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

Q: 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

a (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.

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$A$ has more loss of weight
B$B$ has more loss of weight
C
Both have same loss of weight
DLoss of weight will be in the ratio $1:2$

Easy

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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