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प्रश्न
A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shoreline. How fast is the beam moving along the shoreline when it makes an angle of 45° with the shore?
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उत्तर
Time for one revolution = 10 sec
Now, angular velocity `"dv"/"dt" = (2pi)/10 = pi/5`
From the figure, tan 45° = `"AB"/"OA"`
1 = `x/5`
⇒ x = 5
Again, tan θ = `x/5`
x = 5 tan θ
Differentiating w.r.t. ‘t’
`("d"x)/"dt" = 5 sec^2theta ("d"theta)/"dt"`
= `5 sec^2 (45^circ) (pi/5)`
= `(sqrt(2))^2pi`
= 2π
∴ The beam is moving at the rate of 2π km/sec.
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