Question

Question 5
Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. Determine the radius of the sphere, if the diameter of the smaller one is 5 cm.

Answer 1

Given, weight of one solid sphere,
$m_1=5920g$
Weight of another solid sphere,
$m_2=740g$
Diameter of the smaller sphere = 5 cm
$\therefore$ Radius of the smaller sphere,
$r_2=\frac{5}{2},m_2=740g$

We know that,
$Density=\frac{Mass\left(M\right)}{Volume\left(D\right)}$

$\Rightarrow Volume,V=\frac{M}{D}$

$\Rightarrow V_1=\frac{5920}{D}c{m}^3\dots \left(i\right)$

$V_2=\frac{740}{D}c{m}^3\dots \left(ii\right)$

On dividing eq. (i) by eq. (ii), we get,

$=\frac{V_1}{V_2}=\left(\frac{5920}{D}\right)(\frac{D}{740})$ [Density of the both spheres is same]
$\because$ Volume of a sphere $=\frac{4}{3}\pi r^3$
$\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}=\frac{5920}{740}\Rightarrow {\left(\frac{r_1}{r_2}\right)}^3=\frac{592}{74}$
$\Rightarrow {\left(\frac{r_1}{\frac{5}{2}}\right)}^3=\frac{592}{74}$ $[\because r_2=\frac{5}{2}cm]$
$\Rightarrow \frac{r_1^3}{\frac{125}{8}}=\frac{592}{74}\Rightarrow \frac{8r_1^3}{125}=\frac{592}{74}$
$\Rightarrow r_1^3=\frac{592}{74}\times \frac{125}{8}=\frac{74000}{592}=125$
$\therefore r_1=5cm$
Hence, the radius of the larger sphere is 5 cm.