The coefficient of apparent expansion of merc | vedclass

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

Question

(A) The relationship between real expansion,apparent expansion,and vessel expansion is given by: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$.Sinc

Vedclass · Accepted Answer

$9 	imes 10^{-6}/^{\circ}C$

Vedclass · Answer

(A) The relationship between real expansion,apparent expansion,and vessel expansion is given by: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$.Since the real expansion coefficient of mercury $(\gamma_{real})$ is constant,we have: $(\gamma_{app} + \gamma_{vessel})_{glass} = (\gamma_{app} + \gamma_{vessel})_{steel}$.Given $\gamma_{app, glass} = 153 	imes 10^{-6}/^{\circ}C$ and $\gamma_{app, steel} = 144 	imes 10^{-6}/^{\circ}C$.For a solid vessel,the volume expansion coefficient is $\gamma_{vessel} = 3\alpha$.For steel,$\gamma_{vessel, steel} = 3 	imes (12 	imes 10^{-6}) = 36 	imes 10^{-6}/^{\circ}C$.Substituting these values into the equation: $153 	imes 10^{-6} + \gamma_{vessel, glass} = 144 	imes 10^{-6} + 36 	imes 10^{-6}$.$153 	imes 10^{-6} + \gamma_{vessel, glass} = 180 	imes 10^{-6}$.$\gamma_{vessel, glass} = (180 - 153) 	imes 10^{-6} = 27 	imes 10^{-6}/^{\circ}C$.Since $\gamma_{vessel, glass} = 3\alpha_{glass}$,we have $3\alpha_{glass} = 27 	imes 10^{-6}/^{\circ}C$.Therefore,$\alpha_{glass} = 9 	imes 10^{-6}/^{\circ}C$.

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6}/^{\circ}C$ and in a steel vessel is $144 \times 10^{-6}/^{\circ}C$. If $\alpha$ for steel is $12 \times 10^{-6}/^{\circ}C$,then $\alpha$ for glass is:

Similar Questions

Which one of the following is not a biopolymer?

Power dissipated across the $8\,\Omega$ resistor in the circuit shown here is $2\,W$. The power dissipated in watt units across the $3\,\Omega$ resistor is

The $Ca^{2+}$ and $F^{-}$ ions are located in a $CaF_2$ crystal,respectively at face-centered cubic lattice points and in:

In the circuit shown,the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance $= 1\,\Omega$ and the ideal cell has an $e.m.f. = 10\,V$. Then,the Ammeter has a coil resistance equal to .............. $\Omega$.

The ratio of $C_p$ and $C_v$ of a gas $X$ is $1.4$. The number of atoms of the gas $X$ present in $11.2 \ L$ of it at $NTP$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module