Question

plz answer quetion no (a)

Answer 1

Heya User, Here is Your Answer :

Given, Radius of Sphere (r) = 0.21 cm

Given, Density of Iron = 7.8 g/m^3

Volume of Sphere =
$\frac{4}{3} \pi \times r {}^{3} \\ = > \: \frac{4}{3} \times 3.14 \times (0.21) {}^{3} \\ = > \: \frac{4}{3} \times \frac{157}{50} \times ( \frac{21}{100} ) {}^{3} \\ = > \: \frac{2}{3} \times \frac{157}{25} \times \frac{ {21}^{3} }{ {100}^{3} } \\ = > \: \frac{314 \times 9261}{75 \times 100 {}^{3} } \\ = > \: \frac{969318}{25 \times 100 {}^{3} } \\ = > \: 0.03877 \: cm {}^{ 3}$

Now, As we Know that Density = Mass per unit Volume or Mass ÷ Volume

Therefore,

$density \: = \frac{mass}{volume} \\ = > \: mass \: = density \: \times volume \\ = > \: mass \: = 7.8 \: \times 0.03877 \\ mass \: = 0.3024 \: grams$


Hope it would help you