ABC26GN5011 · Volume and Surface Area
Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
A solid piece of iron of dimensions $49 \times 33 \times 24 \mathrm{cm}$ is moulded into a sphere. The radius of the sphere is
(a)21 cm
(b)28 cm
(c)35 cm
(d)None of these
Answer
Explanation
Volume of the solid $=(49 \times 33 \times 24) \mathrm{cm}^{3}$. Let the radius of the sphere be $r$. $$\begin{array}{r} \text { Then, } \frac{4}{3} \pi r^{3}=(49 \times 33 \times 24) \Leftrightarrow r^{3}=\left(\frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22}\right) \\ =(21)^{3} \Leftrightarrow r=21 . \end{array}$$
Explanation as extracted from the printed page; notation may be imperfect.
Open in whiteboard · Browse this chapter in the app