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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 3 · Set 1 · Q38
A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is $\displaystyle 21$ cm. Cylindrical pool - Outer diameter is $\displaystyle 50$ cm and inner diameter is $\displaystyle 40$ cm. Height of solid base is $\displaystyle 14$ cm. Height of water filled is $\displaystyle 7$ cm.
Observe the figure and answer the following questions :(i)Determine the total height of the fountain.(ii)Find the volume of the ball.(iii)If one-third of the ball is submerged in the water, find the volume of the water filled in the pool.Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is $\displaystyle 21$ cm. Cylindrical pool - Outer diameter is $\displaystyle 50$ cm and inner diameter is $\displaystyle 40$ cm. Height of solid base is $\displaystyle 14$ cm. Height of water filled is $\displaystyle 7$ cm.
Observe the figure and answer the following questions :
(i)
Determine the total height of the fountain.
(ii)
Find the volume of the ball.
(iii)
If one-third of the ball is submerged in the water, find the volume of the water filled in the pool.
Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
Marking-scheme solution
(i) Total height of the fountain \(\displaystyle =14+21=35 \mathrm{~cm}\)
(ii) Volume of the ball \(\displaystyle =\frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}\)
\[=4851 \mathrm{~cm}^{3}
\]
(iii) (a) Volume of water = Volume of inner upper part \(\displaystyle -\frac{1}{3} \times\) Volume of the ball
\[\begin{aligned}
& =\frac{22}{7} \times 20 \times 20 \times 7-\frac{1}{3} \times 4851 \\
& =7183 \mathrm{~cm}^{3}
\end{aligned}
\]
OR
(iii) (b) Required area = Outer CSA of cylinderical part + Surface area of the ball
\[\begin{aligned}
& =2 \times \frac{22}{7} \times 25 \times(14+7)+4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \\
& =4686 \mathrm{~cm}^{2}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.