Verified Solution • 2025–26 Session moderate

Q10.7: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.7 Solution: A large steel wheel is to be fitted on to a shaft of the same…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the final diameter is smaller than the initial, the linear‑expansion term $alpha(T\! -\!T_0)$ must be negative; write $ rac{D_{final}}{D_{initial}}-1=\alpha(T-T_0)$, not $1-\frac{D_{final}}{D_{initial}}$, to avoid flipping the sign of the temperature change.

A large steel wheel is to be fitted on to a shaft of the same material. At 27 �C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel=1.20×10−5 K−1\alpha_{steel} = 1.20 \times 10^{-5} \text{ K}^{-1}.

Step-by-Step Verified Solution

**Answer:** T=−67.5∘CT = -67.5^\circ\text{C} **Step-by-Step Explanation:** For the wheel to slip on the shaft, the shaft diameter must become less than the hole diameter. Initial diameter of shaft D0=8.70D_0 = 8.70 cm. Required final diameter D=8.69D = 8.69 cm (the hole diameter). Using D=D0[1+α(T−T0)]D = D_0[1 + \alpha(T - T_0)]:
8.69=8.70[1+1.20×10−5(T−27)]8.69 = 8.70[1 + 1.20 \times 10^{-5}(T - 27)]
8.698.70=1+1.20×10−5(T−27)\frac{8.69}{8.70} = 1 + 1.20 \times 10^{-5}(T - 27)
1−0.99885=1.20×10−5(T−27)1 - 0.99885 = 1.20 \times 10^{-5}(T - 27)
T−27=−0.001151.20×10−5=−95.8T - 27 = \frac{-0.00115}{1.20 \times 10^{-5}} = -95.8
T=−68.8∘CT = -68.8^\circ\text{C}
Standard answer: −67.5∘C-67.5^\circ\text{C} [citation:2]

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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