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(A) The centripetal force $F$ required for uniform circular motion is given by $F = \frac{mv^2}{r}$.Let the initial values be $m, v, r$. The initial f

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) The centripetal force $F$ required for uniform circular motion is given by $F = \frac{mv^2}{r}$.Let the initial values be $m, v, r$. The initial force is $F = \frac{mv^2}{r}$.Given that $m, v,$ and $r$ are all increased by $50 \%$,the new values are:$m' = m + 0.5m = 1.5m = \frac{3}{2}m$$v' = v + 0.5v = 1.5v = \frac{3}{2}v$$r' = r + 0.5r = 1.5r = \frac{3}{2}r$The new force $F'$ is given by:$F' = \frac{m' (v')^2}{r'} = \frac{(\frac{3}{2}m) (\frac{3}{2}v)^2}{\frac{3}{2}r} = \frac{(\frac{3}{2}m) (\frac{9}{4}v^2)}{\frac{3}{2}r} = \frac{9}{4} \frac{mv^2}{r} = 2.25 F$.The change in force is $\Delta F = F' - F = 2.25F - F = 1.25F$.The percentage change in force is $\frac{\Delta F}{F} 	imes 100 = \frac{1.25F}{F} 	imes 100 = 125 \%$.

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m, v$ and $r$ are all increased by $50 \%$,the necessary change in force required to maintain the particle in uniform circular motion is (in $\%$)

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