Calculating Idol Mass from Water Displacement

The problem asks for the mass of the idol based on the surface area of water in a well and the water level rise upon immersion. The key principle is that the volume of water displaced is equal to the volume of the submerged idol.

Volume of Displaced Water

The volume of water displaced ($V_{displaced}$) is calculated by multiplying the surface area of the water ($A$) by the rise in water level ($h$).

• Surface Area, $A = 30 \ m^2$.
• Water level rise, $h = 10 \ cm$. Convert this to meters: $h = 10 \ cm = 0.1 \ m$.

Therefore, the volume of displaced water is:

$ V_{displaced} = A \times h $

$ V_{displaced} = 30 \ m^2 \times 0.1 \ m = 3 \ m^3 $

Since the idol is fully immersed, the volume of the idol ($V_{idol}$) is equal to the volume of displaced water:

$ V_{idol} = V_{displaced} = 3 \ m^3 $

Calculating Idol Mass

The mass of the idol ($m_{idol}$) can be found using its density ($\rho_{idol}$) and its volume ($V_{idol}$).

• Density of idol material, $\rho_{idol} = 2 \ g/cm^3$.
• Volume of idol, $V_{idol} = 3 \ m^3$.

First, convert the density from $g/cm^3$ to $kg/m^3$ for consistency:

$ \rho_{idol} = 2 \ \frac{g}{cm^3} = 2 \times \frac{10^{-3} \ kg}{(10^{-2} \ m)^3} = 2 \times \frac{10^{-3}}{10^{-6}} \ \frac{kg}{m^3} = 2 \times 10^3 \ kg/m^3 = 2000 \ kg/m^3 $

Now, calculate the mass using the formula $Mass = Density \times Volume$:

$ m_{idol} = \rho_{idol} \times V_{idol} $

$ m_{idol} = 2000 \ kg/m^3 \times 3 \ m^3 $

$ m_{idol} = 6000 \ kg $

The mass of the idol is 6000 kg.