Question 50 of 70
A load of mass is suspended from a steel wire of length and radius in Searle's apparatus experiment. The increase in length produced in the wire is Now the load is fully immersed in a liquid of relative density The relative density of the material of load is The new value of increase in length of the steel wire is:
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Answer: (D)
{\delta }^{'}=\dfrac{\left({\rho }_{0}-{\rho }_{L}\right)vg\times L}{Ay}\Rightarrow \dfrac{\delta ^'}{\delta }=\dfrac{{\rho }_{0}-{\rho }_{L}}{{\rho }_{0}}=\dfrac{8-2}{8}