Conversion One Shape Another: Explanation

Ans. A plastic drum resembles a frustum of a cone. The Height (h) of the drum=63 cm. Upper Radius (R) of the drum = 25 cm. Lower Radius (r) of the drum= 15 cm. Using the formula for the volume of the frustum of the cone the Volume (V) of the drum shall be: 1/3πH (R² + Rr + r²)

= 1/3 × 22/7 × 63(25²  + 25×15 +15²)
= 66 (625 + 375 + 225)
= 80,850 cm³
= 80 L 850 ml
One litre of milk costs Rs 45, and 80 L & 850 ml shall cost = 80.850×45 = Rs 3,638.25

The Drum stores milk of cost = Rs 3,638.25. Total Surface Area of the Drum = π l (R+r) + πR²  + πr²
l =√H2 +(R-r)²
l = 63² + (25-15) ²
= √3969+100
= 63.79 cm

Using the formula πl (R+r) +πR² +πr²
=22/7 × 63.79 (25+15) + 22/7 (25)² + 22/7(15)²
= 22/7 × 63.79 × 40 + 22/7 × 625 + 22/7 × 225
= 22/7[2551.6+625+225] = 10,690.74 cm²
Cost of painting per sq cm is 50 ps.
So for 10,690.74 sq cm the cost of painting shall be 10,690.74 × 50 = Rs 5,345 approximately.

Ans. Given that, Copper rod diameter = 1 cm.

Length of the copper rod = 8 cm

Length of new wire = 18 m = 18 × 100 = 1800 cm.

We know that the rod should be in the cylindrical shape. 

Hence, the volume of the rod = π × (½)² × 8 

= 2π cm³

Therefore, the volume of the copper rod = 2π cm³

If “r” is the radius of cross-section of the wire, then the volume of the wire is given as:

The volume of the wire =  π × r² × 1800

Since the volume of the copper rod and the volume of the new wire should be equal, then we can write

⇒ π × r² × 1800 = 2π

⇒ r² = 2π/1800π

⇒ r² = 1/900

⇒ r = 1/30

Hence, the thickness of the wire should be the diameter of the cross-section of the new wire.

Thickness = (1/30) x 2 = 1/15 cm.

Thus, the thickness of wire is approximately equal to 0.067 cm

Ans. Radius of cylindrical candle = 1.75/2 = 0.875 cm

Volume of one cylindrical candle = πr²h = π x (0.875)² x (0.02) cm³

= π x 0.0153125 = 0.048125 cm³

Volume of cuboid candle = 5.5 x 10 x 3.5 = 192.5 cm³

Thus, number of cylindrical candles = Volume of cuboid candle/Volume of one cylindrical candle

= 192.5/0.048125 = 4000

Ans. Volume of the iron ball = (4/3) πr³ = (4/3) x (22/7 )x 21 x 21 x 21 = 38,808 cm³

Let us assume that the radius of the smaller balls is r.

So volume of 27 smaller balls = 27 x (4/3) πr³ = 36πr³

Thus we have, the volume of the big iron ball = volume of 27 smaller balls

⇒38,808 cm3 = 36πr³

⇒ r³ = 343 cm³

So r = 7 cm

Therefore, the radius of each spherical ball is 7 cm.

Ans. Radius of the sphere  = \(= \frac{21}{2}\) cm.

Volume of the sphere \(\frac{4}{3} \pi r^3 = (\frac{4}{3}\pi \times \frac{21}{2}\times \frac{21}{2}\times \frac{21}{2})\)  cm³.

 \((\frac{3087 \pi}{2})\)cm³.

Radius of each \(=\frac{7}{2}\) cone cm and its height = 3 cm.

Volume of each cone \(\frac{1}{3} \pi r^2 h = (\frac{1}{3} \pi \times \frac{7}{2} \times \frac{7}{2}\times 3)\)  cm³.

= \((\frac{49 \pi}{4})\)cm³.

Required number of cones \(= \frac{\text{volume of the sphere}}{\text {volume of each cone}}\)

= \((\frac{3087 \pi}{2} \times \frac{4}{49 \pi})\) = 126.

Ans. External radius of the sphere = 5 cm.

Internal radius of the sphere = 3 cm.

Volume of metal in the hollow sphere\(\frac{4}{3} \pi (R^3 - r^3) = \frac{4}{3} \pi \times [(5)^3 - (3)^3] =(\frac{392 \pi}{3})\)  cm³.

Let the radius of the solid cylinder be r cm.

Height of the solid cylinder = \(\frac{8}{3}\) cm.

Volume of the solid cylinder= \((\pi r^2 h) = (\pi r^2 \times \frac{8}{3}) \text{cm}^3= (\frac{8 \pi r^2}{3})\)  cm³.

Volume of cylinder = Volume of metal in hollow sphere.

∴  \(\frac{8 \pi r^2}{3} \times \frac{392 \pi}{3}\)⇒ r2 = 49 ⇒ r = 7.

Hence, the diameter of the cylinder formed (2 x 7) cm = 14 cm.

Curved surface area of the cylinder

 \(2 \pi r h = (2 \times \frac{22}{7} \times 7 \times \frac{8}{3})\)cm²  = \(\frac{352}{3}\)cm² \(117 \frac{1}{3}\) cm².

Ans: Radius of the well = \(\frac{7}{2}\)  m m.

Depth of the well  22.5 m.

∴ Volume of the earth dugout = \pi \times (3.5) \times 22.5 m³

\(= \pi \times \frac{7}{2} \times \frac{7}{2} \times \frac{45}{2}\)m³.

Let the width of the embankment be r metres.

Embankment forms a cylindrical shell whose inner and outer radii are 3.5 m and (r + 3.5) m respectively and height 1.5 m.

∴ Volume of the embankment = \(\pi\)[(r + 3.5)² – (3.5)²] x 1.5 m³

=\(\pi\)r(r +7) x \(\frac{3}{2}\)m3.

But, Volume of the embankment  Volume of the earth dugout

=\(\pi\)r(r +7) x \(\frac{3}{2} =\pi \times \frac{7}{2} \times \frac{7}{2}\times \frac{45}{2}\)

⇒ \(\pi\)r(r +7) = \(\frac{49}{4} \times 15\)

⇒ 4r² + 28r = 735

⇒4r² + 28r –  735 = 0

⇒ \(r = {-28 \pm \sqrt{784 +11760} \over 8} \)

⇒ \(r = {-28 \pm \sqrt{12544} \over 8}\) =\({-28 \pm {112} \over 2a}\) = \(\frac{84}{5}\) = 10.5. [r > 0]

Hence, the width of the embankment is 10.5 m.

Ans- Volume of a solid sphere of radius r = Volume of cone of base  and height r

⇒ \(\frac{4}{3} \pi r^3 = {1 \over 3} \pi r^2_1(r)\)

⇒ 4r² = r²₁

⇒ r = 2r

Thus, the radius of the base of the cone is 2r.