Suresh, who runs a bakery, uses a conical shaped equipment to write decorative labels (e.g., Happy Birthday etc.) using cream. The height of this equipment is 7 cm and the diameter of the base is 5 mm. A full charge of the equipment will write 330 words on an average. How many words can be written using three fifth of a litre of cream?
Show Hint
Always convert dimensions to consistent units before calculating volumes.
For conical volumes, small radii drastically reduce the volume, so be careful with conversions (mm to cm).
Words per unit volume scaling is a direct proportionality application.
Step 1: Volume of conical equipment. The formula is
\[
V = \tfrac{1}{3}\pi r^2 h.
\]
Given diameter \(=5\ \text{mm}=0.5\ \text{cm}\), so radius \(r=0.25\ \text{cm}\). Height \(h=7\ \text{cm}\). Therefore
\[
V=\tfrac{1}{3}\pi (0.25)^2 (7)=\tfrac{1}{3}\pi(0.0625)(7)=\tfrac{1}{3}\pi(0.4375).
\]
\[
V\approx0.458\ \text{cm}^3.
\]
Step 2: Words written per unit volume. One charge (volume \(0.458\ \text{cm}^3\)) writes 330 words. So per cubic cm:
\[
\frac{330}{0.458}\approx720.5\ \text{words per cm}^3.
\]
Step 3: Available cream volume. Three fifth of a litre \(= \frac{3}{5}\times1000=600\ \text{cm}^3\). So total words:
\[
600\times720.5\approx432,300.
\]
Step 4: Comparing with options, none match exactly (closest is option D 43200, which is a factor of 10 off). Correct choice is
\[
\boxed{\text{None of the above}}.
\]
