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Answers
Answer:
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Question:
An elastic belt is placed around the rim of a pulley of radius 5cm. Form one poit. C on the belt elaaticbelt is pulled directly away from the centre O of the pulley unit it is at P, 10 cm from the point O. What is the length of the belt that is still in contact with the pulley. (Use π = 3.14 and √3 = 1.72)
Answer:
20.93 cm
Step-by-step explanation:
Angle A is 90° because tangent at any point of a circle is perpendicular to the radius through the point of contact.
Now, in ∆OAP (by Pythagoras theorem)
(Hypotenuse)² = (Base)² + (Perpendicular)²
→ OP² = OA² + AP²
(where the value of OP is 10 cm and OA is 5 cm)
→ (10)² = (5)² + AP²
→ 100 - 25 = AP²
→ 75 = AP²
→ AP = 53√3 cm
Now,
cosØ = Base/Hypotenuse
cosØ = 5/10 = 1/2
Ø = 60°
From above value of angle AOP is 60°. Similarly, value of angle BOP is also 60° because ∆AOP ≈ ∆BOP. So, angle AOB is 120° (60° + 60°).
Reflex of angle AOB = 360° - 120° = 240°
So,
Length of major arc = 2πr (θ/360°)
= πr (θ/180°)
= 3.14 × 5 × 240°/180°
= 20.93
Therefore, the length of the belt that is still in contact with the pulley is 20.93 cm.