Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

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Irrigation

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Fluid Mechanics and Hydraulic Machines

Hydrology

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Engineering Mechanics

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Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

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

2

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

3

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

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

4

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10^{–12} m, the minimum electron energy required is close to -

A

25 keV

B

500 keV

C

100 keV

D

1 keV

$$\lambda $$ = $${h \over p}$$ {$$\lambda $$ = 7.5 $$ \times $$ 10^{$$-$$12}}

P = $${h \over \lambda }$$

KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$$

$$ = {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$$ J

KE = 25 Kev

P = $${h \over \lambda }$$

KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$$

$$ = {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$$ J

KE = 25 Kev

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