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

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B $$(x,z, t) is given by $$-$$ (c is the velocity of light)

A

$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$

B

$${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$

C

$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$

D

$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$

$$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$$

$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$

$$ \therefore $$ $$\overrightarrow K = 6\widehat i + 8\widehat k;$$ direction of waves travel.

i.e., direction of 'c'.

$$ \therefore $$ Direction of $$\widehat B$$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B $$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B = {E \over C} = {{10} \over C}$$

$$ \therefore $$ $$\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$$

$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$

$$ \therefore $$ $$\overrightarrow K = 6\widehat i + 8\widehat k;$$ direction of waves travel.

i.e., direction of 'c'.

$$ \therefore $$ Direction of $$\widehat B$$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B $$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B = {E \over C} = {{10} \over C}$$

$$ \therefore $$ $$\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$$

2

An electromagnetic wave of intensity 50 Wm^{–2} enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

A

$$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$$

B

$$\left( {\sqrt n ,\sqrt n } \right)$$

C

$$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$$

D

$$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$$

C $$ = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}$$

V = $$ = {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$$ [For transparent medium $$\mu $$_{r} $$ \approx $$ $$\mu $$_{0}]

$$ \therefore $$ $${C \over V}$$ $$=$$ $$\sqrt k = $$ n

$${1 \over 2} \in {}_0\,E_0^2$$C $$=$$ intensity $$=$$ $${1 \over 2}$$$$ \in $$_{0} kE^{2}v

$$ \therefore $$ E$$_0^2$$C $$=$$ kE^{2}v

$$ \Rightarrow $$ $${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n $$

similarly

$${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$$

V = $$ = {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$$ [For transparent medium $$\mu $$

$$ \therefore $$ $${C \over V}$$ $$=$$ $$\sqrt k = $$ n

$${1 \over 2} \in {}_0\,E_0^2$$C $$=$$ intensity $$=$$ $${1 \over 2}$$$$ \in $$

$$ \therefore $$ E$$_0^2$$C $$=$$ kE

$$ \Rightarrow $$ $${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n $$

similarly

$${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$$

3

A 27 mW laser beam has a cross-sectional area of 10 mm^{2}. The magnitude of the maximum electric field in this electromagnetic wave is given by :

[Given permittivity of space $$ \in $$_{0} = 9 $$ \times $$ 10^{–12} SI units, Speed of light c = 3 $$ \times $$ 10^{8} m/s]

[Given permittivity of space $$ \in $$

A

2 kV/m

B

1 kV/m

C

1.4 kV/m

D

0.7 kV/m

Intensity of EM wave is given by

$${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$$

$$ = {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$$ $$ = {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$$

$$E = \sqrt 2 \times {10^3}kv/m$$

$$ = 1.4$$ kv/m

$${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$$

$$ = {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$$ $$ = {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$$

$$E = \sqrt 2 \times {10^3}kv/m$$

$$ = 1.4$$ kv/m

4

A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 $$ \times $$ 10^{–6} J/ T when a magnetic intensity of 60 $$ \times $$ 10^{3} A/m is applied. Its magnetic susceptibility is

A

3.3 $$ \times $$ 10^{–4}

B

2.3 $$ \times $$ 10^{–2}

C

4.3 $$ \times $$ 10^{–2}

D

3.3 $$ \times $$ 10^{–2}

x = $${1 \over H}$$

I = $${{Magnetic\,moment} \over {Volume}}$$

I = $${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$$ = 20 N/m^{2}

x = $${{20} \over {60 \times {{10}^{ + 3}}}}$$ = $${1 \over 3} \times {10^{ - 3}}$$

= 0.33 $$ \times $$ 10^{$$-$$3} = 3.3 $$ \times $$ 10^{$$-$$4}

I = $${{Magnetic\,moment} \over {Volume}}$$

I = $${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$$ = 20 N/m

x = $${{20} \over {60 \times {{10}^{ + 3}}}}$$ = $${1 \over 3} \times {10^{ - 3}}$$

= 0.33 $$ \times $$ 10

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