Current Electricity — Page 8 | JEE Physics

Q71

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

A 20

\Omega

B 30

\Omega

C 5

\Omega

D 10

\Omega

Correct Answer

Option A

Solution

500 watt at 100 v P = Vi 500 = Vi i = 5 Amp V = i $\times$ R R = 20

Q72

Five identical cells each of internal resistance 1

\Omega

and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?

A 1

\Omega

B 25

\Omega

C 5

\Omega

D 10

\Omega

Correct Answer

Option A

Solution

{i_1} = {{25} \over {5 + R}}

{i_2} = {5 \over {R + {1 \over 5}}}

{i_1} = {i_2} \Rightarrow 5\left( {R + {1 \over 5}} \right) = 5 + R

4R = R R = 1

\Omega

Q73

For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :

A 1

\Omega

B 5

\Omega

C 4

\Omega

D 2

\Omega

Correct Answer

Option D

Solution

{I_{\max }} = {{50} \over 2} = 25

mA

R = {V \over I} = {{50mV} \over {25mA}} = 2\Omega

Q74

A cell of emf 90 V is connected across series combination of two resistors each of 100

\Omega

resistance. A voltmeter of resistance 400

\Omega

is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

A 40 V

B 90 V

C 45 V

D 80 V

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{400 \times 100}{500}+100 \\\\ & =180 \Omega \\\\ & \mathrm{i}=\frac{90}{180}=\frac{1}{2} \mathrm{~A} \\\\ & \text{ Reading }=\frac{1}{2} \times \frac{400 \times 100}{500} \\\\ & =40 \text{ volt } \end{aligned}

Q75

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at

-

10

^\circ

C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 k

\Omega

resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33

\times

105 J kg

-

1, specific heat of ice = 2

\times

103 J kg

-

1 and density of ice = 103 kg/m3

A 0.353 s

B 35.3 s

C 3.53 s

D 70.6 s

Correct Answer

Option B

Solution

Given, the length of the water pipe, L = 1 m The cross-sectional area of the water pipe, A = 1 cm2 = 10 $-$ 4 m2 The temperature of the ice = $-$ 10

^\circ

C Current passing in the conductor, I = 0.5 A Resistance of the conductor, R = 4 k

\Omega

The latent heat of fusion for ice, Lf = 3.33 $\times$ 105 J/kg The density of the ice, d = 1000 kg/m3 The specific heat of the ice, cp, ice = 2 $\times$ 103 J/kg Heat required to melt the ice at 10

^\circ

C to 0

^\circ

C Q = mcp

\Delta

T + mLf $\Rightarrow$ Q = dVcp

\Delta

T + dVLf = 1000 $\times$ 10 $-$ 4 $\times$ 2 $\times$ 103 $\times$ (10) + 1000 $\times$ 10 $-$ 4 $\times$ 3.33 $\times$ 105 ( $\because$ V = A $\times$ L) = 35300 J According to the Joule's law of heating, H = I2Rt $\Rightarrow$ 35300 = (0.5)2(4000) (t) $\Rightarrow$ t = 35.3 s Thus, the minimum time required to melt the ice is 35.3 s.

Q76

Two resistors R1 = (4

\pm

0.8)

\Omega

and R2 = (4

\pm

0.4)

\Omega

are connected in parallel. The equivalent resistance of their parallel combination will be :

A (4

\pm

0.4)

\Omega

B (2

\pm

0.4)

\Omega

C (2

\pm

0.3)

\Omega

D (4

\pm

0.3)

\Omega

Correct Answer

Option C

Solution

Given, R1 = (4 $\pm$ 0.8)

\Omega

R2 = (4 $\pm$ 0.4)

\Omega

Equivalent resistance when the resistors are connected in parallel is given by

{1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} \Rightarrow {1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 4}

{R_{eq}} = 2\,\Omega

Now,

{{\Delta {R_{eq}}} \over {R_{eq}^2}} = {{\Delta {R_1}} \over {R_1^2}} + {{\Delta {R_2}} \over {R_2^2}}

Substituting the values in the above equation, we get

{{\Delta {R_{eq}}} \over 4} = {{0.8} \over {16}} + {{0.4} \over {16}} \Rightarrow \Delta {R_{eq}} = 0.3\,\Omega

$\therefore$ The equivalent resistance in parallel combination is

{R_{eq}} = (2 \pm 0.3)\Omega

.

Q77

If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:

A

{{KS} \over {(S + G)}}

B

{{(G + S)} \over {nKS}}

C

{{nKS} \over {(G + S)}}

D

{{nK(G + S)} \over S}

Correct Answer

Option D

Solution

According to the information, current through galvanometer = nK

\Rightarrow {S \over {S + G}}i = nK

\Rightarrow i = {{nK(S + G)} \over S}

Q78

A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for

{1 \over 3}

deflection in the galvanometer. Which of the below is true for measuring value of G?

A

{1 \over 3}

deflection method cannot be used for determining the resistance of the galvanometer.

B

{1 \over 3}

deflection method can be used and in this case the G equals to twice the value of shunt resistances.

C

{1 \over 3}

deflection method can be used and in this case, the G equals to three times the value of shunt resistances.

D

{1 \over 3}

deflection method can be used and in this case the G value equals to the shunt resistances.

Correct Answer

Option B

Solution

The circuit for the given situation is: Since G and S are in parallel, $\Rightarrow$

{i \over 3}

$\times$ G =

{2i \over 3}

$\times$ S $\Rightarrow$ G = 2S $\Rightarrow$ G equals twice the value of shunt resistance.

Q79

What will be the most suitable combination of three resistors A = 2

\Omega

, B = 4

\Omega

, C = 6

\Omega

so that

\left( {{{22} \over 3}} \right)

\Omega

is equivalent resistance of combination?

A Parallel combination of A and C connected in series with B.

B Parallel combination of A and B connected in series with C.

C Series combination of A and C connected in parallel with B.

D Series combination of B and C connected in parallel with A.

Correct Answer

Option B

Solution

{R_{eq}} = {{2 \times 4} \over {2 + 6}} + 6 = {{22} \over 3}

$\Rightarrow$ A and B are in parallel and C is in series.

Q80

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20

\Omega

. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

A 2.5

\Omega

B 4

\Omega

C 5

\Omega

D 10

\Omega

Correct Answer

Option C

Solution

{{1.5 \times 10} \over {10 + {r \over 2}}} = 1.2

$\Rightarrow$ r = 5

\Omega