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Question

Physics Question on mechanical properties of fluid

A wooden block floating in a bucket of water has 45\frac{4}{5} of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :

A

0.5

B

0.7

C

0.6

D

0.8

Answer

0.6

Explanation

Solution

In 1st1^{st} situation
Vbρbg=VsρwgV_{b} \rho_{b}g = V_{s}\rho_{w}g
VsVb=ρbρw=45\frac{V_{s}}{V_{b}} =\frac{\rho_{b}}{\rho_{w}} = \frac{4}{5} .....(i)
here VbV_b is volume of block
VsV_s is submerged volume of block
ρb\rho_b is density of block
ρw\rho_w is density of water
& Let ρo\rho_o is density of oil
finally in equilibrium condition
Vbρbg=Vb2ρog+Vb2ρwgV_{b} \rho_{b} g = \frac{V_{b}}{2} \rho_{o} g + \frac{V_{b}}{2} \rho_{w}g
2ρb=ρ0+ρw2\rho_{b} = \rho_{0} + \rho_{w}
ρoρw=35=0.6\Rightarrow \frac{\rho_{o}}{\rho_{w}} = \frac{3}{5} = 0.6