Question

A toothed wheel of diameter $50\text{ cm}$is attached to a smaller wheel of diameter $30\text{ cm}$.Then the number of revolutions the smaller wheel will make when the larger one makes $30$revolution is
A) $30$
B) $50$
C) $40$
D) $60$

Answer 1

Hint:When two toothed wheels are attached with each other, then the distance covered by one wheel for $n$ revolutions is equal to the distance covered by the second wheel for $m$ revolutions. The distance covered by the wheel for $1$ revolution is given by the circumference of that circular wheel. We know that the circumference of the circle is $\pi d$ where $d$ is the diameter of the circular wheel.

Complete step by step answer:
Given that, the toothed wheel of $50cm$ diameter is attached to the wheel of $30cm$ tooth wheel.
The distance covered by the circular wheel of the diameter $50cm$ is given by the circumference of that circular wheel. Hence the circumference of the circle of diameter $50cm$ is
$\begin{align}
  & {{D}_{1}}=\pi d \\
 & =50\pi
\end{align}$
The distance covered by the circular wheel of the diameter $30cm$ is given by the circumference of that circular wheel. Hence the circumference of the circle of diameter $30cm$ is
$\begin{align}
  & {{D}_{2}}=\pi d \\
 & =30\pi
\end{align}$
Given that the large circle which is having the diameter of $50cm$ will make $30$ revolutions, then the distance covered by the large circle is
$\begin{align}
  & {{D}_{n}}=30\left( {{D}_{1}} \right) \\
 & =30\left( 50\pi \right)
\end{align}$
Now let us assume that the small tooth wheel of diameter $30cm$ cover the distance ${{D}_{n}}$ which is made by larger wheel of diameter $50cm$, in $k$ revolutions. Then the distance covered by the small wheel of diameter $30cm$ for $k$ revolutions is
$\begin{align}
  & {{D}_{k}}=k\left( {{D}_{2}} \right) \\
 & =k\left( 30\pi \right)
\end{align}$
Now we have ${{D}_{n}}={{D}_{k}}$, substituting the both the values in the above equation then
$\begin{align}
  & 30\left( 50\pi \right)=k\left( 30\pi \right) \\
 & k=\dfrac{30\left( 50\pi \right)}{30\pi } \\
 & k=50
\end{align}$
Hence the smaller wheel of diameter $30cm$ will make $50$ revolutions while the larger wheel of diameter $50cm$ makes $30$ revolutions.

Note:
Don’t find the value of the circumference by multiplying the diameter with the value of $\pi =3.14$. Remain the $\pi $ as it is till the end of the problem because at the ending, we can cancel the $\pi $. You can also solve the problem by using the formula $\dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$ where ${{d}_{1}},{{d}_{2}}$ are the diameters of the two circles and ${{n}_{1}},{{n}_{2}}$ are the number of revolutions. For this problem ${{d}_{1}}=50,{{d}_{2}}=30cm,{{n}_{1}}=30,{{n}_{2}}=?$ so from the formula
$\begin{align}
  & \dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}} \\
 & \dfrac{30}{50}=\dfrac{30}{{{n}_{2}}} \\
 & {{n}_{2}}=\dfrac{30\times 50}{30} \\
 & {{n}_{2}}=50
\end{align}$
Here also we get number of revolutions for smaller circle is $50$.