A 1 m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.
- 1:4
- 1:2
- 4:1
- 2:1
The Correct Option is B
Solution and Explanation
To solve this problem, we need to understand the relationship between the lengths of wires and their resistances.
Let's denote the length of wire \(X\) as \(x\) meters and the length of wire \(Y\) as \(y\) meters. The total length of the wire is 1 meter, so:
\(x + y = 1 \, \text{meter}\)
Next, the wire \(X\) is stretched into a new wire \(W\) such that the length of \(W\) is twice the length of \(X\):
\(L_W = 2x\)
According to the problem, the resistance of \(W\) is twice that of \(Y\). Resistance \(R\) for a wire is given by the formula:
\(R = \rho \frac{L}{A}\)
where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. When a wire is stretched, its volume remains constant:
\(xA_X = 2xA_W\)
This implies that the cross-sectional area of \(W\) is half of \(X\):
\(A_W = \frac{A_X}{2}\)
The resistance of \(W\) becomes:
\(R_W = \rho \frac{2x}{A_W} = \rho \frac{2x}{\frac{A_X}{2}} = \rho \frac{4x}{A_X}\)
Given \(R_W = 2R_Y\), where:
\(R_Y = \rho \frac{y}{A_Y}\)
Equating the resistances:
\(\rho \frac{4x}{A_X} = 2 \times \rho \frac{y}{A_Y}\)
Assuming the original cross-sectional area \(A_X = A_Y\), we find:
\(\frac{4x}{A_X} = 2 \times \frac{y}{A_Y}\)
Which simplifies to:
\(4x = 2y\)
Therefore:
\(2x = y\)
Substitute back into the equation for total length:
\(x + 2x = 1\)
\(3x = 1\)
\(x = \frac{1}{3}\) and \(y = \frac{2}{3}\)
The ratio of \(x\) to \(y\) is:
\(\frac{x}{y} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}\)
Hence, the correct ratio of the length of \(X\) to \(Y\) is 1:2.
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