Question

A roller 1.5m long has a diameter of 70cm. how many revolutions will it make to level a play ground measuring 50m × 33m

Answer 1

lenght of roller or height = 1.5 m
diameter of roller =0.7 m
radius of roller = 0.7/2 m

CSA of cylinder = 2πrh
= 2 × 22/7 × 0.7/2 × 1.5
= 3.3 m sq
area of field = 50×33 = 1650 m sq
no. of revolution taken = 1650/3.3 = 500revolu.

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Answer 2

Answer:

Hence the no. of revolutions will it make to level a playground is

500 revolutions.

Step-by-step explanation:

As per the data provided in the given question.

The given data is as follows,

The length of a roller i.e. height is (h) is 1.5 m and its diameter (d) is

70 cm.

The area of a playground is $50\;m\times33\;m$.

We have to find the no. of revolutions will it make to level a playground.

At first, we will convert the diameter to 70cm into m as below.

We know that,

$1\;cm=0.01\;m$

So we will multiply 70 into 0.01 we get,

$=70\times0.01\\=0.7\;m$

Now we will calculate the radius (r).

A radius of a circle is half of the diameter.

$r=\frac{1}{2}\times d \\r=\frac{1}{2}\times0.7\\ r=\frac{0.7}{2}\;m$

We will calculate the curved surface area of the roller by using a formula.

$The\;curved\;surface\;area\;of\;roller=2\pi rh$

Now we will substitute the given value and value of $\pi$ in the above formula.

$=2\times\frac{22}{7}\times\frac{0.7}{2}\times1.5\\=\frac{44}{7}\times\frac{0.7}{2}\times1.5$

We will divide 44 by 2 we get 22,

$=\frac{22}{7}\times0.7\times1.5\\ =22\times0.1\times1.5\\ =22\times0.15\\=3.3\;m^{2}$

The area of a playground is,

$=50\times33\\=1650\;m^{2}$

Now we will calculate no. of revolution as below,

$No.\;of\;revolutions=\frac{1650}{3.3}$

We will divide 1650 by 3.3

Further, we get

$=500\;Revolutions$

Hence the no. of revolutions will it make to level a playground is

500 revolutions.