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(B) The tension $T$ in the string is due to the mass $M = 1 \ kg$ hanging from it. Since the system is accelerating upwards with $a = 2 \ m/s^2$,the e

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$0.01$

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(B) The tension $T$ in the string is due to the mass $M = 1 \ kg$ hanging from it. Since the system is accelerating upwards with $a = 2 \ m/s^2$,the effective acceleration is $g_{eff} = g + a = 10 + 2 = 12 \ m/s^2$.The tension $T$ at any point in the string (neglecting the mass of the string itself as it is very small compared to $1 \ kg$) is $T = M(g + a) = 1 	imes 12 = 12 \ N$.The speed of the transverse pulse $v$ is given by $v = \sqrt{\frac{T}{\mu}}$,where $\mu = 1.2 \ g/m = 1.2 	imes 10^{-3} \ kg/m$.$v = \sqrt{\frac{12}{1.2 	imes 10^{-3}}} = \sqrt{10^4} = 100 \ m/s$.Since the speed is constant,the time taken $t$ to travel the length $L = 1 \ m$ is $t = \frac{L}{v} = \frac{1}{100} = 0.01 \ s$.

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