D

Let the height of the cone be h [latex]\frac {}{}[/latex]
Volume of the cone
= [latex]\frac {1}{3} \times \pi \times {(12)}^{2} \times h[/latex]
= 48πh [latex]{cm}^{3}[/latex]
Volume of the sphere
= [latex]\frac {4}{3} \times \pi \times {(r)}^{3}[/latex]
= [latex]\frac {4}{3} \times \pi \times {(12)}^{3}[/latex]
= 2304 [latex]{cm}^{3}[/latex]
Since the volumes are equal
48πh = 2304π
Solving for h
h = [latex]\frac {2304π}{48π}[/latex]
= 48 cm

A

Clearly, we have r = 3 cm and h = 4 cm.
Volume = [latex]\frac {1}{3} \times \pi \times {r}^{2} \times h[/latex]
= Volume of the cone
= [latex]\frac {1}{3} \times \pi \times {(3)}^{2} \times 4[/latex][latex]{cm}^{3}[/latex] = 12π [latex]{cm}^{3}[/latex]

B

1 hectare = 10,000 [latex]{m}^{2}[/latex]
So, Area = (1.5 x 10000) [latex]{m}^{2}[/latex] = 15000 [latex]{m}^{2}[/latex].
Depth = [latex]\frac {5}{100} m = \frac {1}{20}m[/latex]
Volume = (Area x Depth) = [latex](15000 \times \frac {1}{20}) {m}^{3} = 750 {m}^{3} [/latex]

C

2(15 + 12) x h = 2(15 x 12)
[latex]\Rightarrow h = \frac {180}{27} m = \frac {20}{3} m [/latex]
i.e,Volume = [latex](15 \times 12 \times \frac {20}{3}){m}^{3} = 1200 {m}^{3}[/latex]

B

External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron = [latex](\frac {22}{7} \times [{(4)}^{2}) - {(3)}^{2}]\times 21) {cm}^{3}[/latex]
= [latex](\frac {22}{7} \times 7 \times 1 \times 21) {cm}^{3}[/latex]
= 462 [latex]{cm}^{3}[/latex]