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Physics Question on Thermal Expansion

The variation of density of a solid with temperature is given by the formula

A

d2=d11+γ(t2−t1)d_2=\frac{d_1}{1+\gamma(t_2-t_1)}

B

d2=d11−γ(t2−t1)d_2=\frac{d_1}{1-\gamma(t_2-t_1)}

C

d2=d11−2γ(t2−t1)d_2=\frac{d_1}{1-2\gamma(t_2-t_1)}

D

d2=d11+2γ(t2−t1)d_2=\frac{d_1}{1+2\gamma(t_2-t_1)}

Answer

d2=d11+γ(t2−t1)d_2=\frac{d_1}{1+\gamma(t_2-t_1)}

Explanation

Solution

From the definition of γ\gamma, we know it is defined as change in volume per volume per unit temperature difference. So, γ=V2−V1V1(t2−t1)\gamma=\frac{ V _{2}- V _{1}}{ V _{1}\left( t _{2}- t _{1}\right)} Or V2=V1[1+γ(t2−t1)]V _{2}= V _{1}\left[1+\gamma\left( t _{2}- t _{1}\right)\right] Now, density == mass/ volume, So, replacing V1V _{1} with Md1\frac{ M }{ d _{1}} and V2V _{2} with Md2\frac{ M }{ d _{2}} we have, d2=d1[1+γ(t2−t1)]d _{2}=\frac{ d _{1}}{\left[1+\gamma\left( t _{2}- t _{1}\right)\right]}