A) \[0.01\text{ }{}^\circ C\]
B) \[0.1\text{ }{}^\circ C\]
C) \[7\text{ }{}^\circ C\]
D) \[1.1\text{ }{}^\circ C\]
Correct Answer: B
Solution :
Energy dissipated in the process\[=m\,g({{h}_{1}}-{{h}_{2}})\] This energy is absorbed by the ball, if rise in temperature of the ball is t, then\[mg({{h}_{1}}-{{h}_{2}})=m\,\,s\,t\] s = specific heat of the material of the ball \[0.1\times 10\times (10-5.4)=0.1\times 460\times t\] \[t=\frac{10\times 4.6}{460}\] \[=0.1{{\,}^{o}}C\]You need to login to perform this action.
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