A) \[{{0.01}^{0}}C\]
B) \[{{0.1}^{0}}C\]
C) \[{{1}^{0}}C\]
D) \[{{1.1}^{0}}C\]
Correct Answer: B
Solution :
According to energy conservation, change in potential energy of the ball, appears in the form of heat which raises the temperature of the ball. ie, \[mg({{h}_{1}}-{{h}_{2}})=mc.\,\,\Delta \theta \] \[\Rightarrow \] \[\Delta \theta =\frac{g({{h}_{1}}-{{h}_{2}})}{c}\] \[=\frac{10(10-5.4)}{460}={{0.1}^{o}}C\]You need to login to perform this action.
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