An aluminium rod of length 1 m and a steel rod of length 2 m both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is `200 Js^(-1)m K^(-1)` and `50 Js^(-1) m^(-1) K^(-1)` respectively. The temperatures of the free ends are maintained at 300 K and 500 K. What is the temperature of the junction?

Options

A. 322K
B. 350K
C. 367K
D. 400K

Answer

A

Explanation

Let junction temperature is TK, then in steady state heat flow rate is same through both rods.

So using ,
`Q/t = (KA (T_2 - T_1) )/(l) `, we have
` Q/t = (k_s A_s (500 - T ) )/(2) = (k_(Al) A_(Al) (T - 300 ) )/(1)`
As areas of both rods is same, so
`(k_t (500 - T ) )/(2) = (k_(Al) (T - 300 ) )/(1)`
`rArr (50 (500 - T ) )/(2) = (200 (T - 300) )/(1)`
`rArr 25 xx 500 - 25 T = 200 T - 200 xx 300`
`rArr 225 T = 12500 + 60000`
` rArr T = (72500)/(225) = 322.2 ~~ 322 K`