A mass of $2 k g$ is whirled in a horizontal circle by means of a string at an initial speed of $5$ r.p.m. Keeping the radius constant the tension in the string is doubled. The new speed is nearly:

7 r.p.m

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14 r.p.m

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10 r.p.m

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20 r.p.m

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Solution

The correct option is A 7 r.p.m
$C a s e 1 C a s e 2 m = 2 k g m = 2 k g w = \frac{5 \times 2 π}{60} = \frac{π}{6} w_{2} r a d i u s = r r a d i u s = r T_{2} = T a s T_{2} = T o r m w_{2}^{2} r = 2 m w^{2} r w_{2} = \sqrt{2} w = \sqrt{2} \frac{π}{6} = 7 r p m$

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A mass of 2kg is whirled in a horizontal circle by means of a string at an initial speed of 5 r.p.m. Keeping the radius constant the tension in the string is doubled. The new speed is nearly:

The correct option is A 7 r.p.mCase1Case2m=2kgm=2kgw=5×2π60=π6w2radius=rradius=rT2=TasT2=Tormw22r=2mw2rw2=√2w=√2π6=7rpm

A mass of $2 k g$ is whirled in a horizontal circle by means of a string at an initial speed of $5$ r.p.m. Keeping the radius constant the tension in the string is doubled. The new speed is nearly:

The correct option is A 7 r.p.m
$C a s e 1 C a s e 2 m = 2 k g m = 2 k g w = \frac{5 \times 2 π}{60} = \frac{π}{6} w_{2} r a d i u s = r r a d i u s = r T_{2} = T a s T_{2} = T o r m w_{2}^{2} r = 2 m w^{2} r w_{2} = \sqrt{2} w = \sqrt{2} \frac{π}{6} = 7 r p m$