Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

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Reinforced Cement Concrete

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General Aptitude

1

A resistance is shown in the figure. Its value and tolerance are given respectively by :

A

270 $$\Omega $$, 10 %

B

27 k$$\Omega $$, 10 %

C

27 k$$\Omega $$, 20 %

D

270 $$\Omega $$, 5 %

From color code table :

For Red value is 2

For Violet value is 7

For Orange multiplier is 10^{3}

For Silver tolarence is 10%

$$ \therefore $$ Resistance and tolerance is

= 27 $$ \times $$ 10^{3} $$ \pm $$ 10%

= 27 k$$\Omega $$ $$ \pm $$ 10%

For Red value is 2

For Violet value is 7

For Orange multiplier is 10

For Silver tolarence is 10%

$$ \therefore $$ Resistance and tolerance is

= 27 $$ \times $$ 10

= 27 k$$\Omega $$ $$ \pm $$ 10%

2

Two point charges q_{1}$$\left( {\sqrt {10} \mu C} \right)$$ and q_{2}($$-$$ 25 $$\mu $$C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,

[take $${1 \over {4\pi { \in _0}}}$$ = 9 $$ \times $$ 10^{9} Nm^{2}C^{$$-$$2}]

[take $${1 \over {4\pi { \in _0}}}$$ = 9 $$ \times $$ 10

A

$$\left( {63\widehat i - 27\widehat j} \right) \times {10^2}$$

B

$$\left( { - 63\widehat i + 27\widehat j} \right) \times {10^2}$$

C

$$\left( {81\widehat i - 81\widehat j} \right) \times {10^2}$$

D

$$\left( { - 81\widehat i + 81\widehat j} \right) \times {10^2}$$

Electric field due to $$\sqrt {10} \,\mu C$$ charge :

$$\overrightarrow {{E_1}} = - $$ E

Where,

E

$$ = 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}$$

$$ = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m$$

sin $$\theta $$

and cos$$\theta $$

$$ \therefore $$ $$\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)$$

$$ = 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)$$

Electric field due to $$-$$ 25 $$\mu $$C charge,

$$\overrightarrow {{E_2}} = $$ E

where

E

$$ = 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}$$

$$ = 9 \times {10^3}$$ V/m

sin$$\theta $$

and cos$$\theta $$

$$ \therefore $$ $$\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)$$

$$ = 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)$$

$$ \therefore $$ Net electric field,

$$\overrightarrow E $$ = $${\overrightarrow E _1}$$ + $${\overrightarrow E _2}$$

$$ = \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m$$

3

Charge is distributed within a sphere of radius R with a volume charge density $$\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

A

a log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$

B

$${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$

C

a log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$

D

$${a \over 2}$$ log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$

Volume of this spherical layer,

dv = (4$$\pi $$r

charge present in this layer,

dq = $$\rho $$ (4$$\pi $$r

= $${A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)$$

= $$A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)$$

$$ \therefore $$ Total charge in the sphere,

Q= $$\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr$$

= 4$$\pi $$A$$\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr$$

= 4$$\pi $$A$$\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R$$

= 4$$\pi $$A $$\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)$$

$$ \therefore $$ Q = 2$$\pi $$aA $$\left( {1 - {e^{ - {{2R} \over a}}}} \right)$$

$$ \Rightarrow $$ $${1 - {e^{ - {{2R} \over a}}}}$$ = $${Q \over {2\pi aA}}$$

$$ \Rightarrow $$ $${{e^{ - {{2R} \over a}}}}$$ = 1 $$-$$ $${Q \over {2\pi aA}}$$

$$ \Rightarrow $$ $${e^{{{2R} \over a}}}$$ = $${1 \over {1 - {Q \over {2\pi aA}}}}$$

$$ \Rightarrow $$ $${{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$

$$ \Rightarrow $$ R = $${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$

4

In the given circuit the the internal resistance of the 18 V cell is negligible. If R_{1} = 400 $$\Omega $$, R_{3} = 100 $$\Omega $$ and R_{4} = 500 $$\Omega $$ and the reading of an ideal voltmeter across R_{4} is 5V, then the value of R_{2} will be :

A

300 $$\Omega $$

B

450 $$\Omega $$

C

550 $$\Omega $$

D

230 $$\Omega $$

Voltage accross resistance R_{4} = 5 V

$$ \therefore $$ IR_{4} = 5 V

$$ \Rightarrow $$ 500 $$ \times $$ I = 5

$$ \Rightarrow $$ I = $${1 \over {100}}$$ A

$$ \therefore $$ Voltage across resistor R_{3} = $${1 \over {100}}\left( {100} \right)$$ = 1 A

$$ \therefore $$ Total drop in resistance R_{3} and R_{4} = 5 + 1 = 6V

So, voltage accross R_{2} resistance is also 6V as R_{3}, R_{4} and R_{2} are in parallel

$$ \therefore $$ Voltage accross R_{1} resistor R_{1} resistor = 18 $$-$$ 6 = 12 V

$$ \therefore $$ Current through R_{1} resistor = $${{12} \over {400}}$$ = $${3 \over {100}}$$ A

$$ \therefore $$ Current through R_{2} resistor

= $${3 \over {100}} - {1 \over {100}}$$

= $${2 \over {100}}$$ A

$$ \therefore $$ $$\left( {{2 \over {100}}} \right)$$ R_{2} = 6

$$ \Rightarrow $$ R_{2} = 300 $$\Omega $$

$$ \therefore $$ IR

$$ \Rightarrow $$ 500 $$ \times $$ I = 5

$$ \Rightarrow $$ I = $${1 \over {100}}$$ A

$$ \therefore $$ Voltage across resistor R

$$ \therefore $$ Total drop in resistance R

So, voltage accross R

$$ \therefore $$ Voltage accross R

$$ \therefore $$ Current through R

$$ \therefore $$ Current through R

= $${3 \over {100}} - {1 \over {100}}$$

= $${2 \over {100}}$$ A

$$ \therefore $$ $$\left( {{2 \over {100}}} \right)$$ R

$$ \Rightarrow $$ R

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