AP EAMCET · PHYSICS · Laws of Motion
Two masses \(m_1\) and \(m_2\) are connected by a light string passing over smooth pulley. When set free \(m_1\) moves downwards by 3 m in 3 s . The ratio of \(\frac{m_1}{m_2}\) is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)
- A \(\frac{9}{7}\)
- B \(\frac{8}{7}\)
- C \(\frac{10}{7}\)
- D \(\frac{15}{13}\)
Answer & Solution
Correct Answer
(B) \(\frac{8}{7}\)
Step-by-step Solution
Detailed explanation
\(a=\left(\frac{m_1-m_2}{m_1+m_2}\right) g\) Now, \(S=u t+\frac{1}{2} a t^2\) \(\Rightarrow 3=0+\frac{1}{2}\left(\frac{m_1-m_2}{m_1+m_2}\right) \times 10 \times(3)^2\)…
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