Joint Entrance Examination

Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 $$\Omega $$. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 $$\Omega $$. Then we can conclude :

A

Resistance of galvanometer is 200 $$\Omega $$

B

Full scale deflection current is 2 mA.

C

Current sensitivity of galvanometer is 20 $$\mu $$A/division.

D

Resistance required on R.B. for a deflection of 10 divisions is 9800 $$\Omega $$.

Let full scale deflection of current = I

In case 1, when R = 2400 $$\Omega $$ and deflection of 40 divisions present.

$$ \therefore $$ $${{40} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${4 \over 5}{\rm I}$$ = $${2 \over {G + 2400}}$$ . . .(1)

Incase 2, when R = 4900 $$\Omega $$ and deflection of 20 divisions present

$$ \therefore $$ $${{20} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${2 \over 5}{\rm I}$$ = $${2 \over {G + 4900}}$$ . . .(2)

From (1) and (2) we get,

$${4 \over 2} = {{G + 4900} \over {G + 2400}}$$

$$ \Rightarrow $$ 2G + 4800 $$=$$ G + 4900

$$ \Rightarrow $$ G $$=$$ 100 $$\Omega $$

Putting value of G in equation (1), we get,

$${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$$

$$ \Rightarrow $$ $${\rm I} = 1$$ mA

Current sensitivity $$=$$ $${{\rm I} \over {number\,\,of\,\,divisions}}$$

$$=$$ $${1 \over {50}}$$

$$=$$ 0.02 mA / division

$$=$$ 20 $$\mu $$A / division

Resistance required for deflection of 10 divisions

$${{10} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${1 \over 5} \times 1 \times {10^{ - 3}} = {2 \over {100 + R}}$$

$$ \Rightarrow $$ R $$=$$ 9900 $$\Omega $$

In case 1, when R = 2400 $$\Omega $$ and deflection of 40 divisions present.

$$ \therefore $$ $${{40} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${4 \over 5}{\rm I}$$ = $${2 \over {G + 2400}}$$ . . .(1)

Incase 2, when R = 4900 $$\Omega $$ and deflection of 20 divisions present

$$ \therefore $$ $${{20} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${2 \over 5}{\rm I}$$ = $${2 \over {G + 4900}}$$ . . .(2)

From (1) and (2) we get,

$${4 \over 2} = {{G + 4900} \over {G + 2400}}$$

$$ \Rightarrow $$ 2G + 4800 $$=$$ G + 4900

$$ \Rightarrow $$ G $$=$$ 100 $$\Omega $$

Putting value of G in equation (1), we get,

$${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$$

$$ \Rightarrow $$ $${\rm I} = 1$$ mA

Current sensitivity $$=$$ $${{\rm I} \over {number\,\,of\,\,divisions}}$$

$$=$$ $${1 \over {50}}$$

$$=$$ 0.02 mA / division

$$=$$ 20 $$\mu $$A / division

Resistance required for deflection of 10 divisions

$${{10} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$ $${1 \over 5} \times 1 \times {10^{ - 3}} = {2 \over {100 + R}}$$

$$ \Rightarrow $$ R $$=$$ 9900 $$\Omega $$

2

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$$\Omega $$, it shows full
scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a
voltmeter of range 0 – 10V is:

A

4.005 × 10^{3} $$\Omega $$

B

1.985 × 10^{3} $$\Omega $$

C

2.535 × 10^{3} $$\Omega $$

D

2.045 × 10^{3} $$\Omega $$

Given : Current through the galvanometer,

i_{g} = 5 × 10^{–3} A

Galvanometer resistance, G = 15 $$\Omega $$

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = i_{g} (R + G)

10 = 5 × 10^{–3} (R + 15)

$$ \therefore $$ R = 2000 – 15 = 1985

= 1.985 × 10^{3} $$\Omega $$

i

Galvanometer resistance, G = 15 $$\Omega $$

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = i

10 = 5 × 10

$$ \therefore $$ R = 2000 – 15 = 1985

= 1.985 × 10

3

A magnetic needle of magnetic moment 6.7 $$\times$$ 10^{-2} A m^{2} and moment of inertia 7.5 $$\times$$ 10^{-6} kg m^{2} is
performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

A

8.76 s

B

6.65 s

C

8.89 s

D

6.98 s

Given : Magnetic moment, M = 6.7 × 10^{–2} Am^{2}

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10^{–6} Kgm^{2}

Using, T = $$2\pi \sqrt {{I \over {MB}}} $$

= $$2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $$

= $${{2\pi } \over {10}} \times 1.06$$ s

Time taken for 10 complete oscillations

t = 10T = 2$$\pi $$ × 1.06

= 6.6568 $$ \simeq $$ 6.65 s

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10

Using, T = $$2\pi \sqrt {{I \over {MB}}} $$

= $$2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $$

= $${{2\pi } \over {10}} \times 1.06$$ s

Time taken for 10 complete oscillations

t = 10T = 2$$\pi $$ × 1.06

= 6.6568 $$ \simeq $$ 6.65 s

4

In a certain region static electric and magnetic fields exist. The magnetic field is given by $$\overrightarrow B = {B_0}\left( {\widehat i + 2\widehat j - 4\widehat k} \right)$$ . If a test charge moving with a velocity $$\overrightarrow \upsilon = {\upsilon _0}\left( {3\widehat i - \widehat j + 2\widehat k} \right)$$ experiences no force in that region, then the electric field in the region, in SI units, is :

A

$$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {3\widehat i - 2\widehat j - 4\widehat k} \right)$$

B

$$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {\widehat i + \widehat j + 7\widehat k} \right)$$

C

$$\overrightarrow E = {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$$

D

$$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$$

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

$$\therefore\,\,\,$$ F_{e} + F_{m} = 0

$$\therefore\,\,\,$$ F_{e} = $$-$$ q ($$\overrightarrow v $$ $$ \times $$ $$\overrightarrow B)$$

= $$-$$ qB_{0} $$\upsilon $$_{0} [(3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$) $$ \times $$ ($$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$)]

= $$-$$ q$$\upsilon $$_{0} B_{0} (14$$\widehat j$$ + 7$$\widehat k$$)

Electric field produced by the charge q,

$$\overrightarrow E $$ = $${{\overrightarrow {{F_e}} } \over q}$$

= $${{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}$$

= $$-$$ $$\upsilon $$_{0} B_{0} (14 $${\widehat j}$$ + 7 $${\widehat k}$$)

$$\therefore\,\,\,$$ F

$$\therefore\,\,\,$$ F

= $$-$$ qB

= $$-$$ q$$\upsilon $$

Electric field produced by the charge q,

$$\overrightarrow E $$ = $${{\overrightarrow {{F_e}} } \over q}$$

= $${{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}$$

= $$-$$ $$\upsilon $$

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