The coefficients of apparent expansion of a l | vedclass

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

Question

(D) The coefficient of real expansion of a liquid is constant and is given by the relation: $\gamma_{	ext{real}} = \gamma_{	ext{app}} + \gamma_{	ex

Vedclass · Accepted Answer

$\frac{\gamma_1-\gamma_2}{3}+\alpha$

Vedclass · Answer

(D) The coefficient of real expansion of a liquid is constant and is given by the relation: $\gamma_{	ext{real}} = \gamma_{	ext{app}} + \gamma_{	ext{vessel}}$.For vessel $A$,the coefficient of apparent expansion is $\gamma_1$ and the coefficient of linear expansion is $\alpha$. Thus,the coefficient of volume expansion is $\gamma_A = 3\alpha$.Therefore,$\gamma_{	ext{real}} = \gamma_1 + 3\alpha$.For vessel $B$,the coefficient of apparent expansion is $\gamma_2$ and let the coefficient of linear expansion be $\alpha_B$. Thus,the coefficient of volume expansion is $\gamma_B = 3\alpha_B$.Therefore,$\gamma_{	ext{real}} = \gamma_2 + 3\alpha_B$.Since $\gamma_{	ext{real}}$ is the same for the liquid in both cases,we equate the two expressions:$\gamma_1 + 3\alpha = \gamma_2 + 3\alpha_B$.Rearranging for $\alpha_B$:$3\alpha_B = \gamma_1 - \gamma_2 + 3\alpha$.$\alpha_B = \frac{\gamma_1 - \gamma_2}{3} + \alpha$.

The coefficients of apparent expansion of a liquid when determined using two different vessels $A$ and $B$ are $\gamma_1$ and $\gamma_2$,respectively. If the coefficient of linear expansion of the vessel $A$ is $\alpha$,the coefficient of linear expansion of the vessel $B$ is

Similar Questions

If the temperature of the sun were to increase from $T$ to $2T$ and its radius from $R$ to $2R$,then the ratio of the radiant energy received on earth to what it was previously will be

Limestone is present in the blast furnace production of iron in order to:
$(I)$ provide a source of $CaO$
$(II)$ remove some impurities
$(III)$ supply $CO_2$

Which of the following is mismatched?

The sensitivity of a galvanometer that measures current is decreased by $\frac{1}{40}$ times by using a shunt resistance of $10 \Omega$. Then,the value of the resistance of the galvanometer is (in $\Omega$)

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module