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A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. - Physics

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प्रश्न

A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. Specific gravity of wood is 0.8 and that of lead is 11.3. 

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उत्तर

Given:
Density of wood, ρw = 0.8 gm/cc
Density of lead, ρpb = 11.3 gm/cc
Weight of the cubical wood block, mw  = 200 g  

The cubical block floats in water.
Now,
(mw+ mpb) × g = (Vw + Vpb)ρ × g
Here,
ρ = Density of water
Vw = Volume of wood
Vpb = Volume of lead

\[\Rightarrow ( \text{m}_\text{w} + \text{m}_{\text{pb}} ) = \left( \frac{\text{m}_\text{w}}{\rho_\text{w}} + \frac{\text{ m}_{\text{pb}}}{\rho_{\text{pb}}} \right)\rho\]

\[ \Rightarrow (200 + \text{m}_{\text{pb}} ) = \left( \frac{200}{0 . 8} + \frac{\text{m}_{\text{pb}}}{11 . 3} \right) \times 1\]

\[ \Rightarrow \text{m}_{\text{pb}} - \frac{\text{m}_{\text{pb}}}{11 . 3} = 250 - 200\]

\[ \Rightarrow \frac{10 . 3 \text{m}_{\text{pb}}}{11 . 3} = 50\]

\[ \Rightarrow \text{m}_{\text{pb}} = \frac{50 \times 11 . 3}{10 . 3} = 54 . 8 \text{ gm }\]

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अध्याय 13: Fluid Mechanics - Exercise [पृष्ठ २७४]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 13 Fluid Mechanics
Exercise | Q 16 | पृष्ठ २७४

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