Solve this.

36. The arrangement is kept in a vertical plane and all masses are released from rest with strings  taut.
In column 1, different values of masses are given. Match with corresponding parameters in column 2.
Symbols have their ususal  meanings. pulleys and strings are ideal.


    Column 1                                                      column 2
      m₁ m₂  m₃  m4                                           (P) |a₁|  = |a₃|
(A)  2m  m   3m  4m                                        (Q) T₁ =  T₂
(B)  m   2m  4m  3m                                        (R) T₂ > T₁
(C)  m   2m  3m  4m                                        (S) |a₁| > |a₃|

(D)  4m  3m  2m  m                                         (T)  T₃= O
 

Dear Student,

Please find below the solution to the asked query:

If we write the equations for the four blocks,

$$\begin{gathered} Here, all the blocks are having the same acceleration \\ So, always \left|a_1\right|=\left|a_3\right| \\ Equation for the first block is, \\ {m}_{1}a=T_1-m_{1}g ----\left(1\right) \\ Equation for the second block is, \\ {m}_{2}a=m_{2}g+T_3-T_1 ----\left(2\right) \\ Equation for the third block, \\ {m}_{3}a=T_2-m_{3}g-T_3 ----\left(3\right) \\ Equation for the fourth b \\ {m}_{4}a=m_{4}g-T_2 ----\left(4\right) \\ So, \\ \left.\begin{array}{c}{m}_{3}a=T_2-m_{3}g-T_3\\ m_{2}a=m_{2}g+T_3-T_1\\ m_{4}a=m_{4}g-T_2 \end{array}\right\}\Rightarrow \left(m_2+m_3+m_4\right)a=\left(m_2-m_3+m_4\right)g-T_1 \\ Now, \\ \left.\begin{array}{r}\left(m_2+m_3+m_4\right)a=\left(m_2-m_3+m_4\right)g-T_1\\ m_{1}a=T_1-m_{1}g\end{array}\right\} \\ \Rightarrow \left(m_1+m_2+m_3+m_4\right)a=\left(-m_1+m_2-m_3+m_4\right)g \\ a=\frac{\left(-m_1+m_2-m_3+m_4\right)g}{\left(m_1+m_2+m_3+m_4\right)} \\ if m_1=2m; m_2=m; m_{3 }=3m and m_4=4m \\ a=0, \\ {T}_1=m_{1}g=2mg and T_2=m_{4}g=4mg \\ {T}_2 > T_1 \\ {T}_3=T_1-m_{2}g=2mg-mg=mg \\ if m_1=m; m_2=2m; m_{3 }=4m and m_4=3m \\ a=0, \\ {T}_1=m_{1}g=mg and T_2=m_{4}g=3mg \\ {T}_2 > T_1 \\ {T}_3=T_1-m_{2}g=mg-2mg=-mg \\ if m_1=m; m_2=2m; m_{3 }=3m and m_4=4m \\ a=\frac{g}{5}, \\ {T}_1=m_1\left(a+g\right)=mg+\frac{mg}{5}=\frac{6mg}{5} and \\ {T}_2=m_{4}g-m_{4}a=4mg-\frac{4mg}{5}=\frac{16mg}{5} \\ {T}_2 > T_1 \\ {T}_3=T_1-m_{2}g+m_{2}a=\frac{6mg}{5}-2mg+\frac{2mg}{5}=-\frac{2mg}{5} \\ if m_1=4m; m_2=3m; m_{3 }=2m and m_4=m \\ a=\frac{-g}{5}, \\ {T}_1=m_1\left(a+g\right)=4mg-\frac{4mg}{5}=\frac{16mg}{5} and \\ {T}_2=m_{4}g-m_{4}a=mg+\frac{mg}{5}=\frac{6mg}{5} \\ S \\ {T}_1 > T_2 \\ {T}_3=T_1-m_{2}g+m_{2}a=\frac{16mg}{5}-3mg-\frac{3mg}{5}=-\frac{2mg}{5} \end{gathered}$$

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