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10.1, thermonetry,thermal expansion and calorimetry — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics10.1, thermonetry,thermal expansion and calorimetry1 Mark
MCQ
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
    $\frac{t_Q}{150}=\frac{t_P-180}{100}$
  • ✓
    $\frac{t_Q}{100}=\frac{t_P-30}{150}$
  • C
    $\frac{t_P}{180}=\frac{t_Q-40}{100}$
  • D
    $\frac{t_P}{100}=\frac{t_Q-180}{150}$
✓

Answer

Correct option: B.
$\frac{t_Q}{100}=\frac{t_P-30}{150}$
b
$\frac{\text { reading on scale }-\text { Lower fixed point }}{\text { upper fixed point }-\text { lower fixed point }}=\text { constant }$

$\frac{t_P-30}{180-30}=\frac{t_Q-0}{100-0}$

$\frac{t_P-30}{150}=\frac{t_Q}{100}$

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