A beaker of height H is made up of a material | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the cross-sectional area of the beaker be $A$. The initial volume of the liquid is $V_0 = A H$.When heated by temperature $T$,the new volume o

Vedclass · Accepted Answer

$8 \alpha TH$

Vedclass · Answer

(D) Let the cross-sectional area of the beaker be $A$. The initial volume of the liquid is $V_0 = A H$.When heated by temperature $T$,the new volume of the liquid is $V_L = V_0(1 + \gamma_L T) = AH(1 + \alpha T)$,where $\gamma_L = \alpha$ is the volume expansion coefficient of the liquid.The new cross-sectional area of the beaker is $A' = A(1 + 2\alpha_s T)$,where $\alpha_s = 3\alpha$ is the linear expansion coefficient of the beaker material. Thus,$A' = A(1 + 2(3\alpha)T) = A(1 + 6\alpha T)$.The new height of the beaker is $H' = H(1 + \alpha_s T) = H(1 + 3\alpha T)$.The new height of the liquid $H_L$ in the beaker is $H_L = V_L / A' = AH(1 + \alpha T) / [A(1 + 6\alpha T)]$.Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $H_L \approx H(1 + \alpha T)(1 - 6\alpha T) \approx H(1 + \alpha T - 6\alpha T) = H(1 - 5\alpha T)$.The reduction in the level of the liquid is $\Delta H = H' - H_L = H(1 + 3\alpha T) - H(1 - 5\alpha T) = H(1 + 3\alpha T - 1 + 5\alpha T) = 8\alpha TH$.

Similar Questions

$A$ boy is able to chew $20 \, g$ of ice cream in $5 \, minutes$. Calculate the power of the boy in horse power $(HP)$. (Assume the latent heat of fusion of ice is $80 \, cal/g$ and $1 \, cal = 4.2 \, J$).

Water falls from a height of $210 \, m$. Assuming the whole of the energy due to the fall is converted into heat,the rise in temperature of water would be .......... $^\circ C$ $(J = 4.3 \, J/cal)$.

Find the difference in temperature between the water at the top and the bottom of a $20 \ m$ high waterfall, assuming $10 \%$ of the energy of the fall is spent in heating the water. [Use specific heat capacity of water $= 4000 \ J \ kg^{-1} K^{-1}$ and $g = 10 \ m/s^2$]. (in $^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module