Question

The diameter of a roller 120cm long is 84cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling it at the rate of Rs. 2.50 per square metre.

Answer 1

Hint: We use the formula of curved surface area of the cylinder which is given by $CSA=2\pi rh$, where r is the radius of the base and h is the height of the cylinder.

Complete Step-by-Step Solution:
Given that the diameter of the roller is $d$=84cm and the height of the roller is $h$=120cm. We have to determine the cost of levelling it at the rate of Rs. 2.50 per square metre.

Let the radius of the roller be r then radius becomes $r=\dfrac{d}{2}=\dfrac{84}{2}=42$cm
Height of the roller is h =120cm.

Now because the roller is cylindrical in shape, we consider the curved surface area of the roller is equal to the curved surface area of the cylinder.

Now we see the area covered by the roller in 1 revolution is equal to the curved surface area of the roller which is anyway equal to the curved surface area of the cylinder given by $CSA=2\pi rh$ Substituting the values of radius r and height h in the formula we get,

The area covered by the roller in 1 revolution= $2\pi (42)(120)$

Taking the value of $\pi =\dfrac{22}{7}$ and solving the above equation we get,
The area covered by the roller in 1 revolution =$2\left( \dfrac{22}{7} \right)\left( 5040 \right)=31680c{{m}^{2}}$
Hence, we obtain the area covered by the roller in 1 revolution = $31680c{{m}^{2}}$

Now we determine the area covered by the roller in 500 revolutions, this can be done by multiplying 500 by the area covered by the roller in 1 revolution

Then, the area covered by the roller in 500 revolutions will be$=(31680)(500)=15840000c{{m}^{2}}$
Converting $15840000c{{m}^{2}}$into ${{m}^{2}}$we have to remove 4 zeroes at the end because we are dealing with ${{m}^{2}}$
Hence, the area covered by the roller in 500 revolutions will be 1584$$$${{m}^{2}}
Last step is to determine the cost of levelling it at the rate of Rs. 2.50 per square metre. Costs of levelling is determined by rate per square metre multiplied to the area covered by the roller in 500 revolutions.
Then Cost of levelling= $(2.50)(1584)=Rs.3960$
Hence, the cost of levelling the roller at the rate of Rs. 2.50 per square metre is Rs. 3960

Note: The possibility of error in the question can be at the point where you do not convert the area covered by the roller in 500 revolutions from $c{{m}^{2}}$to ${{m}^{2}}$which is a big mistake because, we do all calculations in Standard units. Proceeding the question in different units may lead to wrong answers.