Q187 / 200 · Manipal (MET) 2010 · Physics

In the figure shown, $m_1=10 \mathrm{~kg}, m_2=6 \mathrm{~kg}$, $m_3=4 \mathrm{~kg}$. If $T_3=40 \mathrm{~N}, T_2=?$ Figure

  • A
    13 N
  • Correct answer
    B
    32 N
  • C
    25 N
  • D
    35 N

Solution

Let $a$ be the acceleration of each block. Then, $\begin{array}{ll} & T_3=\left(m_1+m_2+m_3\right) a...(i) \\ \text { and } & T_2=\left(m_1+m_2\right) a...(ii) \end{array}$ From Eqs. (i) and (ii), we get $\begin{aligned} T_2 & =\left(\frac{m_1+m_2}{m_1+m_2+m_3}\right) \times T_3 \\ & =\left(\frac{10+6}{10+6+4}\right) \times 40=32 \mathrm{~N} \end{aligned}$

Hence, the correct answer is option: 'B'.

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