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The wheels of a car of diameter 80cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66km per hour?

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Last updated date: 16th Jul 2024
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Answer
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Hint – In this question the diameter of the car's wheel is given to us. The complete revolutions which each tire will be making is simply the total distance traveled by the car divided by the circumference of the tire of the car. Use this concept to get the answer.

Complete step-by-step answer:
It is given that the diameter of the wheel is $ = 80$ cm.
So, the radius (r) of the wheel is half of diameter.
$ \Rightarrow r = \dfrac{{80}}{2} = 40$cm
So the circumference (C) of the wheel is $ = 2\pi r$
$ \Rightarrow C = 2\pi r = 2\left( {\dfrac{{22}}{7}} \right)\left( {40} \right)$ cm.
Now as we know 1cm = (1/100) m.
$ \Rightarrow C = 2\pi r = \dfrac{{2\left( {\dfrac{{22}}{7}} \right)\left( {40} \right)}}{{100}}$ m………………………… (1)
Now the distance covered by the car at the speed of 66 km/hr. in 10 minutes will be distance (d) is equal to multiplication of speed and time.
$ \Rightarrow d = 66\dfrac{{{\text{km}}}}{{{\text{hr}}}}\left( {10{\text{ min}}} \right)$
Now as we know $60{\text{ min = 1 hr}}$
$ \Rightarrow 10{\text{ min = }}\dfrac{{10}}{{60}}{\text{ hr}}$
$ \Rightarrow d = 66 \times \dfrac{{10}}{{60}} = 11$ km.
Now as we know that 1 km = 1000 m
Therefore 11 km = 11000 m.
$ \Rightarrow d = 11000$ m.
So, the complete revolution (S) each wheel makes is the ratio of total distance covered by the wheel divided by circumference of the wheel.
\[ \Rightarrow S = \dfrac{d}{C} = \dfrac{{11000}}{{\dfrac{{2\left( {\dfrac{{22}}{7}} \right)\left( {40} \right)}}{{100}}}} = \dfrac{{11000 \times 7 \times 100}}{{2 \times 22 \times 40}} = 4375\] Revolutions.
So, the complete revolutions (S) does each wheel make is 4375 revolutions.
So, this is the required answer.

Note – Whenever we face such type of problems the key concept is simply to have the understanding of the basic formula like circumference of circular tire, distance is product of speed and time. The note point is to convert the units of time and distance in synchronized pattern all throughout as it helps in getting on the right track to reach the answer.