45 gm of alcohol are needed to completely fill up a weight thermometer at 15°.C. The weight of alcohol which will overflow when the weight thermometer is heated to 33°C is (Take γa=121×10−5/∘C)

γa=m15−m33m33T2−T1=45−m33m33(33−15)or     121×10−5=45−m33m33×18Solving, we get m33=45/(1⋅02178)gmWeight of alcohol overflowing =m15−m33=45−451⋅02178=45(1⋅02178−1)1⋅02178=45×0⋅021781⋅02178=0⋅96gm∴Weight of alcohol overflowing = 0.96 gm