As per the question,
The ice point on the thermometer =96.0mm
recorded length of the mercury =33.6mm
Let the required temperature is x,
x−0100−0=33.6−096−0\dfrac{x-0}{100-0}=\dfrac{33.6-0}{96-0}100−0x−0=96−033.6−0
⇒x=33.6×10096∘C\Rightarrow x=\dfrac{33.6\times 100}{96}^\circ C⇒x=9633.6×100∘C
x=35∘Cx=35^\circ Cx=35∘C
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Sir this really help me Exactly same answer
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Sir this really help me Exactly same answer
Thanks so much