10-1.Thermometry, Thermal Expansion and Calorimetry
normal

એક અચળ કદ થર્મોમીટર દબાણનું માપ $50 \,cm$ અને $90 \,cm$ (પારાનું) એ $0^{\circ} C$ અને $100^{\circ} C$ બતાવે છે જે ક્રમશ તો જ્યારે $P=60 \,cm$ (પારાનું) હોય ત્યારે તાપમાન ............ $^{\circ} C$

A

$25$

B

$40$

C

$15$

D

$12.5$

Solution

(a)

$T _0=0^{\circ}\,C , T _1=100^{\circ}\, C , h_0=50\, cm$

and $h _1=90\,cm$

Rate of change in height wrt $T =\frac{ h _1- h _0}{ T _1- T _0}$

$\frac{\Delta h }{\Delta t }=\frac{40}{100}$

for height $(h)=60\, cm$, let the

corresponding temperature be $T$.

Then $\,\frac{ h – h _0}{ T – T _0}=\frac{\Delta h }{\Delta t }=\frac{40}{100}$

$\Rightarrow \frac{60-50}{T-0^{\circ} C }=\frac{4}{10}$

$\Rightarrow \frac{10 \times 10}{4}= T \Rightarrow T =25^{\circ}\,C$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.