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

Which of the following option correctly describes the variation of the speed *v* and acceleration *‘a’* of a point mass falling vertically in a viscous medium that applies a force F = − *kv,* where ‘k’ is a constant, on the body ? (Graphs are schematic and not drawn to scale)

A

B

C

D

Equation of motion for the mass,

ma = mg $$-$$ kv

$$ \Rightarrow $$ $${{dv} \over {dt}} = {{mg - kv} \over m}$$

$$ \Rightarrow $$ $$\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt} $$

$$ \Rightarrow $$ $$ - {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$$

$$ \Rightarrow $$ $$\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$$

$$ \Rightarrow $$ $$1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$$

$$ \Rightarrow $$ $${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$$

$$ \Rightarrow $$ $$v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$$

ma $$=$$ mg $$-$$ k $$ \times $$ $${{mg} \over k}$$ (1 $$-$$ e$$^{ - {{kt} \over m}}$$)

$$=$$ mg $$-$$ mg + mge$$^{ - {{kt} \over m}}$$

a $$=$$ g e$$^{ - {{kt} \over m}}$$

ma = mg $$-$$ kv

$$ \Rightarrow $$ $${{dv} \over {dt}} = {{mg - kv} \over m}$$

$$ \Rightarrow $$ $$\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt} $$

$$ \Rightarrow $$ $$ - {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$$

$$ \Rightarrow $$ $$\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$$

$$ \Rightarrow $$ $$1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$$

$$ \Rightarrow $$ $${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$$

$$ \Rightarrow $$ $$v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$$

ma $$=$$ mg $$-$$ k $$ \times $$ $${{mg} \over k}$$ (1 $$-$$ e$$^{ - {{kt} \over m}}$$)

$$=$$ mg $$-$$ mg + mge$$^{ - {{kt} \over m}}$$

a $$=$$ g e$$^{ - {{kt} \over m}}$$

2

Consider a water jar of radius R that has water filled up to height H and is kept on astand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :

A

$$x = r\left( {{H \over {H + h}}} \right)$$

B

$$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 2}}}$$

C

$$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 4}}}$$

D

$$x = r{\left( {{H \over {H + h}}} \right)^{{2}}}$$

v_{1} = velocity of water when it leak from hole

v_{2} = velocity of water when it reach the ground.

From Bernoulli's principle,

$${1 \over 2}\rho {v_1}^2 + \rho gh$$ = $${1 \over 2}\rho {v_2}^2$$

$$ \Rightarrow $$ $${v_1}^2$$ + 2gh = $${v_2}^2$$

From Torricelli's theorem,

v_{1} = $$\sqrt {2gH} $$

$$ \therefore $$ $${v_2}^2$$ = 2gh + 22gH

From continuity equation,

$${A_1}{v_1}$$ = $${A_2}{v_2}$$

$$ \Rightarrow $$ $$\pi {r^2} \times \sqrt {2gH} $$ = $$\pi {x^2}\sqrt {2g\left( {h + H} \right)} $$

$$ \Rightarrow $$ x^{2} = r^{2}$$\sqrt {{H \over {H + g}}} $$

$$ \Rightarrow $$ x = r $${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$$

v

From Bernoulli's principle,

$${1 \over 2}\rho {v_1}^2 + \rho gh$$ = $${1 \over 2}\rho {v_2}^2$$

$$ \Rightarrow $$ $${v_1}^2$$ + 2gh = $${v_2}^2$$

From Torricelli's theorem,

v

$$ \therefore $$ $${v_2}^2$$ = 2gh + 22gH

From continuity equation,

$${A_1}{v_1}$$ = $${A_2}{v_2}$$

$$ \Rightarrow $$ $$\pi {r^2} \times \sqrt {2gH} $$ = $$\pi {x^2}\sqrt {2g\left( {h + H} \right)} $$

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x = r $${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$$

3

A thin 1 m long rod has a radius of 5 mm. A force of 50 $$\pi $$kN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is **false** ?

A

$${{\Delta \gamma } \over \gamma }$$ gets minimum contribution
from the uncertainty in the length.

B

The figure of merit is the largest for the length of the rod.

C

The maximum value of $$\gamma $$ that can be determined is 2 $$ \times $$ 10^{14} N/m^{2}

D

$${{\Delta \gamma } \over \gamma }$$ gets its maximum contribution
from the uncertainty in strain

4

A bottle has an opening of radius a and length b. A cork of length b and radius (a + $$\Delta $$a) where ($$\Delta $$a < < a) is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is *B* and frictional coefficient between the bottle and cork is $$\mu $$ then the force needed to push the cork into the bottle is :

A

($$\pi $$ $$\mu $$ B b) $$\Delta $$a

B

(2$$\pi $$ $$\mu $$ B b) $$\Delta $$a

C

($$\pi $$ $$\mu $$ B b) a

D

(4$$\pi $$ $$\mu $$ B b) $$\Delta $$a

Bulk modulus, B = $${{\Delta P} \over {{{\Delta V} \over V}}}$$

V_{i} = $$\pi $$ (a + $$\Delta $$a)^{2}b

V_{f} = $$\pi $$a^{2}b

$$ \therefore $$ $$\Delta $$V = 2$$\pi $$ab$$\Delta $$a

$$ \therefore $$ $${{{\Delta V} \over V}}$$ = $${{2\pi ab\Delta a} \over {\pi {a^2}b}}$$ = $${{2\Delta a} \over a}$$

$$ \therefore $$ $$\Delta $$P = B $$ \times $$ $${{{\Delta V} \over V}}$$

= B $$ \times $$ $${{2\Delta a} \over a}$$

Normal Force (N) = $$\Delta $$P $$ \times $$ A

= B $$ \times $$ $${{2\Delta a} \over a}$$ $$ \times $$ (2$$\pi $$a)b

= 4$$\pi $$B$$\Delta $$ab

$$ \therefore $$ Frictional force = $$\mu $$N = (4$$\pi $$$$\mu $$Bb)$$\Delta $$a

V

V

$$ \therefore $$ $$\Delta $$V = 2$$\pi $$ab$$\Delta $$a

$$ \therefore $$ $${{{\Delta V} \over V}}$$ = $${{2\pi ab\Delta a} \over {\pi {a^2}b}}$$ = $${{2\Delta a} \over a}$$

$$ \therefore $$ $$\Delta $$P = B $$ \times $$ $${{{\Delta V} \over V}}$$

= B $$ \times $$ $${{2\Delta a} \over a}$$

Normal Force (N) = $$\Delta $$P $$ \times $$ A

= B $$ \times $$ $${{2\Delta a} \over a}$$ $$ \times $$ (2$$\pi $$a)b

= 4$$\pi $$B$$\Delta $$ab

$$ \therefore $$ Frictional force = $$\mu $$N = (4$$\pi $$$$\mu $$Bb)$$\Delta $$a

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (1) *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*