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