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A cane is taken out from a refrigerator ...

A cane is taken out from a refrigerator at `0^(@)"C"`. The atmospheric temperature is `25^(@)"C"`. If t1 is the time taken to heat from `0^(@)"C"` to `5^(@)"C"` and `t_(2)` is the time taken from `10^(@)"C"` to `15^(@)"C"`, then the wrong statements are
(1) `t_(1)gtt_(2)`
(2)`t_(1)=t_(2)`
(3) There is no relation
(4)`t_(1)ltt_(2)`

A

`t_1 gt t_2`

B

`t_1 lt t_2`

C

`t_1=t_2`

D

There is no relation

Text Solution

Verified by Experts

The correct Answer is:
B
`(T_1-T_2)/(t)=K((T_1 +T_2)/(2) -T_0)`
`implies (0-5)/(t_1)=K(5/2 - 25)`
Now,
`(10-15)/(t_2)=K((25)/(2)-25)`
by (i) and (ii) -
`implies (t_2)/(t_1)=(5-50)/(25-50) implies (t_2)/(t_1)=(-45)/(-25)`
`implies 5 t_2=9t_1`
`implies t_2 = 9/5 t_1` Hence, `t_2 gt t_1`.
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