The resistance of a platinum resistance thermometer of ice point- the steam point and the boiling point of sulphur are 2-50- 3-50 and 6-50- respectively- Find the boiling point of sulphur on the platinum scale- The ice point and steam point measure 0- and 100- respectively-

T=R_t R_0/R_100 R_0× 100=6.5 2.50/3.5 2.50× 100^∘ ⇒ t=400^∘ Temperature is equal to % increase of the quantity it is related to. ...

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Standard XI

Physics

Resistance Thermometers

The resistanc...

Question

The resistance of a platinum resistance thermometer of ice point, the steam point and the boiling point of sulphur are 2.50, 3.50 and 6.50 $Ω$ respectively. Find the boiling point of sulphur on the platinum scale. The ice point and steam point measure $0^{\circ}$ and $100^{\circ}$ respectively.

400^{\circ}

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500^{\circ}

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600^{\circ}

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700^{\circ}

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Solution

The correct option is A $400^{\circ}$
$t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100 = \frac{6.5 - 2.50}{3.5 - 2.50} \times 100^{\circ}$
$\Rightarrow t = 400^{\circ}$
Temperature is equal to $%$ increase of the quantity it is related to.

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The resistance of a platinum resistance thermometer of ice point, the steam point and the boiling point of sulphur are 2.50, 3.50 and 6.50Ω respectively. Find the boiling point of sulphur on the platinum scale. The ice point and steam point measure 0∘ and 100∘ respectively.

The correct option is A 400∘t=Rt−R0R100−R0×100=6.5−2.503.5−2.50×100∘ ⇒t=400∘ Temperature is equal to % increase of the quantity it is related to.

The resistance of a platinum resistance thermometer of ice point, the steam point and the boiling point of sulphur are 2.50, 3.50 and 6.50 $Ω$ respectively. Find the boiling point of sulphur on the platinum scale. The ice point and steam point measure $0^{\circ}$ and $100^{\circ}$ respectively.

The correct option is A $400^{\circ}$
$t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100 = \frac{6.5 - 2.50}{3.5 - 2.50} \times 100^{\circ}$
$\Rightarrow t = 400^{\circ}$
Temperature is equal to $%$ increase of the quantity it is related to.