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Question

On a temperature scale Y, water freezes at $$-160^o$$ Y and boils at $$-50^o$$ Y. On this Y scale, a temperature of $$340 \ K$$ is :


A
106.3o Y
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B
96.3o Y
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C
86.3o Y
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D
76.3o Y
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Solution

The correct option is A $$-86.3^o$$ Y
Since the temperature scale is assumed to be linear, slope in two cases will be same. Hence,

$$\dfrac { Y-(-160) }{ -50-(-160) } =\dfrac { K-273 }{ 373-273 } $$,

$$\dfrac { Y+160 }{ 110 } =\dfrac { K-273 }{ 100 } $$

$$Y=\dfrac { 11 }{ 10 } (K-273)-160$$

$$Y=\dfrac { 11 }{ 10 } (340-273)-160=-{ 86.3 }^{ 0 }Y$$

Physics

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