The graph between two temperature scales P and Q is shown in the figure. Between the upper fixed point and the lower fixed point, there are 150 equal divisions on scale P and 100 divisions on scale Q. [2023]

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Q.
The graph between two temperature scales P and Q is shown in the figure. Between the upper fixed point and the 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 [2023]

1 $\frac{t_{P}}{100} = \frac{t_{Q} - 180}{150}$

2 $\frac{t_{Q}}{150} = \frac{t_{P} - 180}{100}$

3 $\frac{t_{P}}{180} = \frac{t_{Q} - 40}{100}$

4 $\frac{t_{Q}}{100} = \frac{t_{P} - 30}{150}$

Ans.
(4)

$\frac{reading on scale - lower fixed point}{upper fixed point - lower fixed point} = constant$

$\frac{t_{p} - 30}{180 - 30} = \frac{t_{0} - 0}{100 - 0}$

$\frac{t_{p} - 30}{150} = \frac{t_{0}}{100}$

Solution

Q.
The graph between two temperature scales P and Q is shown in the figure. Between the upper fixed point and the 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 [2023]

1 $\frac{t_{P}}{100} = \frac{t_{Q} - 180}{150}$

2 $\frac{t_{Q}}{150} = \frac{t_{P} - 180}{100}$

3 $\frac{t_{P}}{180} = \frac{t_{Q} - 40}{100}$

4 $\frac{t_{Q}}{100} = \frac{t_{P} - 30}{150}$

Ans.
(4)

$\frac{reading on scale - lower fixed point}{upper fixed point - lower fixed point} = constant$

$\frac{t_{p} - 30}{180 - 30} = \frac{t_{0} - 0}{100 - 0}$

$\frac{t_{p} - 30}{150} = \frac{t_{0}}{100}$

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Solution

Q. The graph between two temperature scales P and Q is shown in the figure. Between the upper fixed point and the 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 [2023]

1 tP100=tQ-180150

2 tQ150=tP-180100

3 tP180=tQ-40100

4 tQ100=tP-30150

Ans. (4) reading on scale-lower fixed pointupper fixed point-lower fixed point=constant tp-30180-30=t0-0100-0 tp-30150=t0100

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Q.
The graph between two temperature scales P and Q is shown in the figure. Between the upper fixed point and the 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 [2023]

1 $\frac{t_{P}}{100} = \frac{t_{Q} - 180}{150}$

2 $\frac{t_{Q}}{150} = \frac{t_{P} - 180}{100}$

3 $\frac{t_{P}}{180} = \frac{t_{Q} - 40}{100}$

4 $\frac{t_{Q}}{100} = \frac{t_{P} - 30}{150}$

Ans.
(4)

$\frac{reading on scale - lower fixed point}{upper fixed point - lower fixed point} = constant$

$\frac{t_{p} - 30}{180 - 30} = \frac{t_{0} - 0}{100 - 0}$

$\frac{t_{p} - 30}{150} = \frac{t_{0}}{100}$