An elastic belt is placed round the rim of a pulley of radius 5 cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from O.find the length of the belt that is in contact with the rim of the pulley.

Refer the following figure:

Now in Triangle OAP

the angle at centre can be calculated by trigonometry sin $\theta $= 5/10= 1/2 --------> θ= 60

^{o}

total angle made by AB = 2 $\theta $ = 120

^{o}

so length of belt in contact = (360-120) 2$\mathrm{\pi}$(5)/360 = 20*3.14/3=

**20.933 cm**
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