4. A pillar in the shape of a cylinder has radius 21 cm and height 3 m . Find the curved surtace area and the volume of the pillar. 5. The volume of a 1 metre iron rod is \( 1386 \mathrm{cu} . \mathrm{cm} \). Find its diameter. 6. A rectangular piece of paper, 44 cm long and 16 cm broad, is rolled along its length to form a cylinder. What is the radius of the cylinder? What is its volume? 7. A tank is in the form of a right circular cylinder. Its height is 48 m and the base area is 616 sq . m . Find the lateral surface area and volume of the cylinder.

Certainly! Let's tackle each of the problems one by one. --- ## **Problem 4** **A pillar in the shape of a cylinder has a radius of 21 cm and a height of 3 m. Find the curved surface area and the volume of the pillar.** **Given:** - Radius, \( r = 21 \) cm - Height, \( h = 3 \) m \( = 300 \) cm (since \(1 \, \text{m} = 100 \, \text{cm}\)) ### **a. Curved Surface Area (CSA)** The formula for the curved surface area of a cylinder is: \[ \text{CSA} = 2\pi r h \] **Calculations:** \[ \text{CSA} = 2 \times \pi \times 21 \, \text{cm} \times 300 \, \text{cm} \] \[ \text{CSA} = 2 \times 3.1416 \times 21 \times 300 \] \[ \text{CSA} \approx 2 \times 3.1416 \times 21 \times 300 \approx 39,\!583.2 \, \text{cm}^2 \] **Alternatively, expressing in terms of \( \pi \):** \[ \text{CSA} = 2 \times \pi \times 21 \times 300 = 12,\!600\pi \, \text{cm}^2 \] ### **b. Volume of the Cylinder** The formula for the volume of a cylinder is: \[ \text{Volume} = \pi r^2 h \] **Calculations:** \[ \text{Volume} = \pi \times (21)^2 \times 300 \] \[ \text{Volume} = \pi \times 441 \times 300 \] \[ \text{Volume} = 132,\!300\pi \, \text{cm}^3 \approx 415,\!480 \, \text{cm}^3 \] --- ## **Problem 5** **The volume of a 1-meter iron rod is \( 1386 \, \text{cm}^3 \). Find its diameter.** **Given:** - Volume, \( V = 1386 \, \text{cm}^3 \) - Length (height), \( h = 1 \) m \( = 100 \) cm Assuming the rod is cylindrical: \[ V = \pi r^2 h \] **Solving for Radius (\( r \)):** \[ 1386 = \pi r^2 \times 100 \] \[ r^2 = \frac{1386}{100\pi} = \frac{13.86}{\pi} \approx \frac{13.86}{3.1416} \approx 4.416 \, \text{cm}^2 \] \[ r \approx \sqrt{4.416} \approx 2.1 \, \text{cm} \] **Diameter (\( d \)) is twice the radius:** \[ d = 2r = 2 \times 2.1 \, \text{cm} = 4.2 \, \text{cm} \] **Answer:** The diameter of the iron rod is approximately **4.2 cm**. --- ## **Problem 6** **A rectangular piece of paper, 44 cm long and 16 cm broad, is rolled along its length to form a cylinder. What is the radius of the cylinder? What is its volume?** **Given:** - Length of paper, \( L = 44 \) cm (becomes the circumference of the cylinder) - Breadth of paper, \( B = 16 \) cm (becomes the height of the cylinder) ### **a. Radius of the Cylinder** The circumference (\( C \)) of the cylinder is equal to the length of the paper: \[ C = 2\pi r = 44 \, \text{cm} \] \[ r = \frac{44}{2\pi} = \frac{22}{\pi} \approx \frac{22}{3.1416} \approx 7 \, \text{cm} \] **Answer:** The radius of the cylinder is approximately **7 cm**. ### **b. Volume of the Cylinder** The formula for the volume of a cylinder is: \[ \text{Volume} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height. **Given:** - \( r \approx 7 \) cm - \( h = 16 \) cm **Calculations:** \[ \text{Volume} = \pi \times (7)^2 \times 16 = \pi \times 49 \times 16 = 784\pi \, \text{cm}^3 \approx 2,\!462.4 \, \text{cm}^3 \] **Answer:** The volume of the cylinder is approximately **2,\!462.4 cm³**. --- ## **Problem 7** **A tank is in the form of a right circular cylinder. Its height is 48 m and the base area is \( 616 \, \text{m}^2 \). Find the lateral surface area and volume of the cylinder.** **Given:** - Height, \( h = 48 \) m - Base area, \( A = 616 \, \text{m}^2 \) ### **a. Lateral Surface Area (LSA)** First, we need to find the radius of the base. \[ A = \pi r^2 = 616 \, \text{m}^2 \] \[ r^2 = \frac{616}{\pi} \approx \frac{616}{3.1416} \approx 196 \, \text{m}^2 \] \[ r = \sqrt{196} = 14 \, \text{m} \] Now, the lateral surface area: \[ \text{LSA} = 2\pi r h \] \[ \text{LSA} = 2 \times \pi \times 14 \times 48 = 2 \times 3.1416 \times 14 \times 48 \approx 4,\!224 \, \text{m}^2 \] \[ \text{Alternatively, expressing in terms of } \pi: \] \[ \text{LSA} = 2 \times \pi \times 14 \times 48 = 1,\!344\pi \, \text{m}^2 \approx 4,\!224 \, \text{m}^2 \] ### **b. Volume of the Cylinder** The formula for the volume is: \[ \text{Volume} = \text{Base Area} \times \text{Height} = A \times h \] \[ \text{Volume} = 616 \, \text{m}^2 \times 48 \, \text{m} = 29,\!568 \, \text{m}^3 \] **Answer:** - **Lateral Surface Area:** \( 1,\!344\pi \, \text{m}^2 \) or approximately **4,\!224 m²** - **Volume:** **29,\!568 m³** --- If you have any further questions or need additional explanations, feel free to ask!