Advertisements
Advertisements
प्रश्न
A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?
Advertisements
उत्तर
Piece of metal weighs in air = 44.5 f
Piece of metal weighs in liquid = 39.5 f.
Loss of weight of metal in liquid = 44.5 - 39.5 = 5f.
Relative density = weight of solid in air/ loss of weight of solid in water.
Relative density of liquid =44.5f/5f =8.9
Relative density of liquid = 8.9
APPEARS IN
संबंधित प्रश्न
A solid of density 5000 kg m-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m-3. Calculate the apparent weight of the solid in water.
A metal cube of edge 5 cm and density 9 g cm-3 is suspended by a thread so as to be completely immersed in a liquid of density 1.2 g cm-3. Find the tension in thread. (Take g = 10 m s-2)
How are the (i) Mass, (ii) Volume and (iii) Density of a metallic piece affected, if at all, with an increase in temperature?
The unit of relative density is :
A piece of stone of mass 113 g sinks to the bottom in water contained in a measuring cylinder and water level in cylinder rises from 30 ml to 40 ml. Calculate R.D. of stone.
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.
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.
Calculate the mass of a body whose volume is 2m3 and relative density is 0.52.
