A) \[\frac{mg}{4h}\sqrt{{{d}^{2}}+4{{h}^{2}}}\]
B) \[\frac{mg}{4h}\times d\]
C) \[\frac{mg}{4h}\times \sqrt{{{d}^{2}}+4{{h}^{2}}}\]
D) \[\frac{mgh}{d}\]
Correct Answer: A
Solution :
Let \[T\] be the force exerted by each friend on the roe. The component of each tension acting vertically upwards as clear from the figure below is Thus, \[T\cos \theta =mg\] or, \[T=\frac{mg}{2\cos \theta }\] ? (i) Let us determine the value of \[\cos \theta \] from the figure below: \[\cos \theta =\frac{base}{hypotenuse}=\frac{h}{\sqrt{{{\left( \frac{d}{2} \right)}^{2}}+{{h}^{2}}}}\] Substituting, it in equation (i), we haveYou need to login to perform this action.
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