Current Electricity — Page 7 | NEET Physics

Q61

A ring is made of a wire having a resistance R 0 = 12

\Omega

. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub circuit between these point is equal to

{8 \over 3}\Omega

.

A

{{{l_1}} \over {{l_2}}} = {5 \over 8}

B

{{{l_1}} \over {{l_2}}} = {1 \over 3}

C

{{{l_1}} \over {{l_2}}} = {3 \over 8}

D

{{{l_1}} \over {{l_2}}} = {1 \over 2}

Correct Answer

Option D

Solution

Let x is the resistance per unit length then equivalent resistance

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

{{\left( {{x_1}{l_1}} \right)\left( {{x_2}{l_2}} \right)} \over {x{l_1} + x{l_2}}}

\Rightarrow {8 \over 3} = x{{{l_1}} \over {{{{l_1}} \over {{l_2}}} + 1}}

...(i) also

R_0 = xl_1 + xl_2

12 = x(l_1 + l_2)

12 = x{l_2}\left( {{{{l_1}} \over {{l_2}}} + 1} \right)

...(ii) Now

{{\left( i \right)} \over {\left( {ii} \right)}} \Rightarrow {{{8 \over 3}} \over {{{12} \over 1}}} = {{{{x{l_1}} \over {\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}}} \over {x{l_2}\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}}

= {{{l_1}} \over {{l_2}{{\left( {{{{l_1}} \over {{l_2}}} + 1} \right)}^2}}}

{\left( {{{{l_1}} \over {{l_2}}} + 1} \right)^2} \times {8 \over {36}} = {{{l_1}} \over {{l_2}}}

\left( {{y^2} + 1 + 2y} \right) \times {8 \over {36}} = y

(where y =

{{{{l_1}} \over {{l_2}}}}

) 8y 2 + 8 + 16y = 36y $\Rightarrow$ 8y 2 – 20y + 8 = 0 $\Rightarrow$ 2y 2 – 5y + 2 = 0 $\Rightarrow$ 2y 2 – 4y – y + 2 = 0 $\Rightarrow$ 2y (y – 2) – 1(y – 2) = 0 $\Rightarrow$ (2y – 1) (y – 2) = 0 $\Rightarrow$

y = {{{l_1}} \over {{l_2}}} = {1 \over 2}

or 2

Q62

In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is

A +1 V

B

-

1 V

C +2 V

D

-

2 V

Correct Answer

Option A

Solution

Current from D to C = 1A $\therefore$ V D – V C = 2 × 1 = 2V V A = 0 $\therefore$ V C = 1V, $\therefore$ V D – V C = 2 $\Rightarrow$ V D – 1 = 2 $\therefore$ V D = 3V $\therefore$ V D – V B = 2 $\therefore$ 3 – V B = 2 $\therefore$ V B = 1V

Q63

A thermocouple of negligible resistance produces an e.m.f. of 40 µV/ºC in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 µA/division, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be

A 1ºC

B 0.5 ºC

C 0.1ºC

D 0.25ºC

Correct Answer

Option D

Solution

For minimum deflection of 1 division required current = 1 µA $\Rightarrow$ Voltage required = IR = (1µA) (10) = 10 µV $\therefore$ 40 µV ≡ 1ºC $\Rightarrow$ 10 µV ≡

{1 \over 4}

ºC = 0.25ºC

Q64

A current of 2 A flows through a 2

\Omega

resistor when connected across a battery a 2

\Omega

resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9

\Omega

resistor. The internal resistance of the battery is

A 0.5

\Omega

B 1/3

\Omega

C 1/4

\Omega

D 1

\Omega

Correct Answer

Option B

Solution

Let

\varepsilon

be the emf and r be internal resistance of the battery. In the first case,

2 = {\varepsilon \over {2 + r}}

In the second case,

0.5 = {\varepsilon \over {9 + r}}

Divide (i) by (ii), we get

{2 \over {0.5}} = {{9 + r} \over {2 + r}} \Rightarrow 4 + 2r = 4.5 + 0.5r

1.5r = 0.5 \Rightarrow r = {{0.5} \over {1.5}} = {1 \over 3}\Omega

Q65

If power dissipated in the 9

\Omega

resistor in the circuit shown is 36 watt, the potential difference across the 2

\Omega

resistor is

A 4 volt

B 8 volt

C 10 volt

D 2 volt

Correct Answer

Option C

Solution

We have,

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

\Rightarrow 36 = {{{V^2}} \over 9}

\Rightarrow V = 18

Current passing through the 9

\Omega

resistor is

{i_1} = {V \over R} = {{18} \over 9} = 2A

The 9

\Omega

and 6

\Omega

resistors are in parallel, therefore

{i_1} = {6 \over {9 + 6}} \times i

where i is the current delivered by the battery.

\therefore i = {{2 \times 15} \over 6} = 5A

Thus, potential difference across 2

\Omega

resistor is V = iR = 5 × 2 = 10V

Q66

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at length

l

1 cm and

l

2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

A

k({l_2} - {l_1})

and

k{l_2}

B

k{l_1}

and

k({l_2} - {l_1})

C

k\left( {{l_2} - {l_1}} \right)

and

k{l_1}

D

k{l_1}

and

k{l_2}

Correct Answer

Option B

Solution

(i) When key between the terminals 1 and 2 is plugged in, P.D. across R = IR =

k l_1

$\Rightarrow$ R =

k l_1

as I = 1A (ii) When key between terminals 1 and 3 is plugged in, P.D. across (X + R) = I(X + R) =

k l_2

$\Rightarrow$ X + R =

k l_2

$\therefore$ X =

k (l_2 – l_1)

$\therefore$ R =

kl_1

and X =

k (l_2 – l_1)

Q67

Consider the following two statements. (A) Kirchoff's junction law follows from the conservation of charge. (B) Kirchhoff's loop law follows from the conservation of energy. Which of the following is correct?

A Both (A) and (B) are wrong

B (A) is correct and (B) is wrong

C (A) is wrong and (B) is correct

D Both (A) and (B) are correct

Correct Answer

Option D

Solution

Kirchhoff’s junction law or Kirchhoff’s first law is based on the conservation of charge.

Kirchhoff’s loop law or Kirchhoff’s second law is based on the conservation of energy.

Hence both statements (A) and (B) are correct.

Q68

A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points, A and B as shown in the figure is

A

3\,\Omega

B

6\pi \,\Omega

C

6\,\Omega

D

0.6\pi \,\Omega

Correct Answer

Option D

Solution

The resistance of length 2 $\pi$ R is 12

\Omega

. Hence the resistance of length $\pi$ R is 6

\Omega

. Thus two resistances of 6

\Omega

can be represented as shown in fig. $\therefore$ Equivalent resistance R =

{{6 \times 6} \over {12}} = 3

Q69

The mean free path of electrons in a metal is 4

\times

10

-

8 m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units V/m

A

5 \times {10^{ - 11}}

B 8

\times

10

-

11

C 5

\times

10 7

D 8

\times

10 7

Correct Answer

Option C

Solution

Given, Energy = 2 eV = eE $\lambda$ $\therefore$ E =

{{2eV} \over {e\lambda }} = {2 \over {4 \times {{10}^{ - 8}}}} = 5 \times {10^7}

V/m

Q70

See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it ?

A

{\varepsilon _2} - {i_2}{r_2} - {\varepsilon _1} - {i_1}{r_1} = 0

B

- {\varepsilon _2} - \left( {{i_1} + {i_2}} \right)R + {i_2}{r_2} = 0

C

{\varepsilon _1} - \left( {{i_1} + {i_2}} \right)R + {i_1}{r_1} = 0

D

{\varepsilon _1} - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} = 0

Correct Answer

Option D

Solution

Applying Kirchhoff’s equation to the loop ABFE, – (i 1 + i 2 )R – i 1 r 1 +

{\varepsilon _1}

= 0 $\Rightarrow$

{\varepsilon _1}

– (i 1 + i 2 )R – i 1 r 1 = 0