JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
Two masses \(M _{1}\) and \(M _{2}\) are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass \(M _{2}\) is twice that of \(M_{1}\). the acceleration of the system is \(a_{1}\). When the mass \(M_{2}\) is thrice that of \(M_{1}\). The acceleration of The system is \(a_{2}\). The ratio \(\frac{a_{1}}{a_{2}}\) will be.

- A \(\frac{1}{3}\)
- B \(\frac{2}{3}\)
- C \(\frac{3}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(a=\frac{m_{2} g-m_{1} g}{m_{1}+m_{2}}\) \(Case 1\) \(M_{2}=2 m_{1}\) \(a_{1}=\frac{2 m_{1} g-m_{1} g}{3 m_{1}}\) \(a_{1}=g / 3\) Case \(-2\) \(M_{2}=3 m_{1}\) \(a_{2}=\frac{3 m_{1} g-m_{1} g}{4 m_{1}}\) \(a_{2}=\frac{g}{2}\)…
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