The radii of the circular ends of a bucket of height 15 cm

Solution:

We have,

Height, $h=15 \mathrm{~cm}$,

Radius of the upper end, $R=14 \mathrm{~cm}$,

Radius of lower end $=r$,

As,

Volume of the bucket $=5390 \mathrm{~cm}^{3}$

$\Rightarrow \frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)=5390$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 15 \times\left(14^{2}+r^{2}+14 r\right)=5390$

$\Rightarrow \frac{110}{7} \times\left(196+r^{2}+14 r\right)=5390$

$\Rightarrow 196+r^{2}+14 r=\frac{5390 \times 7}{110}$

$\Rightarrow 196+r^{2}+14 r=343$

$\Rightarrow r^{2}+14 r-147=0$

$\Rightarrow r^{2}+21 r-7 r-147=0$

$\Rightarrow r(r+21)-7(r+21)=0$

$\Rightarrow(r+21)(r-7)=0$

$\Rightarrow r+21=0$ or $r-7=0$

$\Rightarrow r=-21$ or $r=7$

As, $r$ cannot be negative

$\therefore r=7 \mathrm{~cm}$

So, the value of r is 7 cm.