Question #d2bd0

Let `V` be the volume of the given cylinder.
Let `d` be density of turpentine.

When the cylinder is weighed in air forces acting on the cylinder`="Weight"darr+"Buoyant force equal to weight of air displaced by it"uarr`
As density of air very small, so is weight of air displaced by the cylinder.
`:."Weight"darr=67g` ......(1)

We also know that weight of cylinder `="Volume"xx"Density"xxg_r`
Where `g_r` is acceleration due to gravity and is `9.81ms^-2`.
`:.` weight of cylinder `=Vxx2700xxg_r` .....(2)

Converting in `kg` and equating (1) and (2) we obtain volume of the cylinder
`67xx10^-3=Vxx2700xxg_r`
`=>V=(67xx10^-3)/(2700xx9.81)m^3` .........(3)

When the cylinder is weighed in turpentine forces acting on the cylinder`="Weight"darr+``"Bouyant force equal to weight of turpentine displaced by it"uarr`
Since density of solid aluminum is much higher than that of turpentine, it is assumed that complete cylinder is immersed in turpentine.
`"Measured weight"="Weight"darr+``"Buoyant force equal to weight of turpentine displaced by it"uarr`

`45xx10^-3=67xx10^-3-Vxxd xxg_r`
`=>45xx10^-3=67xx10^-3-(67xx10^-3)/2700xxd`
Dividing both sides by `10^-3` and rearranging we get
`67/2700xxd =22`
`=>d =22xx2700/67`
`=>d =886.6kgm^-3`

Given
`"The density of Al " d_(Al)=2700kgm^-3`

`=2700*10^3*10^-6gcm^-3=2.7gmcm^-3`

`"The mass of Al Cylinder " m_(Al)=67gm`

`"The volume of Al Cylinder " v_(Al)=m_(Al)/d_(Al)=67/2.7cm^3`

• The apparent Weight of the cylnder in turpentine is `=45" gwt"`

• So the apparent loss in weght of the cylinder when immersed fully in turpentine`=(67-45)=22" gwt"`

• Let density of turpentine be `=d_t gcm^-3`

• So the volume of turpentine displaced by cylinder`=67/2.7cm^3`

• So the weight of turpentine displaced by cylinder`=67/2.7*d_t" gwt"`

• This weight must be equal to the apparent loss in weight

• Hence

`67/2.7*d_t=22`

`=>d_t=(22xx2.7)/67gcm^-3~~0.886gcm^-3`
`=0.886xx10^-3xx10^6kgm^-3=866kgm^-3`