Current Electricity — Page 3 | NEET Physics

Q21

The reciprocal of resistance is :

A conductance

B reactance

C mobility

D conductivity

Correct Answer

Option A

Solution

R = {1 \over G}

Thus reciprocal of resistance (R) is conductance (G)

Q22

A cell of emf 4 V and internal resistance 0.5

\Omega

is connected to a 7.5

\Omega

external resistance. The terminal potential difference of the cell is

A 0.375 V

B 3.75 V

C 4.25 V

D 4 V

Correct Answer

Option B

Solution

From Kirchhoff's loop law :

V - Ir - IR = 0

\Rightarrow I = \left( {{V \over {r + R}}} \right)

Terminal potential difference across cell,

{V_{AB}} = IR

{V_{AB}} = {{RV} \over {r + R}}

= {{7.5 \times 4} \over {0.5 + 7.5}} = 3.75\,V

Q23

The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is R 0 , then the potential drop (V) across the resistor R is

A

{{2{V_0}R} \over {2{R_0} + 3R}}

B

{{4{V_0}R} \over {3{R_0} + 16R}}

C

{{4{V_0}R} \over {3{R_0} + R}}

D

{{2{V_0}R} \over {4{R_0} + R}}

Correct Answer

Option B

Solution

Equivalent resistance across point AC

{R_{AC}} = {{{{{R_0}} \over 4} \times R} \over {{{{R_0}} \over 4} + R}} = {{R{R_0}} \over {{R_0} + 4R}}

From voltage divider rule

{V_{AC}} = {{{R_{AC}}} \over {{R_{AC}} + {R_{CB}}}}{V_0} = {{{{R{R_0}} \over {{R_0} + 4R}}{V_0}} \over {{{R{R_0}} \over {{R_0} + 4R}} + {{3{R_0}} \over 4}}}

= {{4R{R_0}{V_0}} \over {4R{R_0} + 3R_0^2 + 12R{R_0}}} = {{4R{V_0}} \over {3{R_0} + 16R}}

Q24

Two resistors of resistance, 100

\Omega

and 200

\Omega

are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100

\Omega

to that in 200

\Omega

in a given time is

A 1 : 2

B 2 : 1

C 1 : 4

D 4 : 1

Correct Answer

Option B

Solution

For parallel combination

P = {{{V^2}} \over R}

{{{P_1}} \over {{P_2}}} = {{{R_2}} \over {{R_1}}}

\Rightarrow {{{P_1}} \over {{P_2}}} = {{200} \over {100}} = {2 \over 1}

Q25

A copper wire of length 10 m and radius

\left( {{{{{10}^{ - 2}}} \over {\sqrt \pi }}} \right)

m has electrical resistance of 10

\Omega

. The current density in the wire for an electric field strength of 10 (V/m) is

A 10 4 A/m 2

B 10 6 A/m 2

C 10

-

5 A/m 2

D 10 5 A/m 2

Correct Answer

Option D

Solution

Resistance,

R = \rho {L \over A} = {L \over {\sigma A}}

\Rightarrow \sigma = {L \over {RA}}

Also current density

j = \sigma E = {{LE} \over {RA}}

J = {{10 \times 10} \over {10 \times \pi {{\left( {{{{{10}^{ - 2}}} \over {\sqrt \pi }}} \right)}^2}}} = {{100} \over {10 \times \pi \times \left( {{{{{10}^{ - 4}}} \over \pi }} \right)}}

= {10^5}

A/m 2

Q26

A wheatstone bridge is used to determine the value of unknown resistance X by adjusting the variable resistance Y as shown in the figure. For the most precise measurement of X, the resistance P and Q

A Should be approximately equal to 2X

B Should be approximately equal and are small

C Should be very large and unequal

D Do not play any significant role

Correct Answer

Option B

Solution

We know, a wheatstone bridge is said to be most precise when it is most sensitive.

This can be done by making ratio arms equal.

Thus (2) is correct option.

Q27

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

A 62 cm

B 60 cm

C 21.6 cm

D 64 cm

Correct Answer

Option B

Solution

$\phi$ = constant E unknown = $\phi$ I b $\Rightarrow$ E unknown $\propto$ I b

{{{E_1}} \over {{E_2}}} = {{{I_1}} \over {{I_2}}} \Rightarrow {{1.5} \over {2.5}} = {{36} \over x}

x = {{36 \times 5} \over 3} = 60

cm

Q28

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25

\Omega

. What will be the effective resistance if they are connected in series?

A 4

\Omega

B 0.25

\Omega

C 0.5

\Omega

D 1

\Omega

Correct Answer

Option A

Solution

{R \over 4}

= .25 R = 1 Now these four resistances are arranged in series R S = R + R + R + R = 4R = 4 $\times$ 1 = 4

\Omega

Q29

Column - I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match column-I and column-II with appropriate relations.<br><br><table> <thead> <tr> <th></th> <th>Column - I</th> <th></th> <th>Column - II</th> </tr> </thead> <tbody> <tr> <td>(A)</td> <td>Drift Velocity</td> <td>(P)</td> <td>

{m \over {n{e^2}p}}

</td> </tr> <tr> <td>(B)</td> <td>Electrical Resistivity</td> <td>(Q)</td> <td>

ne{\upsilon _d}

</td> </tr> <tr> <td>(C)</td> <td>Relaxation Period</td> <td>(R)</td> <td>

{{eE} \over m}\tau

</td> </tr> <tr> <td>(D)</td> <td>Current Density</td> <td>(S)</td> <td>

{E \over J}

</td> </tr> </tbody> </table>

A (A)-(R); (B)-(Q); (C)-(S); (D)-(P)

B (A)-(R); (B)-(S); (C)-(P); (D)-(Q)

C (A)-(R); (B)-(S); (C)-(Q); (D)-(P)

D (A)-(R); (B)-(P); (C)-(S); (D)-(Q)

Correct Answer

Option B

Solution

Current density, J =

{I \over A}

= nev d Drift velocity, V d =

{{eE} \over m}

$\tau$ ; Electrical resistivity, $\rho$ =

{m \over {n{e^2}\tau }}

or $\rho$ =

{E \over J}

Relaxation period,

\tau = {m \over {n{e^2}\rho }}

A $\to$ R B $\to$ S D $\to$ Q C $\to$ P

Q30

A resistance wire connected in the left gap of a metre bridge balances a 10

\Omega

resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m, then the length of 1

\Omega

of the resistance wire is :

A

1.0 \times {10^{ - 1}}m

B

1.5 \times {10^{ - 1}}m

C

1.5 \times {10^{ - 2}}m

D

1.0 \times {10^{ - 2}}m

Correct Answer

Option A

Solution

Initially,

{P \over {10}} = {{{l_1}} \over {{l_2}}} = {3 \over 2}

$\Rightarrow$

P = {{30} \over 2} = 15\Omega

Now Resistance,

R = {{{\rho l}} \over A}

{{{R_1}} \over {{R_2}}} = {{{l_1}} \over {{l_2}}}

$\Rightarrow$

{{15} \over 1} = {{1.5} \over {{l_2}}}

{l_2} = 0.1m

= 1.0 \times {10^{ - 1}}m