Free Surface Areas and Volumes 02 Practice Test - 10th Grade 

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. 

The hemisphere surmounts the cube, the maximum diameter the hemisphere can have = side of cube as shown 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  =  Surface area of 5 Faces of cube +  Surface area of Hemisphere + Surface area left uncovered 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 = $332.5c{m}^2$

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

$=\pi .(r_1+r_2)l$

$=(\pi .r_1+\pi .r_2)l$

$=(2\pi .r_1+2\pi .r_2)\frac{l}{2}$

$=(18+6)\frac{4}{2}=48c{m}^2$

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

$=1128.75-(300.00+1.60)=827.15m^3$ 

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 $\pi )=540\pi$

$n=\frac{540}{54}=10$

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.

 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 = $10266.66{\text{cm}}^3$

The cylinder is a 3-D object. Here the 3-D surface of the cylinder can be unrolled into a 2-D rectangular sheet of paper. The area of the 3-D surface and that of the 2-D plane are the same. It is clear that the area of rectangle = Length $\times$ Breadth= LB

Height of frustum of cone = h = 14 cm

Radius of one end = $r_1=1$ cm

Radius of the other end = $r_2=2$ cm 

Volume of frustum of cone =
$\frac{1}{3}$ $\pi .h\left(\right(r_1{)}^2+(r_2{)}^2+(r_1\left)\right(r_2\left)\right)$ 

$=\frac{1}{3}\times \frac{22}{7}\times 14(1^2+2^2+(1\left)\right(2\left)\right)$ 

$=\frac{1}{3}\times \frac{22}{7}\times 14\times 7=\frac{308}{3}=102\frac{2}{3}c{m}^3$ 

If the object is 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 both the circular ends.