A) 3600
B) 1800
C) 1200
D) 1000
Correct Answer: B
Solution :
Heat energy required to just melt the bullet \[Q={{Q}_{1}}+{{Q}_{2}}\] Here, \[{{Q}_{1}}=ms\Delta \theta \] \[=m\times 125\times (327-27)\] \[=3.75\times {{10}^{4}}m\] \[{{Q}_{2}}=mL=m\times 2.5\times {{10}^{4}}\] \[=2.5\times {{10}^{4}}m\] \[\therefore \] \[Q=6.25\times {{10}^{4}}\,m\] If v be the speed of bullet, then 50% of\[\frac{1}{2}m{{v}^{2}}\] should be equal to Q. Thus, \[0.5\times \frac{1}{2}m{{v}^{2}}=6.25\times {{10}^{4}}m\] \[\Rightarrow \] \[v=\sqrt{\frac{6.25\times {{10}^{4}}\times 2}{0.5}}\] \[=500\,m/s=\,1800\,km/h\]You need to login to perform this action.
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