A cylindrical pipe of uniform cross section and of length 120 cmis closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water-column increases by 0.7 cm. Calculate the Poisson's ratio of the material of the pipe.

A cylindrical pipe of uniform cross section and of length 120 cmis clo

1.	A cylindrical pipe of uniform cross section and of length 120 cmis closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water-column increases by 0.7 cm. Calculate the Poisson's ratio of the material of the pipe.
Answer» Solution :Poisson.s ratio , `sigma=(LATERAL strai n)/(longitudi nal stra i n)` Here, longtitudinal STRAIN=`l/L=1/120` Let the initial difference of the pipe be D, the DECREASE in its diameter due to elongation be d. Initial volume of water =`pi/4 D^2 TIMES 120` and final volume of water `=pi/4(D-d)^2times(120+0.7)` `=pi/4(D-d)^2times120.7` Since the volume of water inside the cylinder remains unchanged. `pi/4D^2times120=pi/4times(D-d)^2times120.7` or,`(D-d)/D=sqrt(120/120.7)or,1-d/D=sqrt0.9942 =0.997` or,`d/D=1-0.997=0.003` `thereforesigma=(d//D)/(l//L)=0.003/(1/120)=0.003times120=0.36`