Question

Left arm has area 10.0 cm2, right has area 5.00 cm2. 100 g of water isadded to right side. a) Determine the length of the water column. b)density of mercury is 13.6 g/cm3, what distance h does the mercury rise onthe left

Answer 1

0.49 cm

Step-by-step explanation:

We are given that

`A_1=10 cm^2`

`A_2=5 cm^2`

Mass of water,m=100 g

a.
`\rho_w=1 g/cm^3`

Volume,
`V=(mass)/(density)=(100)/(1)=100 cm^3`

Length of water column in the right arm=
`L=(V)/(A_2)=(100)/(5)=20 cm`

b.
`\rho_m=13.6 g/cm^3`

In equilibrium condition

Pressure at point A=Pressure at point B

`P+\rho_mg(h+h_2)=P+\rho_wgL`

`hA_1=h_2A_2`

`h_2=(hA_1)/(A_2)`

`13.6* 9.8(h+h((A_1)/(A_2)))=1* 9.8* 20`

`13.6* h((A_2+A_1)/(A_2))=20`

`h=(20A_2)/(13.6(A_2+A_1))`

`h=(20* 5)/(13.6* (10+5))=0.49 cm`