A conical flask has base radius $a$ cm and height $h$ cm. it is completely filled with milk. The milk is poured into a cylindrical thermos flask whose base radius is $p$ cm. What will be the height of the milk level in the flask?

### Concept & Conservation of Volume When a liquid is poured from one container to another without spilling, its total volume remains constant. Equate the volume of the cone to the volume of the cylinder. $$ V_{\text{cone}} = \frac{1}{3} \pi r^2 h $$ $$ V_{\text{cylinder}} = \pi r^2 h $$ ### Step-by-Step Solution - The volume of milk in the conical flask is $V = \frac{1}{3} \pi a^2 h$. - This milk is transferred into a cylindrical flask with base radius $p$. Let the height of the milk level be $H$. - The volume of milk in the cylinder is $V = \pi p^2 H$. - Equating the volumes: $\pi p^2 H = \frac{1}{3} \pi a^2 h$. - Canceling $\pi$ from both sides: $p^2 H = \frac{1}{3} a^2 h$. - Solving for $H$: $H = \frac{a^2 h}{3 p^2}$. ### Exam Strategy & Shortcut Set up the equation instantly by remembering the ratio of volumes. A cone is exactly one-third the volume of a cylinder with the same base and height. Since bases differ, adjust directly: $\frac{1}{3}a^2h = p^2H$. ### Common Pitfall Forgetting the $\frac{1}{3}$ factor in the cone's volume formula, which incorrectly leads to option (d). ### Final Answer Therefore, the correct answer is **$\frac{a^2h}{3p^2}$ cm**.