Question Number 54825 by Tawa1 last updated on 12/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19
$${v}_{\mathrm{0}} ={volume}\:{of}\:{glass}\:{bottle}\:{before}\:{heating} \\ $$$${v}_{\mathrm{0}} \left(\mathrm{1}+\gamma_{{g}} \bigtriangleup{T}\right)={volume}\:{of}\:{glass}\:{bottle}\:{after}\:{heating} \\ $$$${v}_{\mathrm{0}} \left(\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}\right)={volume}\:{of}\:{mercury}\:{after}\:{heating} \\ $$$${m}_{{e}} ={mass}\:{of}\:{mercury}\:{expelled} \\ $$$${d}_{\mathrm{0}} ={density}\:{of}\:{mercury}\:{before}\:{heating} \\ $$$${mass}\:{of}\:{mercury}\:{remain}\:{same}… \\ $$$${v}_{\mathrm{0}} \left(\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}\right)\left(\frac{{d}_{\mathrm{0}} }{\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}}\right)={v}_{\mathrm{0}} \left(\mathrm{1}+\gamma_{{g}} \bigtriangleup{T}\right)×\left(\frac{{d}_{\mathrm{0}} }{\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}}\right)+{m}_{{e}} \\ $$$${v}_{\mathrm{0}} \left(\frac{{d}_{\mathrm{0}} }{\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}}\right)\left(\gamma_{{m}} −\gamma_{{g}} \right)\bigtriangleup{T}={m}_{{e}} \\ $$$$\left(\frac{{m}_{\mathrm{0}} }{\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}}\right)\left(\gamma_{{m}} −\gamma_{{g}} \right)\bigtriangleup{T}={m}_{{e}} \\ $$$${m}_{\mathrm{0}} =\frac{{m}_{{e}} \left(\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}\right)}{\left(\gamma_{{m}} −\gamma_{{g}} \right)\bigtriangleup{T}} \\ $$$${m}_{\mathrm{0}} =\frac{{m}_{{e}} \left(\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}\right)}{\left(\gamma_{{m}} −\mathrm{3}\alpha_{{g}} \right)\bigtriangleup{T}} \\ $$$${required}\:{answer}\:{is} \\ $$$${m}_{\mathrm{0}} −{m}_{{e}} \\ $$$${m}_{{e}} \left[\frac{\mathrm{1}+\gamma_{{m}} \bigtriangleup{T}}{\left(\gamma_{{m}} −\mathrm{3}\alpha_{{g}} \right)\bigtriangleup{T}}−\mathrm{1}\right] \\ $$$${pls}\:{check}…{and}\:{put}\:{values}… \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 13/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$