Question

For exerting a pressure of 4.8 kg/cm2, the depth of oil (specific gravity 0.8)

Answer 1

Answer:60m

Explanation:

Pressure = W * H.

4.8 * 10^3 = sp. grvty * dnsty * H.

4800 = 0.8 * 1* H.

H = 6000cm = 60m.

Answer 2

The depth of oil column exerting the pressure of $4.8\frac{kg}{cm^{2} }$ is 60 m.

Given,

pressure=$4.8 \frac{kg}{cm^{2} }$

specific gravity=0.8.

To find,

the depth of oil.

Solution:

Firstly the pressure needs to be converted into its S.I. units. The S.I. unit of pressure is $\frac{N}{m^{2} }$ or pascal (Pa).

Pressure=force/area

force=mg

where,

m-mass of the body(oil)

g-acceleration due to gravity $(g=10\frac{m}{s^{2} } )$

force=4.8 x 10 N

force=48 N.

Now, the area needs to be converted into its S.I unit. The S.I. unit of area is $m^{2}$.

area=$\frac{cm^{2} }{10^{4}cm^{2} } m^{2}$

area=$10^{-4} m^{2}$

area=$0.0001m^{2}$

pressure=force/area

pressure=$\frac{48}{0.0001} \frac{N}{m^{2} }$

pressure=$48X10^{4} Pa$

Specific gravity also known as the relative density is the ratio of the density of a liquid to the density of water at 4 degree celsius. It has no value. (Density of water at 4 degree celsius=$1000 \frac{kg}{m^{3} }$

Specific gravity=$\frac{density of liquid(d)}{density of water at 4 degree celsius}$

0.8=d/1000

d=800 $\frac{kg}{m^{3} }$

Pressure exerted by a liquid column is given as,

p=hdg

where,

h-height of liquid column

d-density of liquid

g-acceleration due to gravity.

$48X10^{4} =hX800X10\\\frac{48X10^{4}}{8000} =h\\h=\frac{480}{8} m\\h=60m$

The depth of oil column is 60 m.

#SPJ2