For $T = 2\\pi\\sqrt{\\frac{L}{g}}$, the slope of $T^2$ vs. $L$ is:

University of Ilorin PHY107 2020 past question

1. $\\frac{\\pi^2}{g}$
2. $\\frac{2\\pi^2}{g}$
3. $\\frac{4\\pi^2}{g}$
4. $\\frac{2\\pi}{g}$

Register free to see the answer and step-by-step explanation

More University of Ilorin PHY107 2020 past questions

Related questions

• Increase in viscosity causes:
• How to reduce friction between two surfaces in contact?
• The closeness of a set of values is called:
• Experimental readings are repeated to avoid:
• The copper lid in a calorimeter primarily stops heat loss by:
• Values that are close to one another are said to be: