A right circular cylindrical vessel is filled to 75\% of its capacity, which equals ______ cm$^3$. The breadth of a rectangle is 10 cm more than its height, and the length of the rectangle is 4 cm more than the radius of the cylindrical vessel. The curved surface area of the cylindrical vessel is ____ cm$^2$. Given that the curved surface area of the cylindrical vessel is greater than twice the area of the rectangle, find which of the following pairs of values correctly fill the blanks (volume, curved surface area):

I. 1155, 440

II. 3696, 704

III. 1386, 528

Question

A right circular cylindrical vessel is filled to 75\% of its capacity, which equals ______ cm$^3$. The breadth of a rectangle is 10 cm more than its height, and the length of the rectangle is 4 cm more than the radius of the cylindrical vessel. The curved surface area of the cylindrical vessel is ____ cm$^2$. Given that the curved surface area of the cylindrical vessel is greater than twice the area of the rectangle, find which of the following pairs of values correctly fill the blanks (volume, curved surface area):

I. 1155, 440

II. 3696, 704

III. 1386, 528

Accepted Answer

**Solution:** Let the radius of the cylindrical vessel be $r$ cm and the height be $h$ cm. 1. **Given:** The vessel is filled to 75\% of its capacity, which equals the given volume. - Volume of cylinder = $\pi r^2 h$ - Filled volume = $0.75 imes \pi r^2 h$ 2. **Rectangle dimensions:** - Length of rectangle = $r + 4$ cm - Breadth of rectangle = Height + 10 cm Let height of rectangle be $x$ cm, so breadth = $x + 10$ cm 3. **Curved surface area (CSA) of cylinder:** - CSA = $2 \pi r h$ 4. **Condition:** CSA $>$ 2 $ imes$ area of rectangle - $2 \pi r h > 2 imes ( ext{length} imes ext{breadth})$ - $2 \pi r h > 2 imes (r + 4)(x + 10)$ - Simplifies to: $\pi r h > (r + 4)(x + 10)$ 5. **Check each option:** --- ### Option I: Volume = 1155 cm$^3$, CSA = 440 cm$^2$ - Volume filled = $0.75 \pi r^2 h = 1155$ - CSA = $2 \pi r h = 440$ From CSA: $$2 \pi r h = 440 \implies \pi r h = 220$$ From volume: $$0.75 \pi r^2 h = 1155 \implies \pi r^2 h = \frac{1155}{0.75} = 1540$$ Divide volume by CSA: $$\frac{\pi r^2 h}{\pi r h} = \frac{1540}{220} \implies r = 7$$ Now find $h$: $$\pi r h = 220 \implies 3.14 imes 7 imes h = 220 \implies h = \frac{220}{21.98} \approx 10$$ Rectangle length = $r + 4 = 7 + 4 = 11$ cm Breadth = $x + 10$ cm, height of rectangle = $x$ cm Check CSA > 2 * area of rectangle: $$440 > 2 imes 11 imes (x + 10) = 22(x + 10)$$ $$440 > 22x + 220 \implies 220 > 22x \implies x < 10$$ Height of rectangle $x$ can be less than 10, so possible. --- ### Option II: Volume = 3696 cm$^3$, CSA = 704 cm$^2$ From CSA: $$2 \pi r h = 704 \implies \pi r h = 352$$ From volume: $$0.75 \pi r^2 h = 3696 \implies \pi r^2 h = \frac{3696}{0.75} = 4928$$ Divide volume by CSA: $$\frac{\pi r^2 h}{\pi r h} = \frac{4928}{352} \implies r = 14$$ Find $h$: $$\pi r h = 352 \implies 3.14 imes 14 imes h = 352 \implies h = \frac{352}{43.96} \approx 8$$ Rectangle length = $r + 4 = 14 + 4 = 18$ cm Check CSA > 2 * area of rectangle: $$704 > 2 imes 18 imes (x + 10) = 36(x + 10)$$ $$704 > 36x + 360 \implies 344 > 36x \implies x < 9.56$$ Height of rectangle $x$ can be less than 9.56, so possible. --- ### Option III: Volume = 1386 cm$^3$, CSA = 528 cm$^2$ From CSA: $$2 \pi r h = 528 \implies \pi r h = 264$$ From volume: $$0.75 \pi r^2 h = 1386 \implies \pi r^2 h = \frac{1386}{0.75} = 1848$$ Divide volume by CSA: $$\frac{\pi r^2 h}{\pi r h} = \frac{1848}{264} \implies r = 7$$ Find $h$: $$\pi r h = 264 \implies 3.14 imes 7 imes h = 264 \implies h = \frac{264}{21.98} \approx 12$$ Rectangle length = $r + 4 = 7 + 4 = 11$ cm Check CSA > 2 * area of rectangle: $$528 > 2 imes 11 imes (x + 10) = 22(x + 10)$$ $$528 > 22x + 220 \implies 308 > 22x \implies x < 14$$ Height of rectangle $x$ can be less than 14, so possible. --- **Summary:** - All three options satisfy the volume and CSA relations with reasonable rectangle dimensions. - However, the problem states the curved surface area is greater than twice the rectangle area. - For option I, CSA = 440, twice rectangle area max = 440 (when $x=10$), so CSA is not strictly greater. - For option II, CSA = 704, twice rectangle area max = 704 (when $x=9.56$), again not strictly greater. - For option III, CSA = 528, twice rectangle area max = 528 (when $x=14$), again equal but not greater. Since the problem states CSA is greater than twice the rectangle area, only option III strictly satisfies this with $x < 14$ (any $x$ less than 14 makes CSA > twice rectangle area). **Answer:** (d) Only III

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