Given the relation \( T = 2\pi\sqrt{\frac{I}{C}} \), where \( C = \frac{n\pi r^4}{2L} \), if \( T^2 \) is plotted against \( L \), what is the moment of inertia \( I \) in terms of the slope \( m \)?
University of Ilorin PHY107 2022 past question
- \( \frac{I}{m} = \frac{\pi r^4}{8\pi} \)
- \( \frac{I}{m} = \frac{\pi r^4}{4\pi} \)
- \( \frac{I}{m} = \frac{\pi r^4}{2\pi} \)
- \( \frac{I}{m} = \frac{\pi r^4}{\pi} \)
Register free to see the answer and step-by-step explanation
More University of Ilorin PHY107 2022 past questions
Related questions
- In the equation \( v = \frac{2ga^2(\rho - \sigma)}{9\eta} \), what quantities are usually measured experimentally?
- What is the aim of the viscosity experiment?
- In a simple pendulum experiment, the period (T) does NOT depend on:
- From the equation \( T = 2\pi\sqrt{\frac{m}{k}} \), if \( m^2 \) is plotted against \( T \), what is the spring constant \( k \) when the s…
- Given the expression \( I_e = \lambda T^n \), what is the linearized form?
- Given the expression \( T = k e^{bv} \), what is \( b \) when the slope is 3.32?
