JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Elasticity & SHM

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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 cm^{2}

Y = 2 \times 1 0^{11} Nm^{- 2}

and

g = 10 ms^{- 2}

]

Answer: 3

Solution

The upper wire carries the weight of both loads,

F_{1} = (2 + 1) g = 30

N, while the lower wire carries only the 1 kg load,

F_{2} = 1 \times g = 10

N. Longitudinal strain is

ϵ = \frac{F}{A Y}

; since both wires have the same A and Y, the ratio of strains equals the ratio of tensions:

\frac{ϵ _{1}}{ϵ _{2}} = \frac{F _{1}}{F _{2}} = \frac{30}{10} = 3

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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