The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.
Question:

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

Solution:

Let the inner radius of the hollow cylinder be r1 cm

The outer radius of the hollow cylinder be r2 cm

Then,

$2 \pi r_{1} * h+2 \pi r_{2} * h+2 \pi r_{2}^{2}-2 \pi r_{1}^{2}=4620 \ldots$ (a)

$\pi r_{1}^{2}-\pi r_{2}^{2}=115.5 \ldots$

Now solving eq (a)

$2 \pi r_{1} h+2 \pi r_{2} h+2 \pi r_{2}^{2}-2 \pi r_{1}^{2}=4620$

$\Rightarrow 2 \pi h\left(r_{1}+r_{2}\right)+2 \pi\left(r_{2}^{2}-r_{1}^{2}\right)$

$\Rightarrow 2 \pi h\left(r_{1}+r_{2}\right)+2\left(\pi r_{2}^{2}-\pi r_{1}^{2}\right)$

Now putting the value of (b) in (a) we get

⟹ 2πh(r1 + r2) + 231 = 4620

⟹ 2π ∗ 7(r1 + r2) = 4389

⟹ π(r+ r2) = 313.5 … (c)

Now solving eq (b)

$\pi r_{1}^{2}-\pi r_{2}^{2}=115.5$

$\Rightarrow \pi r_{2}^{2}-\pi r_{1}^{2}=115.5$

$\Rightarrow \pi\left(r_{2}+r_{1}\right)\left(r_{2}-r_{1}\right) \ldots(d)$

Dividing equation (d) by (c) we get

$\frac{\left.\pi r_{1}+r_{2}\left(r_{2}-r_{1}\right)\right)}{\pi\left(r_{1}+r_{2}\right)}=\frac{115}{313.5}$

⟹ r2 − r1 = 0.3684 cm

 

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.