1.

in previous question, if 15 cm of water and spirit each are further poured into the respective arms of the tube. Difference in the level of mercury in the two arms is (Take, relvative density of mercury = 13.6)

Answer»

<P>0.20 CM
0.22 cm
0.27 cm
0.26 cm

Solution :REFER figure. Pressure of mercury level in one arm DUE to water,
`P_(1)=h_(w)rho_(w)g`
`=(10+15)xx1xxg`
`=25 g`

Pressure of mercury level in another arm due to spirit,
`P_(2)=h_(s)rho_(s)g=(12.5+15)xx0.8xxg=22g`
As the pressure in water arm is more, therefore, the mercury will rise in spirit arm.
Let h be the difference in the levels, of mercury in two arms of U TUBE and `rho` be the density of mercury . Then
`P_(1)-P_(2)=hrhog`
`25g-22 g=hxx13.6xxg`
`therefore h=(3)/(13.6)=0.22cm`


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