spring balance reads 200gf when carrying a lump of lead in air . if the lead is now immersed with half of it's volume in brine solution what will be the new reading of the spring balance specific gravity of lead and brine are 11.4 and 1.1; respectively
Answers
The new reading of the spring balance will be 190.4 gf.
Explanation:
=> It is given that,
Weight of lump of lead = 200 gf
Specific gravity of lead = 11.4
Specific gravity of brine = 1.1
Mass of lump of lead = 200 g
Density of lead = 11.4 g cm⁻³
Density of brine = 1.1 sm cm⁻³
=> Volume of lump of lead:
Volume = Mass/Density
= 200 / 11.4
= 17.5 cm³
=> Volume of lump of lead submerged in brine :
According to the question, the lead is immersed with half of it's volume in brine solution
Thus, Volume of lump of lead submerged in brine = 1/2 * Volume of lump of lead
= 1/2 * 17.5
= 8.75 cm³
=> Weight of brine displaced:
= Volume of lump of lead submerged x density of brine x g
= 8.75 * 1.1 * g
= 9.6 gf
Loss in weight is equals to weight of brine displaced
. Thus, Loss in weight is equals to 9.6 gf.
=> Apparent weight:
The new reading of spring balance or apparent weight = True weight - Loss in weight
= 200 gf - 9.6 gf
= 190.4 gf
Hence, the new reading of the spring balance will be 190.4 gf.
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