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(A) $1$. Initial resistance $R_i = 100 \Omega$. When a wire is stretched,its volume remains constant. Since $R = \rho \frac{L}{A} = \rho \frac{L^2}{V}

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$36$

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(A) $1$. Initial resistance $R_i = 100 \Omega$. When a wire is stretched,its volume remains constant. Since $R = ho \frac{L}{A} = ho \frac{L^2}{V}$,$R \propto L^2$. $2$. New length $L' = 1.2 L$. New resistance $R' = R_i 	imes (1.2)^2 = 100 	imes 1.44 = 144 \Omega$. $3$. The wire is bent into a rectangle with length $l$ and breadth $b$ such that $l:b = 3:2$. Let $l = 3x$ and $b = 2x$. The perimeter $2(l+b) = 10x$ corresponds to the total length of the wire. $4$. The resistance of the sides will be proportional to their lengths. Resistance of length $3x$ is $R_l = \frac{3x}{10x} 	imes 144 = 43.2 \Omega$. Resistance of length $2x$ is $R_b = \frac{2x}{10x} 	imes 144 = 28.8 \Omega$. $5$. The rectangle has two sides of $43.2 \Omega$ and two sides of $28.8 \Omega$. Across a diagonal,we have two parallel branches: one branch has $(43.2 + 28.8) = 72 \Omega$ and the other has $(28.8 + 43.2) = 72 \Omega$. $6$. The effective resistance $R_{eq} = \frac{72 	imes 72}{72 + 72} = 36 \Omega$.

$A$ wire of resistance $100 \Omega$ is stretched so that its length increases by $20 \%$. The stretched wire is then bent in the form of a rectangle whose length and breadth are in the ratio $3: 2$. The effective resistance between the ends of any diagonal of the rectangle is (in $\Omega$)

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