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`I_1, I_2, I_3``I_2, I_3, I_1``I_2, I_1, I_3``I_3, I_2, I_1`

Answer :

CSolution :

Let R be the rersistance of the wire and let R' be the resistance <br> of the wire. Energy released in t seconds is `(3V^2)Rxxt.` <br> R' = 2R (as length in twice) <br> Therefore, energy released in t seconds is `((NV^2))/(2R)xxt` <br> But `Q = mc DeltaT` <br> `:. Q' = ((N^2V^2))/(2R)xxt` <br> `:. mc Delta T = ((9V^(2)))/(R )xxt` <br> Applying `Q' = m' c Delta T,` <br> `2mc Delta T = ((N^2V^2))/(2R)xxt` <br> Dividing Eq. (ii) by Eq. (i)j we get <br> `(mc Delta T)/(2mc Delta T) = (9V^2 xxt //R)/(N^2V^2 t/2R)` <br> `:. (1)/(2) = (9xx2)/(N^2) or N = 6`