A) 167ºC
B) 334ºC
C) 500ºC
D) 1000ºC
Correct Answer: C
Solution :
Initial diameter of tyre = (1000 ? 6) mm = 994 mm, so initial radius of tyre \[R=\frac{994}{2}=497\,mm\] and change in diameter DD = 6 mm so \[\Delta R=\frac{6}{2}=3\,mm\] After increasing temperature by Dq tyre will fit onto wheel Increment in the length (circumference) of the iron tyre DL = L ´ a ´ Dq \[=L\times \frac{\gamma }{3}\times \Delta \theta \] [As \[\alpha =\frac{\gamma }{3}]\] \[2\pi \,\Delta R=2\pi \,R\,\left( \frac{\gamma }{3} \right)\,\Delta \theta \]Þ\[\Delta \theta =\frac{3}{\gamma }\frac{\Delta R}{R}=\frac{3\,\times 3}{3.6\times {{10}^{-5}}\times 497}\] Þ\[\Delta \theta ={{500}^{o}}C\]You need to login to perform this action.
You will be redirected in
3 sec