Advertisements
Advertisements
Question
A body weighs W1gf in air and when immersed in a liquid it weighs W2gf, while it weights W3gf on immersing it in water. Find:
- volume of the body
- upthrust due to liquid
- relative density of the solid
- relative density of the liquid
Advertisements
Solution
(i) Volume of the body = W1 − W3 cm3
(ii) Upthrust due to liquid = loss in weight when immersed in liquid = W1 − W2 gf
(iii)
Weight of a body in air = W1gf
Weight of that body in liquid = W2gf
Weight of that body in water = W3gf
RD of solid = `"Weight of solid in air"/"Weight in air - Weight in water"`
= `"W"_1/("W"_1 - "W"_3)`
(iv)
Weight of a body in air = W1gf
Weight of that body in liquid = W2gf
Weight of that body in water = W3gf
RD of Liquid = `(W_1 - W_2)/(W_1 - W_3)`
APPEARS IN
RELATED QUESTIONS
A sphere of iron and another sphere of wood of the same radius are held under water. Compare the upthrust on the two spheres.
[Hint: Both have equal volume inside the water].
A sphere of iron and another of wood, both of same radius are placed on the surface of water. State which of the two will sink? Give a reason for your answer.
How are the (i) Mass, (ii) Volume and (iii) Density of a metallic piece affected, if at all, with an increase in temperature?
Calculate the mass of a body whose volume is 2 m3 and relative density is 0.52.
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in a liquid.
A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
A piece of wax floats in brine. What fraction of its volume will be immersed?
R.D. of wax = 0.95, R.D. of brine = 1.1.
