A wire of length L and three identical cell of negligible internal resistance are connected in series. Due to the current, the temperature of wire is raised by `Delta T` in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross section but of length 2L. The temperature of wire is raised by the same amount `Delta T` in the same time t. The value of N is

Options

A. 4
B. 6
C. 8
D. 9

Answer

B

Explanation

b. Let R be the resistance of rod.
Energy released in t seconds `=(3V^2)/R xx t`
`:. Q = (9V^2)/R xx t`
But `Q = mcDeltaT =(9V^2)/R xx t`................(i)
Let R' be the resistance of rod.
`:. R' = 2R ("as length is twice") `
Therefore, energy released in t seconds
`(NV^2)/(2R) xx t`
`:. Q' (N^2V^2)/(2R) xx t`
Applying `Q' = m'cDeltaT`, we get
`2mcDeltaT = (N^2V^2)/(2R) xx t`........(ii)
Dividing Eq. (ii) by Eq. (i), we get
`(mcDeltaT)/(2mcDeltaT) = (9V^2xxt//R)/(N^2V^2t//2R)`
or `1/2 = (9 xx 2)/N^2 or N^2 = 18 xx 2 pr N = 6 ` .