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# 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 4th June 2014, 1:23 PM
Answered by 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 â€“ 9.6 gf = 190.4 gf

Answered by Expert 4th June 2014, 3:23 PM
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