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(B) Since the tension force is always perpendicular to the velocity vector of the disc,the work done by the tension force is zero. According to the wo

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$2$

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(B) Since the tension force is always perpendicular to the velocity vector of the disc,the work done by the tension force is zero. According to the work-energy theorem,the kinetic energy and thus the speed of the disc remain constant throughout its motion. Therefore,the time taken $t$ is given by $t = \frac{s}{v_0}$,where $s$ is the total distance traversed by the disc. At an arbitrary position,the infinitesimal distance $ds$ covered by the disc is $ds = (l_0 - R	heta) d	heta$. The total distance $s$ is obtained by integrating from $	heta = 0$ to $	heta = \frac{l_0}{R}$: $s = \int_{0}^{l_0/R} (l_0 - R	heta) d	heta = [l_0	heta - \frac{1}{2}R	heta^2]_{0}^{l_0/R} = l_0(\frac{l_0}{R}) - \frac{1}{2}R(\frac{l_0}{R})^2 = \frac{l_0^2}{R} - \frac{l_0^2}{2R} = \frac{l_0^2}{2R}$. Substituting the values $l_0 = 2 \ m$,$R = 1 \ m$,and $v_0 = 1 \ m/s$: $t = \frac{l_0^2}{2Rv_0} = \frac{2^2}{2 	imes 1 	imes 1} = \frac{4}{2} = 2 \ s$.

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