A simple pendulum with a bob of mass m and a | vedclass

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(D) The vertical height $h$ through which the bob rises is given by $h = L(1 - \cos 	heta)$.Using the conservation of mechanical energy,the maximum v

Vedclass · Accepted Answer

$2 BL \sin \left(\frac{	heta}{2}ight)(gL)^{1 / 2}$

Vedclass · Answer

(D) The vertical height $h$ through which the bob rises is given by $h = L(1 - \cos 	heta)$.Using the conservation of mechanical energy,the maximum velocity $v$ at the equilibrium position is given by $v^2 = 2gh$.Substituting $h = L(1 - \cos 	heta)$ and using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$v^2 = 2gL(2 \sin^2(	heta/2)) = 4gL \sin^2(	heta/2)$.Taking the square root,we get $v = 2 \sin(	heta/2) \sqrt{gL}$.The motional e.m.f. induced across a conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$ is given by $V = BvL$.Substituting the value of $v$:$V_{\max} = B \cdot [2 \sin(	heta/2) \sqrt{gL}] \cdot L = 2BL \sin(	heta/2) \sqrt{gL} = 2BL \sin(	heta/2) (gL)^{1/2}$.

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