MATHS M.C.Q

Volume of the new cube formed $=$ total volume of the three cubes
Suppose that a cm is the edge of the new cube, then
$a^3=3^3+4^3+5^3=27+64+125=216=6 \mathrm{~cm}$
$\therefore$ Lateral surface area of the new cube $= 4 a^2=4 \times 6 \times 6=144 \mathrm{~cm}^2$

Let the number of lead shots be $n.$
Volume of the cuboid $= n\ ×$ volume of each lead shot
$\Rightarrow9\times11\times12=\text{n}\times\frac{4}{3}\pi\Big(\frac{0.3}{2}\Big)^3$
$...\Big(\text{since}\ \text{radius}=\frac{0.3}{2}\text{cm}\Big)$
$\Rightarrow9\times11\times12=\text{n}\times\frac{4}{3}\times\frac{22}{7}\times\frac{27}{8000}$
$\Rightarrow\text{n}=84000$
Thus, there are $84000$ lesd shots.

Let the lower limit be $x.$
Here,
Class size $= 10$
$\therefore$ Upper limit $=$ Class size $+$ Lower limit
Upper limit $= (x + 10)$
Mid value of the class interval $= 42$
$\therefore\frac{\text{x}+\text{x}+10}{2}=42$
$\Rightarrow\frac{2\text{x}+10}{2}=42$
$\Rightarrow2\text{x}+10=84$
$\Rightarrow2\text{x}=74$
$\Rightarrow\text{x}=37$
Thus, we have:
Lower limit $= 37$
Upper limit $= 37 + 10 = 47$

Volume of one cube $=$ side$^3$
$= 6^3$
$= 216\ cm^3$
So, volume of cuboid when two cubes are added $= 2 × 216 = 432\ cm^3$

Total surface area of cube $=6 a ^2$ (let a be a side of cube)
So, $6 a^2=96$
$a^2=\frac{96}{6}=16$
$a=4 cm$
Now, lateral surface area of cube is $4 a^2$
So, $4 a^2=4 \times(4)^2$
$=4 \times 16$
$=64 cm^2$

The largest sphere that can be cut from a cube of side $6\ cm$ will have its diameter $=$ side of cube.
i. e. $2r = 6\ cm ⇒ r = 3\ cm$
Volume of that sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\times3\times3\times3=36\pi\text{ cm}^3$
Hence, correct option is $(b)$

Length of resulting cuboid $= x + x = 2x$
Breadth of resulting cuboid $= x$
And Height of the resulting cuboid $= x​$
$\therefore TSA$ of cuboid $= 2(lb + bh + hl)$
$= 2(2x × x + x × × + x × 2x)$
$= 2(5x^2) = 10x^2$

Number of bags $=\frac{\text{Volume of drum}}{\text{volume of one bag}}$
$=\frac{2\pi\text{rh}}{2.1}$
$=\frac{\Big((2\times\frac{22}{7}\times4.2\times3.5)\Big)}{2.1}$
$=92$

$CSA$ of cuboid $= 2(lb + bh + hl)$
$= 2(15 × 10 + 10 × 20 + 20 × 15)$
$= 2(150 + 200 + 300)$
$= 2 × 650$
$= 1300\ cm^2$

Volume of a sphere $=(\frac{4}{3})\pi\text{r}^3$
Volume of a cylinder $=\pi\text{r}^2\text{h}$
Given, cylindrical rod whose height is $8$ times of its radius is melted and recast into spherical balls of same radius.
Let the number of such balls be $‘a’.$
$\Rightarrow\pi\times\text{r}^2\times8\text{r}=\text{a}\times\big(\frac{4}{3}\big)\pi\times\text{r}^3$
$\Rightarrow\text{a}=6$

$\text{Let}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}\ \text{and}\ \frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$
Ratio of the surface area $=\frac{\pi\text{r}_1\text{h}_1}{\pi\text{r}_2\text{h}_2}$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)\times\Big(\frac{\text{h}_1}{\text{h}_2} \Big)$
$=\Big(\frac{2}{3}\Big)^2\times\Big(\frac{5}{3}\Big)^2$
$=\frac{4}{9}\times\frac{5}{3}$
$=\frac{20}{27}$
$=20:27$

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
$=\frac{\text{Increase}\ \text{in}\ \text{the}\ \text{surface}\ \text{area}}{\text{original}\ \text{surface}}\times100$
$=\frac{15\text{a}^2}{2}\times\frac{1}{6\text{a}^2}\times100$
$=125\%$

Given, $d = 628$
$r = 14\ cm$
$h = 20\ cm$
Now,
Curved surface area $=2\pi\text{rh}$
$=2\times\frac{22}{7}\times14\times20$
$=1760\text{cm}^2$

Area covered by the roller in $1$ revolution $=2\pi\text{rh}$
$\frac{2\times22}{7\times42\times100}=100=44\times600=26400\text{cm}^2$
$\therefore$ Area covered in $500$ revolutions $= 26400 × 500\ cm^2$
Area of playground $= 1320\ m^2$

A hemisphere has two surfaces: one top surface and other curved surface.
$T.S.A =2\pi\text{r}^2+(\pi\text{r}^2)$ $\{$Area of Top-face = $\pi\text{r}^2\}$
$=3\pi\text{r}^2$
Hence, correct option is $(c).$

Volume of the wire $ = (2.2 × 10 × 10 × 10)\ cm^3= 2200\ cm^3$
Radius of the wire $= 0.25\ cm =\frac{25}{100}\text{cm}=\frac{1}{4}\text{cm}$
Let the length of wire be $h\ cm.$ Then,
$\pi\times\frac{1}{4}\times\frac{1}{4}\text{h}=2200\Rightarrow\frac{22}{7}\times\frac{1}{16}\times\text{h}=2200$
$\Rightarrow\text{h}=(2200\times\frac{7}{22}\times{16})\text{cm}=\Big(\frac{11200}{100}\Big)\text{m}=112\text{m}$

Diameter of cone $= 10\ cm$
Radius of cone $\text{r}=\frac{10}{2\text{cm}}=5\text{cm}$
Slant height of cone, $l = 13\ cm$
Suppose height of the cone is $h$
$⇒ h^2 = l^2− r^2$
$\Rightarrow\text{h}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}$
$\Rightarrow\text{h}=12\text{cm}$

Let,
$V _1, V_2, V_3 \rightarrow$ Volume of the three cubes
$a_1, a_2, a_3 \rightarrow$ Side of the three cubes
We know that,
$a^3=V$
So,
$a _1^3 = V$
$a _1^3 =\frac{1}{8}$
$a _1 =\frac{1}{2} \ cm$
Similarly,
$a _2^3 =1$
$a _2 =1\ cm$
And;
$a _3^3 =8$
$a _2 =2\ cm$
So the height of the resulting structure,
$=\frac{1}{2}+1+2$
$=0.5+1+2$
$=3.5\ cm$
The height of the structure is $3.5\ cm$ .
Hence, the correct choice is $(a)$.

Cylinderical tunnel will be hollow cylinder of radius $= 1m$
$\big\{\text{r}=\frac{\text{d}}{2}=\frac{2}{2}=1\text{m}\big\}$
Length $= 40m$
Area of iron sheet $=$ Curved surface area of cylinder
$=2\pi\text{rh}$
$=2\pi(1)40$
$=80\pi$

Given surface area of a sphere $= 1386\ cm^2$
$\Rightarrow4\pi\text{r}^2=1386$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=1386$
$\Rightarrow\frac{88}{7}\times\text{r}^2=1386$
$\Rightarrow\text{r}^2=\frac{1386\times7}{88}$
$\Rightarrow\text{r}^2=\frac{441}{88}$
$\Rightarrow\text{r}=\sqrt{\frac{441}{4}}$
$\Rightarrow\text{r}=\frac{21}{2}\text{cm}$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$
$=11\times21\times21$
$=4851\text{cm}^3$

Here the radius of the sphere $= 2\ cm$
$\therefore CSA$ of sphere $=4\pi^2=4\pi(2)^3$
$=16\pi\text{cm}^2$

Surface area of sphere $=4\pi\text{r}^2$
Given surface area $= 1386\ cm^2$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=1386$
$\Rightarrow\text{r}^2=\frac{1386\times7}{88}=\frac{441}{4}$
$\Rightarrow\text{r}=\frac{21}{2}$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
Volume $=\frac{4}{3}\pi\Big(\frac{21}{2}\Big)^3$
$=\frac{4}{3}\times\frac{22}{7}\times\Big(\frac{21}{2}\Big)^3$
$=4851\text{cm}^2$

Let number of ball formed be $n$
Volume of sphere $= n($volume of one spherical ball$)$
$\frac{4}{3}\pi\text{R}^3=\text{n}(\frac{4}{3}\pi\text{r}^3)$
$\text{n}=(\frac{\text{R}}{\text{r}})^3$
$\text{n}=(\frac{3}{0.3})^3$
$\text{n}=1000$

Length of all edges $= 36$
Length of one edge $=\frac{36}{12}=3$
Volumes of cube $= 3 × 3 × 3 = 27$

Let be $r$ the radius of the cylinder
Given: Circumference of cylinder $= 44\ cm$
$=2\pi\text{r}=44$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=44$
$\Rightarrow\text{r}=7\text{r}=7\text{cm}$

Volume of sphere $=$ volume of cylindrical wire
$\frac{4}{3}\pi\text{R}^3=\pi\text{r}^2\text{h}$
$\frac{4}{3}\pi\times21^3=\pi\times(1.4)^2\times\text{h}$
$\text{h}=\frac{4\times21\times21\times21}{3\times1.4\times1.4}$
$\text{h}=6300\text{cm } \text{or }63\text{m}$

Given, dimensions of a room, $l = 10m, b = 10m, h = 5m$
$\therefore$ Length of the longest pole = Diagonal of a cuboid (room)
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$
$=\sqrt{(10)^2+(10)^2+(5)^2}$
$=\sqrt{100+100+25}$
$=\sqrt{225}=15\text{m}$
Hence, the length of the longest pole is $15m.$

Given, Volume of sphere $=$ Volume of cone
$\frac{4}{3}\pi(2\pi)^3=\frac{1}{3}\pi\text{r}^2\text{h}$
$4\times8\text{r}^3=\text{r}^2\text{h}$
$\text{h}=32\text{r}$

Sphere has only one surface i.e. curved surface, so number of faces $= 1$
Hence, correct option is $(a).$