A solid weight 32gf in air and 28.8gf in water. Find the volume and RD

1.	A solid weight 32gf in air and 28.8gf in water. Find the volume and RD of the solid. Find how much it will weigh in a liquid of density 0.9 g/cm³.please give the answer with process in a paper irrelevant answers will be reported please help friends
Answer» ANSWER: Hii..! ${\huge{\tt{\boxed{\fcolorbox{red}{black}{Solution:-}}}}}$ _____________________________________ Given, Weight of solid in air, W, 32 gf Weight of solid when COMPLETELY IMMERSED in water W2 = 28.8 gf (i) VOLUME of solid = Mass / density of solid. $= \frac{32}{10} = 3.2 {m}^{2} \\$ (II) R.D. of solid = $\frac{w1}{w1 - w2} \times r.d \: \: of \: \: water. \\ \\ \\ r.d \: \: of \: \: solid \: = \: \frac{32}{32 - 28.8} \times 1 \\ \\ \\$ So, R.D of water = 10. Hence Solved! _____________________________________ Hope it helps you dear!

A solid weight 32gf in air and 28.8gf in water. Find the volume and RD of the solid. Find how much it will weigh in a liquid of density 0.9 g/cm³.please give the answer with process in a paper irrelevant answers will be reported please help friends

Answer»

ANSWER:

Hii..!

${\huge{\tt{\boxed{\fcolorbox{red}{black}{Solution:-}}}}}$

_____________________________________

Given,

Weight of solid in air, W, 32 gf

Weight of solid when COMPLETELY IMMERSED in water W2 = 28.8 gf

(i) VOLUME of solid = Mass / density of solid.

$= \frac{32}{10} = 3.2 {m}^{2} \\$

(II) R.D. of solid =

$\frac{w1}{w1 - w2} \times r.d \: \: of \: \: water. \\ \\ \\ r.d \: \: of \: \: solid \: = \: \frac{32}{32 - 28.8} \times 1 \\ \\ \\$

So, R.D of water = 10.

Hence Solved!

_____________________________________

Hope it helps you dear!

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