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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassCalculate of COMMedium2 minPYQ_2016

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In the figure shown $A B C$ is a uniform wire. If the center of mass of the wire lies vertically below point $A$ , then $\frac{B C}{A B}$ is close to

Question diagram: In the figure shown A B C is a uniform wire. If the center o

Options:

Answer:

Solution:

$x_{cm} = \frac{x}{2} = \frac{(ρ x) (\frac{x}{2}) \frac{1}{2} + ρ y (\frac{y}{2})}{ρ (x + y)}$

$\Rightarrow \frac{1}{2} + \frac{y}{x} = \frac{y^{2}}{x^{2}}$

$\Rightarrow \frac{y}{x} = \frac{1 + \sqrt{3}}{2} = 1.37$

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Calculate of COM

⏱ 2mℹ️ Source: PYQ_2016

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