Question

What is the radius of spherical ball in inches which is formed by melting a cylinder of base diameter 8 inches and height 160 inches, if the conversion wastage results in a 10% weight loss?

Answer 1

Answer:

radius of sphere is 12 inches

Step-by-step explanation:

volume of cylinder = πr^2h

volume of sphere = 4/3 π r^3

wastage during conversion 10%

Vol of cyl - 10% wastage = volume of sphere

$\pi {r}^{2} h \: - \: \frac{10}{100} \pi {r}^{2} h = \frac{4}{3} \pi {r}^{3} \\ \pi {r}^{2} h(1 - \frac{10}{100} ) = \frac{4}{3} \pi {r}^{3} \\ {4}^{2} \times 160 \times \frac{9}{10} = \frac{4}{3} \times {r}^{3} \\ {r}^{3} = \frac{ {4}^{2} \times 16 \times 9 \times 3}{4} \\ {r}^{3} = {4}^{2} \times 4 \times {3}^{2} \times 3 \\ {r}^{3} = {4}^{3} \times {3}^{3} \\ r = \sqrt[3]{ {4}^{3} \times {3}^{3} } \\ r = 4 \times 3 = 12$

Answer 2

Answer:

The radius of spherical ball is 19 inches.

Step-by-step explanation:

Given a cylinder of base diameter, $r= 8~ inches$

A cylinder of height, $h= 160~ inches$

Loss in wastage = 10%  

Volume of the cylinder, $V_{c} =\pi r^{2} h$

Volume of the sphere, $V_{s} =\frac{4}{3} \pi r^{3}$

Since there is a wastage, $V_{s}=V_{c}-10\% \times V_{c}$  

$\implies \frac{4}{3} \pi r^{3}= \pi r^{2}h\big(1-\frac{10}{100} \big )$

$\implies \frac{4}{3} \pi r^{3}= \pi \times 8^{2}\times 160\big(1-\frac{10}{100} \big )$

$\implies r^{3}= \frac{3}{4} \times 8^{2}\times 160\big(\frac{90}{100} \big )$

$\implies r^{3}= 3 \times 64\times 4 \times 9$

$\implies r^{3}= 6912$  

$\implies r= \sqrt[3]{ 6912}\approx19$

Therefore, the radius of spherical ball is 19 inches.