A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 16656cm3. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of 10 per cm2. (Take π=227)

Given:
Radius of cone = Radius of the hemisphere

\Rightarrow r = 3.5

Total volume

V = 166 \frac{5}{6} c {m^{3}}^{}

∴ V = \frac{1001}{6} c m^{3}

Volume of cone

= \frac{1}{3} π r^{2} h

Where

r =

radius and

h =

height of the cone.
Volume of hemisphere

= \frac{2}{3} π r^{3}

Where

r =

radius of the hemisphere
Let the height of the conical part be h.
Total volume = Volume of cone + Volume of hemisphere

\Rightarrow \frac{1001}{6} = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}

\Rightarrow \frac{1001}{6} = \frac{1}{3} π {(3.5)}^{2} h + \frac{2}{3} π {(3.5)}^{3}

\Rightarrow \frac{1001}{6} = \frac{1}{3} π 12.25 h + 85.75)

Substitute

π = \frac{22}{7}

\Rightarrow \frac{1001 \times 3 \times 7}{6 \times 22} = 12.25 h + 85.75

\Rightarrow 159.25 = 12.25 h + 85.75

\Rightarrow 12.25 h = 159.25 - 85.75

\Rightarrow h = \frac{73.5}{12.25}

\Rightarrow h = 6

Height of the toy = Height of cone + Height of hemisphere

\Rightarrow h = 6 + 3.5

\Rightarrow h = 9.5 cm

Surface area of hemisphere

= 2 π r^{2}

\Rightarrow 2 \times \frac{22}{7} \times 3.5 \times 3.5

\Rightarrow 77 c m^{2}

Cost of painting per

c m^{2} = Rs . 10

\Rightarrow

Cost of painting

77 c m^{2} = 10 \times 77

\Rightarrow

Cost of painting

77 c m^{2} = Rs . 770

Thus, the height of the toy

h = 9.5 cm

and cost of painting the hemisphere

= Rs . 770

.
Hence option 1 is correct.

JEE