Two rods A and B made up of same metal have same length. The ratio of their resistances is 1: 2. If these rods are immeresed in water then loss in weight will be :-

The correct option is A more in A
Resistance $R = ρ \frac{l}{A} = ρ \frac{l^{2}}{V} = ρ \frac{l^{2}}{m} σ$ as $(σ = \frac{m}{V})$
thus, $R \propto \frac{1}{m}$
since $R_{A} : R_{B} = 1 : 2$ so $m_{A} > m_{B}$
Thus, loss in weight in A will be more.

Suggest Corrections

Similar questions

Q. Two rods of the same material have diameters in the ratio

1 : 2

and the lengths in the ratio of

2 : 1

. If the temperature difference between their ends is the same, then the ratio of heats conducted by them in a given time will be :

Q. Two rods having same length and cross-sectional area are connected individually between temperature

T_{1}

and

T_{2}

. If the rate of loss of heat due to condution is equal and specific heat of two rods are in the ratio

1 : 2

, then the ratio of their thermal conductivity will be

Q. Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio

1 : 2

and their lengths are in the ratio

2 : 1

. If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be :

The heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio 1 : 2 and their lengths are in the ratio 2 : 1. If the temperature difference between their ends is the same, the ratio of rate of flow of heat through them will be

Q. Two wires made of same material have lengths in the ratio

1 : 2

and their volumes in the same ratio. The ratio of their resistances is:

Join BYJU'S Learning Program

Two rods A and B made up of same metal have same length. The ratio of their resistances is 1: 2. If these rods are immeresed in water then loss in weight will be :-

The correct option is A more in AResistance R=ρlA=ρl2V=ρl2mσ as(σ=mV)thus, R∝1msince RA:RB=1:2 so mA>mBThus, loss in weight in A will be more.

The correct option is A more in A
Resistance $R = ρ \frac{l}{A} = ρ \frac{l^{2}}{V} = ρ \frac{l^{2}}{m} σ$ as $(σ = \frac{m}{V})$
thus, $R \propto \frac{1}{m}$
since $R_{A} : R_{B} = 1 : 2$ so $m_{A} > m_{B}$
Thus, loss in weight in A will be more.