Consider the following terms:
MV → Material Volume (Control Mass)
V → Control Volume
S → Control Surface
B → Flow property
b → Intensive value of B in any small element of the fluid
ρ → Density of the flow
t → Instantaneous time
\( \vec{v} \) → Velocity of fluid entering or leaving the control volume
\( \vec{n} \) → Outward normal vector to control surface
Which of these equations is the mathematical representation of Reynolds transport theorem in the above terms?
a) \((\frac{dB}{dt})_{MV} = \frac{d}{dt}(\int_sb \rho dS) + \int_vb \rho \vec{v}.\vec{n} dV\)
b) \((\frac{dB}{dt})_{MV} = \frac{d}{dt}(\int_vb \rho dV) + \int_sb \rho \vec{v}.\vec{n} dS\)
c) \((\frac{dB}{dt})_V = \frac{d}{dt}(\int_{MV}b \rho MV) + \int_sb \rho \vec{v}.\vec{n} dS\)
d) \((\frac{dB}{dt})_{MV} = \int_vb \rho dV + \frac{d}{dt}(\int_sb \rho \vec{v}.\vec{n} dS)\)

Answer: b
Explanation:
\((\frac{dB}{dt})_{MV} →\) Instantaneous total change of B in material volume
\(\frac{d}{dt} (\int_vb \rho dV) → \) Instantaneous total change of ” B” within control volume
\( \int_sb \rho \vec{v}. \vec{n}dS → \) Net flow of B into and out of the control volume through control surfaces
Reynolds transport theorem states:
(Instantaneous total change of B in material volume)=(Instantaneous total change of B within control volume + Net flow of B into and out of control volume through control surfaces)
Therefore,
\((\frac{dB}{dt})_{MV} = \frac{d}{dt}(\int_vb \rho dV) + \int_sb \rho \vec{v}.\vec{n}dS\).