c
$\rho v_{2}=10(\text { in air })$
$\rho v_{2}-\rho_{\omega}\left(v_{1}+v_{2}\right)=7.5$ (in water)
$\rho_{\omega}\left(v_{1}+v_{2}\right)=2.5$
$\frac{\rho_{\omega}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}{\rho\left(\mathrm{v}_{2}\right)}=\frac{2.5}{10}=\frac{1}{4}$
$\frac{1000}{8000}\left(\frac{v_{1}}{v_{2}}+1\right)=\frac{1}{4}$
$\frac{v_{1}}{v_{2}}=1$
