Two rods of same material and length have their electric resistance in ratio $1:2$. When both rods are dipped in water, the correct statement will be
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(a) $R = \rho \frac{l}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}}(\rho ,L$ constant) $ \Rightarrow \frac{{{A_1}}}{{{A_2}}} = \frac{{{R_2}}}{{{R_1}}} = 2$
Now, when a body dipped in water, loss of weight $ = V{\sigma _L}g = AL{\sigma _L}g$
So, $\frac{{{{{\rm{(Loss \,of\, weight)}}}_{\rm{1}}}}}{{{{{\rm{(Loss\, of\, wight)}}}_{\rm{2}}}}} = \frac{{{A_1}}}{{{A_2}}} = 2;$ so $A$ has more loss of weight.
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