A particle of mass 200 ,g is whirled in a ver | vedclass

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(B) Given:Mass $m = 200 \,g = 0.2 \,kg$Radius $R = 80 \,cm = 0.8 \,m$Angle $	heta = 60^{\circ}$Speed $v = 1.5 \,m/s$Acceleration due to gravity $g =

Vedclass · Accepted Answer

$1.56$

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(B) Given:Mass $m = 200 \,g = 0.2 \,kg$Radius $R = 80 \,cm = 0.8 \,m$Angle $	heta = 60^{\circ}$Speed $v = 1.5 \,m/s$Acceleration due to gravity $g = 9.8 \,m/s^2$The forces acting on the particle along the radial direction are the tension $T$ (towards the center) and the component of weight $mg \cos 	heta$ (away from the center).The net centripetal force is given by:$T - mg \cos 	heta = \frac{mv^2}{R}$Rearranging for tension $T$:$T = \frac{mv^2}{R} + mg \cos 	heta$Substituting the values:$T = \frac{0.2 	imes (1.5)^2}{0.8} + 0.2 	imes 9.8 	imes \cos(60^{\circ})$$T = \frac{0.2 	imes 2.25}{0.8} + 0.2 	imes 9.8 	imes 0.5$$T = \frac{0.45}{0.8} + 0.98$$T = 0.5625 + 0.98 = 1.5425 \,N$Rounding to two decimal places,we get $T \approx 1.56 \,N$ (using $g = 10 \,m/s^2$ gives $T = 0.5625 + 1 = 1.5625 \,N$).Thus,the correct option is $B$.

$A$ particle of mass $200 \,g$ is whirled in a vertical circle of radius $80 \,cm$ using a massless string. The speed of the particle when the string makes an angle of $60^{\circ}$ with the vertical line is $1.5 \,m/s$. The tension in the string at this position is ........... $N$.

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