A narrow capillary tube when dipped in beaker containing water, the rise is 20 cm . If the area of cross section of the bore is reduced to `(1//4)^(th)` calue , water will rise to a height of

Options

A. 10 cm
B. 20 cm
C. 40 cm
D. 80 cm

Answer

C

Explanation

` A_(2)= 1/4 A_(1)`
`pir_(2)^(2)=1/4pir_(1)^(2)`
`r_(2)=1/2r_(1)`
`h_(2)=(h_(1)r_(1))/(r_(2)) =(h_(1)r_(1))/(1/2r_(2))=2h_(1)=2xx20`
=40 cm