Answer: Option A. -> 10
:
A
Let number of cones which can be filled = n
Diameter of cylinder = d = 12 cm
$\Rightarrow$ Radius of cylinder =
$r=\frac{d}{2}=\frac{12}{2}=6cm$
Height of cylinder = h = 15 cm
Volume of cylinder = $\pi .r^2.h=(\pi \times 6^2\times 15)=540\pi c{m}^3$
Diameter of cone = $d_1=6cm$
Radius of cone = $r_1=\frac{d_1}{2}=\frac{6}{2}=3cm$
Height of cone = $h_1=12cm$
Volume of cone $=\frac{1}{3}\pi (r_1{)}^2{h}_1=\frac{1}{3}\times \pi \times 3^2\times 12=36\pi {\text{cm}}^3$
Radius of hemispherical top of the cone = $r_1=3cm$
Volume of hemisphere top $=\frac{2}{3}\pi (r_1{)}^3=\frac{2}{3}\times \pi \times 3^3=18\pi {\text{cm}}^3$
According to given condition we have:
n× ( Volume of Cone + Volume of Hemispherical top ) = volume of cylinder
$\Rightarrow$ n× (36 $\pi +18\pi )=540\pi$
$\Rightarrow$ n× (54 $\pi )=540\pi$
$n=\frac{540}{54}=10$

Answer: Option D. -> 48cm2
:
D
Slant height of a frustum of a cone =
l = 4 cm
Perimeter of its first circular end =
18 cm
Perimeter of its second circular end =
6 cm
Curved Surface Area of frustum of the cone
$=\pi .(r_1+r_2)l$
$=(\pi .r_1+\pi .r_2)l$
$=(2\pi .r_1+2\pi .r_2)\frac{l}{2}$
$=(18+6)\frac{4}{2}=48c{m}^2$

Answer: Option C. -> Rs. 156.75 , Rs 97.97
:
C
R = 20 cm, r = 8cm, h = 16 cm
l = $\sqrt{h^2+(R-r{)}^2}=\sqrt{256+144}cm=20cm$
Volume of container = $\frac{1}{3}\pi h\left(R^2+r^2+Rr\right)$
$=\frac{1}{3}\times \left(3.14\right)\times 16(400+64+160)c{m}^3$
$=\frac{1}{3}\times 3.14\times 16\times 624c{m}^3$
$=3.14\times 16\times 208c{m}^3$
$=10449.93c{m}^3$
Therefore, the quantity of milk in the container = $\frac{10449.92}{1000}$= 10.45 liters
Cost of milk at the rate of Rs.15 per liters = Rs.{10.45× 15 } = Rs. 156.75
Surface area of the metal sheet used to make the container
$=\pi l(R+r)+\pi r^2=\pi \left[l(R+r)+r^2\right]$
$=3.14\times \left[20\times (20+8)+8^2\right]c{m}^2$
$=3.14\times \left[20\times 28+64\right]c{m}^2$
$=3.14\times 624c{m}^2=1959.36c{m}^2$
Therefore, the cost of the metal sheet at rate of Rs.5 per 100 $c{m}^2$
= Rs. 1959.36 $\times$ $\frac{5}{100}$ = Rs.97.97 approx.

Answer: Option B. -> 100
:
B
Given that diameter of pipe, d = 20 cm
Then, radius of pipe,
$r=\frac{20}{2}=10cm=0.1m$
Speed of water flowing through pipe = 3 $km/h$ = 3000 $m/h$
Pipe is in the form of a cylinder.
So, volume of water flowing through pipe in 1 hour
$=\pi r^{2}h=\pi \times 0.1\times 0.1\times 3000=30\pi m^3$
Let the cylindrical tank be filled in $x$ hours.
Then, volume of water flowing through pipe in $x$ hours
$=30x\pi m^3$
Also it is given that,
diameter of cylindrical tank, $d_1=10m$
$\Rightarrow$ radius of cylindrical tank,$r_1=\frac{10}{2}=5m$
Height of cylindrical tank,$h_1=2m$
Volume of cylindrical tank
$=\pi (r_1{)}^2{h}_1=\pi \times 5\times 5\times 2=50\pi m^3$
Now, volume of water flowing through pipe in $x$ hours = volume of cylindrical tank
$\Rightarrow 30x\pi =50\pi$
$\Rightarrow x=\frac{50}{30}=\frac{5}{3}hours$
$i.e.,x=\frac{5\times 60}{3}minutes=100minutes$
$\therefore$ The tank will be filled in 100 minutes.

Answer: Option B. -> 42 cm2
:
B
Area of a circle
$=\pi r^2$
From Figure, the diameter of circle is 14 cm. Two semi-circlesmake one full circle.
$\therefore$ The area of one full circle is
$=\frac{22}{7}\times 7^2=154c{m}^2$
The total area of square
$={14}^2=196c{m}^2$
The area of shaded portion =[Area of square- Area of full circle]
= 196 - 154 =$42c{m}^2$.
Hence, area of shaded region
$=42c{m}^2$

Answer: Option A. -> True
:
A
Diameter = 14 cm
Perimeter of semi circle = $\pi$ $\times$r =$\frac{22}{7}$x7 = 21.99 cm
Total perimeter of protractor is = 21.99 + 14 = 35.99 cm ~ 36 cm.

Answer: Option C. -> 17.60 cm2
:
C

Height of Cylinder $=h=AO=2.4\text{cm}$
Diameter of Cylinder $=1.4\text{cm}$
Radius of Cone = Radius of Cylinder
$OB=r=\frac{1.4}{2}=0.7\text{cm}$
Lets find Slant Height $l$ of cone, using pythagoras theorem on $△AOB$ , we get
$AB=l=\sqrt{h^2+r^2}=\sqrt{(2.4{)}^2+(0.7{)}^2}$
$l=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5\text{cm}$
Surface Area of remaining Solid
= Surface Area of the Cylinder$+$Inner surface Area of the hollow Cone
$=(2\pi rh+\pi r^2)+\pi rl=\pi r(2h+r+l)=\frac{22}{7}\times 0.7(2\times 2.4+0.7+2.5)=2.2(4.8+0.7+2.5)=2.2\left(8\right)=17.60{\text{cm}}^2$
$\therefore$ Total surface area of the remaining solid is$17.60{\text{cm}}^2$

Answer: Option B. -> 880 cm2
:
B
Surface area = base circle + curved surface area of the cylinder+ curved surface area of the cone.
$=\pi r^2+2\pi rh+\pi rl$
$=\pi (5{)}^2+2\pi (5\left)\right(20)+\pi (5\left)\right(11)$
$=\pi \times 5(5+40+11)$
$=\frac{22}{7}\times 5\times 56$
$=880c{m}^2$.

Answer: Option D. -> 28
:
D
S.A. = 1372 $c{m}^2$ (Given)
$\Rightarrow 2(l\times b+b\times h+l\times h)=1372$
$[\because$ Surface area of a cuboid of dimensions $l\times b\times h$ is given by$2(lb+bh+lh).]$
Also, it is given that$l:b:h=4:2:1$.
Let $l=4k,b=2k$and$h=k$.
$\text{Then,}2(4k\times 2k+2k\times k+4k\times k)=1372$
$⟹28k^2=1372$
$⟹k^2=49$
$⟹k=7$
$\therefore l=4k=4\times 7=28\text{cm}$

Answer: Option A. -> 112
:
A
Volume of iron
$=(440\times 260\times 100)c{m}^3$.
Internal radius of the pipe
$=30cm$.
External radius of the pipe
$=(30+5)cm=35cm$.
Let the length of the pipe be $hcm$.
Volume of iron in the pipe = (External volume) - (Internal volume)
$=\left[\pi \right(35{)}^2\times h-\pi (30{)}^2\times h]c{m}^3$
$=\left(\pi h\right)\left((35{)}^2-(30{)}^2\right)c{m}^3$
$=(65\times 5)\pi hc{m}^3$
Volume of iron in the pipe =Volume of iron block
$\Rightarrow \left(325\pi h\right)c{m}^3=440\times 260\times 100$
$h=\frac{440\times 260\times 100}{325}\times \frac{7}{22}cm$
$h=11200cm$
$h=\frac{11200}{100}=112m$
Hence, the lengthof the pipe is 112 m.