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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 is 7 cm. Find the thickness of the cylinder.

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Solution

Let the inner and outer radii of hollow cylinder be r cm and R cm respectively. Then,

C.S.A = Curved Surface Area

Inner C.S.A of Cylinder + Outer C.S.A + Base area of top + Base area of bottom = Total C.S.A of hollow cylinder

2πrh+2πRh+2(πR2πr2)=4620(i)

and, Base area =πR2πr2=115.5

π(R2r2)=115.5

π(R+r)(Rr)=115.5(ii)

On solving (i), we get

2πh(r+R)+2(πR2πr2)=4620

2πh(r+R)+(2×115.5)=4620

2π×7(r+R)+231=4620

2π×7(r+R)=4620231

π(R+r)=438914=331.5(iii)

On dividing (ii) by (iii), we get

π(R+r)(Rr)π(R+r)=115.5313.5

Rr=719

Hence, Thickness =Rr=719=0.368 cm


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