
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
243.3k+ 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
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

