The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are 8 and 10 respectively. After inserting in a hot bath of temperature 400^ C , the resistance of platinum wire

Correct answer: C. 16 Given R_0=8 , R_ 100 =10 R_ 100 =R_0(1+α Δ T) Also, R_ 400 =R_0(1+α Δ T^1) aligned & 10=8(1+α × 100) 100 α=(1)/(4) & R_ 400 =8(1+400 α)=8(1+1)=16 aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2024 · 4 April · Morning Shift · Physics · Thermal Properties of Matter

The resistances of the platinum wire of a platinum resistance...

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are

8 \Omega

and

10 \Omega

respectively. After inserting in a hot bath of temperature

400^{\circ} \mathrm{C}

, the resistance of platinum wire is :

Options

A $10 \Omega$
B $8 \Omega$
C $16 \Omega$
D $2 \Omega$

Correct answer

C. $16 \Omega$

Step-by-step solution

Given

R_0=8 \Omega, R_{100}=10 \Omega \therefore \mathrm{R}_{100}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T})

Also,

R_{400}=R_0\left(1+\alpha \Delta \mathrm{T}^1\right)

\begin{aligned} & \therefore 10=8(1+\alpha \times 100) \Rightarrow 100 \alpha=\frac{1}{4} \\ & \therefore \mathrm{R}_{400}=8(1+400 \alpha)=8(1+1)=16 \Omega \end{aligned}

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