AP EAMCET · PHYSICS · Laws of Motion
Two situations are shown in fig. (a) and (b).

In each case, \(m_1=3 \mathrm{~kg}\) and \(m_2=4 \mathrm{~kg}\). If \(a_1, a_2\) are the respective accelerations of the blocks in these situations, then the values of \(a_1\) and \(a_2\) are respectively [ \(g=10 \mathrm{~ms}^{-2}\) ]
- A \(\frac{20}{7} \mathrm{~ms}^{-2}, \frac{10}{7} \mathrm{~ms}^{-2}\)
- B \(\frac{10}{7} \mathrm{~ms}^{-2}, \frac{25}{7} \mathrm{~ms}^{-2}\)
- C \(\frac{40}{7} \mathrm{~ms}^{-2}, \frac{10}{7} \mathrm{~ms}^{-2}\)
- D \(\frac{30}{7} \mathrm{~ms}^{-2}, \frac{5}{7} \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(C) \(\frac{40}{7} \mathrm{~ms}^{-2}, \frac{10}{7} \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
\((a)\) \(m_2 g- T =m_2 a_1\) ...\((i)\) \(T=m_1 a_1\) ...\((ii)\) From \((i) \&(i i)\), we get \( a_1=\frac{m_2 g}{m_1+m_2} \) as \(m_1=3 \mathrm{~kg}, m_2=4 \mathrm{~kg}\) \( a_1=\frac{4 \times 10}{3+4}=\frac{40}{7} \mathrm{~m} / \mathrm{s}^2 \) For fig. (b)…
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