A) T/4
B) T/8
C) T/12
D) T/16
Correct Answer: D
Solution :
\[d\text{ }=\text{ }100\text{ }mm,\text{ }l\text{ }=\text{ }1000\text{ }mm\] \[{{\tau }_{\max }}=\frac{16T}{\pi {{d}^{3}}}=60\,\,\text{N/m}{{\text{m}}^{\text{2}}}\] \[T={{T}_{s}}=\frac{\pi {{d}^{3}}}{16}\times 60\] For the hollow shaft, \[{{\tau }_{\max }}=\frac{16T}{\pi {{d}^{3}}(1-{{k}^{4}})}=\frac{16T}{\pi {{d}^{3}}\left[ 1-{{\left( \frac{50}{100} \right)}^{4}} \right]}\] \[=\frac{16}{15}\times \frac{16T}{\pi {{d}^{3}}}\] \[T=\frac{16}{15}\times \frac{\pi {{d}^{3}}}{16\times 60}=\frac{15}{16}{{T}_{s}}=\frac{15}{16}T\] Reduction in torque \[=T\left( 1-\frac{15}{16} \right)=\frac{T}{16}\]You need to login to perform this action.
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