Question

a cylinderical container of 54cm height and 16cm radius is filled with sand.now akl this sand isused to form aconical heap of sand .ifthe height of the conical heap is 18cm, what is the radius of its base

Answer 1

Volume of sand in cylinder = Volume of cone formed by the sand.


Volume of sand in cylinder = πr²h

r = 16cm,
height = 54cm

=> Volume = π × 16² × 54

=> Volume = π × 256 × 54

=> Volume = 13,824π cm³

So this volume of sand is used in making the cone.

Volume of cone = 1/3 × πr²h

h = 18 cm

=> 13,824 π = 1/3 × π × 18 × r²

Cancel π from both sides

=> 13,824 = 6 × r² (Since, 18 × 1/3 = 6)

=> r² = 13824/6

=> r² = 2304

=> r = √2304

=> r = 48 cm

Answer :- 48 cm

Answer 2

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Solution:-

Since, The Sand present in Cylinder is used to make Conical heap.

=> Volume of Cylinder = Volume of Conical Heap.

=> Volume of Cylinder

$= > \pi {r}^{2} h \\ \\ = > \frac{22}{7} \times 16 \times 16 \times 54 ...........(1)$

Now, Volume of Conical heap = Volume of Cone = Volume of Cylinder.

=> Volume of Cone

$= > \frac{22}{7} \times 16 \times 16 \times 54 = \frac{1}{3} \times \pi \times {r}^{2} \times h \\ \\ = > \frac{22}{7} \times 16 \times 16 \times 54 = \frac{1}{3} \times \ \frac{22}{7} \times {r}^{2} \times 18 \\ \\ = > 16 \times 16 \times 54 = \frac{1}{3} \times 18 \times {r}^{2} \\ \\ = > {r}^{2} = \frac{16 \times 16 \times 54 \times 3}{18} \\ \\ = > {r}^{2} = 2304 \\ \\ = > r = \sqrt{2304} \\ \\ = > r = 48cm$

Hence, Radius = 48 cm.