Maths M.C.Q

We know that,
Volume of cone $=\frac{1}{3} \pi r^2 h$
$\Rightarrow 1232=\frac{1}{3} \pi r^2 \mathrm{~h}$
$\Rightarrow 1232=\frac{1}{3} \times \frac{22}{7} \times \mathrm{r}^2 \times 24$
$\Rightarrow \mathrm{r}^2=\frac{7 \times 3 \times 1232}{24 \times 22}$
$\Rightarrow \mathrm{r}^2=49$
$\Rightarrow \mathrm{r}=7 \mathrm{~cm}$
$\Rightarrow \mathrm{r}=7 \mathrm{~cm} \text { and } \mathrm{h}=24 \mathrm{~cm}$
$\mathrm{l}^2=\mathrm{r}^2+\mathrm{h}^2$
$\Rightarrow \mathrm{l}^2=7^2+24^2$
$\Rightarrow \mathrm{l}^2=49+576$
$\Rightarrow \mathrm{l}^2=625$
$\Rightarrow \mathrm{l}=25$
Curved surface area $=\pi \mathrm{rl}$
$=\frac{22}{7} \times 7 \times 25$
$=550 \mathrm{~cm}^2$

Given, $d=6 \mathrm{~cm}$
$\mathrm{r}=3 \mathrm{~cm}$
$\mathrm{h}=14 \mathrm{~cm}$
Now，
Volume of the cylinder $=\pi r^2 h$
$=\frac{22}{7} \times 3 \times 3 \times 14$
$=396 \mathrm{~cm}^3$

Given,
Height $= 14\ cm$
Curved surface area = $264 \mathrm{~cm}^2$
Now,
Curved surface area $=2\pi\text{rh}$
$\Rightarrow264=2\times\frac{22}{7}\times\text{r}\times14$
$\Rightarrow264=88\times\text{r}$
$\Rightarrow\text{r}=\frac{264}{88}$
$\Rightarrow\text{r}=3\text{cm}$
Volume $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times3\times3\times14$
$=396\text{cm}^3$

Volume of a cuboid $=$ length $\times$ breadth $\times$ height
$=15 \times 12 \times 4.5$
$=810 \mathrm{~cm}^3$

Given radius $=10.5=\frac{21}{2}\text{cm}$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$
$=11\times21\times21$
$=4851\text{cm}^3$

Let the radius be $2x$ and the height be $3x\ cm.$
We know that,
Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\pi\times(2\text{x})^2\times3\text{x}$
$=\frac{22}{7}\times12\times\text{x}^3$
$\Rightarrow1617=\frac{22}{7}\times12\text{x}^3$
$\Rightarrow\text{x}^3=\frac{7\times1617}{22\times12}$
$\Rightarrow\text{x}^3=\frac{7\times49}{2\times4}$
$\Rightarrow\text{x}^3=\frac{343}{8}$
$\Rightarrow\text{x}^3=\Big(\frac{7}{2}\Big)^3$
$\Rightarrow\text{x}=\frac{7}{2}$
$\therefore$ radius $=2\times\frac{7}{2}=7\text{cm}$
Height $=3\times\frac{7}{2}=\frac{21}{2}\text{cm}$
$\therefore$ total surface area $=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{35}{2}\Big)$
$=770\text{cm}^2$

Volume of the solid lead ball $=\frac{4}{3}\pi\times\text{r}^3$
$=\frac{4}{3}\pi\times6^3=288\pi\text{cm}^3$
Let the length of the wire be $h.$
Its radius $=\text{r}_1=\frac{0.2}{2}=0.1\text{cm}$
Volume of the wire $=\pi\text{r}_1^2\text{h},$ where $r_1$ is the radius of the wire
Since the wire is drawn into a solid lead ball,
Volume of the solid lead ball $=$ volume of the wire
$\Rightarrow288\pi=\pi(0.1)^2\text{h}$
$\Rightarrow\text{h}=28800\text{cm}=288\text{m}$

Let the slant height be $l$ and $2l$ and their radii be $r_1$ and $r_2$
The ratio of the curved surface area is given by $=\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}$
$\Rightarrow\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}=\frac{2}{1}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2(2)}=\frac{2}{1}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{1}$
$⇒$ Ratio of their radii is $4 : 1.$

Area of the ground $=$ number of person $×$ the amount of space each person occupies
$= 11 × 4$
$\Rightarrow\pi\text{r}^2=44\text{m}^2\ ...(\text{i})$
Given that the volume = $220 \mathrm{~m}^3$
$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=220$
$\Rightarrow\frac{1}{3}\times44\times\text{h}=220 ...($from $(i))$
$\Rightarrow\text{h}=\frac{220\times3}{44}$
$\Rightarrow\text{h}=15\text{m}$
$⇒$ Height of the cone $= 15m.$

Let $r$ be the radius of the cone.
Volume = $1570 \mathrm{~cm}^3$
$1570=\frac{1}{3}\times3.14\times\text{r}^2\times15$
$\Rightarrow1570=3.14\times\text{r}^2\times15$
$\Rightarrow\text{r}^2=100$
$\Rightarrow\text{r}=10\text{cm}$

Required number of persons $=\frac{\text{length}\times\text{breadth}\times\text{height}}{\text{amount}\ \text{of}\ \text{air}\ \text{each}\ \text{person}\ \text{requires}}$
$=\frac{20\times15\times4.5}{5}$
$=270$

Let the radii be $x\ cm$ and $(7 - x)\ cm.$
Volume of the two spheres are in the ratio $64 : 27.$
$\Rightarrow\frac{\frac{4}{3}\pi\text{x}^3}{\frac{4}{3}\pi(7-\text{x})^3}=\frac{64}{27}$
$\Rightarrow\Big(\frac{\text{x}}{7-\text{x}}\Big)^3=\Big(\frac{4}{3}\Big)^3$
$\Rightarrow\frac{\text{x}}{7-\text{x}}=\frac{4}{3}$
$\Rightarrow\text{x}=4\text{cm}$
So, their radii are $4\ cm$ and $3\ cm.$
Difference of their tital surface areas
$=4\pi(4)^2-4\pi(3)^2$
$=4\times\frac{22}{7}(16-9)$
$=88\text{cm}^2$

The diameter of the roller $=84 \mathrm{~cm}=\frac{84}{100} \mathrm{~m}$
So, the radius $=\frac{84}{200} \mathrm{~m}$
The area covered by the roller in $1$ revolution
$=2 \pi \mathrm{rh}$
$=2 \times \frac{22}{7} \times \frac{84}{200} \times 1$
$=2.64 \mathrm{~m}^2$
$\therefore$ Area covered in $500$ complete revolution $=500 \times 2.64=1320 \mathrm{~m}^2$
Thus, the area of the playground is $1320 \mathrm{~m}^2$.

Given that, $r=14 \mathrm{~cm}$
Curved surface area $=1760 \mathrm{~cm}^2$
Now,
Curved surface area $=2\pi\text{rh}$
$\Rightarrow1760=2\times\frac{22}{7}\times14\times\text{h}$
$\Rightarrow1760=88\times\text{h}$
$\Rightarrow\text{h}=\frac{1760}{88}$
$\Rightarrow\text{h}=20\text{cm}$

Let $r$ be the radius of the cone.
$l^2=h^2+r^2$
$\Rightarrow r^2=l^2-h^2$
$=28^2-21^2$
$=49 \times 7$
$\Rightarrow r=7 \sqrt{7} \mathrm{~cm}$
Volume of the cone $=\frac{1}{3} \pi r^2 h$
$\Rightarrow$ Volume of the cone $=\frac{1}{3} \times \frac{22}{7} \times(7 \sqrt{7})^2 \times 21$
$\Rightarrow$ Volume of the cone $=7546 \mathrm{~cm}^3$

The height of the cone is $24\ cm$ and the diameter of its base is $14\ cm$
So, its radius $= 7\ cm$
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{7^2+24^2}$
$\text{l}=\sqrt{49+576}$
$\text{l}=25\text{cm}$
So, curved surface area of the cone $=\pi\text{r}=\frac{22}{7}\times7\times25=550\text{cm}^2$

We know that $1 \mathrm{dm}=10 \mathrm{~cm}$.
Given that $2.2 \mathrm{dm}^3$ is drawn into a cylindrical wire.
That is, $(2.2 \times 1000)=2200 \mathrm{~cm}^3$ is drawn into a cylindrical wire.
Let the radius of the wire be $r$ and the height be $h$.
Volume of the cylindrical $=2200$
$\Rightarrow \pi \mathrm{r}^2 \mathrm{~h}=2200$
$\Rightarrow \frac{22}{7} \times 0.25 \times 0.25 \times \mathrm{h}=2200 \ldots($ Since the diameter $=0.50 \mathrm{~cm})$
$\Rightarrow \mathrm{h}=11200 \mathrm{~cm}$
$\Rightarrow \mathrm{h}=112 \mathrm{~m}$
So, the length of the wire is $112 m .$

Let each edge be $a$
$⇒$ its surface area $= 6a^2$
Now,
New edge $= 150\%$ of a
$=\frac{150}{100}\times\text{a}$
$=\frac{3\text{a}}{2}$
New surface area $=6\times\Big(\frac{3\text{a}}{2}\Big)^2$
$=\frac{27\text{a}^2}{2}$
Increase surface area $=\frac{27\text{a}^2}{2}-6\text{a}^2$
$=\frac{27\text{a}^2-12\text{a}^2}{2}$
$=\frac{15\text{a}^2}{2}$
Percentage increase in its surface area
$=\frac{\text{Increase}\ \text{in}\ \text{the}\ \text{surface}\ \text{area}}{\text{original}\ \text{surface}}\times100$
$=\frac{15\text{a}^2}{2}\times\frac{1}{6\text{a}^2}\times100$
$=125\%$

Let the original side be $' x ' cm$
$\therefore$ Original volume $=x^3 \mathrm{~cm}^3$
Now,
New side $=2 \mathrm{xcm}$
$\therefore \text { New volume }=(2 x)^3$
$=8 x^3 \mathrm{~cm}^3$
So, the volume becomes $8$ times.

Surface area of a sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=4\times22\times3\times21$
$=5544\text{cm}^2$