Answer: The correct answer is 75.6 degree.
Explanation:
Calculate the angular velocity.
\(\omega =\frac{2\pi }{T}\)
Here, T is the time.
Put T=0.50 s.
\(\omega =\frac{2\pi }{0.50}\)
\(\omega =12.6\ rad\ per\ second\)
If T is the tension acting on the sinker then tension equals to the centripetal force due to the circular motion as the sinker is moving in the circular motion.
\(T=mL\omega ^{2}\)
Here, m is the mass of the sinker and L is the length of the sinker.
By resolving the components of the tension, the vertical component of the tension equals to the weight of the sinker.
\(Tcos\theta =mg\)
Calculate the angle that the fishing line makes with the vertical.
\(cos\theta =\frac{mg}{T}\)
\(\theta =cos^{-1}(\frac{mg}{T})\)
Put \(T=mL\omega ^{2}\).
\(\theta =cos^{-1}(\frac{g}{L\omega ^{2}})\)
Put L= 0.25 , g= 9.8 meter per second and \(\omega =12.6\ rad\ per\ second\).
\(\theta =cos^{-1}(\frac{9.8}{(0.25)\(12.6) ^{2}})\)
\(\theta =cos^{-1}(0.25)\)
\(\theta =1.32\)
Convert radian into degree by multiplying 57.3 degree.
\(\theta =75.6\)
Therefore, the angle that the fishing line makes with the vertical is 75.6 degree.
