MATHS M.C.Q

 Let the number of bags be $n$
Volume of drum $= n($volume of each bag$)$
$\pi\text{R}^2\text{h}=\text{n}(2.1)$
$\pi\text{R}^2\text{h}=\text{n}(2.1)$
$\frac{22}{7}\times(4.2)^2\times35=\text{n}(2.1)$
$\text{n}=\frac{\frac{22}{7}\times(4.2)^2\times35}{2.1}$
$=92$

 Given, lateral surface area of a cube $= 256m^2$
We know that, lateral surface area of a cube $= 4 × ($Side$)^2$
$\Rightarrow256=4\times(\text{Side})^2$
$\Rightarrow(\text{Side})^2=\frac{256}{4}=64$
$\Rightarrow\text{Side}=\sqrt{64}=8\text{m}$
$[$taking positive square root because side is always a positive quantity$]$
Now, volume of a cube $= ($Side$)^3= (8)^3= 8 × 8 × 8 = 512m^3$
Hence, the volume of the cube is $512m^3.$

Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
Here height of the carved out cone $=$ Radius of the hemisphere
$\therefore$ Volume of cone $\frac{1}{3}\pi\text{r}^3\times​\text{r}=​\frac{1}{3}\pi\text{r}^3$

Let $r$ be the radius of cylinder and cone and volumes are equal and $h_1$ and $h_2$ be their have $h_2$ is respectively
$\therefore$ Volume of cylinder $=\pi\text{rh}_1$
and volume of cone $=\frac{1}{3\pi\text{r}^2\text{h}_2}$
$\therefore\pi\text{r}^2\text{h}_1=\frac{1}{3\pi\text{r}^2\text{h}_2}$
$\Rightarrow\text{h}_1=\frac{1}{3\text{h}_2}$
$\Rightarrow\frac{\text{h}_1}{\text{h}_2}=\frac{1}{3}$
$\therefore\text{h}_1:\text{h}_2=1:3$

Given $r = 3.5\ cm$
Surface area of sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times3.5\times3.5$
$=154\text{cm}^2$

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

The formula of the volume of a cone with base radius $'r'$ and vertical height $'h'$ is given as
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}=\text{V}$
Since it is given that the radius and height are doubled we have the radius as $'2r'$ and the vertical height as $'2h'$
Volume of modified cone $=\frac{1}{3}\pi(\text{2r})^2(2\text{h})$
$=\frac{8}{3}\pi\text{r}^2\text{h}$
$=8\text{V}$

Volume of a sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi(6)^3=288\pi\text{cm}^3$
On recasting a sphere into cylinder, the volume will remain same.
Volume of a cylinder $=\pi\text{r}^2\text{h}$
Radius $= 0.1\ cm$
$\Rightarrow\pi(0.1)^2\text{h}=288\pi$
$\Rightarrow\text{h}=\frac{288\times1}{0.01}$
$= 28800cm$
$= 288m (\therefore 1m = 100m)$
$h = 288m$

Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$
$=\frac{(20\times100)\times(6\times100)\times50}{(5\times100)\times25\times10}$ $...(1\text{m}=100\text{cm})$
$=\text{480}$

Slant height of cone $=\sqrt{(\text{r}^2+\text{h}^2)}($given $r = 24m\ h = 10m)$
$=\sqrt{(24)^2+(10)^2}$
$=\sqrt{576+100}$
$=\sqrt{676}=26\text{m}$

Let $r_1, r_2$ be the radii of two cylinders and $h_1, h_2$ be the height of two cylinder respectively.
Then $\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}$ and $\frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$
$\therefore$ Required ratio $=\frac{\pi\text{r}^2_1\text{h}_1}{\pi\text{r}^2_2\text{h}_2}$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$
$=\big(\frac{2}{3}\big)^2\big(\frac{5}{3}\big)$
$=\frac{20}{27}=20:27$

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 $=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{35}{2}\Big)$
$=770\text{cm}^2$

We know that,
Length of the longest diagonal $=\sqrt{3}\text{a}$
$\Rightarrow8\sqrt3=\sqrt{3}\text{a}$
$\Rightarrow\text{a}=8$
Now,
Total surface area $=6 \mathrm{a}^2$
$=6 \times(8)^2$
$=6 \times 64$
$=384 \mathrm{~cm}^2$

Let the number of cones be $n.$
Volume of the metallic sphere $= n\ ×$ volume of each cone
$\Rightarrow\frac{4}{3}\pi(10.5)^3=\text{n}\times(3.5)^2(3)$
$\Rightarrow4(10.5)^3=\text{n}(3.5)^2(3)$
$\Rightarrow\text{n}=126$
Thus, the number of such cones is $126.$

Volume of a sphere $=\big(\frac{4}{3}\big)\pi\text{r}^3 $
Volume of a solid cone $=\big(\frac{1}{3}\big)\pi\text{r}^2\text{h}$
Given, solid sphere of radius $r$ is melted and cast into the shape of a solid cone of height $r$
Let the base radius be $A.$
$=\big(\frac{4}{3}\big)\pi\text{r}^3=\big(\frac{1}{3}\big)\pi\times\text{A}^2\times\text{r}$
$\Rightarrow\text{A}=2\text{r}$

Let,
$a →$ Side of each cube
Length of the diagonal $=8\sqrt{3}\text{cm}$
$\sqrt{3}\text{a}=8\sqrt{3}$
$a = 8\ cm$
We have to find the surface area of the cube
Surface area of the cube,
= 6a$^2$
$= 6 × 8^2$
$= 384\ cm^2$
Thus, surface area of the cube is $384\ cm^2$.
Hence, the correct choice is $(b).$

Volume of the wall $=$ length $×$ breadth $×$ height
$= (8 × 100) × (6 × 100) × 22.5 ...(1m = 100\ cm)$
$=800\times600\times\frac{225}{10}$
$=800\times60\times225$
Volume of $1$ brick $=$ length $×$ breadth $×$ height
$=25\times\frac{1125}{100}\times6$
$=\frac{1125}{2}\times3$
Required number of bricks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{wall}}{\text{Volume}\ \text{of}\ 1\ \text{brick}}$
$=\frac{800\times60\times225}{\frac{1125}{2}\times3}$
$=\frac{800\times60\times225\times2}{1125\times3}$
$=6400$

Cost of digging would be $= (4.5m × 2.5m × 2.5m) × 20$
$= ₹ 562.50$

The volume of cuboid $=$ Length $×$ Breadth $×$ Height
$⇒$ volume $= 12 × 8 × 6 = 576\ cu.cm$

Volume of a cube $=$ side$^3$
Volume of a sphere $=\big(\frac{4}{3}\big)\pi\text{r}3$
Given, sphere and a cube are of the same height.
Side $=$ diameter $= 2r$
Ratio of their volumes $=\frac{\frac{4}{3}\times\frac{22}{7}\times\text{r}^2}{(2\text{r})^3}=11:21$

Th ratio of the volume of a right circular cylinder and a right circular cone is given by
$\frac{\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}=\frac{1}{\frac{1}{3}} ...($Same base and same height$)$
$=\frac{3}{1}$
$⇒$ Ratio of the volumes is $3 : 1.$

Let $r\ cm$ be the radius of the cone.
Volume $= 1570\ cm^3$
Then $\frac{1}{3}\times3.14\times1^2\times15=1570$
$\Rightarrow\text{r}^2=\frac{1570}{3.14\times5}=100$
$\Rightarrow\text{r}=10\text{cm}$

Length $(l),$ width $(b)$ and height $(h)$ of the rectangular solid are in the ratio $3 : 2 : 1.$
So, we can take,
$(l) = 3x \ cm$
$(b) = 2x \ cm$
$(h) = x \ cm$
We need to find the total surface area of the box
Volume of the box,
$V = 48\ cm^3$
$lbh = 48$
$(3x)(2x)x = 48$
$6x^3 = 48$
$x^3 = 8$
$x = 2$
Thus,
Surface area of the box,
$= 2(lb + bh + hl)$
$= 2[(3x)(2x) + (2x)x + (x)(3x)]$
$= 2(11x^2)$
$= 22x^2$
$= 22(2)^2$
$= 88\ cm^2$
Thus total surface area of the box is $88\ cm^2.$
Hence, the correct option is $(d).$

$TSA$ of cone $=\pi\text{r}(\text{l}+\text{r})$
$=\frac{22}{7}\times7(10+7)$
$=22\times17$
$=374\text{m}^2$

The beam has a shape of a cuboid.
Volume of a cuboid of length $l,$ breadth $b$ and height $h = l × b × h$
So, the volume of the beam $= 9 × 0.4 × 0.2 = 0.72m^3$
Hence, the weight of the beam if the iron weighs $50\ kg$ per $m^3 = 0.72 × 50 = 36kg$

The formula of the volume of a cone with base radius $'r'$ and vertical height $'h'$ is given as
Volume of cone $=\frac{1}{3\pi\text{r}^2\text{h}}$
Here it is given that the base radius is $'3r'$ and that the vertical height is $'3r'$
Substituting these values in the above equation we get
Volume of cone $=\frac{1}{3\pi}(3\text{r})^2(3\text{r})$
$=9\pi\text{r}^3$

Here, outer curved surface area is asked, so, thickness would be added to radius.
Thus, $r = 3.25 + 0.25 = 3.5\ cm$
Surface area of bowl $=2\pi\text{r}^2=2\times\frac{22}{7}\times(3.5)^2$
$=\frac{2\times22\times3.5\times3.5}{7}$
$=77\text{cm}^2$

The circumference of the base of a right circular cylinder $= 44\ cm$
$2\pi\text{r}=44\text{r}=\frac{44\times7}{(22\times2)}$
$\text{r}=7\text{cm}$

Given,
$2\pi\text{r}=132,\text{r}=\frac{132}{2\pi}$
Volume of cylinder $=\pi\text{r}^2\text{h}$
$=\pi\big(\frac{132}{2\pi\text{r}}\big)^2\times25$
$=\frac{66\times66\times7}{22}\times25$
$=34650\text{cm}^2$

Let the radii of the bases of a cylinder and a cone be $3x\ cm$ and $4x\ cm$ respectively and let their heights be $2y\ cm$ and $3y\ cm$ respectively.
$⇒$ Ratio of the volumes $=\frac{\pi(3\text{x})^2\times2\text{y}}{\frac{1}{3}\pi(4\text{x})^2\times3\text{y}}$
$=\frac{9\times2}{16}$
$=\frac{9}{8}$
$⇒$ Ratio of the volume is $9 : 8.$

We know that,
Lateral surface area of a cube $= 4a^2$
$⇒ 256 = 4a^2$
$\Rightarrow\text{a}^2=\frac{256}{4}$
$⇒ a^2 = 64$
$⇒ a = 8m$
Now,
Volume of the cube $= a^3$
$= 8^3$
$= 512m^3$

Let $I_1$ and $I_2$ be the slant heights of two cones respectively
Given $\frac{\text{l}_1}{\text{l}_2}=\frac{4}{3}$
Now, required ratio $=\frac{\pi\text{rl}_1}{\pi\text{rl}_2}=\frac{\text{l}_1}{\text{l}_2}=\frac{4}{3}$
$\Rightarrow\text{CSA}_1:\text{CSA}_2=4:3$

Volume of sphere $1 :$ volume of sphere $2$
$\frac{4}{3\pi\text{r}_1{^3}}:\frac{4}{3\pi\text{r}_2{^3}}$
$\text{r}_1{^3}:\text{r}_2{^3}=64:27$
$\text{r}_1:\text{r}_2=4:3$

Volume of sphere $1 :$ Volume of sphere $2$
$\frac{4}{3}\pi\text{r}^3_1:\frac{4}{3}\pi\text{r}^3_2$
$\text{r}^3_1:\text{r}^3_2=64:27$
$\text{r}_1:\text{r}_2=4:3$

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 $\text{r}_1$ 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 radius of the cone $= R$ and height $= H$
Then, volume $=\frac{1}{3}\pi\text{R}^2\text{H}$
Now, $R' = R + 20\%$ of $R =\text{R}+\frac{\text{R}}{5}=\frac{\text{6R}}{5}$
$H' = H + 20\%$ of $H =\text{H}+\frac{\text{H}}{5}=\frac{\text{6H}}{5}$
New volume, $\text{v}'=\frac{1}{3}\pi\text{R}'^2\text{H}'$
$=\frac{1}{3}\pi\Big(\frac{6\text{r}}{5}\Big)^2\Big(\frac{6\text{H}}{5}\Big)$
$=\frac{216}{125}\Big(\frac{1}{3}\pi\text{R}^2\text{H}\Big)$
$=\frac{216}{125}\text{v}$
% increase in volume $=\frac{\text{v}'-\text{v}}{\text{v}}\times100$
$=\frac{\frac{216\text{v}}{125}-\text{v}}{\text{v}}\times100$
$=\frac{91}{125}\times100$
$=72.8\%$
$\approx73\%$

Let $r_1$ and $r_2$ be the radius of the two spheres, respectively. Therefore, the ratio of their surface areas,
$\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2}=\frac{125}{64}$
$\Rightarrow\frac{\text{r}^3_1}{\text{r}^3_2}=\frac{125}{64}$
$\Rightarrow\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^3=\Big(\frac{5}{4}\Big)^3$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{5}{4}$
Now, Ratio of their surface area
$\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2}=\frac{\text{r}_1}{\text{r}_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{5}{4}\Big)^3=\frac{25}{16}$
$\therefore SA1 : SA2 = 25 : 16$

Given,
$l + b + h = 19$
Squaring we get
$l^2+b^2+h^2+2\{l b+b h+h l\}=361 \ldots (i)$
Also we know that
$l^2+b^2+h^2=d^2~125 (\text{since d}=5\sqrt{5})$
And Total Surface Area $= 2(lb + bh + hl)$
Using in $(i)$ we get $T.S.A = 361 - 125 = 236\ cm^2$

Let $lb = 2x, bh = 3x, hl = 4x,$ and $lbh = 9000$
multiply all we get
$(lbh)^2 = 24x^3$
$81000000 = 24x^3$
$X = 150.$
Substituting above we get
$b= 15, l = 20$ and $h = 30$
Smallest $= 15$

$n =$ number of cones $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cylinder}}{\text{Volume}\ \text{of}\ \text{one}\ \text{cone}}$
$=\frac{\pi(3)^2\times5}{\frac{1}{3}\pi\Big(\frac{1}{10}\Big)^2\times1}$
$=\frac{9\times5}{\frac{1}{3}\times\frac{1}{100}}$
$=\frac{45}{\frac{1}{300}}$
$=45\times300$
$=13500$

Volume of cylinder $=$ volume of cone
$\Rightarrow\pi\text{r}^2\text{h}_1=\frac{1}{3}\pi\text{r}^2\text{h}_2 ($Let Radius be $r$ for both$)$
$\Rightarrow\frac{\text{h}_1}{\text{h}_2}=\frac{1}{3}$
$\Rightarrow\frac{\text{h}_2(\text{cone})}{\text{h}_1(\text{cylinder})}=\frac{3}{1}$