Current Electricity — Page 2 | JEE Physics

Q11

A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2

\times

10

-

3 ms

-

1. The number of free electrons in each cubic meter of the wire is ___________.

A 625

\times

1025

B 1

\times

1023

C 2

\times

1025

D 2

\times

106

Correct Answer

Option A

Solution

I = neA{V_d}

n = {I \over {eA{V_d}}}

= {{10} \over {1.6 \times {{10}^{ - 9}} \times 5 \times {{10}^{ - 6}} \times 2 \times {{10}^{ - 3}}}}

= {{{{10}^{25}}} \over {16}} = 6.25 \times {10^{27}} = 625 \times {10^{25}}

Q12

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are

r_{1}

and

r_{2}

\left(r_{1}>r_{2}\right)

. If the potential difference across the source of internal resistance

r_{1}

is zero, then the value of R will be :

A

r_{1}-r_{2}

B

\dfrac{r_{1} r_{2}}{r_{1}+r_{2}}

C

\dfrac{r_{1}+r_{2}}{2}

D

r_{2}-r_{1}

Correct Answer

Option A

Solution

\Delta V = 0 \Rightarrow {{2\varepsilon } \over {{r_1} + {r_2} + R}}{r_1} = \varepsilon

\Rightarrow R = {r_1} - {r_2}

Q13

A 72

\Omega

galvanometer is shunted by a resistance of 8

\Omega

. The percentage of the total current which passes through the galvanometer is :

A 0.1%

B 10%

C 25%

D 0.25%

Correct Answer

Option B

Solution

From the given setup

y \times {R_G} = (x - y)({R_S})

\Rightarrow y \times 72 = (x - y) \times 8

\Rightarrow 9y = x - y

\Rightarrow y = {x \over {10}}

or 10% of x

Q14

Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, is :

A r2

-

r1

B r1

-

r2

C r1

D r2

Correct Answer

Option A

Solution

I = {{2\varepsilon } \over {R + {r_1} + {r_2}}}

As per the question,

{{2\varepsilon } \over {R + {r_1} + {r_2}}} \times {r_2} - \varepsilon = 0

\Rightarrow R = {r_2} - {r_1}

Q15

If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a

A low resistance in parallel

B high resistance in parallel

C high resistance in series

D low resistance in series

Correct Answer

Option C

Solution

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.

The same procedure needs to be done if ammeter is to be used as a voltmeter.

Q16

In a large building, three are

15

bulbs of

40

W

,

5

bulbs of

100

W

,

5

fans of

80

W

and

1

heater of

1

kW.

The voltage of electric mains is

220

V.

The minimum capacity of the main fuse of the building will be:

A

8

A

B

10

A

C

12

A

D

14

A

Correct Answer

Option C

Solution

Total power consumed by electrical appliances in the building,

{P_{total}} = 2500W

Watt $=$ Volt $\times$ ampere

\Rightarrow 2500 = V \times {\rm I}

\Rightarrow 2500 = 220

{\rm I}

\Rightarrow I = {{2500} \over {220}} = 11.36 \approx 12A

(Minimum capacity of main fuse)

Q17

The thermo

e.m.f.

of a thermo -couple is

25

\mu V/{}^ \circ C

at room temperature. A galvanometer of

40

ohm

resistance, capable of detecting current as low as

{10^{ - 5}}\,A,

is connected with the thermo couple. The smallest temperature difference that can be detected by this system is

A

{16^0}C

B

{12^0}C

C

{8^0}C

D

{20^0}C

Correct Answer

Option A

Solution

Let $\theta$ be the smallest temperature difference that can be detected by the thermocouple, then

I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta

where

{\rm I}

is the smallest current which can be detected by the galvanometer of resistance

R.

$\therefore$

{10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta

$\therefore$

\theta = {16^ \circ }C.

Q18

The length of a wire of a potentiometer is

100

cm

, and the

e.

m.

f.

of its standard cell is

E

volt. It is employed to measure the

e.m.f.

of a battery whose internal resistance in

0.5\Omega .

If the balance point is obtained at

1=30

cm

from the positive end, the

e.m.f.

of the battery is where

i

is the current in the potentiometer wire.

A

{{30E} \over {100.5}}

B

{{30E} \over {\left( {100 - 0.5} \right)}}

C

{{30\left( {E - 0.5i} \right)} \over {100}}

D

{{30E} \over {100}} - 0.5i

, where i is the current in the potentiometer wire

Correct Answer

Option D

Solution

Potential gradient along wire, K =

{E \over {100}}

volt/cm For battery V = E' – ir, where E' is emf of battery. or K × 30 = E' – ir, where current i is drawn from battery or

{{E \times 30} \over {100}}

= E' + 0.5i or E' =

{{30E} \over {100}} - 0.5i

Q19

An ammeter reads upto

1

ampere. Its internal resistance is

0.81

ohm

. To increase the range to

10

A

the value of the required shunt is

A

0.03\,\Omega

B

0.3\,\Omega

C

0.9\,\Omega

D

0.09\,\Omega

Correct Answer

Option D

Solution

{i_g} \times G = \left( {i - {i_g}} \right)S

$\therefore$

S = {{{i_g} \times G} \over {i - {i_g}}} = {{1 \times 0.81} \over {10 - 1}} = 0.09\Omega

Q20

Thermistors are usually made of

A metal oxides with high temperature coefficient of resistivity

B metals with high temperature coefficient of resistivity

C metals with low temperature coefficient of resistivity

D semiconducting materials having low temperature

Correct Answer

Option A

Solution

Thermistors are usually made of metal-oxides with high temperature coefficient of resistivity.