Let's analyze and solve the questions step-by-step from the image.

Given:

• Height ($h$) = $24 \, \text{cm}$
• Radius ($r$) = $15 \, \text{cm}$

The cylindrical tin is closed on both ends. For a cylinder:

• Volume ($V$) = $\pi r^2 h$
• Surface Area ($A$) = $2\pi r^2 + 2\pi r h$, which includes both the circular ends and the curved side.

7.1: Volume of the tin

$V = \pi r^2 h$ Substitute $r = 15 \, \text{cm}$ and $h = 24 \, \text{cm}$: $V = \pi (15)^2 (24) = \pi (225)(24) = 5400\pi \, \text{cm}^3$ Using $\pi \approx 3.1416$: $V \approx 5400 \times 3.1416 = 16,964.6 \, \text{cm}^3$

7.2: Total Surface Area of the tin

$A = 2\pi r^2 + 2\pi r h$ Substitute $r = 15 \, \text{cm}$ and $h = 24 \, \text{cm}$: $A = 2\pi (15)^2 + 2\pi (15)(24)$ $A = 2\pi (225) + 2\pi (360) = 450\pi + 720\pi = 1170\pi \, \text{cm}^2$ Using $\pi \approx 3.1416$: $A \approx 1170 \times 3.1416 = 3,674.5 \, \text{cm}^2$

7.3: Total Surface Area if the dimensions are doubled

If both $r$ and $h$ are doubled:

• New radius = $2r = 30 \, \text{cm}$
• New height = $2h = 48 \, \text{cm}$

New surface area: $A_{\text{new}} = 2\pi (2r)^2 + 2\pi (2r)(2h)$ $A_{\text{new}} = 2\pi (30)^2 + 2\pi (30)(48)$ $A_{\text{new}} = 2\pi (900) + 2\pi (1440) = 1800\pi + 2880\pi = 4680\pi \, \text{cm}^2$ Using $\pi \approx 3.1416$: $A_{\text{new}} \approx 4680 \times 3.1416 = 14,702.1 \, \text{cm}^2$

Final Answers:

1. Volume of the tin: $\mathbf{16,964.6 \, \text{cm}^3}$
2. Total Surface Area: $\mathbf{3,674.5 \, \text{cm}^2}$
3. Total Surface Area if dimensions are doubled: $\mathbf{14,702.1 \, \text{cm}^2}$

Would you like further details or help with similar problems? Here are some related questions:

1. How does doubling dimensions affect the volume of the cylinder?
2. Can you calculate the ratio of the new surface area to the original surface area?
3. What happens to the curved surface area if only the height is doubled?
4. How does the surface area change if only the radius is doubled?
5. How do real-life factors (e.g., material thickness) affect these calculations?

Tip: Always separate the contributions of the curved surface area and the two circular bases for clarity when solving surface area problems.