A string breaks if its tension exceeds 10 newtons. A stone of mass 250 gm tied to this string of length 10 cm is rotated in a horizontal circle. The maximum angular velocity of rotation can be

Q. a string breaks under a load of 4.8 kg. A mass if 0.5 kg is attached to one end of the string 2m long and is rotated in a horizontal circle. calculate the greatest no of revolutions that the mass can make without breaking the string

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The breaking tension of a string is 10N.A particle of mass 0.1kg tied to it is rotated along a horizontal circle of radius 0.5 metre . The maximum speed with which the particle can be rotated without breaking the string is :

The correct option is A √5m/secThe breaking tension can be equated to the centripetal force. 10N=mv2r 10N=0.1×10×v20.5 v=√5m/s After this velocity the string will break.

The breaking tension of a string is $10 N . A$ particle of mass $0.1$ kg tied to it is rotated along a horizontal circle of radius $0.5$ metre . The maximum speed with which the particle can be rotated without breaking the string is :

The correct option is A $\sqrt{5} m / s e c$
The breaking tension can be equated to the centripetal force.

$10 N = \frac{m v^{2}}{r}$
$10 N = \frac{0.1 \times 10 \times v^{2}}{0.5}$

$v = \sqrt{5} m / s$

After this velocity the string will break.