Question

A hollow glass stopper weighs 23.4gf in air and 3.9 gf in water. If the relative density of glass is 2.6, Calculate:
(a) the volume of the glass.
(b) the volum of the water displaced.
(c) the volume of the hollow space in the stopper.
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=> No Plagiarism
=> Pls help ASAP​

Answer 1

$\sf{\huge{\underline{\orange{Given :-}}}}$

• A hollow glass stopper weights 23.4gf in air and 3.9 gf in water.

• The relative density of glass is 2.6.

$\sf{\huge{\underline{To\:Find :-}}}$

• The volume of the glass.

• The volume of the water displaced.

• The volume of the hollow space in the stopper

$\sf{\huge{\underline{\pink{Formula\:Used:-}}}}$

★ Density = $\dfrac{mass}{volume}$

$\sf{\huge{\underline{\green{Solution:-}}}}$

Case 1 :-

• The volume of the glass.

Volume = Total mass / relative density

➝ 23.4 +3.9/ 2.6

➝ 27.3/2.6

➝ 10.5 cm³

Case 2 :-

• The volume of the water (Its for medium warer ) (V2)

We have, 23.4gf in air

We know,

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

➝ $p = \dfrac{m}{v}$

p = 2.6

m = 3.9 gf

➝ $2.6 = \dfrac{3.9 }{v}$

➝ $v = \cancel {\dfrac{3.9 }{2.6}}$

➝ 1.5 cm³

Case 3 :-

• The volume of the hollow space in the stopper.

➝ Total volume - Volume of water

➝ 10.5 cm³ - 1.5 cm³

➝ 9 cm³

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