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NCERT Exemplar · Q25

Q.(a) A steel wire of mass μ\mu per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains ≪\ll longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m−3^{-3} (Young's modules Y=2×1011Y = 2 \times 10^{11} Nm−2^{-2}).

(b) If the yield strength of steel is 2.5×1082.5 \times 10^{8} Nm−2^{-2}, what is the maximum weight that can be hung at the lower end of the wire?
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The 2525 kg load stretches the wire by ΔL≈3.9×10−3\Delta L\approx 3.9\times10^{-3} m (≈3.9\approx 3.9 mm); the yield strength limits the hangable load to Wmax⁡≈785W_{\max}\approx 785 N (about 8080 kg).

Part (a): Extension under the load

1. Set up Hooke's law for the wire

ΔL=FLAY,\Delta L=\frac{FL}{AY},

with F=Mg=25×9.8=245F=Mg=25\times 9.8=245 N, L=10L=10 m, Y=2×1011Y=2\times10^{11} N m−2^{-2}.

2. Cross-sectional area (radius r=0.1r=0.1 cm =1×10−3=1\times10^{-3} m):

A=πr2=π(1×10−3)2=3.14×10−6 m2.A=\pi r^2=\pi(1\times10^{-3})^2=3.14\times10^{-6}\ \text{m}^2.

3. Compute the extension

ΔL=245×10(3.14×10−6)(2×1011)=24506.28×105=3.9×10−3 m.\Delta L=\frac{245\times 10}{(3.14\times10^{-6})(2\times10^{11})}=\frac{2450}{6.28\times10^{5}}=3.9\times10^{-3}\ \text{m}.

(The wire's own weight, ≈ρALg≈2.4\approx\rho A L g\approx 2.4 N against the 245245 N load, is negligible.)

Part (b): Maximum hangable weight

4. Apply the yield-strength limit …

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