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  • In the figure shown, the pulleys and strings are massless. What will be acceleration of the block of mass 4m just after the system is released from reset. <img alt="\mathrm{\left[\theta=\

In the figure shown, the pulleys and strings are massless. What will be acceleration of the block of mass 4m just after the system is released from reset. \mathrm{\left[\theta=\sin ^{-1}\left(\frac{3}{5}\right)\right]}

Option: 1

\mathrm{\frac{g}{5}}


Option: 2

\mathrm{\frac{5g}{3}}


Option: 3

\mathrm{\frac{5g}{11}}


Option: 4

\mathrm{\frac{11g}{5}}


Answers (1)

best_answer

If a' is the acceleration of block of mass 4m, then 

\mathrm{\begin{gathered} \underbrace{T a \cos 0^{\circ}+T a \cos 0^{\circ}}_{\text {for boTh blocks }}+\underbrace{2 T \cos \theta\left(180^{\circ}-\theta\right) \times a^{\prime}}_{\text {for block of } 4 m}=0 \\ a^{\prime}=\frac{a}{\cos \theta} \downarrow \end{gathered}}

Using Newton's second law-

\mathrm{T-mg=ma} ----------------(1)

\mathrm{\begin{aligned} & 4 m g-2 T \cos \theta=4 m\left(\frac{a}{\cos \theta}\right) \\ & 4 m g-2 T \times \frac{4}{5}=4 m\left(\frac{a}{4 / 5}\right)-\text { (2) } \end{aligned}}                \mathrm{\left\{\begin{array}{l} \because \frac{3}{5}=\sin \theta \\ \cos \theta=\frac{4}{5} \end{array}\right.}

On simplifing above equations,we get

\mathrm{a=\frac{12 q}{33}, \frac{a}{\cos \theta}=\frac{5 g}{41}}

 

Posted by

Divya Prakash Singh

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