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Selina Solutions Class 9 Physics Solutions for Exercise 5C in Chapter 5 - Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation

Question 28 Exercise 5C

A block of wood of mass 24kg floats on water. The volume of wood is 0.032 m3. Find:

(a) The volume of the block below the surface of the water,

(b) The density of the wood.

(density of water = 1000 kg m-3)

Answer:

(a) Given:

Mass of the wood block = 24kg

Volume of the wood block = 0.032 m3

Upthrust = volume of the block below the water surface x density of liquid x g

Upthrust for floatation is equivalent to the weight of the body i.e., 24kgf

 24kgf = v x 1000 x g

 v= 24/1000 = 0.024 m3

(b) To find the density of wood

As per the law of floatation

\frac{\text { Volume of submerged block }}{\text { Net volume of the block }}=\frac{\text { density of wood }}{\text { density of water }}

\begin{array}{l} \Rightarrow \frac{0.024}{0.032}=\frac{d e n s i t y \text { of wood }}{1000} \\ \Rightarrow \text { density of wood }=750 \mathrm{kg} \mathrm{m}^{-3}=7.5 \mathrm{x} 10^{2} \mathrm{kg} \mathrm{m}^{-3} \end{array}

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