On a linear temperature scale $Y$ , water freezes at $- 160^{o} Y$ and boils at $- 50^{o} Y$ . On this Y scale, a temperature of $340 K$ would be read as : (water freezes at $273 K$ and boils at $373 K$ )

{- 73.7}^{o} Y

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{- 233.7}^{o} Y

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{- 86.3}^{o} Y

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{- 106.3}^{o} Y

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Solution

The correct option is D ${- 86.3}^{o} Y$
$Given :$ $L F P = - 160^{o} Y, U F P = - 150^{o} Y$

$Solution :$
$\frac{R e a d i n g o n a n y s c a l e - L F P}{U F P - L F P}$ = constant for all scales
$\frac{340 - 273}{373 - 273} = \frac{^{\circ} Y - (- 160)}{- 50 - (- 160)}$
$⟹ \frac{67}{100} = \frac{y + 160}{110}$
$∴ y = - {86.3}^{\circ}$

$Hence C is the correct option$

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Similar questions

Q. 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 :

On an X temperature scale, water freezes at $- {125.0}^{\circ}$ X and boils at ${375.0}^{\circ}$ X. On a Y temperature scale, water freezes at $- {70.0}^{\circ}$ Y and boils at $- {30.0}^{\circ}$ Y. The value of temperature on X scale equal to the temperature of ${50.0}^{\circ}$ Y on Y scale is: