
A faulty thermometer reads ${5^ \circ }C$ in melting ice and ${99^ \circ }C$ in dry steam. Find the correct temperature in Fahrenheit scale if this faulty thermometer reads ${52^ \circ }C$.
A) ${84^ \circ }C$
B) ${122^ \circ }C$
C) ${56^ \circ }C$
D) ${88^ \circ }C$
Answer
161.7k+ views
Hint: Use the fault reading of the melting ice, dry steam and the given temperature in the formula of the correct reading to find the correct reading in the Celsius scale. Convert the Celsius temperature into the Fahrenheit temperature using the formula.
Formula used:
(1) Formula of the correct reading is given by
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Where $c$ is the correct reading, $f$ is the faulty reading, $d$ is the reading of the dry steam and $m$ is the reading of the melting ice in the faulty thermometer.
(2) The relation between the temperature in the Fahrenheit and the Celsius scale is given by
$\dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Where $C$ is the temperature in the Celsius scale and $F$ is the temperature in the Fahrenheit scale.
Complete step by step solution:
It is given that the
Reading of the melting ice in the faulty thermometer, $m = {5^ \circ }C$
Reading of the dry steam in the faulty thermometer, $d = {99^ \circ }C$
Using the formula of the correct temperature,
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Substituting the known values in the above formula, we get
$\Rightarrow c = \dfrac{{52 - 5}}{{99 - 5}} \times 100$
By performing various arithmetic operations, we get
$\Rightarrow c = {50^ \circ }C$
Using the relation between the Fahrenheit and the Celsius scale,
$\Rightarrow \dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Substitute the temperature in the Celsius,
$\Rightarrow \dfrac{{50}}{5} = \dfrac{{F - 32}}{9}$
By simplification of the above equation, we get
$\Rightarrow F = 90 + 32 = {122^ \circ }F$
Hence the correct reading of the temperature in the Fahrenheit scale is obtained as ${122^ \circ }F$ .
Thus the option (B) is correct.
Note: At a normal condition, the temperature of the melting ice is ${0^ \circ }C$ , it can be said in the Fahrenheit as ${32^ \circ }F$ . Drying steam is obtained when the water is boiled and evaporated. Temperature of the drying steam is ${140^ \circ }F - {180^ \circ }F$ . Remember the formula for the correct reading and the Celsius and the Fahrenheit conversion.
Formula used:
(1) Formula of the correct reading is given by
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Where $c$ is the correct reading, $f$ is the faulty reading, $d$ is the reading of the dry steam and $m$ is the reading of the melting ice in the faulty thermometer.
(2) The relation between the temperature in the Fahrenheit and the Celsius scale is given by
$\dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Where $C$ is the temperature in the Celsius scale and $F$ is the temperature in the Fahrenheit scale.
Complete step by step solution:
It is given that the
Reading of the melting ice in the faulty thermometer, $m = {5^ \circ }C$
Reading of the dry steam in the faulty thermometer, $d = {99^ \circ }C$
Using the formula of the correct temperature,
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Substituting the known values in the above formula, we get
$\Rightarrow c = \dfrac{{52 - 5}}{{99 - 5}} \times 100$
By performing various arithmetic operations, we get
$\Rightarrow c = {50^ \circ }C$
Using the relation between the Fahrenheit and the Celsius scale,
$\Rightarrow \dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Substitute the temperature in the Celsius,
$\Rightarrow \dfrac{{50}}{5} = \dfrac{{F - 32}}{9}$
By simplification of the above equation, we get
$\Rightarrow F = 90 + 32 = {122^ \circ }F$
Hence the correct reading of the temperature in the Fahrenheit scale is obtained as ${122^ \circ }F$ .
Thus the option (B) is correct.
Note: At a normal condition, the temperature of the melting ice is ${0^ \circ }C$ , it can be said in the Fahrenheit as ${32^ \circ }F$ . Drying steam is obtained when the water is boiled and evaporated. Temperature of the drying steam is ${140^ \circ }F - {180^ \circ }F$ . Remember the formula for the correct reading and the Celsius and the Fahrenheit conversion.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
