Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

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

2

Two electric dipoles, A, B with respective dipole moments $${\overrightarrow d _A} = - 4qai$$ and $${\overrightarrow d _B} = - 2qai$$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is -

A

$${{\sqrt 2 R} \over {\sqrt 2 + 1}}$$

B

$${R \over {\sqrt 2 + 1}}$$

C

$${{\sqrt 2 R} \over {\sqrt 2 - 1}}$$

D

$${R \over {\sqrt 2 - 1}}$$

V $$ = {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$$

$$\sqrt 2 x = R + x$$

$$x = {R \over {\sqrt 2 - 1}}$$

dist $$ = {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$$

$$\sqrt 2 x = R + x$$

$$x = {R \over {\sqrt 2 - 1}}$$

dist $$ = {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$$

3

Charges –q and +q located at A and B, respectively, constitude an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP' = $$\left( {{y \over 3}} \right)$$, the force on Q will be close to - $$\left( {{y \over 3} > > 2a} \right)$$

A

9F

B

3F

C

F/3

D

27F

Electric field of equitorial plane of dipole

$$ = - {{K\overrightarrow P } \over {{r^3}}}$$

$$ \therefore $$ At P, F $$ = - {{K\overrightarrow P } \over {{r^3}}}$$Q.

At P^{1} , F^{1} $$ = - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$$

$$ = - {{K\overrightarrow P } \over {{r^3}}}$$

$$ \therefore $$ At P, F $$ = - {{K\overrightarrow P } \over {{r^3}}}$$Q.

At P

4

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R_{1} has the colour code (Orange, Red, Brown). The resistors R_{2} and R_{4} are 80$$\Omega $$ and 40$$\Omega $$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R_{3}_{}, would be -

A

Brown, Blue, Brown

B

Grey, Black, Brown

C

Red, Green, Brown

D

Brown, Blue, Black

R_{1} = 32 $$ \times $$ 10 = 320

for wheat stone bridge

$$ \Rightarrow $$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$

$${{320} \over {{R_3}}} = {{80} \over {40}}$$

$${R_3} = 160$$

$$ \therefore $$ Correct answer is Brown Blue Brown

for wheat stone bridge

$$ \Rightarrow $$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$

$${{320} \over {{R_3}}} = {{80} \over {40}}$$

$${R_3} = 160$$

$$ \therefore $$ Correct answer is Brown Blue Brown

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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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