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Question: A spring balance reacts \(200gF\) when carrying a lump of lead in air. If the lead is now immersed w...

A spring balance reacts 200gF200gF when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine solution, what will be the new reading of the spring balance? Specific gravity of lead and brine 11.411.4 and 1.11.1 respectively.
A. 190.4gF190.4gF
B. 180.4gF180.4gF
C. 210gF210gF
D. 170.4gF170.4gF

Explanation

Solution

The initial reading of the spring balance will be equal to the weight of the lump of lead. After the lead is half immersed in brine solution, the weight of the brine solution will decrease. Using mass equals density multiplied by volume, calculate the mass of lead immersed in solution and thus calculate the resultant mass.

Complete step by step answer:
The initial weight of the lump of lead will be:
W1=mg{W_1} = mg
Here, W1{W_1} is the weight of the lump of lead in the initial given condition.
mm is the initial mass of the lump of lead
gg is the acceleration due to gravity;
As the initial reading on the spring is 200gF200gF thus, the initial weight of the lump of lead is
W1=200gF{W_1} = 200gF --equation 11
The total volume of the lump of lead will be:
V=mρlV = \dfrac{m}{{{\rho _l}}}
Here, ρl{\rho _l} is the density of the lump.
When the lump of lead is half immersed in the brine solution, the solution will exert a force in the upward direction on the lump. Hence, the effective weight will decrease. The effective weight W2{W_2} now will be:
W2=W1−ρgV2{W_2} = {W_1} - \dfrac{{\rho gV}}{2}
Here, ρ\rho is the density of the brine solution.
Since, only half of the volume is immersed.
Substituting the values from above equations, we get
⇒W2=mg−ρgm2ρl\Rightarrow {W_2} = mg - \dfrac{{\rho gm}}{{2{\rho _l}}}
⇒W2=mg(1−ρ2ρl)\Rightarrow {W_2} = mg\left( {1 - \dfrac{\rho }{{2{\rho _l}}}} \right)
⇒W2=200(1−1.12×11.4)\Rightarrow {W_2} = 200\left( {1 - \dfrac{{1.1}}{{2 \times 11.4}}} \right)
⇒W2=190.35gF\Rightarrow {W_2} = 190.35gF
⇒W2≃190.4gF\Rightarrow {W_2} \simeq 190.4gF
This will be the new reading in the spring.

So, the correct answer is “Option A”.

Note:
Only half of the lump of lead was immersed in the brine solution.
Volume, mass and density are related to each other and one can be expressed in terms of the other two.
The lump was immersed in the liquid hence, it will have a feeling of weightlessness like we feel when going down in the lift.