A) \[g(1-\eta )\]
B) \[\eta g\]
C) \[\frac{g}{1+\eta }\]
D) \[\frac{g}{1-\eta }\]
Correct Answer: A
Solution :
\[mg-R=ma\] \[mg=\eta mg=ma\] \[[\because \,R=\eta \,mg]\] \[mg(1-\eta )=ma\] \[a=g(1-\eta )\]
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