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    				| 1. | If a body cools down from `80^(@) C`to `60^(@) C` in 10 min when the temperature of the surrounding of the is `30^(@) C` . Then, the temperature of the body after next 10 min will beA. `50^(@) C`B. `48^(@) C`C. `30^(@) C`D. None of the above | 
| Answer» Correct Answer - D `(theta_(1) - theta_(2))/t = alpha[(theta_(1) + theta_(2))/2 - theta_(0)]` two times we get, Where `alpha` is constant. `(80-60)/(10) = alpha[(80+60)/2 -30]`............(i) `(60-theta)/10 = alpha[(60+theta)/2 -30]`..............(ii) Solving these two equations, we get `theta = 48^(@)`C | |