| String is massless and pulley is smooth in the adjoining figure total mass or left hand side of the pulley side of the pulley is \[{{m}_{1}}\] and on right hand side is \[{{m}_{2}}\] . Friction coefficient between block B and the wedge is \[\mu =\frac{1}{2}\] and \[\theta =30{}^\circ \] Select the wrong answer |
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A) Block B will slide down if \[{{m}_{1}}={{m}_{2}}\]
B) Block B may remain stationary with respect to wedge, for suitable values of \[{{m}_{1}}\] and \[{{m}_{2}}\] with \[{{m}_{1}}>{{m}_{2}}\]
C) Block B cannot remain stationary with respect to wedge in any case
D) Block B will slide down if \[{{m}_{1}}>{{m}_{2}}\]
Correct Answer: B
Solution :
| If \[{{m}_{1}}={{m}_{2}}\], block will slide down when |
| \[\mu \,g\,\cos \theta <g\,\sin \,\theta \] or \[\frac{1}{2}\left( \frac{\sqrt{3}}{2} \right)<\frac{1}{2}\] |
| If \[{{m}_{1}}={{m}_{2}}\], block will slide down when |
| \[\mu \left( g+a \right)\cos \theta <\left( g-a \right)\sin \,\theta \] |
| or \[\frac{1}{2}\left( \frac{\sqrt{3}}{2} \right)<\frac{1}{2}\] where \[a=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,\,g\] |
| therefore in all cases block will slide down |
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