A thermometer has wrong calibration- It reads the melting point of ice as -10-circ- - If it reads 60-circ- in place of 50-circ- C - what is the temperature of the boiling point of water on this scale-90-circ-110-circ-130-circ-120-circ-

Question

A thermometer has wrong calibration- It reads the melting point of ice as  -10-circ- - If it reads 60-circ- in place of 50-circ- C - what is the temperature of the boiling point of water on this scale-90-circ-110-circ-130-circ-120-circ-

Toppr Admin · Accepted Answer

For any two thermometric scales-T1-x2212-LFP1UFP1-x2212-LFP1-T2-x2212-LFP2UFP2-x2212-LFP2-x2234-Tcelsius-x2212-0100-T-x2212-x2212-10-UFP-x2212-x2212-10-x21D2-UFP-10-140-x2234-UFP-130o

A thermometer has wrong calibration. It reads the melting point of ice as $- 10^{\circ}$ . If it reads $60^{\circ}$ in place of $50^{\circ} C$ , what is the temperature of the boiling point of water on this scale.
$90^{\circ}$
$110^{\circ}$
$130^{\circ}$
$120^{\circ}$

For any two thermometric scales,
$\frac{T_{1} - L F P_{1}}{U F P_{1} - L F P_{1}} = \frac{T_{2} - L F P_{2}}{U F P_{2} - L F P_{2}}$

$∴ \frac{T_{c e l s i u s} - 0}{100} = \frac{T - (- 10)}{U F P - (- 10)}$

$\Rightarrow U F P + 10 = 140$
$∴ U F P = 130^{o}$

A thermometer has wrong calibration. It reads the melting point of ice as −10∘ . If it reads 60∘ in place of 50∘C , what is the temperature of the boiling point of water on this scale.90∘110∘130∘120∘

For any two thermometric scales,T1−LFP1UFP1−LFP1=T2−LFP2UFP2−LFP2∴Tcelsius−0100=T−(−10)UFP−(−10)⇒UFP+10=140∴UFP=130o

A thermometer has wrong calibration. It reads the melting point of ice as $- 10^{\circ}$ . If it reads $60^{\circ}$ in place of $50^{\circ} C$ , what is the temperature of the boiling point of water on this scale.
$90^{\circ}$
$110^{\circ}$
$130^{\circ}$
$120^{\circ}$

For any two thermometric scales,
$\frac{T_{1} - L F P_{1}}{U F P_{1} - L F P_{1}} = \frac{T_{2} - L F P_{2}}{U F P_{2} - L F P_{2}}$

$∴ \frac{T_{c e l s i u s} - 0}{100} = \frac{T - (- 10)}{U F P - (- 10)}$

$\Rightarrow U F P + 10 = 140$
$∴ U F P = 130^{o}$