Question

The total surface area of a hollow cylinder which is open from both sides is $4620sq.cm$, area of the base ring is $115.5sq.cm$ and height $7cm$. Find the thickness of the cylinder.

Answer 1

Total surface area of a hollow cylinder opened from both sides $=4620c{m}^2$
Area of base ring $=115.5c{m}^2$
Height of cylinder$\left(h\right)=7cm$
Let $R$ be the outer radius and $r$ be the inner radius

Area of base ring $=$ Area of outer circle $-$ Area of inner circle $=115.5c{m}^2$

$\therefore \pi (R^2-r^2)$$=115.5c{m}^2$

$\Rightarrow \frac{22}{7}(R^2-r^2)=\frac{1155}{10}$

$\Rightarrow (R^2-r^2)=\frac{1155}{10}\times \frac{7}{22}$

$\Rightarrow (R^2-r^2)=\frac{147}{4}$

$\Rightarrow (R-r)(R+r)=\frac{147}{4}$ ...(i)

Area of outer and inner curved surfaces $=$ Total Surface area $-$ area of base ring $=$ $4620-2\times 115.5=4389$

$\Rightarrow 2\pi h(R+r)=4389$

$\Rightarrow 2\times \frac{22}{7}\times 7(R+r)=4389$

$\Rightarrow 44(R+r)=4389$

$\Rightarrow R+r=\frac{4389}{44}$

$\therefore R+r=\frac{399}{4}$..(ii)

From (i) and (ii),

$\Rightarrow R-r=\frac{147}{4}\times \frac{4}{399}=\frac{7}{19}$

$\therefore$ Thickness $=0.36cm$