A) 50 sec
B) 90 sec
C) 60 sec
D) 48 sec
Correct Answer: D
Solution :
Rate of cooling \[\propto \] excess of temperature \[\frac{80{}^\circ -60{}^\circ }{60{}^\circ }=K\left( \frac{80{}^\circ +60{}^\circ }{2}-30{}^\circ \right)\] \[\frac{1}{3}=K(40)\] ?(1) \[\frac{60{}^\circ -50{}^\circ }{t}=K\left( \frac{60{}^\circ +50{}^\circ }{2}-30{}^\circ \right)\] \[\frac{10}{t}=K(25)\] ...(2) Dividing equation (1) by (2), we get \[\frac{t}{30}=\frac{40}{25}\] or \[t=\frac{40}{25}\times 30=48\,\sec \]You need to login to perform this action.
You will be redirected in
3 sec