Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.
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
$1 \frac{2}{3}$
$5 \frac{1}{3}$
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