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The area of cross-section of the wider tube shown in fig., is `800cm^(2)` . If a mass of 12 kg is placed on the massless piston, what is the difference in the level of water in two tubes.
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Here, `A= 800 cm^(2) ` ,
`F= mg = (12 xx 1000 xx 980) dyn es`.
Thus, pressure on the liquid,
`p= F/A = (12 xx 1000 xx 980)/(800) dyn e//cm^(2)`
If h is the difference in level of liquid in the two tubes then,
`p = h rho g = hxx 1 xx 980`
`:. h xx 1 xx 980 = (12 xx10 xx 980)/(800)`
or, ` h=15.0 cm`.

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