A) \[10{}^\circ C\]
B) \[0{}^\circ <<{{T}_{m}}<20{}^\circ C\]
C) \[20{}^\circ C\]
D) Above \[20{}^\circ C\]
Correct Answer: A
Solution :
According to galvanometer principle, heat given = heat taken \[\begin{align} & {{m}_{1}}{{s}_{1}}(100-t)=mL+{{m}_{2}}{{s}_{2}}(t-0) \\ & 50\times 1(100-t)=50\times 80+50\times 1(t-0) \\ \end{align}\] \[t=10{}^\circ C\]
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