A block of wood is floating on oil with half of its volume submerged. If the density of oil $840 k g m^{- 3}$ , the relative density of wood (relative to water) is:

Question

A block of wood is floating on oil with half of its volume submerged- If the density of oil 840kg m-3- the relative density of wood -relative to water- is-0-840-420-211

Toppr Admin · Accepted Answer

-x3F1-wVw-x3F1-oil-xD7-Vsubmerged-x3F1-w-xD7-Vw2-x3F1-oil-xD7-Vw1-x3F1-w-8402-420kg-m3So relative density -xA0-densityofwooddensityofwater-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- - 0-42

A block of wood is floating on oil with half of its volume submerged. If the density of oil $840 k g m^{- 3}$ , the relative density of wood (relative to water) is:
0.84
0.42
0.21
1

$ϱ_{w} V_{w} = ϱ_{o i l} \times V_{s u b m e r g e d}$
$ϱ_{w} \times \frac{V_{w}}{2} = ϱ_{o i l} \times \frac{V_{w}}{1}$
$ϱ_{w} = \frac{840}{2} = 420 k g / m^{3}$
So relative density = $\frac{d e n s i t y o f w o o d}{d e n s i t y o f w a t e r}$
= 0.42

A block of wood is floating on oil with half of its volume submerged. If the density of oil 840kg m−3, the relative density of wood (relative to water) is:0.840.420.211

ϱwVw=ϱoil×Vsubmergedϱw×Vw2=ϱoil×Vw1ϱw=8402=420kg/m3So relative density = densityofwooddensityofwater = 0.42

A block of wood is floating on oil with half of its volume submerged. If the density of oil $840 k g m^{- 3}$ , the relative density of wood (relative to water) is:
0.84
0.42
0.21
1

$ϱ_{w} V_{w} = ϱ_{o i l} \times V_{s u b m e r g e d}$
$ϱ_{w} \times \frac{V_{w}}{2} = ϱ_{o i l} \times \frac{V_{w}}{1}$
$ϱ_{w} = \frac{840}{2} = 420 k g / m^{3}$
So relative density = $\frac{d e n s i t y o f w o o d}{d e n s i t y o f w a t e r}$
= 0.42