A) \[\frac{\mu p}{mg}\]
B) \[\frac{\mu mg}{p}\]
C) \[\mu mgp\]
D) \[\frac{p}{\mu mg}\]
Correct Answer: D
Solution :
[d] \[\therefore P=F.V=constant\] |
When \[F=\mu mg\] net force on block becomes zero, i.e., it has maximum or terminal velocity |
\[\therefore \] \[p=(\mu mg){{V}_{\max }}\] |
\[\therefore \] \[{{V}_{\max }}=\frac{P}{\mu mg}\] |
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