Find the approximate volume of metal in a hollow spherical

Solution:

Given,

The internal radius r = 3 cm

And, external radius R = r + ∆r =3.0005 cm

∆r = 3.0005 – 3 = 0.0005 cm

Let y = r3 ⇒ y + ∆y = (r + ∆r)3 = R3 = (3.0005)3

Differentiating both sides w.r.t., r, we get

$\frac{d y}{d r}=3 r^{2}$

So, $\Delta y=\frac{d y}{d r} \times \Delta r=3 r^{2} \times 0.0005$

$=3 \times(3)^{2} \times 0.0005=27 \times 0.0005=0.0135$

$\therefore(3.0005)^{3}=y+\Delta y$

Volume of the shell $=\frac{4}{3} \pi\left[R^{3}-r^{3}\right]$

$=\frac{4}{3} \pi[27.0135-27]=\frac{4}{3} \pi \times 0.0135$

 

$=4 \pi \times 0.005=4 \times 3.14 \times 0.0045=0.018 \pi \mathrm{cm}^{3}$

Therefore, the approximate volume of the metal in the shell is 0.018π cm3

$1 \frac{2}{3}$

 

$5 \frac{1}{3}$