a glass stopper weights 40 grams in air .when immersed in water. it appears to weight 25 grams .find the relative density of the stopper
Answers
Answer:
V 1 = Volume of cavity
V 2 = Volume of stopper
(V 2 −V 1 )×2.5
=V 2 ×1V2
−V
1
=
2.5
V
2
(
V
2
V
2
−V
1
)=
2.5
10
(
V
2
V
2
−
V
2
V
1
)=
5
2
(1−
V
2
V
1
)−1=
5
2
V
2
V
1
=
5
2
+1
V
2
V
1
=−
5
2
+1
V
2
V
1
=
5
3
⇒3:5
Answer:
Given,
Mass of glass stopper in air = 40 g
Weight of glass stopper in air = m×g
= 40 × 9.8 = 392 N
Mass of glass stopper in water = 25 g
Weight of glass stopper in water = m × g
= 25 × 9.8 = 245 N
Weight of water displaced =
Weight of glass stopper in air - Weight
of glass stopper in water
=> Weight of water displaced
= 392 N - 245 N
=> Weight of water displaced = 147 N
Relative Density of glass stopper =
Weight of glass stopper in air / weight
of water displaced
=> R. D. of glass stopped = 392 N / 147 N
=> R. D of glass stopper = 2.67