Question

A solid floats with (1/4)th of its volume above the surface of water .

calculate the density of solid

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Answer 1

$\huge \red{ \boxed{ \ulcorner{ \mid{ \overline{ \underline{ \bf{ANSWER}}}}} \mid}}$

= 750 kg/m³

$\bf{step \: by \: step \: explanation}$

Let v be volume of solid

and

d be density of solid.

W = Vdg

$Volume \:inside \:water$

$\bf\huge\implies{ v - \frac{v}{4} }$

$\bf\huge\implies{\frac{4v-v}{4}}$

$\bf\huge\implies{ \frac{3v}{4}}$

Weight displaced

$\bf\huge\implies{ \frac{3v}{4} \times {10}^{3} \times g }$

Using principle of floatation

$\bf\implies\: vdg = \frac{3v}{4} \times {10}^{3}$

$\bf\implies\: Hence \: d= \frac{3}{4} ×10^{3}$

$\bf\implies{750 \:kg \: per \: m ^{3}}$

Answer 2

Answer:-

= 750 kg/m³

Explanation:-

Let the volume = v

Let the Density of solid = p

W = Vpg

Volume inside water = V - V/4

= 3V/4

Weight displaced by immersed portion = 3V/4 × 10³ × g

Using principle of floatation, we get

⇒ Vpg = 3V/4 × 10³ ×g

⇒ p = 3/4×10³

⇒ p = 750 kg/m³

Hence, the density of solid is 750 kg/m³.