Volume And Surface Area

Volume of the cylinder :
= Volume of the cube
= (11)³ cm³
= 1331 cm³
Let the radius of the base be r cm
Then,
$$\frac{{22}}{7} \times {r^2} \times 14 = 1331$$
$$ ⇒ {r^2} = \frac{{1331}}{{44}} = \frac{{121}}{4}$$
$$ ⇒ r = \frac{{11}}{2}$$
$$ ⇒ r = 5.5$$

Let their heights be 2h and 3h and radii be r and R respectively
Then,
$$\pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right)$$
$$ ⇒ \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2}$$
$$ ⇒ \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 $$

$$\frac\text{πr}{2}h}{2}\text{πr}{2}h}\ \=\ \frac{924}{264}$$  

⇒ r = (24/364)×2 = 7m 

$${2}\text{πrh}\ \=\ {264}\ \ \⇒\ h}\ \=\ \({264}\ \×\ \frac{7}{22}\×\frac{1}{2}\ \×\frac{1}{7}\)\ \=\ {6}m}$$   

Required Rato =2r/h  =14/6 = 7:3

h = 14 cm, r = 7 cm.  

So, l =$${7}{2}+{14}{2}\ \=\ {245}\ \=\ {7}{5}\text{c}\text{m}$$   

Total surface area = $$\text{πrl}+\text{πr}{2}$$ = $$\frac{22}{7}\×{7}\×{7}{5}\ +\ \frac{22}{7}\×{7}\×{7}$$ 

=$${154}{5}+{1}$$ 

= (3.236×154) sq.cm   

= 498.35 sq.cm

Curved surface area :
$$ = 2\pi rh$$
$$ = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 22.5} \right){{\text{m}}^2}$$
$$ = 495{\text{ }}{{\text{m}}^2}$$
∴ Cost of plastering :
= ₹ (495 × 3)
= ₹ 1485

Volume :
$$ = \pi {r^2}h$$
$$ = \left( {\frac{{22}}{7} \times 1 \times 1 \times 14} \right){m^3}$$
$$ = 44\,{m^3}$$

Let original radius = R
Then, new radius = $$\frac{{\text{R}}}{2}$$
$$\therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}}$$
$$ = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}}$$
$$ = \frac{1}{4}\,Or\,1:4$$

Let the height of the cone be h cm
Then,
$$\pi \times {r^2} \times 6 = \frac{1}{3} \times \pi \times {r^2} \times h$$
$$ ⇒ h = 18\,cm$$

Let the thickness of the bottom be×cm
Then,
$$\left[ {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} \right]$$       $$ = 8000 \times 1000$$
$$ ⇒ 320 \times 250 \times \left( {110 - x} \right) = $$       $$8000 \times 1000$$
$$ ⇒ \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}}$$
$$ ⇒ \left( {110 - x} \right) = 100$$
$$ ⇒ x = 10\,cm$$
$$ ⇒ x = 1\,dm$$

$${\text{Number of bricks :}}$$
$$ = \frac{{{\text{Volume of the wall}}}}{{{\text{Volume of 1 brick}}}}$$
$$ = \left( {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \right)$$
$$ = 6400$$