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You are given a solid iron ball A of mass $15 g$ and a hollow iron ball B of volume $20 {cm}^{3}$ and mass $15 g$ . Both are placed on the surface of water. [[1]] will sink.

[Density of iron $= 7.8 g / {cm}^{3}$ ]

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Detailed Solution

Volume of hollow iron ball B,

V_{1} = 20 {cm}^{3}

Mass of hollow iron ball B,

M_{1} = 15 g

Mass of solid iron ball A,

M_{2} = 15 g

Density of iron,

ρ = 7.8 g / {cm}^{3}

Volume of solid iron ball A

= \frac{Mass}{Density}

\frac{15 g}{7.8 g / c m^{3}}

= 1.92 {cm}^{3}

Upthrust on ball A

=

Volume of iron ball A

\times

density

\times g

= 1.92 \times 1 \times g

= 1.92 g

Upthrust on ball A

=

Volume of iron ball A

\times

density

\times g

= 20 \times 1 \times g

= 20 g

Ball B will sink because the upthrust on ball A (

1.92 g

) is less than its weight

15 g

, while the upthrust on ball B

20 g

is greater than its weight

15 g

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