What is the weight of water contained in a conical vessel $21$ cm deep and $16$ cm in diameter? (R.R.B., 2006)

### Concept & Volume to Weight Conversion First, find the volume of the cone. For water, the volume in cubic centimeters ($cm^3$ or $cc$) is numerically equal to its weight in grams because the density of water is $1 \text{ g/cm}^3$. $$ V = \frac{1}{3}\pi \times r^2 \times h $$ ### Step-by-Step Solution - **Given:** - Depth (height) $h = 21$ cm. - Diameter = $16$ cm $\implies$ Radius $r = 8$ cm. - **Calculation:** - $V = \frac{1}{3} \times \frac{22}{7} \times 8^2 \times 21$ - Cancel the $3$ and $7$ in the denominator with the $21$ in the numerator ($\frac{1}{3} \times \frac{1}{7} \times 21 = 1$). - $V = 22 \times 64$ - $V = 1408$ cm$^3$. - **Weight Conversion:** - Weight of $1$ cm$^3$ of water = $1$ gram. - Total weight = $1408$ grams. - Convert grams to kilograms by dividing by $1000$: - $1408 \text{ g} = 1.408$ kg. ### Exam Strategy & Shortcut Recognize that $\frac{1}{3}$ and $\pi$ (with a denominator of $7$) form a denominator of $21$, which perfectly cancels out the height of $21$. You are left simply multiplying $22$ by $r^2$. ### Common Pitfall Using the diameter ($16$) instead of the radius ($8$) in the volume formula. ### Final Answer Therefore, the correct answer is **1.408 kg**.