Question

23. The diameter of a roller 1.5 m. long is 84 cm. It
takes 100 revolutions to level a playground. The
cost of leveling this playground at the rate of 1 per
square metre will be-​

Answer 1

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Q) The diameter of a roller 1.5 m long is 84 cm . It takes 100 revolutions to level a playground . The cost of levelling this playground at the arate of 1 per square metre will be ?

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★ Given :–

• Diameter (of roller) = 84 cm
• Length (of roller) = 1.5 m
• It takes 100 revolutions to level a ground .
• Rate of levelling = Rs 1 per m²

★ To Find :–

• Cost of levelling .

★ Solution :–

• Radius of the roller = $\dfrac{84}{2}$ = 42 cm = 0.42 m

The CSA will come in contact of the ground.

$\rm \: area _{1 \: revoution} = csa$

$\rm area \: covered \: in \: 100 \: revolutions = \: csa \times 100 = area _{playground}$

$\mapsto \rm \: area _{playground} = 2 \pi \: rh \times 100$

$\mapsto \rm \: area _ {playground} = 200 \times \frac{22}{ \cancel7} \times \cancel{0.42 }\times 1.5 \: {m}^{2}$

$\mapsto \rm \: area _{playground} = 200 \times 22 \times 0.6 \times 1.5 \: {m}^{2}$

$\mapsto \underline{ \boxed{ \rm{area _{playground} = 3960 \: {m}^{2} }}}$

Now ,

$\rightarrow \rm \: cost = rate \times area$

$\rightarrow \rm \: cost = \frac{1}{ \cancel{m}^{2} } \times 3960 \: \cancel {m}^{2}$

$\longrightarrow \underline{ \boxed{ \rm{ \pink{cost = rs \: 3960}}}}$

So ,

The Cost of levelling the plot is Rs 3960 .

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★ Know More :–

• Area of Cylinder = πr²h

• CSA of Cylinder = 2πrh

• TSA of Cylinder = 2πr(h+r)

• Area of base / top = πr²

Where ,

$\green{ \ddot{ \smile}} \: \sf \: r = radius \: of \: base$

$\purple{ \ddot{ \smile}} \: \sf \: h = height \: from \: base \: to \: top$

$\blue{ \ddot{ \smile}} \: \sf \: \pi = \frac{22}{7} \: or \: 3.14$