The corfficient of linear expansion for a certain metal varies with temperature as `alpha(T)`. If `L_0` is the initial elgnth of the metal and the temperature of metal is changed from `T_0` to `T(T_0gtT)`, then

Options

A. `L=L_0int_(T_(0))^(T) alpha(T)dT`
B. `L=L_0[1+int_(T-0)^(T)alpha(T)dT]`
C. `L=L_0[1-int_(T_(0))^(T)alpha(T)dT]`
D. `LgtL_0`

Answer

B

Explanation

As, `(dL)/(L_0)=alpha(T)dT`
`int_(L_0)^(L)dL=L_0int_(T_0)^(T)alpha(T)dT`
`L-L_0=L_0int_(T_0)^(T)alpha(T)dT`
`L=L_0[1+int_(T_0)^(T)alpha(T)dT]`