MATHS M.C.Q

So here, radius would be half of edge of cube, or diameter of rod $=$ edge of cube
$r = 2\ cm$
$h = 4\ cm$
So, volume $=\pi\text{r}^2\text{h}$
$=\pi\times2\times2\times4$
$=16\pi\text{cm}^2$

Volume of the inflated balloon : Volume of the original balloon
$=\frac{4}{3}\pi(2\pi\text{r})^3:\frac{4}{3}\pi(\text{r})^3$
$=8:1$

We know that,
Total surface area of a cube $= 6a^2$
$⇒ 96 = 6a^2$
$\Rightarrow\text{a}^2=\frac{96}{6}$
$⇒ a^2 = 16$
$⇒ a = 4\ cm$
Now,
Volume of the cube $= a^3$
$= 4^3$
$= 64\ cm^3$

Let $a, b, c.$ be the sides of three cubes
Then $a^3 =\frac{1}{8}\Rightarrow\text{a}=\frac{1}{2}$
$b^3= 1 ⇒ b = 1$
$c^3 = 8 ⇒ c = 2,$
Now height of resulting cube
$0.5 + 1 + 2 = 3.5\ cm$

Dimensions of the wall are,
Length $(l) = 8m$
Breadth $(b) = 20\ cm$
$= 0.2m$
Height $(h) = 4\ cm$
Volume of the hall,
$V = lbh$
$= 8 × 4 × 0.2$
$= 6.4m^3$
Cost of building the wall is $₹ 160.$
Hence, the correct option is $(c).$

Volume of pit $= (16 × 12 × 4)m^3$
Volume of a plank $= (4 × 0.5 × 0.2)m^3$
Required number of planks $=\frac{\text{Volume of pit}}{\text{Volume of a plank}}$
$=\frac{16\times12\times4}{4\times0.5\times0.2}=1920$
Hence, $(b)$ is the correct answer.

$C.S.A$ of a cone $=\pi\text{rl}$
If $l' = 1 + 10\%$ of $l$
$=\text{l}+\frac{10}{100}\times\text{l}$
$=\text{l}+\frac{\text{l}}{10}$
And, $r' = r$
$\text{C.S.A.}=\pi\text{r}\Big(\text{l}+\frac{\text{l}}{10}\Big)=\frac{11}{10}\pi\text{rl}$
So, increase in $C.S.A. =\frac{\frac{11}{10}\pi\text{rl}-\pi\text{rl}}{\pi\text{rl}}\times100\%=10\%$

Given: Side of the cube $(a) = 3m$
\therefore Surface Area of Cube $= 6a^2 = 6 × 3 × 3 = 54\ sq. cm$
Now, Cost of painting the cubical box of $1\ sq. m = Rs. 2$
\therefore Cost of painting the cubical box of $54\ sq. m = 54 × 2 = Rs. 108$

We know that,
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow1232=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow1232=\frac{1}{3}\times\frac{22}{7}\times\text{r}^2\times24$
$\Rightarrow\text{r}^2=\frac{7\times3\times1232}{24\times22}$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}=7\text{cm}$
$\Rightarrow r=7 \mathrm{~cm} \text { and } h=24 \mathrm{~cm}$
$I^2=r^2+h^2$
$\Rightarrow I^2=7^2+24^2$
$\Rightarrow I^2=49+576$
$\Rightarrow I^2=625$
$\Rightarrow I=25$
Curved surface area $=\pi\text{rl}$
$=\frac{22}{7}\times7\times25$
$=550\text{cm}^2$

The surface or faces that a cone has are:
$(i)$ Base, $(ii)$ Slanted surface
So, the number of surfaces that a cone has is $2.$

Given, $d = 6\ cm$
$r = 3\ cm$
$h = 14\ cm$
Now,
Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times3\times3\times14$
$=396\text{cm}^3$

$CSA$ of cylinder $=2\pi\text{rh}$
$88=2\times\frac{22}{7}\times\text{r}\times14$
$\text{r}=\frac{88\times7}{2\times22\times14}$
$\text{r}=1\text{cm}$
Hence Diameter of base $= 2\ cm.$

Let,
$l →$ Length of the first cuboid
$b →$ Breadth of the first cuboid
$h →$ Height of the first cuboid
And,
$L → $ Length of the new cuboid
$B →$ Breadth of the new cuboid
$H →$ Height of the new cuboid
We know that,
$L = 2l$
$B = 2b$
$H = 2h$
Surface area of the first cuboid,
$S' = 2(LB + BH + HL)$
$= 2[(2l)(2b) + (2b)(2h) + (2h)(2l)]$
$= 2(4lb + 4bh + 4hl)$
$= 4[2(lb + bh + hl)]$
$= 4S$
The surface area of the new cuboid is $4S.$
So, the correct choice is $(b).$

Volume of a cuboid $=$ Length $×$ Breadth $×$ Height
Volume of pit $= 40m × 12m × 16m = 7680m^3$
Volume of plank $= 4m × 5m × 2m = 40m^3$
No. of planks $=\frac{\text{Volume of pit }}{\text{Volume of plank }}=\frac{7680}{40}=192$

Volume of the cube $= a^3$
$⇒ 512 = a^3$
$⇒ a = 8\ cm$
Now,
Total surface area of a cube $= 6a^2$
$= 6 × (8)^2$
$= 6 × 64$
$= 384\ cm^2$

Volume of a hemisphere $=\frac{2}{3}\pi\text{r}^3=\frac{2}{3}\pi(9)^3\text{cm}^3$
Volume of cylinder $=\pi\text{r}^2\text{h}$
Volume of a cylinderical bottle $=\pi(1.5)^2\times4$
No. of bottles required $=\frac{\text{Volume of a hemisphere}}{\text{Volume of a cylinderical bottle}}$
$=\frac{\frac{2}{3}\pi(9)^2}{\pi(1.5)^2\times4}$
$=\frac{81\times3}{2.25\times1.5\times2}=54$
Thus, total $54$ bottals are required.

curved surface area of a cylinder of radius $'r'$ and heightn $'h'$ is given by $\text{A}=2\pi\text{rh}$
Now, if $r' = 2r$ and $\text{h}'=\frac{\text{h}}{2}$
Then $\text{A}'=2\pi\times(2\text{r})\times\frac{\text{h}}{2}$
$=2\pi\text{rh}=\text{A}$
$⇒ C.S.A.$ remains the same.

Radius of cone $= r\ cm$
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow1232=\frac{1}{3}\times\frac{22}{7}\times\text{r}^2\times24$
$\Rightarrow\text{r}^2=\frac{1232\times3\times7}{22\times24}=49$
$\Rightarrow\text{r}=\sqrt{49}=7\text{cm}$
Area of curved surface area $=\pi\text{rl}$
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25$
$\therefore\frac{22}{7}\times7\times25=550\text{cm}^2$

Given,
Height $= 14\ cm$
Curved surface area $= 264\ cm^2$
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 $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times3\times3\times14$
$=396\text{cm}^3$

Perimeter of one face of a cube $= 4a = 40\ cm ($where $a =$ side of the cube$)$ sides of face of a cube $= 4$
Therefore, side of the cube $=\frac{4\text{a}}{4}=\frac{40}{4}=10\text{cm}$
Therefore, volume of the cube $= a^3 = 10^3 = 1000$ cubic cm

Surface area of a sphere $=4\pi\text{r}^2$
Given, surface area of a sphere is $144\pi\text{m}^2$
$\Rightarrow4\pi\text{r}^2=144\pi$
$\Rightarrow\text{r}=6\text{m}$
Volume of the sphere $=\big(\frac{4}{3}\big)\times\pi\times6^3$
$\Rightarrow$ Volume of the sphere $=288\pi\text{m}^3$

Volume of conical vessel $=\frac{1}{3}\times\pi\times\text{r}^2\times\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times24\times24\times42$
$= 25344\ cm^3$
$= 25.344\ dm^3 (1\ dm^3 =1000\ cm^3)$
$= 25.344\ Kg\ wt. (1\ kg - wt = 1\ dm^3)$

Lateral surface area of the cuboid $= [2(l + b) × h)]$
$=\big(2(15+6)\times\frac{5}{10}\big)\text{m}^2=21\text{m}^2$

Let $V_1$ and $V_2$ be the volume of the two cylinders with radius $r_1$ and height $h_1$, and radius $r_2$ and height $h_2$,
Where, $\frac{2\text{r}_1}{2\text{r}_2}=\frac{3}{1},\frac{\text{h}_1}{\text{h}_2}=\frac{1}{3}$
So,
$\text{V}_1=\pi\text{r}^2_1\text{h}_1\ ...(\text{i})$
Now,
$\text{V}_2=\pi\text{r}^2_2\text{h}_2\ ...(\text{ii})$
From equation $(i)$ and $(ii),$ we have
$\frac{\text{V}_1}{\text{V}_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\Big(\frac{\text{2r}_1}{\text2{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\big(3\big)^2\Big(\frac{1}{3}\Big)=\frac{3}{1}$

Volume of the beam $=$ length $×$ breadth $×$ height
$=9\times\frac{40}{100}\times\frac{20}{100}$
$...\Big(1\text{cm}=\frac{1}{100}\text{m}\\\Rightarrow40\text{cm}=\frac{40}{100}\text{m}\ \text{and}\ 20\text{cm}=\frac{20}{100}\text{m}\Big)$
$=\frac{18}{25}\text{cm}^3$
Weight of the beam $=\frac{18}{25}\times50$
$=36\text{kg}$

The longest rod that can be fitted in the cubical vessel is its diagonal.
side of the cube $(l) = 10\ cm$
So, the diagonal of the cube,
$=\sqrt{3}\text{l}$
$=10\sqrt{3}\text{cm}$
So, the length of the longest rod that can be fitted in the cubical box is $10\sqrt{3}\text{cm}.$
Hence, the correct choice is $(c).​​​​​​$

Volume of cone $=\frac{1}{3}\times\pi\times\text{r}^2\times\text{h}$
$1570=\frac{1}{3}\times\pi\times\text{r}^2\times15$
$\pi\text{r}^2=314\text{cm}^2$
Base area is $314\ cm^3$

The formula of the curved surface area of a cone with base radius $'r'$ and slant height $'1'$ us given as
Curved Surface Area $=\pi\text{rl}$
Hence the base radius is given as $'2r'$ and the slant height is given as $\frac{1}{2}$
Substituting these values in the above equation we have
Curved Surface Area $=\frac{(\pi)(2)(\text{r})(l))}{2}$
$=\pi\text{rl}$

Total surface area of the cuboid $= 2(lb + bh + lh)\ cm^2$
$= 2(12 × 9 + 8 × 9 + 12 × 8)\ cm^2$
$= 2(108 + 72 + 96)\ cm^2$
$= 552\ cm^2$

Volume of water it can hold $=$ Volume of water tank
$= l × b × h$
$= 6 × 5 × 4.5$
$= 135m^3$
$= 135000$ litres $(1m^3 = 1000$ litres$)$

Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
Slant height $=\text{L}=\sqrt{\text{r}^2+\text{h}^2}$
Subsitituting values we get $=\text{r}=\sqrt{343}\text{cm}$
Volume of cones $=\frac{1}{3}\times\frac{22}{7}\times343\times21=7546\text{cm}^3$

Volume of the clay to be speed,
$V = 10m^3$
Area on which the clay is speed
$A = 10arc$
$= 1000m^2$
Let,
$h →$ Rise in the level of the ground
We know that,
$V = A × h$
$\text{h}=\frac{\text{V}}{\text{A}}$
$=\frac{10}{1000}$
$=0.01\times100\text{cm}$
Rise in the level of the ground is $1\ cm.$
Hence, the correct optin is $(a).$

Let the length of the required cloth be $L\ m$ and its breadth be $B\ m.$
Here, $B = 2.5m$
Radius of the conical tent $= 7m$
Height of the tent $= 24m$
Area of cloth required $=$ curved surface area of the conical tent
$\Rightarrow\text{L}\times\text{B}=\pi\text{rl}​​$
$\Rightarrow\text{L}\times\text{B}=\pi\text{rl}​​$
$\Rightarrow\text{L}\times2.5=\frac{22}{7}\times7\times\sqrt{7^2+24^2}$
$\Rightarrow2.5\text{L}=22\times\sqrt{49}+576$
$\therefore\text{L}=\frac{22\times25}{2.5}=220\text{m}$

Diagonal $=\text{s}\sqrt{3}$ Where $s =$ side
Here $s = 8$
Surface area of cube $= 6a^2 = 6 × 8 × 8 = 384\ cm^2$

Volume of reduced cylinder $:$ Volume of original cylinder
$\pi\big(\frac{\text{r}}{2}\big)^2\text{h}:\pi(\text{r})^2\text{h}$
$1:4$

Volume of a cuboid $=$ Length $×$ Breadth $×$ Height
Volume of hall $ = 20m × 15m × 4.5m = 1350m^3$
Volume of air required by $1$ person $= 5m^3$
No. of persons $=\frac{\text{Volume of hall}}{\text{Volume of air required by 1 person }}=\frac{1350}{5}=270$

Volume of a cuboid $=$ length $×$ breadth $×$ height
$= 15 × 12 × 4.5$
$= 810\ cm^3$