Question

The resistance of a certain platinum resistance thermometer is found to be $2.56\Omega$ at $0^{\circ }C$ and $3.56\Omega$ at ${100}^{\circ }C$. When the thermometer is immersed in a given liquid, its resistance is observed to be $5.06\Omega$. The temperature of the liquid is:

• A. ${45}^{\circ }C$
• B. ${250}^{\circ }C$
• C. ${225}^{\circ }C$
• D. ${120}^{\circ }C$

Answer 1

The correct option is B ${250}^{\circ }C$

Solution

We know that,

$R_T=R_0+\alpha (T_1-T_0)$

If $T_0=0^{\circ }C,T_1={100}^{\circ }C,R_0=2.56\Omega$

and $R_T=3.56\Omega$. Then,

$3.56=2.56+\alpha (100-0)$

$\frac{1}{100}=\alpha$

Let $T_L$ be the tan P of liquid,

Then, $5.06=2.56+\frac{1}{100}(T_L-0)$

$\Rightarrow 2.5\times 100=T_L$

$T_L={250}^{\circ }C$

Option - B is correct.