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A U-tube having a liquid of density rho ...

A U-tube having a liquid of density `rho` is accelerated at a `m//s^(2)` , so as to create be the height difference between two columns of l/2 (as shown if figure) . If/is the length of the base of U-tube the value of acceleration given to the system is

A

`4.9 m//s^(2)`

B

`9.8 m//s^(2)`

C

`5.6 m//s^(2)`

D

`6.4 m//s^(2)`

Text Solution

Verified by Experts

(a) Pressure difference between two column,
`Deltap=pg(h_(1)-h_(2))=pg.(1)/(2)`
`:.` Force on the liquid contained in the horizontal portional of the tube. `=(Deltap)xx"area"`. Now this force must be equal to the product of mass of liquid (in bas tube) and acceleration a lof the system
`rArr " " pg.(1)/(2)xxarea=(volume xxdensity)xxa`
`rArr " " pg.(1)/(2)xxarea=arealxxpxxa`
`rArr" " a=g//2=4.9 m//s^(2)`
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