Question

Mercury is poured into a U-tube in which the cross-sectional area of the left-hand limb is three
times smaller than that of the right one. The level of the mercury in the narrow limb is a distance
I=30 cm from the upper end of the tube. How much will the mercury level rise in the right-hand
limb if the left one is filled to the top with water?
110.58cm
270.18cm
3)0.36cm
4)0.46cm
the nf uniform pross-section is filled with two​

Answer 1

Answer:

0.58 cm, none of the options given are correct.

Explanation:

REFER TO THE ATTACHMENT

If there is a rise of ' x ' cm in the right hand limb, there will be a drop of 3x cm in the left hand limb since their cross-sectional areas vary by 3 times while volume of water poured remains constant. => Total in the right hand limb

from point 2 = 3x + x = 4x

The pressure at point 1 = The pressure at point 2. ( since these points are at the same horizontal plane

connecting the same fluid so pascals law applies)

=> atmospheric pressure + p1h1g = atmospheric pressure + p2h2g

after removing Atmospheric pressure and ' g ' on both sides since they are common we get,

p1h1 = p2h2

Density of water = 1 g/cm^3

Density of mercury = 13.6 g/cm^3

=> 1(30 + 3x) = 13.6(4x)

=> 30 + 3x = (54.4)x

=> 30 = (51.4)x

=> x = 0.58 cm