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PhysicsElasticityMiscellaneousMedium2 minPYQ_2024
PhysicsMediumnumerical

One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be

[Area of cross section of wire =0.005 cm2, Y=2×1011 N m-2 and g=10 m s-2]

Question diagram: One end of a metal wire is fixed to a ceiling and a load of
Answer:
3.00
Solution:

As we know, change in length is given by ΔL=FLAY

Therefore, strain required ΔLL=FAY

Now the ratio will be, ΔL1 L1ΔL2 L2=F1 F2=T1 T2=3010=3

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous
2mℹ️ Source: PYQ_2024

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