The total weight of a piece of wood is 6 kg. In the floating state in water its $\frac{1}{3}$ part remains inside the water. On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water?

Question

The total weight of a piece of wood is 6 kg. In the floating state in water its $\frac{1}{3}$ part remains inside the water. On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water?>

Dr. Abubacker · Accepted Answer

(a) Given, $6g = \frac{V}{3} 	imes 10^{3} 	imes g$ …(i)
And $(6 + m)g = V 	imes 10^{3} 	imes g$ …(ii)
Dividing Eq.(ii) by Eq. (ii) by Eq. (i), we get
$\frac{6 + m}{6} = 3$
Or $m = 18 - 6 = 12$ kg

Q. The total weight of a piece of wood is 6 kg. In the floating state in water its $\frac{1}{3}$ part remains inside the water. On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water?>

Explanation

Reviewed by Dr. Abubacker

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