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JEE Main 20241 FebruaryEvening Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2024 · 1 February · Evening Shift · Physics · Mechanical Properties of Solids

One end of a metal wire is fixed to a ceiling and a load of 2kg...

One end of a metal wire is fixed to a ceiling and a load of 2kg2kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1kg1kg 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 =0.005cm2,Y=2×1011Nm−2= 0 . 005{cm}^{2}, Y = 2 \times {10}^{11}N{m}^{- 2} and g=10ms−2g = 10m{s}^{- 2}]

One end of a metal wire is fixed to a ceiling and a load of 2kg hangs from the other end. A similar wire is attached to

Correct answer

3

Step-by-step solution

One end of a metal wire is fixed to a ceiling and a load of 2kg hangs from the other end. A similar wire is attached to

As we know, change in length is given by \DeltaL=FLAY\DeltaL = \dfrac{FL}{AY}

Therefore, strain required \DeltaLL=FAY\dfrac{\DeltaL}{L} = \dfrac{F}{AY}

Now the ratio will be, ΔL1L1ΔL2L2=F1F2=T1T2=3010=3\dfrac{\dfrac{\Delta{L}_{1}}{{L}_{1}}}{\dfrac{\Delta{L}_{2}}{{L}_{2}}} = \dfrac{{F}_{1}}{{F}_{2}} = \dfrac{{T}_{1}}{{T}_{2}} = \dfrac{30}{10} = 3

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