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Question

A 0.50N metal sinker appears (as measured using a spring scale) to have a weight of 0.45N when submerged in water.Find out the relative density of the metal .

A
6
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B
8
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C
9
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D
10
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Solution

The correct option is C 10
Answer is D.

Let us consider that Dw and D be the density of water and solid respectively.
The actual weight of the solid is 0.50 N
Apparent weight of the body = Actual weight - Buoyant force (V×Dw×g).
Let V be the volume of the solid.
Density of water Dw=1g/cm3.
In this case, 0.45=mgV×Dw×g
That is, 0.45=0.50V×1
Therefore, V = 0.05cm3.
The density of the body is given as Mass/Volume, that is, 0.50/0.05=10gcm3.
The relative density of a solid is given as the ratio of the density of a substance to the density of a standard substance under specified conditions.
That is, Relativedensity=DensityofsolidDensityofwater=10gcm31gcm3=10
Hence, the relative density of the solid is 10.

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