A block of mass 2 kg is attached with one end of a light string of length 0.5 m and is moving in circular path in the plane of a smooth inclined of angle $30^{0}$ as shown in figure. If the speed of particle at lowest point is 6 m/s, then the tension in string at the lowest point is, $(g = 10 m / s^{2})$

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answer is 154.

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Detailed Solution

$T - mgsin θ = \frac{{mv}^{2}}{r} T = mgsin θ + \frac{{mv}^{2}}{r} = 2 \times 10 \times \frac{1}{2} + \frac{2 \times 36}{0 .5} = 10 + 144 = 154 N$

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