Question

The figure shows a block of mass $m$ placed on a smooth wedge of mass $M$. Calculate the value of ${}^{\prime }{M}^{\prime }$ and tension in the string, so that the block of mass $m$ will move vertically downwards with acceleration $10$ $m/s^2$. Take $g=10m/s^2$.

• A. The value of $M^{{}^{\prime }}$ is $\frac{Mcot\theta }{1-cot\theta }$
• B. The value of $M^{{}^{\prime }}$ is $\frac{Mtan\theta }{1-tan\theta }$
• C. The value of the tension in the string is $\frac{Mg}{tan\theta }$
• D. The value of tension is $\frac{Mg}{cos\theta }$

Answer 1

The correct option is C The value of the tension in the string is $\frac{Mg}{tan\theta }$

$\begin{array}{c}g\cos \theta =a\sin \theta constrained\\ a=g\cot \theta \\ M^{\prime }g=\left(M+M^{\prime }\right)a\\ \Rightarrow M^{\prime }g=\left(M+M^{\prime }\right)g\cot \theta \\ \Rightarrow M^{\prime }=M\cot \theta +M^{\prime }\cos \theta \\ \Rightarrow M^{\prime }=\frac{M\cos \theta }{1-\cos \theta }\\ M^{\prime }g-T=M^{\prime }.g\cos \theta \\ T=M^{\prime }g\left(1-\cos \theta \right)\\ =\frac{Mg\cos \theta \left(1-\cos \theta \right)}{\left(1-\cos \theta \right)}=\frac{Mg}{\tan \theta }\\ Hence,\\ optionCiscorrectanswer.\end{array}$