Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Two electric bulbs marked $$25W$$ $$-$$ $$220$$ $$V$$ and $$100W$$ $$-$$ $$220V$$ are connected in series to a $$440$$ $$V$$ supply. Which of the bulbs will fuse?

A

Both

B

$$100$$ $$W$$

C

$$25$$ $$W$$

D

Neither

The current upto which bulb of marked $$25W$$-$$220V,$$ will

not fuse $${I_1} = {{{W_1}} \over {{V_1}}} = {{25} \over {220}}Amp$$

Similarly, $${I_2} = {{{W_2}} \over {{V_2}}} = {{100} \over {220}}\,Amp$$

The current flowing through the circuit

$$I = {{440} \over {{{\mathop{\rm R}\nolimits} _{eff}}}},\,\,{{\mathop{\rm R}\nolimits} _{eff}} = {R_1} + {R_2}$$

$${R_1} = {{V_1^2} \over {{P_1}}} = {{{{\left( {220} \right)}^2}} \over {25}};\,\,\,$$

$${R_2} = {{V_2^2} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}$$

$$I = {{440} \over {{{{{\left( {220} \right)}^2}} \over {25}} + {{{{\left( {220} \right)}^2}} \over {100}}}}$$

$$ = {{440} \over {{{\left( {220} \right)}^2}\left[ {{1 \over {25}} + {1 \over {100}}} \right]}}$$

$$I = {{40} \over {220}}\,\,Amp$$

as $${I_1}\left( { = {{25} \over {220}}A} \right) < I\left( { = {{40} \over {220}}A} \right) < {I_2}\left( { = {{100} \over {200}}A} \right)$$

Thus the bulb marked $$25W$$-$$220$$ will fuse.

not fuse $${I_1} = {{{W_1}} \over {{V_1}}} = {{25} \over {220}}Amp$$

Similarly, $${I_2} = {{{W_2}} \over {{V_2}}} = {{100} \over {220}}\,Amp$$

The current flowing through the circuit

$$I = {{440} \over {{{\mathop{\rm R}\nolimits} _{eff}}}},\,\,{{\mathop{\rm R}\nolimits} _{eff}} = {R_1} + {R_2}$$

$${R_1} = {{V_1^2} \over {{P_1}}} = {{{{\left( {220} \right)}^2}} \over {25}};\,\,\,$$

$${R_2} = {{V_2^2} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}$$

$$I = {{440} \over {{{{{\left( {220} \right)}^2}} \over {25}} + {{{{\left( {220} \right)}^2}} \over {100}}}}$$

$$ = {{440} \over {{{\left( {220} \right)}^2}\left[ {{1 \over {25}} + {1 \over {100}}} \right]}}$$

$$I = {{40} \over {220}}\,\,Amp$$

as $${I_1}\left( { = {{25} \over {220}}A} \right) < I\left( { = {{40} \over {220}}A} \right) < {I_2}\left( { = {{100} \over {200}}A} \right)$$

Thus the bulb marked $$25W$$-$$220$$ will fuse.

2

MCQ (Single Correct Answer)

If a wire is stretched to make it $$0.1\% $$ longer, its resistance will:

A

increase by $$0.2\% $$

B

decrease by $$0.2\% $$

C

decrease by $$0.05\% $$

D

increase by $$0.05\% $$

Resistance of wire

$$R = {{\rho l} \over A} = {{\rho {l^2}} \over C}$$ (where $$Al=C$$ )

$$\therefore$$ Fractional charge in resistance

$${{\Delta R} \over R} = 2{{\Delta l} \over l}$$

$$\therefore$$ Resistance will increase by $$0.2\% $$

$$R = {{\rho l} \over A} = {{\rho {l^2}} \over C}$$ (where $$Al=C$$ )

$$\therefore$$ Fractional charge in resistance

$${{\Delta R} \over R} = 2{{\Delta l} \over l}$$

$$\therefore$$ Resistance will increase by $$0.2\% $$

3

MCQ (Single Correct Answer)

Two conductors have the same resistance at $${0^ \circ }C$$ but their temperature coefficients of resistance are $${\alpha _1}$$ and $${\alpha _2}.$$ The respective temperature coefficients of their series and parallel combinations are nearly

A

$${{{\alpha _1} + {\alpha _2}} \over 2},\,{\alpha _1} + {\alpha _2}$$

B

$${\alpha _1} + {\alpha _2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$

C

$${\alpha _1} + {\alpha _2},\,{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}$$

D

$${{{\alpha _1} + {\alpha _2}} \over 2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$

$${R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];$$

$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$

$$R = {R_1} + {R_2}$$

$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$

$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$

$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$

$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$

$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$

$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$

$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$

$$R = {R_1} + {R_2}$$

$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$

$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$

$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$

$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$

$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$

$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$

$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$

4

MCQ (Single Correct Answer)

Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $${t_2}$$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $${t_1}/{t_2}$$ will be

A

$$1$$

B

$${1 \over 2}$$

C

$${1 \over 4}$$

D

$$2$$

Initial energy of capacitor, $${E_1} = {{q_1^2} \over {2C}}$$

Final energy of capacitor, $${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$$

$$\therefore$$ $${t_1}=$$ time for the charge to reduce to $${1 \over {\sqrt 2 }}$$ of its initial value

and $${t_2} = $$ time for the charge to reduce to $${1 \over 4}$$ of its initial value

We have, $${q_2} = {q_1}{e^{ - t/CR}}$$

$$ \Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$$

$$\therefore$$$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$$

and $$\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$$

By $$(1)$$ and $$(2),$$ $${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$$

$$ = {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$$

Final energy of capacitor, $${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$$

$$\therefore$$ $${t_1}=$$ time for the charge to reduce to $${1 \over {\sqrt 2 }}$$ of its initial value

and $${t_2} = $$ time for the charge to reduce to $${1 \over 4}$$ of its initial value

We have, $${q_2} = {q_1}{e^{ - t/CR}}$$

$$ \Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$$

$$\therefore$$$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$$

and $$\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$$

By $$(1)$$ and $$(2),$$ $${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$$

$$ = {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$$

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