A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended, respectively, from its free end. Then the value of L is equal to
- \(\sqrt{L_1L_2}\)
- \(\frac{L_1+L_2}{2}\)
- 2L1 – L2
- 3L1 – 2L2
The Correct Option is C
Solution and Explanation
The correct option is(C): 2L1 – L2
\(y=\frac{FL}{A△L}\)
\(⇒△L=\frac{FL}{Ay}\)
\(⇒L_1=L+\frac{(1g)L}{Ay}....(i)\)
and
\(⇒L_2=L+\frac{(2g)L}{Ay}....(ii)\)
⇒ L = 2L1 – L2
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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.