# NCERT Exemplar Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.1 in Chapter 13 - Surface Areas and Volumes

Question 3 Surface Areas and Volumes - Exercise 13.1

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere.

The radius of the sphere is:

(A) 4.2 cm (B) 2.1 cm (C) 2.4 cm (D) 1.6 cm

(B) 2.1 cm

Explanation:

Height of cone, h = 8.4cm

Radius of base, r = 2.1cm

Thus volume of a cone = (1/3) × π × h × r2

=(1/3) × π × 8.4 × (2.1)2

Now when it is melted to form a sphere, say of radius r1 cm, the volumes of both are going to be equal.

Volume of sphere = (4/3) × π × r13

\begin{array}{l} \therefore \frac{4}{3} \pi \times r_{1}^{3}=\frac{1}{3} \pi \times 2.1^{2} \times 8.4 \\ \quad=2.1^{2} \times 8.4 \\ r_{1}^{3}=2.1^{2} \times 2.1 \\ r_{1}^{3}=2.1^{3} \\ \therefore r_{1}=2.1 \mathrm{cm} \end{array}

Thus the radius of the sphere is 2.1cm which is option b.

Hence, option B is the correct answer.

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