Physics students use a spring scale to measure the weight of a piece of lead. The experiment was performed two times: once in th

Question

Jobisdone

1 month ago

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Physics students use a spring scale to measure the weight of a piece of lead. The experiment was performed two times: once in th

e air and once in water. If the volume of lead is 50 cm3, what is the difference between the two readings on the scale?

Physics

1 answer:

Yuliya22 [3.3K] · Answer 1 · 2026-02-17T16:32:14+03:00

Result:

The scale reading in air exceeds the reading taken in water by 50 grams.

Reasoning:

Since lead has a known density of $11,3 g/cm^{3}$ , and knowing its volume, we can determine the actual weight:

$weight_{lead}=\rho _{lead}*V_{lead}=11,3 g/cm^{3} *50 cm^{3} = 565 g$

When measured in air, the buoyant force exerted by air is negligible due to its low density, so the scale approximately reflects the true weight.

Conversely, in water, buoyancy caused by displaced water can't be ignored.

Given water's density is 1 gram per cubic centimeter and the lead piece displaces a volume equal to its own (as it sinks because of its greater density), the buoyant force acting upward is equivalent to 50 grams.

$Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2} -50 g*9.8 m/s^{2})}{9,8m/s^{2} } = 515 g$

Calculating the difference between the two measurements yields:

$Difference= |Weight _{in water}- Weight _{in air}|=|515 g-565 g|=50 g$

Hence, the air measurement is 50 grams heavier than the water measurement.