 # The difference between the outer and inner Question:

The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in making the cylinder is 176 cm3, find the outer and inner diameters of the cylinder. (Use π = 22/7)

Solution:

The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder are r cm and R cm respectively. The difference between the outer and inner surface area of the hollow cylinder is

$=2 \pi R \times 14-2 \pi r \times 14$

$=28 \pi(R-r) \mathrm{cm}^{2}$

By the given condition, this difference is 88 square cm. Hence, we have

$28 \pi(R-r)=88$

$\Rightarrow R-r=\frac{44 \times 7}{14 \times 22}$

$\Rightarrow R-r=\frac{4 \times 7}{14 \times 2}$

$\Rightarrow R-r=1$

The volume of the metal used in making the cylinder is

$V_{1}=\pi\left\{(R)^{2}-(r)^{2}\right\} \times 14 \mathrm{~cm}^{3}$

By the given condition, the volume of the metal is 176 cubic cm. Hence, we have

$\pi\left\{(R)^{2}-(r)^{2}\right\} \times 14=176$

$\Rightarrow R^{2}-r^{2}=\frac{176 \times 7}{14 \times 22}$

$\Rightarrow R^{2}-r^{2}=4$

$\Rightarrow(R-r)(R+r)=4$

$\Rightarrow 1 \times(R+r)=4$

$\Rightarrow R+r=4$

Hence, we have two equations with unknowns R and r

$R-r=1$

$R+r=4$

Adding the two equations, we have

$(R-r)+(R+r)=1+4$

$\Rightarrow 2 R=5$

$\Rightarrow R=2.5$

Then from the second equation, we have

$r=4-2.5=1.5$

Therefore, the outer and inner diameters of the hollow cylinder are 5cm and 3cm respectively.