Question 3(ii)In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use π=227\pi = \dfrac{22}{7}π=722 )

Question

Question 3(ii)In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use π=227\pi = \dfrac{22}{7}π=722​ )

Accepted Answer

When vertically placed,Height of liquid (h) = 1+62=72\dfrac{1 + 6}{2} = \dfrac{7}{2}21+6​=27​ cmDiameter of base = 7 cmRadius (r) = 72\dfrac{7}{2}27​ cmArea of wet surface=πr2+2πrh=πr(r+2h)=227×72(72+2×72)=11×(3.5+7)=11×10.5=115.5 cm2.\text{Area of wet surface} = πr^2 + 2πrh \[1em] = πr(r + 2h) \[1em] = \dfrac{22}{7} \times \dfrac{7}{2}\Big(\dfrac{7}{2} + 2\times \dfrac{7}{2}\Big) \[1em] = 11 \times (3.5 + 7) \[1em] = 11 \times 10.5 \[1em] = 115.5 \text{ cm}^2.Area of wet surface=πr2+2πrh=πr(r+2h)=722​×27​(27​+2×27​)=11×(3.5+7)=11×10.5=115.5 cm2.Hence, area of wet surface = 115.5 cm2.

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 20

Question 3(ii)