The total surface area of a hollow cylinder which is open from both sides is 4620 sq.

Question:

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

Solution:

Given:

Total surface area of the cylinder $=4620 \mathrm{~cm}^{2}$

Area of the base ring $=115.5 \mathrm{~cm}^{2}$

Height, $h=7 \mathrm{~cm}$

Let $R$ be the radius of the outer ring and $r$ be the radius of the inner ring.

Area of the base ring $=\pi R^{2}-\pi r^{2}$

$115.5=\pi\left(R^{2}-r^{2}\right)$

$\mathrm{R}^{2}-\mathrm{r}^{2}=115.5 \times \frac{7}{22}$

$(R+r)(R-r)=36.75 \quad \ldots \ldots \ldots \ldots$ (i)

Total surface area $=$ Inner curved surface area $+$ Outer curved surface area $+$ Area of bottom and top rings

$4620=2 \pi r h+2 \pi R h+2 \times 115.5$

$2 \pi h(R+r)=4620-231$

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

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

Substituting the value of $R+r$ from the equation (ii) in (i):

$\frac{399}{4}(R-r)=36.75$

$(R-r)=36.75 \times \frac{4}{399}=0.368 \mathrm{~cm}$

$\therefore$ Thickness of the cylinder $=(R-r)=0.368 \mathrm{~cm}$

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