ICSE Class 9 Answered
A spring balance reads 200 gf when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine, what will be the new reading of the spring balance? R.D of lead is 11.4 and R.D of brine is 1.1.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
Expert Answer
Given, weight of lump of lead = 200 gf, R.D of lead = 11.4 and R.D of brine =1.1
Mass of lump of lead = 200 g
Density of lead = 11.4 g 
Density of brine = 1.1 g
Volume of lump of lead = 
Volume of lump of lead submerged in brine = half its volume = 
Weight of brine displaced = Volume of lump of lead submerged x density of brine x g
= 8.75 
1.1 g dyne
= 9.6 gf
But loss in weight = weight of brine displaced
Therefore, Loss in weight = 9.6 gf
Now the apparent weight (or the new reading of spring balance) = True weight - Loss in weight = 200 gf – 9.6 gf = 190.4 gf
Answered by | 04 Jun, 2014, 03:23: PM
ICSE 9 - Physics
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