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

For the P-V diagram given for an ideal gas,

out of the following which one correctly represents the T-P diagram ?

out of the following which one correctly represents the T-P diagram ?

A

B

C

D

We know,

PV = nRT

here R is a constant,

assuming n = number of moles does not change, we get,

PV = KT, K = constant.

Given that, PV = constant,

So, T = constant, Hence the process is isothermol.

From the graph you can see,

Pressure at point 1 is higher than that at point 2. So correct answer is (c)

PV = nRT

here R is a constant,

assuming n = number of moles does not change, we get,

PV = KT, K = constant.

Given that, PV = constant,

So, T = constant, Hence the process is isothermol.

From the graph you can see,

Pressure at point 1 is higher than that at point 2. So correct answer is (c)

2

Two moles of an ideal monatomic gas occupies a volume V at 27^{o}C. The gas expands adiabatically to a
volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

A

(a) 195 K (b) 2.7 kJ

B

(a) 189 K (b) 2.7 kJ

C

(a) 195 K (b) –2.7 kJ

D

(a) 189 K (b) – 2.7 kJ

For adiabatic process,

pv^{$$\gamma $$} = constant.

and we know, pv = nRT

$$\therefore\,\,\,$$ p = $${{nRT} \over v}$$

$$\therefore\,\,\,$$ $${{nRT} \over v} \times {v^\gamma }$$ = constant

$$ \Rightarrow $$$$\,\,\,$$ T v^{$$\gamma $$$$-$$1} = constant.

$$\therefore\,\,\,$$ T_{1} v_{1}^{$$\gamma $$$$-$$1} = T_{2} v_{2}^{$$\gamma $$$$-$$1}

Given that,

This is a monoatomic gas.

So, degree of frequency f = 3

$$\therefore\,\,\,$$ $$\gamma $$ = $${{{c_p}} \over {{c_v}}}$$ = 1 + $${2 \over f}$$ = 1 + $${2 \over 3}$$ = $${5 \over 3}$$

T_{1} = 27 + 273 = 300 K

v_{1} = v

and v_{2} = 2v

$$\therefore\,\,\,$$ T_{2} (2V)$$^{{2 \over 3}}$$ = 300(v)$$^{{2 \over 3}}$$

$$ \Rightarrow $$$$\,\,\,$$ T_{2} = $${{300} \over {{2^{{2 \over 3}}}}}$$ = 189 K

Change in internal energy,

$$\Delta $$U = $${1 \over 2}$$ nfR$$\Delta $$T

= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ 3 $$ \times $$ 8.31 $$ \times $$ (189 $$-$$ 300)

= $$-$$ 2.7 KJ

pv

and we know, pv = nRT

$$\therefore\,\,\,$$ p = $${{nRT} \over v}$$

$$\therefore\,\,\,$$ $${{nRT} \over v} \times {v^\gamma }$$ = constant

$$ \Rightarrow $$$$\,\,\,$$ T v

$$\therefore\,\,\,$$ T

Given that,

This is a monoatomic gas.

So, degree of frequency f = 3

$$\therefore\,\,\,$$ $$\gamma $$ = $${{{c_p}} \over {{c_v}}}$$ = 1 + $${2 \over f}$$ = 1 + $${2 \over 3}$$ = $${5 \over 3}$$

T

v

and v

$$\therefore\,\,\,$$ T

$$ \Rightarrow $$$$\,\,\,$$ T

Change in internal energy,

$$\Delta $$U = $${1 \over 2}$$ nfR$$\Delta $$T

= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ 3 $$ \times $$ 8.31 $$ \times $$ (189 $$-$$ 300)

= $$-$$ 2.7 KJ

3

The mass of a hydrogen molecule is 3.32 $$\times$$ 10^{-27} kg. If 10^{23} hydrogen molecules strike, per second, a fixed wall of area 2 cm^{2} at an angle of 45^{o} to the normal, and rebound elastically with a speed of 10^{3} m/s, then the pressure on the wall is nearly:

A

2.35 $$\times$$ 10^{3} N m^{-2}

B

4.70 $$\times$$ 10^{3} N m^{-2}

C

2.35 $$\times$$ 10^{2} N m^{-2}

D

4.70 $$\times$$ 10^{2} N m^{-2}

Considering one hydrogen molecule :

As collision is elastic so, e = 1

Initial momentum,

$$\overrightarrow {{P_i}} $$ = $${{mv} \over {\sqrt 2 }}$$ $$\widehat i$$ $$-$$ $${{mv} \over {\sqrt 2 }}$$ $$\widehat j$$

Final momentum,

$$\overrightarrow {{P_f}} $$ = $${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$ $$-$$ $${{mv} \over {\sqrt 2 }}\widehat j$$

$$\therefore\,\,\,$$ Change in momentum for single H molecule,

$$\Delta $$P = $$\overrightarrow {{P_f}} $$ $$-$$ $$\overrightarrow {{P_i}} $$

= $${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$

$$\therefore\,\,\,$$ $$\left| {\Delta P} \right|$$ = $${{2mv} \over {\sqrt 2 }}$$

Now for n hydrogen molecule total momentum changes per second,

= $$\left( {{{2mv} \over {\sqrt 2 }}} \right)$$ $$ \times $$ n

As we know,

Force (F) = $${{\Delta P} \over {\Delta t}}$$

= $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ As direction of $$\Delta $$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$$\therefore\,\,\,$$ Force on the wall = $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ Pressure on the wall, P = $${F \over A}$$

= $${{2mv\,n} \over {\sqrt 2 A}}$$

= $${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$$

= 2.35 $$ \times $$ 10^{3} N m^{$$-$$2}

As collision is elastic so, e = 1

Initial momentum,

$$\overrightarrow {{P_i}} $$ = $${{mv} \over {\sqrt 2 }}$$ $$\widehat i$$ $$-$$ $${{mv} \over {\sqrt 2 }}$$ $$\widehat j$$

Final momentum,

$$\overrightarrow {{P_f}} $$ = $${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$ $$-$$ $${{mv} \over {\sqrt 2 }}\widehat j$$

$$\therefore\,\,\,$$ Change in momentum for single H molecule,

$$\Delta $$P = $$\overrightarrow {{P_f}} $$ $$-$$ $$\overrightarrow {{P_i}} $$

= $${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$$

$$\therefore\,\,\,$$ $$\left| {\Delta P} \right|$$ = $${{2mv} \over {\sqrt 2 }}$$

Now for n hydrogen molecule total momentum changes per second,

= $$\left( {{{2mv} \over {\sqrt 2 }}} \right)$$ $$ \times $$ n

As we know,

Force (F) = $${{\Delta P} \over {\Delta t}}$$

= $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ As direction of $$\Delta $$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$$\therefore\,\,\,$$ Force on the wall = $${{{2mv} \over {\sqrt 2 }}}$$ $$ \times $$ n

$$\therefore\,\,\,$$ Pressure on the wall, P = $${F \over A}$$

= $${{2mv\,n} \over {\sqrt 2 A}}$$

= $${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$$

= 2.35 $$ \times $$ 10

4

A Carnot's engine works as a refrigerator between $$250$$ K and $$300$$ K. It receives $$500$$ cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

A

$$420$$ $$J$$

B

$$772$$ $$J$$

C

$$2100$$ $$J$$

D

$$2520$$ $$J$$

Given,

Cold body temperature (T_{2}) = 250 K

Hot body temperature (T_{1}) = 300 K

Received heat (Q_{2}) = 500 cal

Let, required works done = $$\omega $$

$$\therefore\,\,\,\,$$ For the refrigerator,

Efficiency = $$1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ 1 $$-$$ $${{250} \over {300}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $${{{Q_2} + \omega } \over \omega }$$ = $${{300} \over {50}}$$ = 6

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{{Q_2}} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{500 \times 4.2} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = 420\,J$$

Cold body temperature (T

Hot body temperature (T

Received heat (Q

Let, required works done = $$\omega $$

$$\therefore\,\,\,\,$$ For the refrigerator,

Efficiency = $$1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ 1 $$-$$ $${{250} \over {300}} = {\omega \over {{Q_2} + \omega }}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $${{{Q_2} + \omega } \over \omega }$$ = $${{300} \over {50}}$$ = 6

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{{Q_2}} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = {{500 \times 4.2} \over 5}$$

$$ \Rightarrow $$$$\,\,\,\,$$ $$\omega = 420\,J$$

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

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