A spring balance reads 200 ,gF when carrying | vedclass

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(A) The weight of the lead lump in air is $W_1 = mg = 200 \,gF$.The volume of the lead lump is $V = \frac{m}{\rho_{lead}}$,where $\rho_{lead}$ is the

Vedclass · Accepted Answer

$190.4$

Vedclass · Answer

(A) The weight of the lead lump in air is $W_1 = mg = 200 \,gF$.The volume of the lead lump is $V = \frac{m}{ho_{lead}}$,where $ho_{lead}$ is the density of lead.When the lead is immersed with half of its volume in brine,the buoyant force $F_B$ acting on it is equal to the weight of the displaced brine:$F_B = 	ext{Volume immersed} 	imes ho_{brine} 	imes g = \frac{V}{2} 	imes ho_{brine} 	imes g$.The new reading of the spring balance $W_2$ is the apparent weight:$W_2 = W_1 - F_B = mg - \frac{V}{2} ho_{brine} g$.Substituting $V = \frac{m}{ho_{lead}}$:$W_2 = mg - \frac{m}{2 ho_{lead}} ho_{brine} g = mg \left(1 - \frac{ho_{brine}}{2 ho_{lead}}ight)$.Given the specific gravity $\sigma = \frac{ho_{substance}}{ho_{water}}$,we have $\frac{ho_{brine}}{ho_{lead}} = \frac{1.1}{11.4}$.$W_2 = 200 	imes \left(1 - \frac{1.1}{2 	imes 11.4}ight) = 200 	imes \left(1 - \frac{1.1}{22.8}ight) = 200 	imes \left(1 - 0.048245ight) = 200 	imes 0.951755 = 190.351 \,gF$.Rounding to one decimal place,the reading is approximately $190.4 \,gF$.

$A$ 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 solution,what will be the new reading of the spring balance? The specific gravity of lead and brine are $11.4$ and $1.1$ respectively. (Result in $gF$)

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