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Physics Question on mechanical properties of fluid

An object weights m1m_1, in a liquid of density d1d_1 and that in liquid of density d2 d_2\, is  m2.\, m_2. The density dd of the object is

A

d=m2d2−m1d1m2−m1d = \frac{m_2 d_2 - m_1 d_1}{m_2 - m_1}

B

d=m1d1−m2d2m2−m1d = \frac{m_1 d_1 - m_2 d_2}{m_2 - m_1}

C

d=m2d1−m1d2m1−m2d = \frac{m_2 d_1 - m_1 d_2}{m_1 - m_2}

D

d=m1d2−m2d1m1−m2d = \frac{m_1 d_2 - m_2 d_1}{m_1 - m_2}

Answer

d=m1d2−m2d1m1−m2d = \frac{m_1 d_2 - m_2 d_1}{m_1 - m_2}

Explanation

Solution

Given that an object weighs m1m_{1} in a liquid of density d1d_{1} and m2m_{2} in a liquid of density d2d_{2}, so when the density of the object is dd, then we get
V(d−d1)=m1V\left(d-d_{1}\right)=m_{1}
And V(d−d2)=m2V\left(d-d_{2}\right)=m_{2}
Thus, d−d1d−d2=m1m2\frac{d-d_{1}}{d-d_{2}}=\frac{m_{1}}{m_{2}}
So, we get d=m1d2−m2d1m1−m2d=\frac{m_{1} d_{2}-m_{2} d_{1}}{m_{1}-m_{2}}.