A galss flask is filled up to a mark with 50 cc of mercury at `18^@C`. If the flask and contents are heated to `38^@C`, how mech mercury will be above the mark (`alpha` for glass is `9xx10^(-6)//^@C` and coeffiecient of real expansion of mercury is `180xx10^(-6)//^@C`)?

Options

A. 0.85 cc
B. 0.46 cc
C. 0.153 cc
D. 0.05 cc

Answer

C

Explanation

Volume of mercury at `18^@C`
`V_0=50c c`
Volume of mercury at `38^(@)C`
`(V_(38))_(r)=V_(0)(1+gamma_(m) Delta theta)`
Volume of flask at `38^@C`
`(V_(38))_(r)=V_0(1gamma_m Delta theta)`
Volume of mercury at `38^@C` above the tank
`=V_0(1+gamma_mDeltatheta)-V_0(1+gamma_fDeltatheta)`
`V_0(gamma_m-gamma_t)Deltatheta=50[180xx10^-6-3xx9xx10^-6](38-18)`
`=0.153 c c`