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Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

A particle P of mass m is attached to a vertical axis by two strings AP and BP of length I each as shown in the figure. The separation AB rotates around the axis with an angular velocity\[\omega \]. The tension in the two strings are \[{{T}_{1}}\]and\[{{T}_{2}}\]. Then (1) \[{{T}_{1}}-{{T}_{2}}=2mg\] (2) \[{{T}_{1}}+{{T}_{2}}=m{{\omega }^{2}}l\] (3) \[{{T}_{1}}={{T}_{2}}\] (4) \[{{T}_{1}}+{{T}_{2}}=mg\]

A)  1, 2 and 3 are correct

B)  1 and 2 are correct

C)   2 and 4 are correct

D)  1 and 3 are correct

Correct Answer: B

Solution :

                 Let\[{{T}_{1}}\]and\[{{T}_{2}}\]be the tensions in PA and PB respectively. From the figure,                 \[{{T}_{1}}\sin {{30}^{o}}={{T}_{2}}\sin {{30}^{o}}+mg\] or            \[{{T}_{1}}-{{T}_{2}}=2mg\]                      ...(i) and        \[{{T}_{1}}\cos {{30}^{o}}+{{T}_{2}}\cos {{30}^{o}}=m{{\omega }^{2}}l\] or            \[{{T}_{1}}+{{T}_{2}}=m{{\omega }^{2}}l\]                             ...(ii) From Eqs. (i) and (ii)                 \[{{T}_{1}}=\frac{m}{2}[{{\omega }^{2}}l+2g]\] and         \[{{T}_{2}}=\frac{m}{2}[{{\omega }^{2}}l-2g]\] Hence, option  is correct.