A rectangular block of mass m and area of cross-section A floats in a liquid of density \(\rho\) . If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then :
- T \(\propto\) \(\sqrt \rho\)
- T \(\propto\) \(\bigg(\frac{1}{\sqrt A}\bigg)\)
- T \(\propto\bigg(\frac{1}{\rho}\bigg)\)
- T \(\propto\bigg(\frac{1}{\sqrt m}\bigg)\)
The Correct Option is B
Solution and Explanation
Time period of SHM is given by
T= \(2\pi\frac{\sqrt I}{g}\)
according to law of floatation
weight of the block = weight of the liquid displaced
mg=Al\(\rho\)\(g\)
l= \(\frac{m}{A\rho}\)
Then, T=\(2\pi\frac{\sqrt m}{A\rho g}\)
Now we can say that-
T \(\propto\) \(\bigg(\frac{1}{\sqrt A}\bigg)\)
Therefore, the correct option is (B): T \(\propto\) \(\bigg(\frac{1}{\sqrt A}\bigg)\)
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Concepts Used:
Center of Mass
The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.
The formula for the Centre of Mass:
Center of Gravity
The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.