Question

please solve with proper steps ​

Answer 1

Solution

given...

m=12kg=12000gm

area of crossection(a)=800cm²

now... pressure created on the wider tube due to the mass of 12 kg or 12000gm is..

$p = \frac{f}{a} = \frac{mg}{a} \\ \: \: \: = \frac{12000 \times 980}{800} \\ \: \: \: = 14700 \: dyne.cm {}^{ - 2}$

now for the narrower tube...

pressure due to weight of the liquid of height h is equal to 14700 dune/cm ²......

$hdg = 14700 \\ = > h = \frac{14700}{dg} \\ = > h = \frac{14700}{1 \times 980} \\ = > h = 15 \: \: cm$

Hope this helps you.....