Question

Two solid spheres are made of the same metal weighing 5920g and 740g respectively.
Determine the radius of larger sphere , if the diameter of the smaller one is 5cm.​

Answer 1

$\huge\bf\orange{ANSWER:-}$

Let the radius of the bigger sphere be r cm, then

$its \: volume = \frac{4}{3} \pi \: r {}^{3} cm {}^{3}$

Radius of the smaller sphere = ½×5

= 5/2 cm

Therefore,

$Volume \: of \: the \: smaller \: sphere$

$= \frac{4}{3} \pi \: (\frac{5}{2} ) {}^{3} cm {}^{3}$

As the two spheres are made up of the same metal, their volumes are proportional to their weights

Therefore,

$\frac{volume \: of \: bigger \: sphere}{volume \: of \: smaller \: sphere} = \frac{weight \: of \: bigger \: sphere}{weight \: of \: smaller \: sphere \: }$

$= > \frac{4}{3} \pi \: r {}^{3} \div \frac{4}{3}\pi ( \frac{5}{2} ) {}^{3} = \frac{5920}{740}$

$= > \frac{ \frac{r {}^{3} }{125} }{8} = 8$

$= > r {}^{3} = 125$

$= > r = 5$

Hence, the radius of the bigger sphere is $\huge\bold{5cm}$

Answer 2

Answer :

The radius of the larger sphere is 5 cm.

Step by step solution :

Let, R be the radius of the larger sphere.

Then, it's volume be V = $\frac{4}{3}\pi R^{3}$

Similarly, given smaller sphere with 5 cm diameter's volume be

v = $\frac{4}{3}\pi (\frac{5}{2})^{3}\:cm^{3}$

Since, both the spheres are made of same metal, their density be same (say ρ).

We know that,

Density of metal = $\frac{Mass}{Volume}$

For larger sphere,

ρ = $\frac{5920}{\frac{4}{3}\pi R^{3}}$ ...(i)

For smaller sphere,

ρ = $\frac{740}{\frac{4}{3}\pi (\frac{5}{2})^{3}}$ ...(ii)

Equating (i) and (ii), we get

$\frac{5920}{\frac{4}{3}\pi R^{3}}=\frac{740}{\frac{4}{3}\pi (\frac{5}{2})^{3}}$

$\to R^{3} =\frac{5920\times (\frac{5}{2})^{3}}{740}$

$\to R^{3} = 125$

$\to R = 5$

Thus, the radius of the larger sphere is 5 cm