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

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

A

16 : 9

B

25 : 9

C

4 : 1

D

5 : 3

Given that,

$${{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}$$

We know,

I_{max} $$=$$ $${\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}$$

and I_{min} $$ = {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}$$

$$ \therefore $$ $${{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}$$

$$ \Rightarrow $$ $${{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}$$

$$ \Rightarrow $$ $$4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$ $$3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$ $${{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}$$

$$ \Rightarrow $$ $${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}$$

$${{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}$$

We know,

I

and I

$$ \therefore $$ $${{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}$$

$$ \Rightarrow $$ $${{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}$$

$$ \Rightarrow $$ $$4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$ $$3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$ $${{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}$$

$$ \Rightarrow $$ $${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}$$

2

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

A

0.77

B

0.57

C

0.37

D

0.17

Initially :

After putting 2 masses of each 'm' at a distance $${L \over 2}$$ from center :

We know,

Time period (T) = 2$$\pi $$ $$\sqrt {{{\rm I} \over C}} $$

$$ \therefore $$ T $$ \propto $$ $$\sqrt {\rm I} $$

$$ \therefore $$ Frequency (f) $$ \propto $$ $$\sqrt {{1 \over {\rm I}}} $$

$$ \therefore $$ $${{{f_1}} \over {{f_2}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

Also given that,

After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$$ \therefore $$ f_{2} = 0.8f_{1}

$$ \therefore $$ $${{{f_1}} \over {0.8{f_1}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

$$ \therefore $$ $${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$$ = 0.64

Initial moment of inertia of the system,

$${{{\rm I}_1}}$$ = $${{M{{\left( {2L} \right)}^2}} \over {12}}$$

Final moment of inertia of the system,

I_{2} = $${{M{{\left( {2L} \right)}^2}} \over {12}}$$ + 2$$\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)$$

$$ \therefore $$ $${{M{{\left( {2L} \right)}^2}} \over {12}}$$ = 0.64 $$\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$$

$$ \Rightarrow $$ $${{M{L^2}} \over {3 \times 0.64}}$$ = $${{M{L^2}} \over 3}$$ + $${{M{L^2}} \over 2}$$

$$ \Rightarrow $$ $${M \over {1.92}} - {M \over 3} = {m \over 2}$$

$$ \Rightarrow $$ $${{1.08M} \over {3 \times 1.92}}$$ = $${m \over 2}$$

$$ \Rightarrow $$ $${m \over M}$$ = $${{1.08 \times 2} \over {3 \times 1.92}}$$ = 0.37

After putting 2 masses of each 'm' at a distance $${L \over 2}$$ from center :

We know,

Time period (T) = 2$$\pi $$ $$\sqrt {{{\rm I} \over C}} $$

$$ \therefore $$ T $$ \propto $$ $$\sqrt {\rm I} $$

$$ \therefore $$ Frequency (f) $$ \propto $$ $$\sqrt {{1 \over {\rm I}}} $$

$$ \therefore $$ $${{{f_1}} \over {{f_2}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

Also given that,

After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$$ \therefore $$ f

$$ \therefore $$ $${{{f_1}} \over {0.8{f_1}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

$$ \therefore $$ $${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$$ = 0.64

Initial moment of inertia of the system,

$${{{\rm I}_1}}$$ = $${{M{{\left( {2L} \right)}^2}} \over {12}}$$

Final moment of inertia of the system,

I

$$ \therefore $$ $${{M{{\left( {2L} \right)}^2}} \over {12}}$$ = 0.64 $$\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$$

$$ \Rightarrow $$ $${{M{L^2}} \over {3 \times 0.64}}$$ = $${{M{L^2}} \over 3}$$ + $${{M{L^2}} \over 2}$$

$$ \Rightarrow $$ $${M \over {1.92}} - {M \over 3} = {m \over 2}$$

$$ \Rightarrow $$ $${{1.08M} \over {3 \times 1.92}}$$ = $${m \over 2}$$

$$ \Rightarrow $$ $${m \over M}$$ = $${{1.08 \times 2} \over {3 \times 1.92}}$$ = 0.37

3

A musician using an open flute of length 50 cm producess second harmonic sound waves. A person runs towards the musician from another end of hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :

A

666 Hz

B

753 Hz

C

500 Hz

D

333 Hz

Frequency of sound wave produce by flute

= $${{2{V_S}} \over {2\ell }}$$

= $${{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}$$

= 660 Hz

Speed of the observer

= 10km/hr

= 10 $$ \times $$ $${5 \over {18}}$$ m/s

= $${{25} \over 9}$$ m/s

Frequency heard by the observer,

f' = $$\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f$$

= $$\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660$$

= 666 Hz

= $${{2{V_S}} \over {2\ell }}$$

= $${{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}$$

= 660 Hz

Speed of the observer

= 10km/hr

= 10 $$ \times $$ $${5 \over {18}}$$ m/s

= $${{25} \over 9}$$ m/s

Frequency heard by the observer,

f' = $$\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f$$

= $$\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660$$

= 666 Hz

4

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is ƒ_{1}. If the speed of the train is reduced to 17 m/s, the frequency registered is ƒ_{2}. If speed of sound is 340 m/s, then the ratio ƒ_{1}/ƒ_{2} is -

A

19/18

B

20/19

C

21/20

D

18/17

f_{app} = f_{0} $$\left[ {{{{v_2} \pm {v_0}} \over {{v_2} \pm {v_s}}}} \right]$$

f_{1} = f_{0} $$\left[ {{{340} \over {340 - 34}}} \right]$$

f_{2} = f_{0} $$\left[ {{{340} \over {340 - 17}}} \right]$$

$${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$$

f

f

$${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$$

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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*

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Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

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Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

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