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Оп a new scale of temperature (which is linear) and called the $W$ scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?
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Verified Answer
The correct answer is:
$117^{\circ} \mathrm{W}$
Given,
\(\begin{aligned}
& \mathrm{T}_{\mathrm{f}}=39^{\circ} \mathrm{W} \\
& \mathrm{T}_{\mathrm{b}}=230^{\circ} \mathrm{W}
\end{aligned}\)
We know that,
\(\begin{aligned}
& \frac{\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{f}}}{100^{\circ} \mathrm{C}-\mathrm{o}^{\circ} \mathrm{C}}=\frac{\mathrm{T}-39^{\circ} \mathrm{C}}{39^{\circ} \mathrm{C}} \\
& \frac{239-39}{100}=\frac{\mathrm{T}-39}{39} \\
& \frac{200}{100}=\frac{\mathrm{T}-39}{39} \\
& \mathrm{~T}-39=2 \times 39=78^{\circ} \mathrm{W} \\
& \mathrm{T}=39+78=117^{\circ} \mathrm{W}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{T}_{\mathrm{f}}=39^{\circ} \mathrm{W} \\
& \mathrm{T}_{\mathrm{b}}=230^{\circ} \mathrm{W}
\end{aligned}\)
We know that,
\(\begin{aligned}
& \frac{\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{f}}}{100^{\circ} \mathrm{C}-\mathrm{o}^{\circ} \mathrm{C}}=\frac{\mathrm{T}-39^{\circ} \mathrm{C}}{39^{\circ} \mathrm{C}} \\
& \frac{239-39}{100}=\frac{\mathrm{T}-39}{39} \\
& \frac{200}{100}=\frac{\mathrm{T}-39}{39} \\
& \mathrm{~T}-39=2 \times 39=78^{\circ} \mathrm{W} \\
& \mathrm{T}=39+78=117^{\circ} \mathrm{W}
\end{aligned}\)
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