Volume And Surface Area

Details

Details

Details

Let,
$$OP = OQ = OR = r$$
$$\therefore OR = h = r$$
∴ Curved surface area of the hemisphere = $$2\pi {r^2}$$
Curved surface area of a cone = $$\pi rl$$
Where,
$$l = \sqrt {{h^2} + {r^2}} $$
$$\,\,\,\,\, = \sqrt {{r^2} + {r^2}} $$
$$\,\,\,\,\, = r\sqrt 2 $$
∴ Required ratio :
$$ = \frac{{2\pi {r^2}}}{{\pi rl}}$$
$$ = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }}$$
$$ = \frac{2}{{\sqrt 2 }}$$
$$ = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$$
$$ = \frac{{2\sqrt 2 }}{2}$$
$$ = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1$$

Volume of sphere :
$$ = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}}$$
$$ = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}}$$
Volume of each cone :
$$ = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}}$$
$$ = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}}$$
∴ Number of cone :
$$ = \frac{{288\pi }}{{12\pi }}$$
$$ = 24$$

Since the tank is open at the top, we have :
Area of sheet required = Surface area of the tank
$$ = lb + 2\left( {bh + lh} \right)$$
$$ = \left[ {30 \times 20 + 2\left( {20 \times 12 + 30 \times 12} \right)} \right]{{\text{m}}^2}$$
$$ = \left( {600 + 1200} \right){{\text{m}}^2}$$
$$ = 1800{\text{ }}{{\text{m}}^2}$$
Length of sheet required :
$$ = \left( {\frac{{{\text{Area}}}}{{{\text{Width}}}}} \right)$$
$$ = \frac{{1800}}{3}m$$
$$ = 600\,m$$
∴ Cost of the sheet
= ₹ (600 × 10)
= ₹ 6000

Since a cube has 4 diagonals, we have :
Length of a diagonal
$$ = \left( {\frac{{12}}{4}} \right)cm$$
$$ = 3\,cm$$
Let the length of each edge of the cube be a cm
Then,
$$\sqrt 3 a = 3$$
$$or,a = \sqrt 3 $$
∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

Let the length of each pipe be $$l$$ inches
Then, volume of water in thinner pipe :
$$ = \left[ {\pi \times {{\left( {\frac{1}{2}} \right)}^2} \times 1} \right] \text{cu.inch}$$
$$ = \left( {\frac{{\pi l}}{4}} \right)\text{cu.inch}$$
Volume of water in thinker pipe :
$$ = \left( {\pi \times {3^2} \times l} \right)\text{cu.inch}$$
$$ = \left( {9\pi l} \right)\text{cu.inch}$$
∴ Required number of pipes :
$$ = \frac{{9\pi l}}{{\left( {\frac{{\pi l}}{4}} \right)}}$$
$$ = 36$$

Clearly, we have r = 3 cm and h = 4 cm
∴ Volume :
$$ = \frac{1}{3}\pi {r^2}h$$
$$ = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right)\pi {r^3}$$
$$ = 12\pi {\text{ c}}{{\text{m}}^3}$$