Current Electricity — Page 8 | NEET Physics

Q71

A student measures the terminal potential difference (V) of a cell (of emf

\varepsilon

and internal resistance r) as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I, then respectively, equal

A

-

r and

\varepsilon

B r and

-

\varepsilon

C

-

\varepsilon

and r

D

\varepsilon

and

-

r

Correct Answer

Option A

Solution

The terminal potential difference of a cell is given by V + Ir =

\varepsilon

V = V A – V B $\Rightarrow$ V =

\varepsilon

– Ir

\Rightarrow {{dV} \over {dI}} = - r

, Also for, i = 0 then V =

\varepsilon

$\therefore$ slope = – r, intercept =

\varepsilon

Q72

In the circuit shown, the current through the 4

\Omega

resistor is 1 amp when the points P and M are connected to a d.c. voltage source. The potential difference betwen the points M and N is

A 0.5 volt

B 3.2 volt

C 1.5 volt

D 1.0 volt

Correct Answer

Option B

Solution

As the P.D. between 4

\Omega

and 3

\Omega

(in parallel), are the same, 4 × 1 amp = 3 × i 1

\Rightarrow {i_1} = {4 \over 3}A

Total resistance of 4

\Omega

and 3

\Omega

= 12/7

\Omega

. Current in MQP (upper one) =

1 + {4 \over 3} = {7 \over 3}A

$\therefore$

P.D. = {{12} \over 7} \times {7 \over 3} = 4V

Current in MNP =

{4 \over {1.25}} = {{4 \times 4} \over 5} = {{16} \over 5}A

$\therefore$ P.D. across

1\Omega = {{16} \over 5}A \times 1\Omega = {{16} \over 5}volt

$\Rightarrow$ P.D. across 1

\Omega

= 3.2 volt.

Q73

A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively

A both remain the same

B 1.1 times, 1.1 times

C 1.2 times, 1.1 times

D 1.21 times, same

Correct Answer

Option D

Solution

{{\Delta l} \over l} = 0.1

$\therefore$

l = 1.1

but the area also decreases by 0.1. mass =

\rho lA = V\rho

,

\ln l + \ln A = \ln

mass. $\therefore$

{{\Delta l} \over l} + {{\Delta A} \over A} = 0 \Rightarrow {{\Delta l} \over l} = {{ - \Delta A} \over A}

Length increases by 0.1, resistance increases, area decreases by 0.1, then also resistance will increase.

Total increase in resistance is approximately 1.2 times, due to increase in length and decrease in area.

But specific resistance does not change.

Q74

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10

\Omega

. Its internal resistance is

A 2.0 ohm

B zero

C 1.0 ohm

D 0.5 ohm

Correct Answer

Option C

Solution

Here

E > {{ER} \over {R + r}}

, hence the lengths 110 cm and 100 cm are interchanged.

Without being short-circuited through R, only the battery E is balanced.

E = {V \over L} \times {l_1} = {V \over L} \times 110

...(i) When R is connected across E,

Ri = {V \over L} \times {l_2}

\Rightarrow R\left( {{E \over {R + r}}} \right) = {V \over L} \times 100

...(ii) Dividing (i) by (ii), we get

{{R + r} \over R} = {{110} \over {100}}

$\Rightarrow$ 100 R + 100 r = 110 R $\Rightarrow$ 10 R = 100 r $\therefore$

r = {{10R} \over {100}} = {{10 \times 10} \over {100}}

( $\because$ R = 10

\Omega

) $\Rightarrow$ r = 1

\Omega

.

Q75

A current of 3 amp. flows through the 2

\Omega

resistor shown in the circuit. The power dissipated in the 5

\Omega

resistor is

A 1 watt

B 5 watt

C 4 watt

D 2 watt

Correct Answer

Option B

Solution

Clearly, 2

\Omega

, 4

\Omega

and ( 1 + 5)

\Omega

resistors are in parallel.

Hence, potential difference is same across each of them.

$\therefore$ I 1 × 2 = I 2 × 4 = I 3 × 6 Given I 1 = 3A $\therefore$ I 1 × 2 = I 3 × 6 Given I 1 = 3A.

$\therefore$ I 1 × 2 = I 3 × 6 provides

{I_3} = {{{I_1} \times 2} \over 6} = {{3 \times 2} \over 6} = 1A

Now, the potential across the 5

\Omega

resistor is V = I 3 × 5 = 1 × 5 = 5V. $\therefore$ the power dissipated in the 5

\Omega

resistor

P = {{{V^2}} \over R} = {{{5^2}} \over 5} = 5\,watt.

Q76

An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20 o C ? The temperature of boiling water is 100 o C

A 12.6 min

B 4.2 min

C 6.3 min

D 8.4 min

Correct Answer

Option C

Solution

Power = 220 V × 4 A = 880 watts. = 880 J/s. Heat needed to raise the temperature of 1 kg water through 80°C = ms.

\Delta

T × 4.2 J/cal = 1000 g × 1 cal/g × 80 × 4.2 J/cal. $\therefore$ Time taken =

{{1000 \times 1 \times 80 \times 4.2} \over {880}} = {{336 \times {{10}^3}} \over {880}}

= 382 s = 6.3 min.

Q77

The total power dissipated in watt in the circuit shown here is

A 40

B 54

C 4

D 16

Correct Answer

Option B

Solution

Power dissipiated = P

= {{{V^2}} \over R} = {{{{\left( {18} \right)}^2}} \over 6} = 54W

Q78

Three resistances, P, Q, R each of 2

\Omega

and an unknown resistance S from the four arms of a Wheatstone bridge circuit. When a resistance of 6

\Omega

is connected in parallel to S the bridge gets balanced. What is the value of S?

A 3

\Omega

B 6

\Omega

C 1

\Omega

D 2

\Omega

.

Correct Answer

Option A

Solution

A balanced wheatstone bridge simply requires

{P \over Q} = {R \over S} \Rightarrow {2 \over 2} = {2 \over S}

Therefore, S should be 2

\Omega

. A resistance of 6

\Omega

is connected in parallel. In parallel combination,

{1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}

{1 \over 2} = {1 \over 6} + {1 \over S}

\Rightarrow S = 3\Omega

Q79

Power dissipated across the 8

\Omega

resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3

\Omega

resistor is

A 3.0

B 2.0

C 1.0

D 0.5

Correct Answer

Option A

Solution

Power = V . I = I 2 R

{i_2} = \sqrt {{{Power} \over R}} = \sqrt {{2 \over 8}}

=

\sqrt {{1 \over 4}} = {1 \over 2}A

Potential over

8\Omega = R{i_2} = 8 \times {1 \over 2} = 4V

This is the potential over parallel branch. So,

{i_1} = {4 \over 4} = 1A

Power of

3\Omega = {i_1}^2R = 1 \times 1 \times 3 = 3W

Q80

Kirchoff's first and second laws of electrical circuits are consequences of

A conservation of energy and electric charge respectively

B conservation of energy

C conservation of electric charhge and energy respectively

D conservation of electric charge.

Correct Answer

Option C

Solution

Kirchhoff’s first law of electrical circuit is based on conservation of charge and Kirchhoff’s second law of electrical circuit is based on conservation of energy.