Match the following columns:

Solution:

(a)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 10 cm and h = 30 cm.
Capacity of the bucket = Volume of the frustum of the cone

$=\frac{\pi h}{3}\left(R^{2}+r^{2}+R r\right)$

$=\frac{22}{7} \times \frac{1}{3} \times 30 \times\left[(20)^{2}+\left(10^{2}\right)+(20 \times 10)\right] \mathrm{cm}^{3}$

$=\frac{220}{7} \times[400+100+200] \mathrm{cm}^{3}$

$=\left(\frac{220}{7} \times 700\right) \mathrm{cm}^{3}$

 

$=22000 \mathrm{~cm}^{3}$

Hence, $(a) \Rightarrow(q)$

(b)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 12 cm and h = 15 cm.

Slant height of the bucket, $l=\sqrt{h^{2}+(R-r)^{2}}$

$=\sqrt{(15)^{2}+(20-12)^{2}}$

$=\sqrt{225+64}$

$=\sqrt{289}$

 

$=17 \mathrm{~cm}$

Hence, $(b) \Rightarrow(s)$

(c)
Let R and r be the top and base of the bucket and let l be its slant height.
Then, R = 33 cm, r = 27 cm and h = 10 cm

Total surface area of the bucket $=\pi\left[R^{2}+r^{2}+l(R+r)\right]$

$=\pi \times\left[(33)^{2}+(27)^{2}+10 \times(33+27)\right]$

$=\pi \times[1089+729+600]$

 

$=2418 \pi \mathrm{cm}^{2}$

Hence, $(c) \Rightarrow(p)$

(d)
Let the diameter of the required sphere be d.

Then, volume of the sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi\left(\frac{\mathrm{d}}{2}\right)^{3}$

Therefore,

$\frac{4}{3} \pi\left(\frac{\mathrm{d}}{2}\right)^{3}=\frac{4}{3} \pi(3)^{3}+\frac{4}{3} \pi(4)^{3}+\frac{4}{3} \pi(5)^{3}$

$\Rightarrow \frac{4}{3} \pi \frac{\mathrm{d}^{3}}{8}=\frac{4}{3} \pi \times\left[(3)^{3}+(4)^{3}+(5)^{3}\right]$

$\Rightarrow \frac{\mathrm{d}^{3}}{8}=216$

$\Rightarrow d^{3}=1728$

$\Rightarrow d^{3}=12^{3}$

 

$\Rightarrow d=12 \mathrm{~cm}$

Hence, $(d) \Rightarrow(r)$