Current Electricity — Page 3 | JEE Physics

Q21

The electrochemical equivalent of a metal is

{3.35109^{ - 7}}

kg

per Coulomb. The mass of the metal liberated at the cathode when a

3A

current is passed for

2

seconds will be

A

6.6 \times {10^{57}}/kg

B

9.9 \times {10^{ - 7}}\,kg

C

19.8 \times {10^{ - 7}}\,kg

D

1.1 \times {10^{ - 7}}\,kg

Correct Answer

Option C

Solution

The mass liberated

m,

electrochemical equivalent of a metal

Z,

are related as

m = Zit

\Rightarrow m = 3.3 \times {10^{ - 7}} \times 3 \times 2

= 19.8 \times {10^{ - 7}}\,kg

Q22

The thermo

emf

of a thermocouple varies with temperature

\theta

of the hot junction as

E = a\theta + b{\theta ^2}

in volts where the ratio

a/b

is

{700^ \circ }C.

If the cold junction is kept at

{0^ \circ }C,

then the neutral temperature is

A

{1400^ \circ }C

B

{350^ \circ }C

C

{700^ \circ }C

D No neutral temperature is possible for this termocouple.

Correct Answer

Option D

Solution

Neutral temperature is the temperature of a hot junction at which

E

is maximum.

\Rightarrow {{dE} \over {d\theta }} = 0

or

a + 2b\theta = 0 \Rightarrow \theta = {{ - a} \over {2b}} = - 350

Neutral temperature can never be negative hence no $\theta$ is possible.

Q23

The nagative

Zn

pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by

0.13g

in

30

minutes. If the electrochemical equivalent of

Zn

and

Cu

are

32.5

and

31.5

respectively, the increase in the mass of the positive

Cu

pole in this time is

A

0.180

g

B

0.141

g

C

0.126

g

D

0.242

g

Correct Answer

Option C

Solution

According to Faraday's first law of electrolysis

m = z \times q

For same

q,

\,\,\,\,\,\,\,\,\,

m \propto Z

$\therefore$

{{{m_{Cn}}} \over {{m_{Zn}}}} = {{{Z_{Cu}}} \over {{Z_{Zn}}}}

\Rightarrow {m_{Cu}} = {{{Z_{Cu}}} \over {{Z_{Zn}}}} \times {m_{Zn}}

= {{31.5} \over {32.5}} \times 0.13 = 0.126\,g

Q24

The resistance of hot tungsten filament is about

10

times the cold resistance. What will be resistance of

100

W

and

200

V

lamp when not in use ?

A

20\Omega

B

40\Omega

C

200\Omega

D

400\Omega

Correct Answer

Option B

Solution

P = Vi = {{{V_2}} \over R}

{R_{hot}} = {{{V^2}} \over P} = {{200 \times 200} \over {100}} = 400\Omega

{R_{cold}} = {{400} \over {10}} = 40\Omega

Q25

Two voltmeters, one of copper and another of silver, are joined in parallel. When a total charge

q

flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are

{Z_1}

and

{Z_2}

respectively the charge which flows through the silver voltmeter is

A

{q \over {1 + {{{Z_2}} \over {{Z_1}}}}}

B

{q \over {1 + {{{Z_1}} \over {{Z_2}}}}}

C

q{{{Z_2}} \over {{Z_1}}}

D

q{{{Z_1}} \over {{Z_2}}}

Correct Answer

Option A

Solution

Mass deposited

m = Zq \Rightarrow Z \propto {1 \over q} \Rightarrow {{{Z_1}} \over {{Z_2}}} = {{{q_2}} \over {{q_1}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Also

q = {q_1} + {q_2}\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

\Rightarrow {q \over {{q_2}}} = {{{q_1}} \over {{q_2}}} + 1\,\,\,\,\,\,\,\,\,\,

(\,Dividing\,\,\left( {ii} \right)

by

\left. {{q_2}\,} \right)

\Rightarrow {q_2} = {q \over {1 + {{{q_1}} \over {{q_2}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

From equations

(i)

and

(iii),

{q_2} = {q \over {1 + {{{Z_2}} \over {{z_1}}}}}

Q26

An electric bulb is rated

220

volt -

100

watt. The power consumed by it when operated on

110

volt will be

A

75

watt

B

40

watt

C

25

Watt

D

50

Watt

Correct Answer

Option C

Solution

The resistance of the bulb is

R = {{{V^2}} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}

The power consumed when operated at

110

V

is

P = {{{{\left( {110} \right)}^2}} \over {{{\left( {220} \right)}^2}/100}} = {{100} \over 4} = 25\,W

Q27

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

A four times

B doubled

C halved

D one fourth

Correct Answer

Option B

Solution

H = {{{V^2}t} \over R}

Resistance of half the coil

= {R \over 2}

$\therefore$ As

R

reduces to half,

'H'

will be doubled.

Q28

Two sources of equal

emf

are connected to an external resistance

R.

The internal resistance of the two sources are

{R_1}

and

{R_2}\left( {{R_1} > {R_1}} \right).

If the potential difference across the source having internal resistance

{R_2}

is zero, then

A

R = {R_2} - {R_1}

B

R = {R_2} \times \left( {{R_1} + {R_2}} \right)/\left( {{R_2} - {R_1}} \right)

C

R = {R_1}{R_2}/\left( {{R_2} - {R_1}} \right)

D

R = {R_1}{R_2}/\left( {{R_1} - {R_2}} \right)

Correct Answer

Option A

Solution

{\rm I} = {{2\varepsilon } \over {R + {R_1} + {R_2}}}

Potential difference across second cell

= V = \varepsilon - {\rm I}{R_2} = 0

\varepsilon - {{2\varepsilon } \over {R + {R_1} + {R_2}}}.{R_2} = 0

R + {R_1} + {R_2} - 2{R_2} = 0

R + {R_1} - {R_2} = 0

$\therefore$

R = {R_2} - {R_1}

Q29

A material

'B'

has twice the specific resistance of

'A'.

A circular wire made of

'B'

has twice the diameter of a wire made of

'A'

. Then for the two wires to have the same resistance, the ratio

{l \over B}/{l \over A}

of their respective lengths must be

A

1

B

{l \over 2}

C

{l \over 4}

D

2

Correct Answer

Option D

Solution

{\rho _B} = 2{\rho _A}

{d_B} = 2{d_A}

{R_B} = {R_A} \Rightarrow {{{\rho _B}{\ell _B}} \over {{A_B}}} = {{{P_A}{\ell _A}} \over {{A_A}}}

$\therefore$

{{{\ell _B}} \over {{\ell _A}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{d_B^2} \over {d_A^2}}

= {{{\rho _A}} \over {2{\rho _A}}} \times {{4d_d^2} \over {d_A^2}} = 2

Q30

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}

Correct Answer

Option D

Solution

{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}