Answer: Option B. -> 463.39  m2
:
B
Given
r = 7m
h = 1m
Surface area of hemisphere = 2$\pi$$r^2$
$=2\times \frac{22}{7}\times (7{)}^2$= 308 $m^2$.
For calculating the surface area of a cone we need to calculate its slant height,
l = $\sqrt{r^2+h^2}$
l = $\sqrt{49+1}=\sqrt{50}$m.
Surface area of cone = $\pi$rl= $\frac{22}{7}\times 7\times \sqrt{50}=155.39m^2$.
So, area of cloth required = (308 + 155.39)$m^2$= 463.39$m^2$

Answer: Option A. -> Volume of cone = Volume of sphere
:
A
If any solid is reshaped into a new solid then volume always remains constant.

Answer: Option B. -> ₹ 7590
:
B

The total surface area of the tent:
$T=2\pi r{h}_1+\pi rl\pi =\frac{22}{7};r=2.1\text{m}{h}_1=\text{height of the cylinder}=4\text{m}{h}_2=\text{height of the cone}=2.8\text{m}l=\text{slant height of the cone}l=\sqrt{{2.8}^2+{2.1}^2}=3.5\text{m}$
So,
$T=(2\times \frac{22}{7}\times 2.1\times 4+\frac{22}{7}\times 2.1\times 3.5)T=75.9{\text{m}}^2$
Therefore, the total cost of canvas at₹100 per sq meter =$75.9\times 100=₹7590$.

Answer: Option C. -> 8074 cm2 
:
C

Total surface area of the bucket
$=TSA=\pi \times l\times (r_1+r_2)+\pi \times ({r_1}^2+{r_2}^2)$
Where $\pi =\frac{22}{7},$
$l=$ Slant height of the bucket
$=\sqrt{(h{)}^2+(r_1-r_2{)}^2}$
$r_1=28cm$ and $r_2$ = 7 cm
$l=$ Slant height of the bucket
$=\sqrt{(45{)}^2+(21{)}^2}cm=49.6cm$
Therefore,
$TSA=[\frac{22}{7}\times 49.6\times (28+7)+\frac{22}{7}\times ({28}^2+7^2)]=8074c{m}^2$

Answer: Option A. -> 10266.66 cm3
:
A
The volume of the toy = volume of hemisphere + volume of a cone
$V=\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^{2}h=\frac{2}{3}\times \frac{22}{7}(14{)}^3+\frac{1}{3}\times \frac{22}{7}(14{)}^2\left(22\right)=\frac{2}{3}\times 22\times 2\times 14\times 14+\frac{1}{3}\times 22\times 2\times 14\times 22=\frac{1}{3}[2\times 22\times 2\times 14\times 14+22\times 2\times 14\times 22]=\frac{1}{3}[17248+13552]=\frac{1}{3}\times 30800V=10266.66{\text{cm}}^3$
$\therefore$ Volume of the toy =$10266.66{\text{cm}}^3$

Answer: Option D. -> Curved surface area + area of both the circular ends.
:
D
If the objectis solid, then all the surfaces have to be painted. It consists of no hollow faces.
Therefore the area to be painted is,
curved surface area of frustum+ area of boththe circular ends.

Answer: Option B. -> 827.15
:
B
The volume of air inside the shed (when there are no people or machinery) is given by the
volume of air inside the cuboid and inside the half cylinder, taken together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.
Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid +$\frac{1}{2}$volume of the cylinder
= $[15\times 7\times 8+\frac{1}{2}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 15]m^3=1128.75m^3$
Next, the total space occupied by the machinery = $300m^3$
And the total space occupied by the workers = $20\times 0.08m^3=1.6m^3$
Therefore, the volume of the air, when there are machinery and workers
$=1128.75-(300.00+1.60)=827.15m^3$

Answer: Option A. -> 332.5 cm2
:
A
The hemisphere surmounts the cube, themaximum diameter the hemisphere can have = side of cube asshown in the figure below. The hemisphere will just touch the sides of top face of the cube.
Therefore, maximum diameter Hemisphere can have = 7 cm
$\Rightarrow \text{Radius of Hemisphere = r =}\frac{7}{2}=3.5cm$

Surface area of solid = Surfacearea of 5 Faces of cube + Surface area of Hemisphere + Surface area leftuncovered on the top face of the cube (shown in blue)
$=5l^2+2.\pi .r^2+\text{Area of Square ABCD - Area of inscribed circle in square ABCD}$
$=(5\times 7\times 7)+(2\times \frac{22}{7}\times 3.5\times 3.5)+(49-(\frac{22}{7}\times 3.5\times 3.5\left)\right)$
$=245+77+(49-38.5)=322+10.5=332.5c{m}^2$
$\therefore$ The surface area of combined solid = $332.5c{m}^2$

Answer: Option A. -> True
:
A
Height of frustum of cone = h = 14 cm
Radius of one end = $r_1=1$ cm
Radius of the otherend = $r_2=2$ cm

Volume of frustum of cone =
$\frac{1}{3}$$\pi .h\left(\right(r_1{)}^2+(r_2{)}^2+(r_1\left)\right(r_2\left)\right)$
$=\frac{1}{3}\times \frac{22}{7}\times 14(1^2+2^2+(1\left)\right(2\left)\right)$
$=\frac{1}{3}\times \frac{22}{7}\times 14\times 7=\frac{308}{3}=102\frac{2}{3}c{m}^3$

Answer: Option A. -> 22176
:
A
Circumference of the circle
$=2\pi r=528cm$
(where ris the radius of the circle)
$\Rightarrow r=\frac{528}{2\pi }=84cm$
Now, area of the circle = $\pi r^2$
=$\frac{22}{7}$ x${84}^2$ = 22176$c{m}^2$