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Q8.11: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.11 Solution: A 14.5 kg mass, fastened to the end of a steel wire of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often calculate the wire tension at the lowest point by using only the centripetal term $m\omega^2L$ and forget the gravitational force $mg$ that also acts downward, so the correct tension is $T=mg+m\omega^2L$. Remember to add both contributions before applying Hooke’s law; otherwise the elongation will be severely underestimated.

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm_. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Step-by-Step Verified Solution

**Answer:** Elongation = 1.87×10−31.87 \times 10^{-3} m =1.87= 1.87 mm **Step-by-Step Explanation:** ω=2\omega = 2 rev/s =4π= 4\pi rad/s =12.57= 12.57 rad/s. At the lowest point, tension:
T=m(g+ω2L)=14.5(9.8+157.9×1.0)=14.5×167.7=2.43×103 NT = m(g + \omega^2 L) = 14.5(9.8 + 157.9 \times 1.0) = 14.5 \times 167.7 = 2.43 \times 10^3 \text{ N}
Area A=0.065A = 0.065 cm_ =6.5×10−6= 6.5 \times 10^{-6} m_. Ysteel=2.0×1011Y_{steel} = 2.0 \times 10^{11} Pa.
ΔL=TLAY=2.43×103×1.06.5×10−6×2.0×1011=1.87×10−3 m\Delta L = \frac{TL}{AY} = \frac{2.43 \times 10^3 \times 1.0}{6.5 \times 10^{-6} \times 2.0 \times 10^{11}} = 1.87 \times 10^{-3} \text{ m}
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All Questions in this Chapter

Q8.1 Q8.2 Q8.3 Q8.4 Q8.5 Q8.6 Q8.7 Q8.8 Q8.9 Q8.10 Q8.11 Q8.12 Q8.13 Q8.14 Q8.15 Q8.16

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