A conical pendulum of length 1,m makes an ang | vedclass

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Question

$\begin{array}{l} Given,	heta  = {45^ \circ },r = 0.4m,g = 10m/{s^2}\ T\,\sin \,	heta  = \frac{{m{v^2}}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\begin{array}{l} Given,	heta  = {45^ \circ },r = 0.4m,g = 10m/{s^2}\ T\,\sin \,	heta  = \frac{{m{v^2}}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\ T\,\cos \,	heta  = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {ii} ight)\ From\,equation\,\left( i ight) and \left( {ii} ight)\ We\,have,\,	an 	heta  = \frac{{{v^2}}}{{rg}}\ {V^2} = rg\,\,\,\,\,\,\,\,\,\,\,\,	heta  = {45^ \circ }\ Hence,\,speed\,of\,the\,pendulum\,in\,its\,circular\ path,\ V = \sqrt {rg}  = \sqrt {0.4 	imes 10}  = 2m/s \end{array}$

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