over 2}$$ = (wsin$$\theta $$ $$-$$ f) $$ \times $$ v

$$ \Rightarrow $$ $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} \right)$$ $$ \times $$ v

$$ \Rightarrow $$ $${{w \over {20}} \times }$$ v = $${{3w} \over 4}$$

$$ \Rightarrow $$ v = 15 m/s

2

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n^{2} R t^{2} where n is a constant. The power delivered to the particle by the force acting on it, is :

A

M n^{2} R^{2} t

B

M n R^{2} t

C

M n R^{2} t^{2}

D

$${1 \over 2}$$ M n^{2} R^{2} t^{2}

We know,

centripetal acceleration = $${{{V^2}} \over R}$$

$$ \therefore $$ According to question,

$${{{V^2}} \over R}$$ = $${n^2}R{t^2}$$

$$ \Rightarrow $$ V^{2} = n^{2} R^{2} t^{2}

$$ \Rightarrow $$ V = nRt

$$ \Rightarrow $$ $${{dV} \over {dt}}$$ = nR

Power (P) = Force (F) $$ \times $$ Velocity (V)

= M $${{dV} \over {dt}}$$(V)

= M (nR) (nRt)

= Mn^{2}R^{2}t

centripetal acceleration = $${{{V^2}} \over R}$$

$$ \therefore $$ According to question,

$${{{V^2}} \over R}$$ = $${n^2}R{t^2}$$

$$ \Rightarrow $$ V

$$ \Rightarrow $$ V = nRt

$$ \Rightarrow $$ $${{dV} \over {dt}}$$ = nR

Power (P) = Force (F) $$ \times $$ Velocity (V)

= M $${{dV} \over {dt}}$$(V)

= M (nR) (nRt)

= Mn

3

A neutron moving with a speed āvā 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

4

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

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*

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*

Dual Nature of Radiation *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

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