Question

The resistance of the wire in the platinum resistance thermometer at ice point is $T_{1}$ and at steam point is $T_{2}$ . When the thermometer is inserted in an unknown hot bath its resistance is found to be $T_{1}$. The temperature of the hot bath is

• A. 100 $T_{1} = T_{2}$C
• B. 200 $T_{1} = \frac{1}{T_{2}}$C
• C. 300 $7.5 \times 10^{- 4}ms^{- 1}$C
• D. 350 $3 \times 10^{- 10}Vm^{- 1}$C

Answer 1

Answer: (B)

200 $T_{1} = \frac{1}{T_{2}}$C

Answer 2

: Here,

$R_{0} = 5\Omega,R_{100} = 5.25\Omega,R_{t} = 5.5\Omega$

As,

$R_{t} = R_{0}(1 + \alpha t)\therefore R_{100} = R_{0}(1 + \alpha 100)$

$\alpha = \frac{R_{100} - R_{0}}{R_{0} \times 100}$ …(i)

Let the temperature of hot bath be $t^{o}C$

$R_{t} = R_{0}(1 + \alpha t)$

$\alpha = \frac{R_{t} - R_{0}}{R_{0} \times t}$

Equating equations (i) and (ii), we get

$\frac{R_{100} - R_{0}}{R_{0} \times 100} = \frac{R_{t} - R_{0}}{R_{0} \times t}$

$t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100 = \frac{5.5 - 5}{5.25 - 5} \times 100$

$= \frac{0.5}{0.25} \times 100 = 200^{o}C$