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

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Engineering Mathematics

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Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B s slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz, the original frequency of B is :

A

430 Hz

B

420 Hz

C

428 Hz

D

422 Hz

Frequency of B, f_{B} = 425 $$ \pm $$ 5 = 420 or 430 Hz

As tension of string B is increased

So, frequency of B, f_{B} should also increase [as f $$ \propto $$ $$\sqrt T $$]

If initially f_{B} = 430 Hz then when f_{B} increases by increasing the tension then f_{B} $$-$$ f_{A} increases that means beat frequency increase.

So, f_{B} can't be 430 Hz

When f_{B} = 420 then when f_{B} increases f_{A} $$-$$ f_{B} decreases means beat frequency decreases. So, correct f_{B} = 420 Hz.

As tension of string B is increased

So, frequency of B, f

If initially f

So, f

When f

2

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :

A

28 cm

B

32 cm

C

36 cm

D

40 c

Given, End correction (e) = 1 cm

For first resonance,

$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm

For second resonance,

$${3\lambda \over 4} = {l_2} + e$$

$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm

For first resonance,

$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm

For second resonance,

$${3\lambda \over 4} = {l_2} + e$$

$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm

3

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

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

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

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

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

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

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

4

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

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