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Question

On a linear temperature scale Y, water freezes at 160Y and boils at 50Y. On this Y scale, a temperature of 340 K would be read as: (water freezes at 273 k and boils at 373 K)

A
73.7Y
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B
233.7Y
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C
86.3Y
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D
106.3Y
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Solution

The correct option is B 86.3Y
Given,

Freezing point on Y scale =1600Y

Boing point om Y scale =500Y

Freezing point on kelvin scale =273K

Boiling point on kelvin scale =373K

Here we have to find the temperature equivalent to 340K on Y scale

Since the temperature scale is assumed to be linear, slope in two cases will be same. Hence,

Y(160)50(160)=K273373273,

Y+160110=K273100

Y=1110(K273160

Y=1110(340273)160=86.30Y

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