Free Surface Areas and Volumes 02 Practice Test - 9th Grade 

We know that, 
$Volumeofacylindricalcontainer$
=$\pi$$r^{2}h$

$1000c{m}^3=1L$

Hence, the volume of each cylindrical tumbler
$=\frac{22}{7}\times 7^2\times 10$

$=1540c{m}^3$

So, the total volume of juice required
$=1540\times 25$
$=38500{cm}^3$
= 38.5 L

Given that,
r:h = 3:4
Let r = $3x$ and h = $4x$

From the relation
$l^2=h^2+r^2$,

The Slant height:
$l=\sqrt{(3x{)}^2+(4x{)}^2}$
$\Rightarrow l=5x$

We know that,
Total surface area of cone
$=\pi r(r+l)$
Curved surface area of cone
$=\pi rl$

Hence, the required ratio
$=\frac{Totalsurfaceareaofcone}{Curvedsurfacearea}$

$=\frac{\pi r(r+l)}{\pi rl}$

$=\frac{\pi \times 3x(3x+5x)}{\pi \times 3x\times 5x}$

$=8:5$

Given that, the ratio of the height of the pillar to that of dome is 4:3 .

$\Rightarrow \frac{h_{cylinder}}{h_{cone}}=\frac{4}{3}$

We know that,
$volumeofcylinder=\pi r^{2}h$.
$volumeofcone=\frac{1}{3}\pi r^{2}h$.

Now, ratio of  volume of cylinder to the volume of cone  
$=\frac{\pi r^2{h}_{cylinder}}{\frac{1}{3}\pi r^2{h}_{cone}}$

$=\frac{\pi r^2\times 4}{\frac{1}{3}\pi r^2\times 3}$

$=\frac{4}{1}$

$=4:1$

Total Surface Area = 2( l x b + b x h + h x l)
 = 2(26 x 14 + 14 x 6.5 + 6.5 x 26)
= 2(364 + 91 + 169
$=1248m^2$
 

The curved surface area of a cone = $\pi rl$
where, $r$ is radius and $l$ is slant height.

$\Rightarrow 308=\pi rl$
$\Rightarrow \frac{22}{7}\times r\times 14=308$
$\Rightarrow r=\frac{308\times 7}{22\times 14}$
$\Rightarrow r=7m$

Base area
$=\pi r^2$
$=\frac{22}{7}\times 7^2$
$=154m^2$

Volume of cube = ​(side)${}^3$

Volume of cuboid = (length) $\times$ (breadth) $\times$ (height)

So height of cuboidal tank can be found by equating the two

$\therefore \text{height}=\frac{{\text{side}}^3}{\text{length}\times \text{breadth}}$

$=\frac{{10}^3}{5\times 10}$

$\therefore$ Height = 20m

The volume will remain constant.

Let's assume $x$ such hemispherical balls will be formed.

So  $\frac{4}{3}\pi R^3=x\frac{2}{3}\pi r^3$

(Where R is the radius of spherical ball and r is the radius of the hemispheres)

$\Rightarrow \frac{4}{3}\pi {\left(21\right)}^3=x\frac{2}{3}\pi {\left(7\right)}^3$

$\Rightarrow x=\frac{2\times 21\times 21\times 21}{7\times 7\times 7}$
$\Rightarrow x=54$

Volume of cylinder $=\pi \times r^2\times h$
$\Rightarrow 231=\frac{22}{7}\times r^2\times 6$
$\Rightarrow r^2=231\times \frac{7}{22}\times \frac{1}{6}$
$\Rightarrow r^2=12.25$
$\Rightarrow r=3.5$
So, a ball will fit inside the cylinder if its radius is less than or equal to 3.5 $cm$.