Question

A cubical block of wood of specific gravity 0.5 and chunk of concrete of specific gravity 2.5 are fastened together. The ratio of mass of concrete to the mass of wood which makes the combination to float with its entire volume submerged in water is 

A. $\dfrac{5}{3}$​

B. $\dfrac{6}{5}$​

C. $\dfrac{5}{2}$​

D. $\dfrac{4}{3}$​

Answer 1

Hint: In order to explain this question, we should know the concept of how the flotation depends on density. If an object has density less than the density of water then the object floats otherwise it sinks. 

Complete step by step answer: According Archimedes principle,

Buoyant force=weight of the object floats 

Buoyant force ${f_b} = \rho vg$

When the two blocks float in water, the weight of two blocks must balance with net force of buoyancy of two blocks. 

$F=mg$

$\Rightarrow {m_1}g + {m_2}g = \rho vg$…… (1)

$\Rightarrow \text{Density} = \dfrac{\text{mass}}{\text{volume}}$

$\Rightarrow {m_1} = 2.5 $

$\Rightarrow {m_2} = 0.5 $

Equation (1) becomes

$2.5{v_1}g + 0.5{v_2}g = ({v_1} + {v_2})g $

$\Rightarrow 2.5{v_1}g - {v_1}g = {v_2}g - 0.5{v_2}g $

$\Rightarrow 1.5{v_1} = 0.5{v_2} $

$\Rightarrow {v_2} = 3{v_1} $

For block one: ${m_1} = {\rho _1}{v_1}$……. (2)

For block two: ${m_2} = {\rho _2}{v_2}$…… (3)

Divide 2 and 3 equation 

$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{\rho _1}{v_1}}}{{{\rho _2}{v_2}}}$

$\Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{2.5{v_1}}}{{0.5{v_2}}}$

$\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{5}{3}$

Hence the correct option is A.

Note: Specific gravity is also known as relative density, it is the ratio of the density of a substance to that of the standard substance. Specific gravity is highest in heavy metals and lowest in water.