A uniform thermometer scale is at steady stat | vedclass

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(B) Let the temperature at the $60 \, cm$ mark be $T$.Since the scale is uniform,the temperature varies linearly with the length.The relationship betw

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$68$

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(B) Let the temperature at the $60 \, cm$ mark be $T$.Since the scale is uniform,the temperature varies linearly with the length.The relationship between temperature $T$ and length $L$ is given by the linear interpolation formula:$\frac{T - T_0}{T_{100} - T_0} = \frac{L - L_0}{L_{100} - L_0}$Here,$T_0 = 20^{\circ} C$ at $L_0 = 0 \, cm$ and $T_{100} = 100^{\circ} C$ at $L_{100} = 100 \, cm$.Substituting the values for $L = 60 \, cm$:$\frac{T - 20}{100 - 20} = \frac{60 - 0}{100 - 0}$$\frac{T - 20}{80} = \frac{60}{100}$$\frac{T - 20}{80} = 0.6$$T - 20 = 0.6 	imes 80$$T - 20 = 48$$T = 48 + 20 = 68^{\circ} C$Thus,the temperature at the $60 \, cm$ mark is $68^{\circ} C$.

$A$ uniform thermometer scale is at steady state with its $0 \, cm$ mark at $20^{\circ} C$ and $100 \, cm$ mark at $100^{\circ} C$. The temperature of the $60 \, cm$ mark is ............. $^{\circ} C$.

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