Q) 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 21 cm.
Cylindrical pool – Outer diameter is 50 cm and inner diameter is 40 cm.
Height of solid base is 14 cm.
Height of water filled is 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.
(iv) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.

(Q 36 – 30/3/3 – CBSE 2026 Question Paper)

Ans:

(i) Total height of the fountain:

Total Height of the fountain = Height of Solid base + Diameter of sphere

= 14 cm + 21 cm = 35 cm

Therefore, the total height of the fountain is 35 cm.

(ii) Volume of the ball:

We know that the volume of a sphere is given by, V_Sph = π r ³

Here, we have diameter of the spherical ball = 21 cm

∴ Radius of the spherical ball = cm

∴ Volume of the spherical ball, V_Sph= π r ³

=

=

= 4 x 22 x  = 8 x 11 x

= x 11 x

= 11 x 21 x 21 = 11 x 441 = 4851 cm ³

Therefore, the volume of the ball is 4,851 cm ³

(iii) Volume of the water filled in the pool:

When the ball is submerged, the water occupies the cylindrical space minus the volume of the submerged part of the ball.

∴ Volume of the water filled in the pool = Volume of water filled – Volume occupied by the ball

Step 1: Now, Volume of water filled = Inner cylinderical volume above solid base

We know that the volume of a cylinder is given by, V_Cyl = π r ² h

Here, we have inner diameter of cylinder = 40 cm

∴ Inner radius of cylinder = = 20 cm

From diagram, height of water level above solid base = 7 cm

∴ Volume of water filled = π r ² h = π (20)² (7)

= (20) ² (7) = 22 x 400 = 8800 cm ³

Step 2: Next, Volume occupied by the ball inside water

= 1/3 of ball’s volume           (∵ 1/3 ^rd ball is submerged)

= x 4851       (∵ ball’s volume = 4851 from part (i))

= 1617 cm ³

Step 3: ∴ Volume of the water filled in the pool

= Volume of water filled – Volume occupied by the ball

= 8800 – 1617 = 7183 cm ³

∴ Therefore, the volume of the water filled in the pool is 7,183 cm ³.

(iv) Sum of the outer curved surface area of the cylindrical part and surface area of the ball:

Step 4: Outer curved surface area of cylindrical part

We know that, curved surface area of the cylinder = 2 π r h

Here, we have outer diameter of cylinder = 50 cm

∴ Outer radius of cylinder = = 25 cm

Height of the cylinder = Height of solid part + Height of water level

= 14 + 7 = 21 cm

∴ curved surface area of the cylinder

= 2 π r h = 2 x (25) (21)

= 2 x 22 x 25 x 3 = 3300 cm ²

Step 5: Surface area of the ball:

We know that, surface area of the sphere = 4 π r ²

Here, we have Diameter of the spherical ball = 21 cm

∴ Radius of the spherical ball = cm

∴ Surface area of the spherical ball = 4 π r ²

=

=

= x 21 x 21

= 22 x 3 x 21 = 66 x 21 = 1386 cm ²

Step 6: Sum of the outer curved surface area of the cylindrical part and surface area of the ball
= 3300 + 1386 = 4686 cm²

Therefore, Sum of the outer curved surface area of the cylindrical part and surface area of the ball is 4,686 cm².

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