l લંબાઈના બે દળ રહિત સામાન્ય બિંદુ પરથી બે સમ | vedclass

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Question

$tan 	heta \,\, = \,\,\frac{F}{{Mg}}\,\,$ ($	heta $ નાનો છે)

$	heta \,\, = \,\,\frac{F}{{Mg}}\,\, \Rightarrow \,\,F\,\, = \,\,Mg	heta $

$\fr

Vedclass · Accepted Answer

$v$ $\propto$ $x^{-1/2}$

Vedclass · Answer

$tan 	heta \,\, = \,\,\frac{F}{{Mg}}\,\,$ ($	heta $ નાનો છે)

$	heta \,\, = \,\,\frac{F}{{Mg}}\,\, \Rightarrow \,\,F\,\, = \,\,Mg	heta $

$\frac{{K{Q^2}}}{{{x^2}}}\, = \,\,Mg\,\,\frac{x}{\ell }$

${Q^2}\,\, = \,\,\frac{{Mg}}{{K\ell }}\,{x^3}\,\, \Rightarrow \,\,Q\,\, = \,\,\sqrt {\frac{{Mg}}{{K\ell }}} \,{x^{\frac{3}{2}}}$

$\left( {\frac{{dQ}}{{dt}}} ight)\,\, = \,\,\sqrt {\frac{{Mg}}{{K\ell }}} \,\,\left( {\frac{3}{2}\,{x^{\frac{1}{2}}}} ight)\,\,\left( {\frac{{dx}}{{dt}}} ight)$

$ = \,\,\left( {\frac{{dQ}}{{dt}}} ight)\,$ અચળ છે તેથી $\frac{{dx}}{{dt}}\, \propto \,{x^{ - \frac{1}{2}}}$ અથવા $V\,\, \propto \,\,{x^{ - \frac{1}{2}}}$

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