Question

Find the volume of a sphere ,if its surface area is 154 sq.cm.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that radius of sphere = r cm

Given that,

$\rm \: Surface Area_{(Sphere)} = 154$

$\rm \: 4\pi {r}^{2} \: = \: 154$

$\rm \: 4 \times \dfrac{22}{7} \times {r}^{2} \: = \: 154$

$\rm \: 4 \times \dfrac{1}{7} \times {r}^{2} \: = \: 7$

$\rm \: {r}^{2} = \dfrac{7 \times 7}{4}$

$\rm \: {r}^{2} = \dfrac{7 \times 7}{2 \times 2}$

$\rm\implies \:r \: = \: \dfrac{7}{2} \: cm$

Now,

$\rm \: Volume_{(Sphere)} \:$

$\rm \: = \: \dfrac{4}{3} \pi {r}^{3}$

$\rm \: = \: \dfrac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$

$\rm \: = \: \dfrac{11 \times 49}{3}$

$\rm \: = \: \dfrac{539}{3}$

$\rm \: \approx \: 179.67 \: {cm}^{3}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Volume_{(Sphere)} = \frac{539}{3} \approx \: 179.67 \: {cm}^{3} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

V≈179.7

Step-by-step explanation: