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A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in water? The density of lead is 11.4 g/cm³.

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Answer:

1020g

Step-by-step explanation:

Volume of can=
1100cm^3=1100* 10^(-6)m^3


1cm^3=10^(-6)m^3

Mass of can=80g=
(80)/(1000)=0.08kg

1Kg=1000g

Density of lead=
11.4g/cm^3=11.4* 10^(3)=11400kg/m^3

By using
1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force =
F_b=Weight of can+weight of lead


\rho_wV_cg=m_cg+m_lg

Where
\rho_w=10^3kg/m^3=Density of water


m_c=Mass of can


m_l=Mass of lead


V_c=Volume of can

Substitute the values then we get


1000* 1100* 10^(-6)=0.08+m_l


1.1-0.08=m_l


m_l=1.02 kg=1.02* 1000=1020g


1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

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