Оп a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39^ W and 239^ W respectively. What will be the temperature on the new scale, corresponding to a

Correct answer: B. 117^ W Given, ( aligned & T_ f =39^ W & T_ b =230^ W aligned ) We know that, ( aligned & T_ b -T_ f 100^ C-o^ C = T-39^ C 39^ C & (239-39)/(100)= T-39 39 & (200)/(100)= T-39 39 & ~T-39=2 × 39=78^ W & T=39+78=117^ W aligned )

NEET 2008

Exam: NEET
Year: 2008
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2008 · Physics · Thermal Properties of Matter

Оп a new scale of temperature (which is linear) and called the W...

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?

Options

A $78^{\circ} \mathrm{W}$
B $117^{\circ} \mathrm{W}$
C $200^{\circ} \mathrm{W}$
D $139^{\circ} \mathrm{W}$

Correct answer

B. $117^{\circ} \mathrm{W}$

Step-by-step solution

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}

)

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