### Q.21:- A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

**Answer:-**Mass of the stone, *m* = 0.25 kg

Radius of the circle, *r* = 1.5 m

Number of revolution per second, n = 40 / 60 = 2 / 3 rps

Angular velocity, ω = *v* / *r* = 2πn

The centripetal force for the stone is provided by the tension *T*, in the string, i.e.,

*T* = *F*_{Centripetal}

= *mv ^{2}* /

*r*=

*mr*ω =

*mr*(2πn)

^{2}

= 0.25 × 1.5 × (2 × 3.14 × (2/3) )

^{2}

= 6.57 N

Maximum tension in the string,

*T*

_{max}= 200 N

T

_{max}=

*mv*

^{2}_{max}/

*r*

∴ v

_{max}= (T

_{max}×

*r*/

*m*)

^{1/2}

= (200 × 1.5 / 0.25)

^{1/2}

= (1200)

^{1/2}= 34.64 m/s

Therefore, the maximum speed of the stone is 34.64 m/s.