Question #d2bd0

2 Answers
Sep 1, 2016

#886.6kgm^-3#

Explanation:

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#

Sep 1, 2016

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#