The total surface area of a hollow cylinder, which is open from both the sides, is 3575 $c m^{2}$ ; area of its base ring is 357.5 $c m^{2}$ and its height is 14 cm, then the thickness of the cylinder is 3.5 cm.

Open in App

Solution

Let us take the outer radius as $R$ and inner radius as $r$ .
Area of ring is $A = π (R^{2} - r^{2})$
$357.5 = π (R^{2} - r^{2})$
$357.5 \div π = (R^{2} - r^{2})$ --- eq1
Surface area of hollow cylinder $= 2 π R h + 2 π r h + 2 π (R^{2} - r^{2})$
$3575 = 2 π R h + 2 π r h + 2 π (R^{2} - r^{2})$
$3575 \div π = 2 R h + 2 r h + 2 (R^{2} - r^{2})$ ---eq2
From eq1 & eq2,
$3575 \div π = 2 R h + 2 r h + 2 \times 357.5 \div π$
$2860 \div π = 2 R h + 2 r h$
$2860 \div π = 28 (R + r)$
$(R + r) = \frac{2860}{28} \div π$
Thickness $(R - r) = \frac{(R^{2} - r^{2})}{(R + r)} = \frac{357.5 \div π}{\frac{2860}{28} \div π} = 3.5 c m$
Thickness of the cylinder $= 3.5 c m$

Suggest Corrections

Similar questions

3575 c m^{2};

357.5 c m^{2}

14 c m

3.5 c m

3575 c m^{2}

357.5 c m^{2}

14 c m

Find the height of a cylinder if its total surface area is 2552 $c m^{2}$ and the radius is 14 cm.

28

20

{c m}^{2}

12 c m

8 c m

2 c m

Join BYJU'S Learning Program