A) \[98{}^\circ C\]
B) \[110{}^\circ C\]
C) \[40{}^\circ C\]
D) \[60{}^\circ C\]
Correct Answer: A
Solution :
Let temperature of thermometer be \[\theta {{\,}^{o}}C\]at \[60{{\,}^{o}}C\]then \[100-60=150-\theta \] \[\Rightarrow \] \[40=150-\theta \] (i) Also \[60-0=\theta -20\] \[60=\theta -20\] (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{40}{60}=\frac{150-\theta }{\theta -20}\] \[\Rightarrow \] \[\frac{2}{3}=\frac{150-\theta }{\theta -20}\] \[\Rightarrow \] \[5\theta =490\] \[\Rightarrow \] \[\theta =98{{\,}^{o}}C\]
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