A) 0.196°C
B) 1.960°C
C) 0.96°C
D) 0.0196°C
Correct Answer: A
Solution :
If energy is converted into heat, then \[ms\Delta \theta =\frac{1}{2}\frac{m{{v}^{2}}}{J}=\frac{mgh}{4.2}\]where s is specific heat of water. \[\therefore \]\[m\times 1\times {{10}^{3}}\times \Delta \theta =\frac{m\times 9.8\times 84}{4.2}\] \[\Rightarrow \] \[\Delta \theta ={{0.196}^{o}}C\] Rise in temperature\[=0.196{}^\circ C\]
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