JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
The graph between two temperature scales \(P\) and \(Q\) is shown in the figure. Between upper fixed point and lower fixed point there are \(150\) equal divisions of scale \(P\) and \(100\) divisions on scale \(Q\). The relationship for conversion between the two scales is given by :
- A \(\frac{t_Q}{150}=\frac{t_P-180}{100}\)
- B \(\frac{t_Q}{100}=\frac{t_P-30}{150}\)
- C \(\frac{t_P}{180}=\frac{t_Q-40}{100}\)
- D \(\frac{t_P}{100}=\frac{t_Q-180}{150}\)
Answer & Solution
Correct Answer
(B) \(\frac{t_Q}{100}=\frac{t_P-30}{150}\)
Step-by-step Solution
Detailed explanation
\(\frac{\text { reading on scale }-\text { Lower fixed point }}{\text { upper fixed point }-\text { lower fixed point }}=\text { constant }\) \(\frac{t_P-30}{180-30}=\frac{t_Q-0}{100-0}\) \(\frac{t_P-30}{150}=\frac{t_Q}{100}\)
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