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
B
96.3o Y
C
86.3o Y
D
76.3o Y

Solution

## The correct option is A $$-86.3^o$$ YSince 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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