Question

an object of mass 0.5 kg is held at end of a string 0.8 long if the string makes three revolution in 1.2 second find the tension of a string​

Answer 1

Answer:

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Step by step solution:

$we \ have \ to \ find \ tension\\\\Let \ us \ do\\\\t=\frac{1.2}{3}=0.4 sec\\\\v=\frac{2\pi r }{t}=\frac{2\pi 0.8 }{0.4}=4\pi \\\\we \ know \ tension \ equals \ to\\\\T=\frac{mv^{2} }{r} =\frac{0.5 \times4\pi \times 4\pi }{0.8}=98.7N$

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Answer 2

Number of revolutions = 3

Time taken = 1.2 sec

Hence, time period for one revolution = 1.2/3 = 0.4 sec

Mass of object = 0.5 Kg

Length of the string = radius of the circle = r = 0.8m

Velocity of the object = 2∏ r /T = (2 * ∏ * 0.8) / 1.2 = 4 ∏m/s

Tension T = mv2 / r = 0.5 * (4 ∏)2 / 0.8 = 98.7 N