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Byju's Answer

Standard XII

Physics

Calorimetry

On a hypothet...

Question

On a hypothetical scale A the ice point is $42^{o}$ and the steam point is $182^{o}$ For another scale B, the ice point is $- 10^{o}$ and steam point in $90^{o}$ ) If B reads $60^{o}$ , the reading of A is,

160^{o}

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140^{o}

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120^{o}

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110^{o}

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Solution

The correct option is B $140^{o}$
Given,
$B = 60^{0} => \frac{A - 42}{182 - 42} = \frac{B - (- 10)}{90 - (- 10)}$
$=> \frac{A - 42}{140} = \frac{B + 10}{100} => \frac{A - 42}{14} = \frac{60 + 10}{10}$
$=> \frac{A - 42}{14} = \frac{70}{10} => A - 42 = 98$
$A = 140^{0}$

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On a hypothetical scale A the ice point is 42o and the steam point is 182o For another scale B, the ice point is −10o and steam point in 90o) If B reads 60o, the reading of A is,

The correct option is B 140oGiven,B=600=>A−42182−42=B−(−10)90−(−10)=>A−42140=B+10100=>A−4214=60+1010=>A−4214=7010=>A−42=98A=1400

On a hypothetical scale A the ice point is $42^{o}$ and the steam point is $182^{o}$ For another scale B, the ice point is $- 10^{o}$ and steam point in $90^{o}$ ) If B reads $60^{o}$ , the reading of A is,

The correct option is B $140^{o}$
Given,
$B = 60^{0} => \frac{A - 42}{182 - 42} = \frac{B - (- 10)}{90 - (- 10)}$
$=> \frac{A - 42}{140} = \frac{B + 10}{100} => \frac{A - 42}{14} = \frac{60 + 10}{10}$
$=> \frac{A - 42}{14} = \frac{70}{10} => A - 42 = 98$
$A = 140^{0}$