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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.

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719

423

354

845

answer is A.

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Detailed Solution

Here,
It is given a hollow cylinder of total surface area

and area of the base ring

and the height of cylinder 7 cm.
Let outer radius be `R’ cm, inner radius be

r

cm.
The thickness of cylinder is

R − r

.
Area of a hollow cylinder =

2 π R 2 − r 2 + 2 π R h + 2 π r h

Area of a hollow cylinder

= 2 π R + r h + R − r

Area of base =

π R 2 − π r 2

Area of base

= π R + r R − r

S u r f a c e a r e a a r e a o f t h e b a s e = 4620 115.5

⇒ 2 π R + r h + R − r π R + r R − r = 4620 115.5

⇒ 2 h + R − r R − r = 4620 115.5

Let us consider

R − r = t

⇒ 2 h + t t = 40

⇒ 2 h + t = 40 t

⇒ 2 h + 2 t = 40 t

⇒ 2 h = 38 t

Putting

⇒ 2 × 7 = 38 t

⇒ 14 = 38 t

⇒ t = 1438 ⇒ t = 719 ⇒ R − r = 719

The thickness of the cylinder is

R − r

.
The thickness of the cylinder is

.
Hence option (1) is correct.

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