In a mercury thermometer- the ice point is marked as 10-o and the steam point is marked as 130-o- At 40-oC temperature- what will this thermometer read -in degrees-78666258

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Toppr Admin · Accepted Answer

Steampoint-x2212-icepointgiventemp-x2212-icepoint-constant100-x2212-040-x2212-0-130-x2212-10x-x2212-10-x21D2-104-120x-x2212-10-x21D2-x-x2212-10-48-x21D2-x-58-x2103-

In a mercury thermometer, the ice point is marked as $10^{o}$ and the steam point is marked as $130^{o}$ . At $40^{o} C$ temperature, what will this thermometer read (in degrees)?
$78$
$66$
$62$
$58$

$\frac{s t e a m p o i n t - i c e p o i n t}{g i v e n t e m p . - i c e p o i n t} = c o n s t a n t \frac{100 - 0}{40 - 0} = \frac{130 - 10}{x - 10} \Rightarrow \frac{10}{4} = \frac{120}{x - 10} \Rightarrow x - 10 = 48 \Rightarrow x = 58 ℃$

In a mercury thermometer, the ice point is marked as 10o and the steam point is marked as 130o. At 40oC temperature, what will this thermometer read (in degrees)?78666258

steampoint−icepointgiventemp.−icepoint=constant100−040−0=130−10x−10⇒104=120x−10⇒x−10=48⇒x=58℃

In a mercury thermometer, the ice point is marked as $10^{o}$ and the steam point is marked as $130^{o}$ . At $40^{o} C$ temperature, what will this thermometer read (in degrees)?
$78$
$66$
$62$
$58$

$\frac{s t e a m p o i n t - i c e p o i n t}{g i v e n t e m p . - i c e p o i n t} = c o n s t a n t \frac{100 - 0}{40 - 0} = \frac{130 - 10}{x - 10} \Rightarrow \frac{10}{4} = \frac{120}{x - 10} \Rightarrow x - 10 = 48 \Rightarrow x = 58 ℃$