A string A of length 0.314 m of Young’s modulus is connected to another string B of length and Young’s modulus both twice those of A

Question

A string A of length 0.314 m of Young’s modulus $2 \times 10^{10} N / m^{2}$ is connected to another string B of length and Young’s modulus both twice those of A. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is ______ mm.

(radius of both the strings is 0.2 mm and acceleration due to gravity = 10 ${m/s}^{2}$ )

(Mass of both strings is to be neglected as compared to the mass of load) [2026]

(radius of both the strings is 0.2 mm and acceleration due to gravity = 10 ${m/s}^{2}$ )

(Mass of both strings is to be neglected as compared to the mass of load) [2026]

Smart Achievers · Accepted Answer

(2)

$Δ L_{total} = Δ L_{1} + Δ L_{2} = \frac{M g L_{1}}{A Y_{1}} + \frac{M g L_{2}}{A Y_{2}}$

$Δ L_{total} = \frac{M g}{A} [\frac{L}{Y} + \frac{2 L}{2 Y}] = \frac{2 M g L}{A Y}$

$Δ L_{total} = \frac{2 \times 0.8 \times 10 \times 0.314}{π \times {(0.2)}^{2} \times 10^{- 6} \times 2 \times 10^{10}} = 2 mm$

Solution

1 3

2 2

3 1.9

4 1

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Solution

1 3

2 2

3 1.9

4 1

Ans. (2) ΔLtotal=ΔL1+ΔL2=MgL1AY1+MgL2AY2 ΔLtotal=MgA[LY+2L2Y]=2MgLAY ΔLtotal=2×0.8×10×0.314π×(0.2)2×10-6×2×1010=2 mm

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