A non-conducting body floats in a liquid at 2 | vedclass

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

Question

(A) Let $V$ be the volume of the body and $\rho_L$ be the density of the liquid.Since the body floats, the weight of the body equals the weight of the

Vedclass · Accepted Answer

$15.6 	imes 10^{-4} \, ^{\circ}C^{-1}$

Vedclass · Answer

(A) Let $V$ be the volume of the body and $ho_L$ be the density of the liquid.Since the body floats, the weight of the body equals the weight of the displaced liquid: $mg = V_{imm} ho_L g$.Thus, $V_{imm} ho_L = 	ext{constant}$.At $t_1 = 20^{\circ}C$, $V_{imm,1} = \frac{2}{3}V$, so $\frac{2}{3}V ho_1 = m$.At $t_2 = 100^{\circ}C$, $V_{imm,2} = \frac{3}{4}V$, so $\frac{3}{4}V ho_2 = m$.Equating the two: $\frac{2}{3} ho_1 = \frac{3}{4} ho_2$, which gives $ho_2 = \frac{8}{9} ho_1$.Using the relation $ho_2 = \frac{ho_1}{1 + \gamma_R \Delta t}$, we have $1 + \gamma_R \Delta t = \frac{ho_1}{ho_2} = \frac{9}{8}$.$\gamma_R \Delta t = \frac{9}{8} - 1 = \frac{1}{8}$.Given $\Delta t = 100 - 20 = 80^{\circ}C$, so $\gamma_R = \frac{1}{8 	imes 80} = \frac{1}{640} = 0.0015625 = 15.6 	imes 10^{-4} \, ^{\circ}C^{-1}$.

Similar Questions

Which of the following $statement(s)$ is/are correct?

What are the oxidation numbers of nitrogen in $NH_4NO_3$?

The current through an ideal $PN$-junction diode shown in the following circuit diagram will be.....$mA$.

Two spherical black bodies have radii $r_1$ and $r_2$. Their surface temperatures are $T_1$ and $T_2$. If they radiate the same power,then $\frac{r_2}{r_1}$ is:

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module