Question

The diameter and thickness of a hollow metals sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere.

Answer 1

A.T.Q
Diameter of hollow sphere is 12 cm
it's thickness is 0.01 m
Density of the metal is 8.88 gm/cm^3
Outer surface area of sphere:
$4\pi {r}^{2}$

Radius=Diameter/2
Radius=12/2= 6 cm
outer surface area
$= 4 \times \frac{22}{7} \times {(6)}^{2} \\ = \frac{4 \times 22 \times3 6}{7} \\ = 452.57 {cm}^{2}$
Now,Mass of the sphere m= density *volume
To calculate volume V=
$\frac{4}{3} \pi {r}^{3}$
for inner V1=
$\frac{4}{3} \times \pi \times {(r1)}^{3}$
inner radius= outer radius - thickness
= 6-1
=5 cm
V1
$= \frac{4}{3} \times \frac{22}{7} \times ( {5)}^{3} \\ = \frac{4 \times 22 \times 125}{21} \\ = 523.81 \: {cm}^{3}$
V2=
$\frac{4 \times 22 \times ( {6)}^{3} }{21} \\ = 905.15 {cm}^{3}$
Volume=V2-V1

$905.15 - 523.81 \\ = 381.34 {(cm)}^{3}$
Mass of sphere=8.88(381.34)
=3386.30 gm
or 3.39 kg

Answer 2

$formul a \: volume \: \: of \: hollow \: sphere \: \\ \frac{4}{3} \pi \: ( {R }^{3} - {r}^{3} ) \\ \frac{4}{3} \times \frac{22}{7} \times ( {6}^{3} - {5}^{3} ) \\ \frac{88}{3 \times 7} \times (216 - 125) \\ \frac{88}{3 \times 7} \times 91 \\ \frac{88 \times 13}{7} = 381.34 {cm}^{3} \\ mass \: of \: sphere = 381.34 \times 8.88 \\ = 3386.30gm \\ = 3.38kg$
■I HOPE ITS HELP■