अर्थं लिखत-
(क) पात्रताम्
(ख) दलमिव
(ग) उत्थाय
(घ) अवतीर्य
Answers
Answer:
\sf{\huge{\underline{\orange{Given :-}}}}
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 :-}}}
ToFind:−
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:-}}}}
FormulaUsed:−
★ Density = \dfrac{mass}{volume}
volume
mass
\sf{\huge{\underline{\green{Solution:-}}}}
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}
volume
mass
➝ p = \dfrac{m}{v} p=
v
m
p = 2.6
m = 3.9 gf
➝ 2.6 = \dfrac{3.9 }{v} 2.6=
v
3.9
➝ v = \cancel {\dfrac{3.9 }{2.6}} v=
2.6
3.9
➝ 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
______________________________________
Answer:
i hope it helps you phir krdo Support
