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Two wires of equal length made of materials of resistivity ratio $1 : 2$ and area of cross-section $3 : 2$ have a potential drop across them is in the ratio $X : Y$ when connected in series. Which of the following is the ratio $X : Y$ ?

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3 : 1

2 : 5

5 : 2

1 : 3

answer is D.

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Detailed Solution

If two wires of equal length made of materials of resistivity ratio

1 : 2

and area of cross-section

3 : 2

have a potential drop across them is in the ratio

X : Y

when connected in series, then the ratio of

X : Y

1 : 3

.
Let the resistance, area of cross-section and resistivity of the wires two be

R_{1}, ρ_{1}, A_{1}

and

R_{2}, ρ_{2}, A_{2}

, respectively.
Hence, it is given that

\frac{ρ_{1}}{ρ_{2}} = \frac{1}{2}

and

\frac{A_{1}}{A_{2}} = \frac{3}{2}

It is known that “the resistance

(R)

of the wire is directly proportional to length of the wire

(l)

and the resistivity

(ρ)

of the material. It is inversely proportional to the area of the cross-section

(A)

.” Mathematically,

R = ρ \frac{l}{A}

Therefore, for wire

1

R_{1} = ρ_{1} \frac{l}{A_{1}} (i)

Similarly, for wire 2,

R_{2} = ρ_{2} \frac{l}{A_{2}} (ii)

On dividing

(i)

and

(ii)

, we get

\Rightarrow \frac{R_{1}}{R_{2}} = \frac{ρ_{1}}{ρ_{2}} \times \frac{A_{2}}{A_{1}}

On substituting the given values,

\Rightarrow \frac{R_{1}}{R_{2}} = \frac{1}{2} \times \frac{2}{3}

\Rightarrow \frac{R_{1}}{R_{2}} = \frac{1}{3}

\Rightarrow \frac{I R_{1}}{{IR}_{2}} = \frac{1}{3}

From Ohm’s law,

\Rightarrow \frac{V_{1}}{V_{2}} = \frac{1}{3}

Hence the ratio of the potential drop

X : Y

1 : 3 .

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