question_answer

Two blocks are placed on a wedge with coefficients of friction being different for two blocks. Choose the correct option (friction is not sufficient to prevent the motion)

A)  if\[{{m}_{1}}<{{m}_{2}}\], then normal reaction between blocks will be non-zero

B)  if\[{{m}_{1}}={{m}_{2}}\]then normal reaction between the blocks will be zero

C)  if\[{{\mu }_{1}}={{\mu }_{2}}\], then normal reaction between the blocks will be zero

D)  None of the above                                                               

Correct Answer: C

Solution :

 Idea The normal reaction between the blocks will be non-zero only if the initial acceleration of block \[{{m}_{2}}\] is more than block\[{{m}_{1}}\]. Then, only the block \[{{m}_{2}}\] will push block \[{{m}_{1}}\] and normal reaction will develop and it is possible only when \[{{\mu }_{2}}<{{\mu }_{1}}\] The normal reaction between the blocks will not depend on the masses \[{{m}_{1}}\]and\[{{m}_{2}}\]. If \[{{\mu }_{1}}={{\mu }_{2}}=\mu \] \[\Rightarrow \] \[{{a}_{1}}=(g\sin \theta -\mu g\cos \theta )\] \[\Rightarrow \]\[{{a}_{2}}=(g\sin \theta -\mu g\cos \theta )\] \[\Rightarrow \]\[{{a}_{1}}={{a}_{2}}\] So, there is no relative motion between the blocks, so normal reaction between them is zero. TEST Edge Both are identical planes, if m does not slip then M will? M will also not slip because in this case motion of block will depend only on (l not on mass of the block.