Question

a silica bulb of negligible expansivity holds 340 grams of mercury at 0 drgree celsius when full some steel balls are introduced and remaining space is occupied at 0 degree celsius by 255 grams of mercury .on heating the bulb and its contents to 100 degree celsius,4.8 grams of mercury over flows .find the coeficient of linear expansion of steel.the coeficient of expansion mercury is 1.8*10 minus 4 degree per centigrade

Answer 1

Answer:

Explanation:

volume of the bulb at 0 degree =340/13.6=24cc

volume of mercury in the bulb after the balls were introduced =255/13.6=18.75cc.

volume os steel ball =25-18.75=6.25cc.

volume of mercury overflowed at 100 degre

=4.800/ρ100=4.800(1+γΔθ)/ρο=4.800(1+1.8*10^-4*100)/13.6=0.3593cc

But the volume of mercury overflowed=Expansion of mercury +Expansion of steel ball

or, 0.3593=18.75*1.8*10^-4*100 +6.25*3α*100 .(γ of steel ball =3α of steel ball)

∴α=11.6*10^-6 C^-1 which is requred linear expansivity of steel ball.