A block of wood floats in a liquid of density 0.8g/cm sq. with one fourth of its volume submerged. In oil the block floats with 60% of its volume submerged. Find the density of (a) wood? (b) oil?

Answer 1

Let the volume of the block of wood be #V cm^3# and its density be #d_w gcm^-3#

So the weight of the block #=Vd_w g# dyne, where g is the acceleration due to gravity #=980cms^-2#

The block floats in liquid of density #0.8gcm^-3# with #1/4 th# of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid#=1/4Vxx0.8xxg# dyne.

Thus, by eliminating floatation

#Vxxd_wxxg=1/4xxVxx0.8xxg#

#=>d_w=0.2gcm^"-3"#,

Now let the density of oil be #d_o gcm^"-3"#

The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil = #60%xxVxxd_o xxg# dyne

Applying the flotation condition now, we obtain

#60%xxVxxd_o xxg=Vxxd_wxxg#

#=>60/100xxcancelVxxd_o xxcancelg=cancelVxx0.2xxcancelg#

#=>d_o=0.2xx10/6=1/3=0.33gcm^-3#

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Answer 2

Benjamin Asiedu

(a) Density of wood: (0.4 , \text{g/cm}^3)

(b) Density of oil: (0.6 , \text{g/cm}^3)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.