Current Electricity — Page 9 | JEE Physics

Q81

An electric cable of copper has just one wire of radius 9 mm. Its resistance is 14

\Omega

. If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius 3 mm connected in parallel, then the new resistance of the combination will be :

A 9

\Omega

B 18

\Omega

C 28

\Omega

D 126

\Omega

Correct Answer

Option B

Solution

Initially, copper wire radius (r1) = 9 mm Resistance (R) = 14

\Omega

We know,

R = {{\rho L} \over A} = {{\rho L} \over {\pi r_1^2}} = 14

Now this copper wire is replaced by 7 parallel copper wire of resistance R1.

$\therefore$ Equivalent resistance of 7 parallel copper wire,

{1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}} + {1 \over {{R_1}}}

\Rightarrow {1 \over {{R_{eq}}}} = {7 \over {{R_1}}}

\Rightarrow {R_{eq}} = {{{R_1}} \over 7}

= {1 \over 7} \times {{\rho L} \over {\pi r_2^2}}

= {1 \over 7} \times {{\rho L} \over {\pi {{\left( {{{{r_1}} \over 3}} \right)}^2}}}

[as

{r_2} = {{{r_1}} \over 3}

;

{r_1} = 9

m and

{r_2} = 3

]

= {1 \over 7} \times {{\rho L} \over {\pi r_1^2}} \times 9

= {1 \over 7} \times 14 \times 9

= 18\,\Omega

Q82

Which of the following physical quantities have the same dimensions?

A Electric displacement

(\overrightarrow{\mathrm{D}})

and surface charge density

B Displacement current and electric field

C Current density and surface charge density

D Electric potential and energy

Correct Answer

Option A

Solution

Electric displacement

(\overrightarrow D ) = {\varepsilon _0}\overrightarrow E

\Rightarrow [\overline D ] = [{\varepsilon _0}][\overline E ]

= [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^1}{A^{ - 1}}{T^{ - 3}}]

[\overline D ] = [{L^{ - 2}}{T^1}{A^1}]

[Surface charge density]

= {{[Q]} \over {[A]}}

[\sigma ] = [AT{L^{ - 2}}]

\Rightarrow \overrightarrow D

and

[\sigma ]

have same dimensions

Q83

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor. (B) The drift velocity is inversely proportional to the area of cross-section of given conductor. (C) The drift velocity does not depend on the applied potential difference to the conductor. (D) The drift velocity of electron is inversely proportional to the length of the conductor. (E) The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below :

A (A) and (B) only

B (A) and (D) only

C (B) and (E) only

D (B) and (C) only

Correct Answer

Option B

Solution

We know, Resistivity $\rho=\dfrac{m}{n e^2 \tau}$ Where $\tau$ is relaxation time As temperature $\uparrow, \tau \downarrow, \rho \uparrow, \mathrm{R} \uparrow, \mathrm{i} \downarrow, \mathrm{v}_{\mathrm{d}} \downarrow$ as $i=$ ne $\mathrm{A} v_{\mathrm{d}}$ Statement (A) is correct and hence statement (E) is incorrect.

\mathrm{R}=\rho \frac{l}{A}, I=\frac{V}{R}=\frac{V A}{\rho l}=n e A v_d \rightarrow v_d=\text{ constant }

For a given $\mathrm{V}, v_{\mathrm{d}}$ is independent of $\mathrm{A}$ .

Hence, statement (B) is incorrect Form above $v_d \propto$ V statement (C) is also incorrect.

Since $v_d \propto \dfrac{1}{\ell}$ as seen from above statement, so statement (D) is correct.

Q84

A wire of resistance R1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :

A 9 : 1

B 1 : 9

C 4 : 1

D 3 : 1

Correct Answer

Option A

Solution

Length is increased by twice of its original length.

So, if original length is $l_1$ then final length is $l_2$ = $l_1$ + 2 $l_1$ = 3 $l_1$ .

Then area becomes, A2 =

{{{A_1}} \over 3}

\begin{aligned} & R_1=\frac{p l_1}{A_1} \\\\ & R_2=\frac{p l_2}{A_2}=\frac{p \times 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\\\ & \therefore R_2: R_1=9: 1 \end{aligned}

Q85

Given below are two statements : Statement I : A uniform wire of resistance

80 \,\Omega

is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be

5 \,\Omega

. Statement II: Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio

3: 2

. In the light of the above statements, choose the most appropriate answer from the option given below

A Both statement I and statement II are correct

B Both statement I and statement II are incorrect

C Statement I is correct but statement II is incorrect

D Statement I is incorrect but statement II is correct

Correct Answer

Option C

Solution

Statement I :

{R_{1\,part}} = {{80} \over 4} = 20\,\Omega

\Rightarrow {R_{eff}} = {{20} \over 4} = 5\,\Omega

Statement II : Ratio

= {{{{{{(\Delta V)}^2}} \over {3R}}} \over {{{{{(\Delta V)}^2}} \over {2R}}}}

= {2 \over 3}

Q86

Two metallic wires of identical dimensions are connected in series. If

\sigma_{1}

and

\sigma_{2}

are the conductivities of the these wires respectively, the effective conductivity of the combination is :

A

\dfrac{\sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}

B

\dfrac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}

C

\dfrac{\sigma_{1}+\sigma_{2}}{2 \sigma_{1} \sigma_{2}}

D

\dfrac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}

Correct Answer

Option B

Solution

R = {R_1} + {R_2}

\Rightarrow {{{l_1} + {l_2}} \over {\sigma A}} = {{{l_1}} \over {{\sigma _1}A}} + {{{l_2}} \over {{\sigma _2}A}}

\Rightarrow {2 \over \sigma } = {1 \over {{\sigma _1}}} + {1 \over {{\sigma _2}}}

\Rightarrow \sigma = {{2{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}

Q87

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils. Reason R: Constantan and manganin have very small value of temperature coefficient of resistance. In the light of the above statements, choose the correct answer from the options given below.

A Both A and R are true and R is the correct explanation of A.

B Both A and R are true but R is NOT the correct explanation of A.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option A

Solution

Since they have low temperature coefficient of resistance, their resistance remains almost constant.

Q88

A

1 \mathrm{~m}

long wire is broken into two unequal parts

\mathrm{X}

and

\mathrm{Y}

. The

\mathrm{X}

part of the wire is streched into another wire W. Length of

W

is twice the length of

X

and the resistance of

\mathrm{W}

is twice that of

\mathrm{Y}

. Find the ratio of length of

\mathrm{X}

and

\mathrm{Y}

.

A 1 : 4

B 1 : 2

C 4 : 1

D 2 : 1

Correct Answer

Option B

Solution

\frac{\mathrm{R}_{\mathrm{X}}}{\mathrm{R}_{\mathrm{Y}}}=\frac{\ell_{\mathrm{X}}}{\ell_{\mathrm{Y}}}

When wire is stretched to double of its length, then resistance becomes 4 times

\begin{aligned} &\mathrm{R}_{\mathrm{W}}=4 \mathrm{R}_{\mathrm{X}}=2 \mathrm{R}_{\mathrm{Y}} \\\\ &\frac{\mathrm{R}_{\mathrm{X}}}{\mathrm{R}_{\mathrm{Y}}}=\frac{1}{2} \end{aligned}

So. $\dfrac{\ell_{\mathrm{x}}}{\ell_{\mathrm{y}}}=\dfrac{1}{2}$

Q89

Given below are two statements : One is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

. Assertion A : For measuring the potential difference across a resistance of

600 \Omega

, the voltmeter with resistance

1000 \Omega

will be preferred over voltmeter with resistance

4000 \Omega

. Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance. In the light of the above statements, choose the most appropriate answer from the options given below.

A

\mathbf{A}

is not correct but

\mathbf{R}

is correct

B Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is the correct explanation of

\mathbf{A}

C

\mathbf{A}

is correct but

\mathbf{R}

is not correct

D Both

\mathbf{A}

and

\mathbf{R}

are correct but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

Correct Answer

Option A

Solution

Assertion A is incorrect because the preferred voltmeter for measuring the potential difference across a resistance of 600 ohm is actually the voltmeter with a higher resistance, not a lower resistance.

The reason for this is that when a voltmeter is connected in parallel with the resistance being measured, it will draw current away from the resistance, reducing the potential difference across it.

A higher resistance voltmeter will draw less current and therefore have a smaller effect on the potential difference being measured.

In this case, the voltmeter with resistance 4000 ohm would be preferred because it would draw less current than the voltmeter with resistance 1000 ohm, leading to a more accurate measurement of the potential difference across the resistance of 600 ohm.

Therefore, assertion A is incorrect because it states that the voltmeter with resistance 1000 ohm is preferred, when in reality the voltmeter with resistance 4000 ohm is preferred.

Reason R states that a voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.

This is correct because Ohm's law states that the current through a resistor is proportional to the voltage across it and inversely proportional to the resistance.

Thus, a voltmeter with higher resistance will draw less current, making it a better choice for measuring the potential difference across another resistance.

Q90

Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :

A

\dfrac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^{2}}

B

\dfrac{n^{2} R}{n-1}

C

\dfrac{(n-1) R}{(2 n-1)}

D

\dfrac{(n-1) R}{n}

Correct Answer

Option A

Solution

When, a uniform wire of resistance $\mathrm{R}$ is shaped into a regular $\mathrm{n}$ -sided polygon, the resistance of each side will be,

\frac{\mathrm{R}}{\mathrm{n}}=\mathrm{R}_1

Let $R_1$ and $R_2$ be the resistance between adjacent corners of a regular polygon $\therefore$ The resistance of $(\mathrm{n}-1)$ sides, $\mathrm{R}_2=\dfrac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}$ Since two parts are parallel, therefore,

\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)+\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}} \\\\ & \Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}^2}{\mathrm{n}^2} \times \frac{\mathrm{n}}{\mathrm{R}+\mathrm{nR}-\mathrm{R}} \\\\ &\Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^2} \end{aligned}