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(A) The resistance of a rod is given by $R = \rho \frac{L}{A}$. Since the material $(\rho)$ and length $(L)$ are constant,$R \propto \frac{1}{A}$.Give

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$A$ has more loss of weight

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(A) The resistance of a rod is given by $R = ho \frac{L}{A}$. Since the material $(ho)$ and length $(L)$ are constant,$R \propto \frac{1}{A}$.Given $\frac{R_1}{R_2} = \frac{1}{2}$,we have $\frac{A_1}{A_2} = \frac{R_2}{R_1} = 2$,which means $A_1 = 2A_2$.When a body is dipped in water,the loss of weight is equal to the buoyant force,given by $F_B = V \sigma_L g = A L \sigma_L g$,where $V$ is the volume,$\sigma_L$ is the density of the liquid,and $g$ is the acceleration due to gravity.Since $L$,$\sigma_L$,and $g$ are the same for both rods,the loss of weight is directly proportional to the cross-sectional area $A$.Therefore,$\frac{	ext{Loss of weight}_1}{	ext{Loss of weight}_2} = \frac{A_1}{A_2} = 2$.Since the ratio is $2$,rod $A$ (the first rod) experiences a greater loss of weight.

Two rods of the same material and length have their electric resistance in the ratio $1:2$. When both rods are dipped in water,the correct statement will be:

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