A piece of wood weighs 10 g in air. When a heavy piece of metal is suspended below it, the metal being submerged in water. The ' weight of wood in air + metal in water is 14 g . The 'weight ' when both the wood and metal are submerged in water is 2g . Find the volume and density of wood .
Asked by
| 11th Sep, 2012,
01:35: PM
Expert Answer:
Let the volume of wood=Vw
volume of metal=Vm
density of wood=dw
density of metal=dm
given,
Vwdw=10
Vmdm-Vm=4
total weight of wood & metal-total bouyant force=apparant weight of both when emerged
(Vwdw+Vmdm)-Vmx(density of water)-Vwx(density of water)=2
(Vwdw-Vw)+(Vmdm-Vm)=2
putting the values from prev given equation:
(Vwdw-Vw)+4=2
Vw-Vwdw=2
Vw=2+10=12c
dw=10/12=0.833g/cc
Let the volume of wood=Vw
volume of metal=Vm
density of wood=dw
density of metal=dm
given,
Vwdw=10
Vmdm-Vm=4
total weight of wood & metal-total bouyant force=apparant weight of both when emerged
(Vwdw+Vmdm)-Vmx(density of water)-Vwx(density of water)=2
(Vwdw-Vw)+(Vmdm-Vm)=2
putting the values from prev given equation:
(Vwdw-Vw)+4=2
Vw-Vwdw=2
Vw=2+10=12c
dw=10/12=0.833g/cc
Answered by
| 11th Sep, 2012,
03:36: PM
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