Two rods of the same material and length have their electric resistance in ratio 1:2. When both rods are dipped in water, the correct statement will be
A. A has more loss of weight
B. B has more loss of weight
C. Both have same loss of weight
D. Loss of weight will be in the ratio 1:2

Hint: Find the relationship between electric resistance and the area of the rod made of the specified material first. Using the radius of the two wires specified in the question, calculate the ratio of the areas of the two rods. The resistance of the wire should then be calculated using the same ratio since resistance is inversely proportional to area.

Formula used:
Electric resistance, $R = \rho \dfrac{L}{A}$
Where, $\rho $ is resistivity, L is length and A is the area of the rod given.
Loss of weight = $V{\sigma _L}g = AL{\sigma _L}g$
where, $\sigma $ is density of the given material and V is volume here.

Complete step by step solution:
Given information:
Let resistance of rod1 = ${R_1}$
Resistance of rod2 = ${R_2}$
Then, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2}$
Now, we know that:
Electric resistance, $R = \rho \dfrac{L}{A}$
As both the rods of the same material so $\rho $ and L will be constant here.
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2} = \dfrac{{{A_2}}}{{{A_1}}}$
When rods are dipped in water there will be loss of weight.
Loss of weight = $V{\sigma _L}g = AL{\sigma _L}g$
Other quantities remain the same as A.
$\dfrac{{{\text{(Loss of weight)}_1}}}{{{\text{(Loss of weight)}_2}}} = \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{2}{1} = 2$
So, A has more loss of weight when rods are dipped in water.

Hence, the correct answer is option A.

Note:The remaining parameters other than area in the formula for electric resistance were taken constant here since the rods provided in the question are of the same material and mass, but this is not always the case. The solution will vary if various materials are utilized for the rods.