In a mercury thermometer, the ice point (lower fixed point) is marked as $10^\circ $ and the stream point (upper fixed point) is marked as$130^\circ $. At $40^\circ $ temperature what will this thermometer read?
A $78^\circ $
B $66^\circ $
C $62^\circ $
D $58^\circ $

Hint: We know that $\dfrac{{sp - ip}}{{temp - ip}}$=constant. Where, sp is the steam point, ip is the ice point and temp is the given temperature. Therefore we can equate this equation for two different thermometers with different stream points and ice points.

Complete step by step answer:
We know that $\dfrac{{sp - ip}}{{temp - ip}}$=constant.
Where, sp= steam point,
 ip= ice point
temp= given temperature.
Therefore we can equate this equation for two different thermometers with different stream points and ice points.
$\dfrac{{SP - IP}}{{t - IP}} = \dfrac{{sp - ip}}{{temp - ip}}$
Here, SP= Steam point of normal thermometer= 100℃
IP= Ice point of normal thermometer= 0℃
t= Given temperature
sp= steam point of new thermometer
ip= ice point of new thermometer
temp= unknown temperature
Substituting the value the equation becomes,
$\dfrac{{100 - 0}}{{40 - 0}} = \dfrac{{130 - 10}}{{temp - 10}}$
$\implies \dfrac{{100}}{{40}} = \dfrac{{120}}{{temp - 10}}$
$
\implies temp - 10 = \dfrac{{4800}}{{100}} \\
\implies temp = 48 + 10 = 58^\circ \\
 $
 At $40^\circ $ temperature this new thermometer will read 58℃.

So, the correct answer is “Option D”.

Note:
Students should learn the standard steam point and ice point of Celsius scale, Fahrenheit scale and Kelvin scale to solve such types of questions.
We can solve this question using alternative method as explained below,
Let the relation between two thermometers be y=ax+b, where x is the initial temperature on Celsius scale and y is the temperature on the new scale.
For ice point x=0 and y=10.
Substituting these values in the equation we get,
10=0×a+b
b =10
For stream point x=100 and y=130.
Substituting these values in the equation we get,
130=100a+10
a =$\dfrac{{130 - 10}}{{100}} = \dfrac{{120}}{{100}}$
a=1.2
Now, we need to find the value of y when the value of x=40.
y=1.2×40+10
y=48+10
y=58
At $40^\circ $ temperature this new thermometer will read 58℃.