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Question Number 56971 by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19

$s o u r c e . . b o o k$
	Answered by mr W last updated on 27/Mar/19

	$m a s s o f w h o l e r o p e :$ $M = \int_{0}^{L} e^{\frac{x}{L}} d x = L (e - 1)$ $m a s s o f l e f t r o p e h a l f :$ $m = \int_{0}^{\frac{L}{2}} e^{\frac{x}{L}} d x = L (\sqrt{e} - 1)$ $a c c e l e r a t i o n o f r o p e :$ $a = \frac{1}{M} = \frac{1}{L (e - 1)}$ $t e n s i o n i n r o p e a t x = \frac{L}{2} :$ $T = m a = \frac{L (\sqrt{e} - 1)}{L (e - 1)} = \frac{1}{\sqrt{e} + 1} N$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19

	$e x c e l l e n t s i r . . . k h u b b h a l o . . d a r u n . . .$
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