Math, asked by bvamtayaru82, 3 months ago

the diameter of the wheels of a bus is 142 cm . how many revolutions per minute must A wheel make in order to move at a speed of 66 km / hr

take pi = 22 / 7

plz explain very very clearly plz I am facing so much difficulty ​

Answers

Answered by SitaramKeLuvKush
290

Given

  • Diameter of the wheels of a bus is 142 cm.
  • Wheel make in order to move at a speed of 66 km / hr.

We Find

Revolutions per minute must a wheel make in order to move at a speed of 66 km / hr

We Know

☆ Circumference of wheel :- 2 × 22/7 × Radius

= 2 × 22/7 × 71

= 446.28 CM

☆ distance traveled in 60 min = 66 × 1000 × 100 cm

So, distance traveled in 1 min = \frac{60}{66} ×1000×100 cm \:  \\  \\ =110000 cm \\ \\

According to the question

Now , We Find the revolution in 1 min, So :-

Revolution in 1 min =  \frac{110000}{446.28} \\  \\

Revolution in 1 min =  \frac{110000}{446.28} \\  \\

Revolution in 1 min =  \cancel\frac{110000}{446.28} \\  \\

Revolution in 1 min =  \boxed {\red{246.48}} \\  \\

Hence, The required answer is 247 CM ( Approximately ).

Answered by Rohitranawatyadav
0

Step-by-step explanation:

Given

Diameter of the wheels of a bus is 142 cm.

Wheel make in order to move at a speed of 66 km / hr.

We Find

Revolutions per minute must a wheel make in order to move at a speed of 66 km / hr

We Know

☆ Circumference of wheel :- 2 × 22/7 × Radius

= 2 × 22/7 × 71

= 446.28 CM

☆ distance traveled in 60 min = 66 × 1000 × 100 cm

So, distance traveled in 1 min = \begin{gathered}\frac{60}{66} ×1000×100 cm \: \\ \\ =110000 cm \\ \\\end{gathered}

66

60

×1000×100 cm

=110000 cm

According to the question

Now , We Find the revolution in 1 min, So :-

Revolution in 1 min = \begin{gathered} \frac{110000}{446.28} \\ \\\end{gathered}

446.28

110000

Revolution in 1 min = \begin{gathered} \frac{110000}{446.28} \\ \\\end{gathered}

446.28

110000

Revolution in 1 min = \begin{gathered} \cancel\frac{110000}{446.28} \\ \\\end{gathered}

446.28

110000

Revolution in 1 min = \begin{gathered} \boxed {\red{246.48}} \\ \\\end{gathered}

246.48

Hence, The required answer is 247 CM ( Approximately ).

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