Question

. A solid body weighs 2.10 N in air. Its relative
density is 8.4 . How much will the body weigh if
placed
(i) in water,
(ii) in a liquid of relative density 1.2 ?​

Answer 1

Answer:

(I)upthrust=weight in air/relative density=0.25

there fore the weight in water is original weight-apparent weight which is 1.85N

(2)1.8N

Answer 2

Answer:

(1). Weight of solid in water = 1.85 N.

(2). Weight of body in liquid = 1.80N

Explanation:

• In context to the given question, we have to find the Weigh  of body if it is placed in

1. in water,
2. in a liquid of relative density 1.2 ?​

Given,

(1).  IN WATER

⇒ Weight of solid in air = 2.10 N

⇒ Relative density of solid = 8.4

Now,

we know that,

⇒ Relative density = weight of solid in air/ loss of weight of solid in water.

Hence,

⇒ Loss of weight of solid in water = weight of solid in air/ Relative density.

Loss of weight of solid in water = 2.10/8.4 = 0.25 N.

⇒ Actual Weight of solid in water = weight in air - loss of weight in water

⇒Actual Weight of solid in water = 2.10 - 0.25 = 1.85 N.

∴ Actual Weight of solid in water = 1.85 N.

(2). IN A LIQUID OF RELATIVE DENSITY 1.2

we know that,

⇒Loss of weight of solid in water = 0.25 N

⇒Up thrust due to liquid of relative density 1.2  = Loss of weight of solid in water× R.D. of liquid

= 0.25 x 1.2 = 0.30 N

⇒Weight of body in liquid = Weight of body in air - Up thrust due to liquid

= 2.10-0.30= 1.80 N

∴Weight of body in liquid = 1.80N