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 rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its
instantaneous angular acceleration will be -

A

$${g \over {13l}}$$

B

$${g \over {2l}}$$

C

$${g \over {3l}}$$

D

$${7g \over {3l}}$$

Applying torque equation about point P.

2M

I = 2M

$$ \therefore $$ $$\alpha $$ = $$-$$ $${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$-$$ $${g \over {13\ell }}$$

$$ \therefore $$ $$\alpha $$ = $${g \over {13\ell }}$$ anticlockwise

2

A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :

A

$$2\sqrt {{k \over p}} $$

B

$$2\sqrt {{p \over k}} $$

C

$$\sqrt {{{2p} \over 2}} $$

D

$$\sqrt {{{2k} \over p}} $$

$${{dp} \over {dt}} = F = kt$$

$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$

$$2p = {{K{T^2}} \over 2}$$

$$T = 2\sqrt {{P \over K}} $$

$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$

$$2p = {{K{T^2}} \over 2}$$

$$T = 2\sqrt {{P \over K}} $$

3

The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

A

$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$

B

$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$

C

$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$

D

$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$

$${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$$

$${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$$

4

A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It
bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by :

A

$${m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$

B

$${m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

C

$$m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$

D

$$m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

v = $$\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $$

v

By momentum conservation

m$$\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} $$

$$= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)} $$

$$ \Rightarrow $$$$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}$$

$$ = $$ MV

and e = 1 = $${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}$$

$$\sqrt {2g\ell } $$ $$\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right) $$

$$= $$ V

m$$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)$$

$$ = $$ MV

Dividing

$${{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}$$

By componendo divided

$${{m - M} \over {m + M}}$$ = $${{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}$$

$$ \Rightarrow $$ $${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$$

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2004 (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*