When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to ${(\frac{2}{5})}^{t h}$ of the distance between the lower and the upper fixed points. The temperature of liquid in K is
$217.15$
$313.15$
$220$
$330$

Question

When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to ${(\frac{2}{5})}^{t h}$ of the distance between the lower and the upper fixed points. The temperature of liquid in K is
$217.15$
$313.15$
$220$
$330$

Toppr Admin · Accepted Answer

Given-Tb-x2212-Tf-100Temperature of liquid- T-25 distance between lower and upper fixed pointTemperature of liquid- T-25-xD7-100-400COn kelvin scale- Tk-273-15-40-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-Tk-313-15KThe correct option is B-

When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to (25)th of the distance between the lower and the upper fixed points. The temperature of liquid in K is217.15313.15220330

Given:Tb−Tf=100Temperature of liquid, T=25 distance between lower and upper fixed pointTemperature of liquid, T=25×100=400COn kelvin scale, Tk=273.15+40 Tk=313.15KThe correct option is B.

When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to ${(\frac{2}{5})}^{t h}$ of the distance between the lower and the upper fixed points. The temperature of liquid in K is
$217.15$
$313.15$
$220$
$330$

Given:
$T_{b} - T_{f} = 100$
Temperature of liquid, $T = \frac{2}{5}$ distance between lower and upper fixed point
Temperature of liquid, $T = \frac{2}{5} \times 100 = 40^{0} C$
On kelvin scale, $T_{k} = 273.15 + 40$
$T_{k} = 313.15 K$
The correct option is B.