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Surface Areas and Volumes — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsSurface Areas and Volumes1 Mark
Question
Match the following columns:
 
Column $I$
 
Column $II$
$a.$ A solid metallic sphere of radius $8\ cm$ is melted and the material is used to make solid right cones with height $4\ cm$ and base radius of $8\ cm$. How many cones are formed? $q.$ $8$
$b.$
A $20 - m -$ deep well with diameter $14m$ is dug up and the earth from digging is evenly spread out to form a platform $44m$ by $14m$. The height of the platform is $........ m.$
 $s.$ $5$
$c.$
A sphere of radius $6 \ cm$ is melted and recast in the shape of a cylinder of radius $4\ cm$. Then, the height of the cylinder is $......... \ cm.$
 $p.$ $18$
$d.$
The volumes of two spheres are in the ratio $64 : 27$. The ratio of their surface areas is $....... .$
 $r.$ $16 : 9$
✓

Answer

 
Column $I$
 
Column $II$
$a.$ A solid metallic sphere of radius $8\ cm$ is melted and the material is used to make solid right cones with height $4\ cm$ and base radius of $8\ cm$. How many cones are formed? $q.$ $8$
$b.$
A $20 - m -$ deep well with diameter $14m$ is dug up and the earth from digging is evenly spread out to form a platform $44m$ by $14m$. The height of the platform is $........ m.$
 $s.$ $5$
$c.$
A sphere of radius $6 \ cm$ is melted and recast in the shape of a cylinder of radius $4\ cm$. Then, the height of the cylinder is $......... \ cm.$
 $p.$ $18$
$d.$
The volumes of two spheres are in the ratio $64 : 27$. The ratio of their surface areas is $....... .$
 $r.$ $16 : 9$
$(a)$ Number of cones formed $=\frac{\text{Volume of the sphere}}{\text{Volume of each cone}}$
$=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{1}{3}\pi\text{r}^2\text{h}}$
$=\frac{4\text{r}}{\text{h}}$
$=\frac{4\times8}{4}$
$=8$
$(b)$ Volume of the earth dug out $=$ Volume of the cylinder
$=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times7^2\times20$
Let the height of the platfrom be $h.$
Volume of the platfrom $=$ volume of the cuboid
$=44\times14\times\text{h}$
$\Rightarrow\frac{22}{7}\times7^2\times20=44\times14\times\text{h}$
$\Rightarrow3080=616\times\text{h}$
$\Rightarrow\text{h}=\frac{3080}{616}$
$\Rightarrow\text{h}=5\text{m}$
$(c)$ Volume of the sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi(6)^3$
Let $ h$ be the height of the cylinder.
Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\pi(4)^2\text{h}$
$\Rightarrow\frac{4}{3}\pi(6)^3=\pi(4)^2\text{h}$
$\Rightarrow\frac{1}{3}(6)^3=(4)^2\text{h}$
$\Rightarrow\text{h}=\frac{228}{16}=18\text{cm}$
$(d)$ Let the radii of the sphere be $R$ and $r.$
Radio of their volume $=\frac{\frac{4}{3}\pi\text{R}^3}{\frac{4}{3}\pi\text{r}^3}$
$\Rightarrow\frac{\frac{4}{3}\pi\text{R}^3}{\frac{4}{3}\pi\text{r}^3}=\frac{64}{27}$
$\Rightarrow\Big(\frac{\text{R}}{\text{r}}\Big)^3=\Big(\frac{4}{3}\Big)^3$
Ratio of there surface areas $=\frac{4\pi\text{R}^2}{4\pi\text{r}^2}$
$=\Big(\frac{\text{R}}{\text{r}}\Big)^2$
$=\Big(\frac{4}{3}\Big)^2$
$=\frac{16}{9}$

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