Current Electricity — Page 16 | JEE Physics

Q151

In a meter bridge experiment null point is obtained at

20

cm

, from one end of the wire when resistance

X

is balanced against another resistance

Y.

If

X < Y

, then where will be the new position of the null point from the same end, if one decides to balance a resistance of

4

X

against

Y

A

40

cm

B

80

cm

C

50

cm

D

70

cm

Correct Answer

Option C

Solution

In the first case

{X \over Y} = {{20} \over {80}} = {1 \over 4}

In the second case

{{4X} \over Y} = {\ell \over {100 - \ell }} \Rightarrow \ell = 50

Q152

In a Wheatstone's bridge, three resistance

P, Q

and

R

connected in the three arms and the fourth arm is formed by two resistances

{S_1}

and

{S_2}

connected in parallel. The condition for the bridge to be balanced will be

A

{P \over Q} = {{2R} \over {{S_1} + {S_2}}}

B

{P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {{S_1}{S_2}}}

C

{P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {2{S_1}{S_2}}}

D

{P \over Q} = {R \over {{S_1} + {S_2}}}

Correct Answer

Option B

Solution

{P \over Q} = {R \over S}

where

S = {{{S_1}{S_2}} \over {{S_1} + {S_2}}}

Q153

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :

A 0.4%

B 0.2%

C 0.8%

D 0.6%

Correct Answer

Option C

Solution

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

Also volume will remain constant i.e., Al = constant

\Rightarrow A \propto {1 \over l}

$\therefore$

R \propto {l^2}

{{\Delta R} \over R} = 2{{\Delta l} \over l} = 0.8

Q154

The length of a given cylindrical wire is increased by

100\%

. Due to the consequent decrease in diameter the change in the resistance of the wire will be

A

200\%

B

100\%

C

50\%

D

300\%

Correct Answer

Option D

Solution

{R_f} = {n^2}{R_1}

Here

n=2

(length becomes twice) $\therefore$

{R_f} = 4{R_i}

New resistance

=400

of

{R_i}

$\therefore$ Increase

= 300\%

Q155

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then :

A P1 = 4W, P2 = 16 W

B P1 = 16W, P2 = 4 W

C P1 = 9W, P2 = 16 W

D P1 = 16W, P2 = 9 W

Correct Answer

Option B

Solution

{R_1} = {{{{220}^2}} \over {25}}

{R_2} = {{{{220}^2}} \over {100}}

L = {{220} \over {{R_1} + {R_2}}}

{P_1} = {i^2}\,{R_1}

{P_2} = {i^2}\,\,({R_2}\, = \,4W)

= {{{{220}^2}} \over {\left( {{{{{220}^2}} \over {25}} + {{{{220}^2}} \over {100}}} \right)}} \times {{{{220}^2}} \over {25}}

= {{400} \over {25}} = 16W

Q156

A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when :-

A R = 1000 r

B R = r

C R = 2r

D R = 0.001 r

Correct Answer

Option B

Solution

Current i =

{E \over {r + R}}

Power generated in R P = i2R

P = {{{E^2}R} \over {{{\left( {r + R} \right)}^2}}}

For maximum power

{{dP} \over {dR}} = 0

{E^2}\left[ {{{{{\left( {r + R} \right)}^2} \times 1 - R \times 2(r + R)} \over {{{\left( {r + R} \right)}^4}}}} \right] = 0

$\Rightarrow$ r = R or R = r

Q157

The resistance of bulb filmanet is

100\Omega

at a temperature of

{100^ \circ }C.

If its temperature coefficient of resistance be

0.005

per

^ \circ C

, its resistance will become

200\,\Omega

at a temperature of

A

{300^ \circ }C

B

{400^ \circ }C

C

{500^ \circ }C

D

{200^ \circ }C

Correct Answer

Option B

Solution

R{}_1 = {R_0}\left[ {1 + \alpha \times 100} \right] = 100\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

{R_2} = {R_0}\left[ {1 + \alpha \times T} \right] = 200\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

On dividing we get

{{200} \over {100}} = {{1 + \alpha T} \over {1 + 100\alpha }}

\Rightarrow 2 = {{1 + 0.005T} \over {1 + 100 \times 0.005}}

\Rightarrow T = {400^ \circ }C

NOTE : We may use this expression as an approximation because the difference in the answers is appreciable.

For accurate results one should use

R = {R_0}{e^{\alpha \Delta T}}

Q158

Two resistors 400

\Omega

and 800

\Omega

are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k

\Omega

across 400

\Omega

resistor is close to :

A 2.05 V

B 1.95 V

C 2 V

D 1.8 V

Correct Answer

Option B

Solution

i = {6 \over {800 + {{400 \times 10000} \over {400 + 10000}}}}

i = {6 \over {800 + {{40000} \over {104}}}}

i = {6 \over {800 + 384.61}} = {6 \over {1184.61}} = 0.00506

V400 = i1 $\times$ 400 =

\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)i \times 400

=

\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)\left( {0.00506} \right) \times 400

= 1.95 V

Q159

Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :

A 240 W

B 60 W

C 30 W

D 120 W

Correct Answer

Option A

Solution

In series condition, equivalent resistance is 2R thus power consumed is 60W =

{{{\varepsilon ^2}} \over {2R}}

In parallel condition, equivalent resistance is R/2 thus new power is P' =

{{{\varepsilon ^2}} \over {\left( {R/2} \right)}}

or P' = 4P = 240W