Three rectangles $A_1$, $A_2$ and $A_3$ have the same area. Their lengths $a_1$, $a_2$ and $a_3$ respectively are such that $a_1 < a_2 < a_3$. Cylinders $C_1$, $C_2$ and $C_3$ are formed from $A_1$, $A_2$ and $A_3$ respectively by joining the parallel sides along the breadth. Then

### Concept & Logical Deduction When a rectangle is formed into a cylinder by joining the sides along its breadth, the length ($a$) forms the circumference of the cylinder's base, and the breadth ($b$) becomes the height ($h$). Volume of cylinder: $$V = \frac{\text{Circumference}^2 \times \text{Height}}{4\pi}$$ ### Step-by-Step Solution * Let the constant area of each rectangle be $K$. So, $a_i \times b_i = K$. * Circumference $C_i = a_i$, Height $h_i = b_i$. * Volume $V_i = \frac{a_i^2 b_i}{4\pi}$. * Since $a_i b_i = K$, we can rewrite the volume as: $$V_i = \frac{a_i (a_i b_i)}{4\pi} = \frac{a_i K}{4\pi}$$ * This shows that for a constant area $K$, the volume is directly proportional to the length $a_i$ ($V_i \propto a_i$). * Given $a_1 < a_2 < a_3$, it directly follows that $V_1 < V_2 < V_3$. ### Exam Strategy & Shortcut Recognize that volume depends on the square of the radius (derived from length) but only linearly on height (breadth). Squaring a larger quantity produces a much larger result. Hence, rolling along the longer side always produces a cylinder with a larger volume. Since $a_3$ is the longest, $C_3$ has the maximum volume. ### Common Pitfall Misinterpreting "joining the parallel sides along the breadth" as using the breadth for the circumference. Even so, confusing whether maximizing radius or height impacts volume more is a common trap. ### Final Answer Therefore, the correct answer is **$C_3$ will enclosed maximum volume**.