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

Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is :

A

$${{ - q} \over {1 + \sqrt 2 }}$$

B

+ q

C

$$-$$ 2q

D

$${{ - \sqrt 2 q} \over {\sqrt 2 + 1}}$$

U = K$$\left[ {{{{q^2}} \over a} + {{Qq} \over a} + {{Qq} \over {a\sqrt 2 }}} \right] = 0$$

$$ \Rightarrow $$ q = $$-$$ Q$$\left[ {1 + {1 \over {\sqrt 2 }}} \right]$$

$$ \Rightarrow $$ Q = $${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$$

2

The resistance of the meter bridge AB in given figure is 4 $$\Omega $$. With a cell of emf $$\varepsilon $$ = 0.5 V and rheostat
resistance R_{h} = 2 $$\Omega $$ the null point is obtained at some point J. When the cell is replaced by another one of emf $$\varepsilon $$ = $$\varepsilon $$_{2} the same null point J is found for R_{h} = 6 $$\Omega $$. The emf $$\varepsilon $$_{2} is, :

A

0.3 V

B

0.6 V

C

0.5 V

D

0.4 V

Potential gradient with R_{h} = 2$$\Omega $$

is $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$$ L $$=$$ 100 cm

Let null point be at $$\ell $$ cm

thus $$\varepsilon $$_{1} $$=$$ 0.5V $$=$$ $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell $$ . . .(1)

Now with R_{h} $$=$$ 6$$\Omega $$ new potential gradient is

$$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$$ and at null point

$$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$$ . . .(2)

dividing equation (1) by (2) we get

$${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$$ thus $${\varepsilon _2} = 0.3$$

is $$\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$$ L $$=$$ 100 cm

Let null point be at $$\ell $$ cm

thus $$\varepsilon $$

Now with R

$$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$$ and at null point

$$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$$ . . .(2)

dividing equation (1) by (2) we get

$${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$$ thus $${\varepsilon _2} = 0.3$$

3

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 $$\Omega $$ resistor is connected in series with R_{1}, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R_{1} + 10) $$\Omega $$ such that the null point shifts back to its initial position is :

A

40 $$\Omega $$

B

30 $$\Omega $$

C

20 $$\Omega $$

D

60 $$\Omega $$

$${{{R_1}} \over {{R_2}}} = {2 \over 3}\,\,$$ . . .(i)

$${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$$ . . .(ii)

$${{2{R_2}} \over 3} + 10 = {R_2}$$

$$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega $$

& $${R_1} = 20\Omega $$

$${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$$

$$R = 60\,\Omega $$

$${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$$ . . .(ii)

$${{2{R_2}} \over 3} + 10 = {R_2}$$

$$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega $$

& $${R_1} = 20\Omega $$

$${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$$

$$R = 60\,\Omega $$

4

Determine the electric dipole moment of the system of the three charges, placed on the vertices of an equilateral triangle, as shown in the figure :

A

$$2q\ell \widehat j$$

B

$$\left( {q\ell } \right){{\widehat i + \widehat j} \over {\sqrt 2 }}$$

C

$$\sqrt 3 \,q\ell {{\widehat j - \widehat i} \over {\sqrt 2 }}$$

D

$$ - \sqrt 3 \,q\ell \widehat j$$

$$\left| {{P_1}} \right| = $$ q(d)

$$\left| {{P_2}} \right| = $$ qd

|Resultant| $$=$$ 2 P cos30

2 qd$$\left( {{{\sqrt 3 } \over 2}} \right)$$ = $$\sqrt 3 $$ qd

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (5) *keyboard_arrow_right*

AIEEE 2003 (6) *keyboard_arrow_right*

AIEEE 2004 (4) *keyboard_arrow_right*

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AIEEE 2011 (2) *keyboard_arrow_right*

AIEEE 2012 (3) *keyboard_arrow_right*

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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*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*