A) \[\frac{g}{m}\]
B) \[\frac{2}{3}g\]
C) \[\frac{3}{2}g\]
D) \[\frac{g}{2}\]
Correct Answer: A
Solution :
Let a be the resultant acceleration with which the fireman slides down, then \[mg-T=ma\] Given, \[{{T}_{m}}=\frac{2}{3}mg\], \[a={{a}_{m}}\] \[\therefore \] \[mg-\frac{2}{3}mg=m{{a}_{m}}\] \[\Rightarrow \] \[{{a}_{m}}=g-\frac{2}{3}g=\frac{g}{3}\]You need to login to perform this action.
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