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A $4 kg$ block is attached to a vertical rod by means of two strings of equal length. When the system rotates about the axis of the rod, the strings are extended as shown in figure.
How many revolutions per minute must the system make in order for the tension in the upper string to be $200 N$ ? What is the tension in the lower string then?

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$69.6 rpm$ , 100 N

$49.6 rpm$ , 110 N

$39.6 rpm$ , 150 N

$59.6 rpm$ , 120 N

answer is D.

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Detailed Solution

$T_{1} \cos θ + T_{2} \cos θ = m r ω^{2}$

$ω = (2 n π)$

Here, n= number of revolutions per second. Substituting the proper values in Eq. (i),

$200 \times (\frac{3}{5}) + T_{2} \times (\frac{3}{5}) = (4) (3) (2 n π)^{2}$

$or 600 + 3 T_{2} = 240 n^{2} π^{2}$

$Further, T_{1} \sin θ = T_{2} \sin θ + m g$

$or 200 \times \frac{4}{5} = T_{2} \times \frac{4}{5} + 4 \times 10$

$or 800 = 4 T_{2} + 200$

$Solving Eqs. (ii) and (iii) we get,$

$T_{2} = 150 N and n = 0.66 rps$

$= 39.6 rpm$

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