MHT CET · Physics · Motion In Two Dimensions
Two objects of masses ' \(m_1\) ' and ' \(m_2\) ' are moving in the circles of radii ' \(r_1\) ' and ' \(r_2\) ' respectively. Their respective angular speeds ' \(\omega_1\) ' and ' \(\omega_2\) ' are such that they both complete one revolution in the same time ' \(t\) '. The ratio of linear speed of ' \(\mathrm{m}_2\) ' to that of ' \(\mathrm{m}_1\) ' is
- A \(\omega_1: \omega_2\)
- B \(\mathrm{T}_2: \mathrm{T}_1\)
- C \(\mathrm{m}_1: \mathrm{m}_2\)
- D \(\mathrm{r}_2: \mathrm{r}_1\)
Answer & Solution
Correct Answer
(D) \(\mathrm{r}_2: \mathrm{r}_1\)
Step-by-step Solution
Detailed explanation
The cars complete one revolution in the same time.
\(\begin{array}{ll}
\therefore & \omega_1=\omega_2 \\
\therefore & \frac{\mathrm{v}_1}{\mathrm{r}_1}=\frac{\mathrm{v}_2}{\mathrm{r}_2} \Rightarrow \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\mathrm{r}_2}{\mathrm{r}_1}
\end{array}\)
\(\begin{array}{ll}
\therefore & \omega_1=\omega_2 \\
\therefore & \frac{\mathrm{v}_1}{\mathrm{r}_1}=\frac{\mathrm{v}_2}{\mathrm{r}_2} \Rightarrow \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\mathrm{r}_2}{\mathrm{r}_1}
\end{array}\)
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