Level of a certain liquid at `0^(@)C` and `100^(@)C` are `0` and `10 mm` on a given fixed scale (as shown in fig.) coefficient of volume expansion this liquid varies with temperature as `gamma = gamma_(0) (1+(T)/(100))` (where `T` in `.^(@)C)` Find the level (in mm) of liquid at `48^(@)C`

Answer

4

Explanation

`dv = Adx`
`V_(0)gamma_(0) ((T)/(100)+1) dT = Adx`
`rArr V_(0)gamma_(0) int_(0)^(100) ((T)/(100)+1) dT =A int_(0)^(10) dx`
`rArr V_(0)gamma_(0) xx ((100^(2))/(2xx100)+100) = A xx 10`
`rArr V_(0)gamma_(0) xx 150 = A xx 10 …(i)`
II case
`rArr V_(0) gamma_(0) int_(0)^(48) ((T)/(100)+1) dT = A int_(0)^(x)dx`
`V_(0) gamma_(0) xx ((48^(2))/(2xx100)+48)=A xx x ...(ii)`
`(1)/(15) xx 48 (0.24 +1) - x`
`x = 1.24 xx 3.2 = 3.968 = 4mm`