MATHS M.C.Q

Length of the river $= 1.5\ m$
Breadth of the river $= 30\ m$
Depth of the river $= 3\ km = 3000m$
Now, volume of water that runs into the sea $= 1. 5 × 30 × 3000\ m^3= 135000\ m^3$
$\therefore$ Volume of water that runs into the sea per minute $=\frac{135000}{60}=2250\text{m}^3$

Let the radii of two cylinders be $2r$ and $3r$ respectively and their heights are in the ratio $5h$ and $3h.$ Their volumes be $\mathrm V_1$ and $\mathrm V_2$. Than,
$\frac{\text{V}_1}{\text{V}_2}=\frac{\pi(2\text{r})^2(5\text{h})}{\pi(3\text{r})^2(3\text{h})}=\frac{4\text{r}^2\times5\text{h}}{9\text{r}^2\times3\text{h}}=\frac{20}{27}$
Hence, $(b)$ is the correct answer.

Let the required number of balls be $n.$
$\Rightarrow\text{n}\times\frac{4}{3}\pi\times(2)^3=\frac{4}{3}\pi\times(8)^3$
$\Rightarrow\text{n}=\frac{8^3}{2^3}$
$\Rightarrow\text{n}=\frac{8\times8\times8}{8}$
$\Rightarrow\text{n}=64$

The ratio between the curved surface area and total surface area given by,
$\frac{2\pi\text{rh}}{2\pi\text{rh}+2\pi\text{r}^2}=\frac{1}{2}$
$\Rightarrow\frac{2\pi\text{r}(\text{h})}{2\pi\text{r}(\text{h}+\text{r})}=\frac{1}{2}$
$\Rightarrow\frac{\text{h}}{\text{h}+\text{r}}=\frac{1}{2}$
$\Rightarrow\text{h}+\text{r}=2\text{h}$
$\Rightarrow\text{h}=\text{r}$
Given total surface area $=2\pi\text{rh}+2\pi\text{r}^2=2\pi\text{r}(\text{h}+\text{r})=616$
$\Rightarrow2\pi\text{r}(\text{h}+\text{r})=616$
$\Rightarrow2\pi\text{r}(\text{r}+\text{r})=616$ $(\because\text{h}=\text{r})$
$\Rightarrow2\pi\text{r}(2\text{r})=616$
$\Rightarrow4\pi\text{r}^2=616$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=616$
$\Rightarrow\text{r}^2=\frac{616\times7}{88}$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}=7\text{cm}$
$\Rightarrow\text{h}=7\text{cm}$ $(\because\text{h}=\text{r})$
Volume of the cylinder$=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times7\times7\times7$
$=1078\text{cm}^3$

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 $= 220m^3$
$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=220$
$\Rightarrow\frac{1}{3}\times44\times\text{h}=220 ...($from $(i))$
$\Rightarrow\text{h}=\frac{220\times3}{44}$
$\Rightarrow\text{h}=15\text{m}$
$⇒$ Height of the cone $= 15m.$

Surface area of a cube $= 96\ cm^2$
Surface area of a cube $= 6 ($Side$)^2 = 96$
$⇒ ($Side$)^2 = 16$
$⇒ ($Side$) = 4\ cm$
$[$taking positive square root because side is always a positive quantity$]$
Volume of cube $= ($Side$)^3 = (4)^3 = 64\ cm^3$
Hence, the volume of the cube is $64\ cm^3.$

$TSA$ of cube $-\ LSA$ of cube
$= 6a^2 - 4a^2$
$= 2a^2$
$= 2 × 4 × 4$
$= 32\ cm^2$

The volume of a spherical shell is given by $\frac{4}{3}\pi(\text{R}^3-\text{r}^3)$ where $R =$ Larger radius and $r =$ smaller radius.

The ratio of the volumes of two sphere is given by $\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}.$
$\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}=\frac{1}{8}$
$\Rightarrow\frac{\text{r}^3}{\text{R}^3}=\frac{1}{8}$
$\Rightarrow\Big(\frac{\text{r}}{\text{R}}\Big)^3=\frac{1}{8}$
$\Rightarrow\frac{\text{r}}{\text{R}}=\frac{1}{2}\ ...(\text{i})$
$⇒$ Ratio of the volumes $= 1 : 2$
Now,
Ratio of their surface area $=\frac{4\pi\text{r}^2}{4\pi\text{R}^2}$
$=\frac{\text{r}^2}{\text{R}^2}$
$=\Big(\frac{\text{r}}{\text{R}}\Big)^2$
$=\Big(\frac{1}{2}\Big)^2 ...($from $(i))$
$=\frac{1}{4}$
$⇒$ Ratio of their surface area $= 1 : 4.$

Let $S$ be the total surface area of the closed cylinder with radius $r$ and height $h,$ then
$\text{S}=2\pi\text{r}^2+2\pi\text{rh}$
$\Rightarrow\text{S}=2\pi\text{r}+2(\text{r}+\text{h})$

Volume of cylinder $=\pi\text{r}^2\text{h}$
$2310=\frac{22}{7}\times7\times7\times\text{h}$
$\text{h}=\frac{2310}{22\times7}$
$\text{h}=15\text{cm}$
$=15\text{cm}$

$\frac{\text{r}}{\text{h}}=\frac{2}{3}$
$\text{r}=\frac{2}{3}\text{h}$
Volume of cylinder $=\pi\text{r}^2\text{h}$
$1617=\frac{22}{7}\cdot\frac{2}{3}\text{h}\cdot\frac{2}{3}\text{h}\cdot\frac{2}{3}\text{h}\cdot\text{h}$
$\text{h}^3=\frac{1617.7.9}{22.4}$
$\text{h}=\frac{21}{2}\text{cm}\text{r}=7\text{cm}$
Total surface area of cylinder $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\cdot\frac{22}{7}\big(7\cdot\frac{21}{2}+7\cdot7\big)$
$=770\text{cm}^2$

The formula of the curved surface area of a cone with base radius $'r'$ and slant height '' is given as
Curved Surface Area $=\pi\text{rl}$
Now, it is said that the slant height has increase by $10\%.$ So the new slant height $'1.1 l'$
So, now
New Curved Surface Area $=1.1\pi\text{rl}$
We see that the percentage increase of the Curved Surface Area is $10\%$

Volume of a cylinder $=\pi\text{r}^2\text{h}$
Diameter $= 6\ cm$
$⇒$ radius $= 3\ cm$
$\Rightarrow\text{Volume}=\frac{22}{7}\times3^2\times14$
$= 22 × 9 × 2 = 396\ cm^3$

$TSA$ of cube $-\ LSA$ of cube
$= 6a^2 - 4a^2$
$= 2a^2$
$= 2 × 4 × 4$
$= 32\ cm^2$

Volume of cone$^1$ : Volume of cone$^2$
$\frac{1}{3}\pi\text{r}^2_1:\frac{1}{3}\pi\text{r}^2_2\text{h}_2$
$=3^2\times1:1^2\times3=9:3$
$3:1$

If all of these have equal bases, then their radii are equal.
Their heights are same. (given)
$\text{r}=\text{h}_1=\text{h}_2$
$\text{V}_\text{cone}=\frac{1}{3}\pi\text{r}^2\text{h}_1=\frac{1}{3}\pi\text{r}^2(\text{r})=\frac{1}{3}\pi\text{r}^3$
$\text{V}_\text{hemisphere}=\frac{2}{3}\pi\text{r}^3$
$\text{V}_\text{cylinder}=\pi\text{r}^2\text{h}_2=\pi\text{r}^2(\text{r})=\pi\text{r}^3$
$\text{V}_\text{cone}:\text{V}_\text{hemisphere}:\text{V}_\text{cylinder}=\frac{1}{2}:\frac{2}{3}:1=1:2:3$
Hence, correct option is $(a).$

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
$=4\pi(4)^2-4\pi(3)^2$
$=4\times\frac{22}{7}(16-9)$
$=88\text{cm}^2$

Let the number of bottles be $n.$
The number of bottles needed to empty the bowl $= n\ ×$ volume of each cylinder
that is, volume of the hemispherical bowl $= n\ ×$ volume of each cylinder
$\Rightarrow\frac{2}{3}\pi(9)^2=\text{n}\times\pi(1.5)^2(4)$
$\Rightarrow\text{n}=54$
Thus, there are $54$ bottles.

Let the dimensions of Cuboid be $a, b, c$ respectively.
Volume, $V = abc = 12\ cm^3$
If $a^{\prime}=2 a, b^{\prime}=2 b, c^{\prime}=2 c,$ then
$V^{\prime}=a^{\prime} b^{\prime} c^{\prime}=8 a b c=8 \times 12=96 \mathrm{~cm}^3$

Maximum length of a pencil $=$ diagonal of the cuboid
Now, the diagonal of cuboid is $=\sqrt{1^2+\text{b}^2+\text{h}^2}$
Thus,
Length of longest rod $=\sqrt{8^2+{6}^2+5^2}$
$=\sqrt{64+36+25}$
$=\sqrt{125}$
$=5\sqrt{5\text{cm}}$
$=5(2.24)\text{cm}$
$=11.2\text{cm}$

Total surface area of a cone $=\pi\text{R}(\text{L}+\text{R})$
Where, $R =$ Radius and $L =$ Slant height
$\therefore\text{T.S.A.}=\pi\Big(\frac{\text{r}}{2}\Big)\Big(\text{2l}+\frac{\text{r}}{2}\Big)$
$=\pi\text{r}\Big(\text{l}+\frac{\text{r}}{4}\Big)$

A cube has total $12$ edges.
Let, $a →$ edge of the cube
Sum of all the edges of the cube $= 12a$
$36 = 12a$
$a = 3\ cm$
Volume of that cube,
$V = a^3$
$= 3^3$
$= 27\ cm^3$
Volume of the cube is $27\ cm^3$.
Hence, the correct option is $(b).$

Volume of a cuboid $= l × b × h = 9 × 11 × 12\ cm^3$
Radius of a lead shot $= 0.15\ cm$
Volume of a lead shot $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi(0.15)^3\text{cm}^3$
No. of lead shot $=\frac{\text{Volume of a cuboid }}{\text{Volume of a lead shot }}$
$=\frac{9\times11\times12}{\frac{4}{3}\pi(0.15)^2}$
$=\frac{9\times11\times3\times3}{\frac{22}{7}\times0.003375}$
$=84000$

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

$\frac{\text{AD}}{\text{AB}}=\frac{\text{DF}}{\text{BC}}$
$\Rightarrow\frac{\frac{\text{h}}2{}}{\frac{\text{h}}{2}+\frac{\text{h}}{2}}=\frac{\text{DF}}{\text{BC}}$
$\Rightarrow\frac{\text{DF}}{\text{BC}}=\frac{1}{2}$
$\Rightarrow\text{DF}=\frac{\text{BC}}{2}=\frac{\text{r}}{2}$
Volume of full cone $=\frac{1}3{}\pi\text{r}^2\text{h}$
Volumeof small cone formed $=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\frac{\text{h}}{2}$
$=\frac{1}{3}\pi\frac{\text{r}^2}{4}\frac{\text{h}}{2}$
$=\frac{1}{8}\Big(\frac{\pi\text{r}^2\text{h}}{3}\Big)$
Ratio of volume of two parts $=\frac{\text{Volume of small cone}}{\text{Volume of full cone}-\text{Volume of small cone}}$
$=\frac{\frac{1}{8}\Big(\frac{\pi\text{r}^2\text{h}}{3}\Big)}{\frac{\pi\text{r}^2\text{h}}{3}-\frac{\pi\text{r}^2\text{h}}{8\times3}}$
$=\frac{1}{7}$

Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$
$=\frac{40\times12\times16}{4\times5\times2}$
$=192$

$\text{V}=\frac{1}{3}\pi\text{R}^2\text{H}$
If $R' = 2R$ and $H' = 2H, $then
$\text{V}'=\frac{1}{3}\pi(\text{2R})^2(\text{2H})$
$=8\Big(\frac{1}{3}\pi\text{R}^2 \text{H}\Big)$
$=\text{8V}$
$\Rightarrow\text{V}'=\text{8V}$

Let $r$ be the radius of the cone.
$\mathrm{I}^2=\mathrm{h}^2+\mathrm{r}^2$
$\Rightarrow \mathrm{r}^2=\mathrm{I}^2-\mathrm{h}^2$
$=28^2-21^2$
$=49 \times 7$
$\Rightarrow\text{r}=7\sqrt{7}\text{cm}$
Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$⇒$ Volume of the cone $=\frac{1}{3}\times\frac{22}{7}\times(7\sqrt{7})^2\times21$
$⇒$ Volume of the cone $=7546\text{cm}^3$

Let,
$a →$ Side of each cube
So, the dimensions of the resulting cuboid are,
Length $(l) = 3a$
Breadth $(b) = a$
Height $(h) = a$
Total surface area of the cuboid,
$= 2(lb + bh + hl)$
$= 2[(3a)a + a × a + a(3a)]$
$= 14a^2$
Sum of the surface area of the three cubes,
$= 3(6a^2$)
$= 18a^2$
Required ratio,
$=\frac{14\text{a}^2}{18\text{a}^2}$
$= 7 : 9$
Thus, the required ratio is $7 : 9.$
Hence, the correct choise is $(a).$

The circumference of its base $= 110\ cm$
$2\pi\text{r}=110$
$\text{r}=\frac{35}{2}$
$CSA = 4400\ cm^2$
$2\pi\text{rh}=4400$
$ \text{h}=\frac{4000}{2\pi\text{r}}$
$=\frac{4400}{110}$.
$=40\text{cm}$

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 $=\pi\text{r}=\frac{22}{7}\times7\times25=550\text{cm}^2$