Question

1.find the volume, area of the curved surface and the total surface area of a cylinder whose height and radius of base are 20 cm and 28 CM respectively.

2.the diameter of a roller is 2.1 metre and its length is 1.75 metre if it takes 500 revolutions to level a field find the cost of levelling it at 75 paise per metre ²
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Answer 1

1)

given:

h = height = 20cm

r = radius = 28cm

to find:

volume , tsa , csa

Sol.

volume of cylinder = $\pi r^{2} h$

= 22/7 * 28 * 28 * 20 cm^3

= 22 * 4 * 28 * 20 cm^3

= 49280 cm^3

or , 49.28 l

csa of cylinder = $2\pi rh$

= 2 * 22/7 * 28 * 20

= 2* 22 * 4 * 20

= 3520cm^2

tsa of cylinder = $2\pi rh + 2\pi r^{2}$

= 3520 + 2 * 22/7 * 28 * 28

= 3520 + 2 * 22 * 4 *28

= 3520 + 4928

= 8448cm^2

2)

given:

d = diameter = 2.1 m

h = length  = 1.75 m

revolutions = 500

and cost of  leveling per m^2 is Rs 0.75/-

to find:

total cost of leveling

Sol.

total area covered = csa * total no. of revolutions

= $2\pi rh * 500$

= $\pi d h *500$

= 22/7 * 2.1 * 1.75 * 500

= 22 * 0.3 * 1,75 *500 m^2

now , total cost  = cost of leveling a m^2 * total area

=  22* 0.3 * 1.75 * 500 * 0.75

= Rs 4331.25/-

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\bigstar \: \: {{ \boxed{{\bold\green{Curved \: Surface \: Area_{(Cylinder)}\: = \:2\pi rh)}}}}}$

$\bigstar \: \: {{ \boxed{{\bold\blue{Total \: Surface Area_{(Cylinder)}\: = \:2\pi r(h +r)}}}}}$

$\bigstar \: \: {{ \boxed{{\bold\red{Volume_{(Cylinder)}\: = \:\pi r^2 h }}}}}$

CALCULATION :-

$\large\underline\blue{\bold{ \sf \: ANSWER \: - \: 1}}$

Given that

• Height of Cylinder, h = 20 cm

• Radius of Cylinder, r = 28 cm

So,

• Volume of cylinder is given by

$\longmapsto \rm \: {{ {\large{\bold\red{Volume_{(Cylinder)}\: = \:\pi r^2 h }}}}}$

$\longmapsto \rm \: Volume_{(Cylinder)}\: = \: \dfrac{22}{7} \times 28 \times 28 \times 20$

$\longmapsto \boxed{ \purple{ \bf \: Volume_{(Cylinder)}\: = \: 49280 \: {cm}^{3} }}$

Now,

• Total Surface Area of cylinder is given by

$\rm :\implies\: \red{TSA_{(cylinder)} = 2\pi \: r(h + r)}$

$\longmapsto \rm \: TSA_{(cylinder)} = 2 \times \dfrac{22}{7} \times 28 \times (28 + 20)$

$\longmapsto \rm \: TSA_{(cylinder)} = 44 \times 4 \times 48$

$\longmapsto \boxed{ \green{ \bf \: TSA_{(cylinder)} = 8448 \: {cm}^{2} }}$

Now,

• Curved Surface Area of cylinder is given by

$\rm :\implies\: \purple{CSA_{(cylinder)} \: = 2\pi \: rh}$

$\longmapsto \rm \: CSA_{(cylinder)} = 2 \times \dfrac{22}{7} \times 28 \times 20$

$\longmapsto \boxed{ \orange{ \bf \:CSA_{(cylinder)} = 3520 \: {cm}^{2} }}$

$\large\underline\blue{\bold{ \sf \: ANSWER \: - \: 2}}$

Given that

• Diameter of cylindrical roller, d = 2.1 m

So,

• Radius of cylindrical roller, r = 1.05 m

And

• Height of cylindrical roller, h = 1.75 m

Now,

We know that,

$\bull \: \: \: \red{ \sf \: Area \: covered \: in \: 1 \: revolution \: = CSA_{(cylinder)}}$

$\rm :\implies\:Area \: covered \: in \: 500 \: revolution = 500 \times 2\pi \: r h$

$\rm :\implies\:Area_{(covered)} \: = \: 500 \times 2 \times \dfrac{22}{7} \times 1.05 \times 1.75$

$\rm :\implies\:Area_{(covered)} \: = 5775 \: {m}^{2}$

Now,

Cost of levelling the field is Rs 0.75 per square metre.

So,

Total Cost of levelling the field is

$\longmapsto \rm \: Cost_{(levelling)} \: = 5775 \times 0.75$

$\longmapsto \boxed{ \green{ \bf \: Cost_{(levelling)} \: = Rs \: 4331.25}}$