The equation for real gas is given by $ \left( P + \frac{a}{V^2} \right)(V - b) = RT $, where $ P $, $ V $, $ T $, and $ R $ are the pressure, volume, temperature and gas constant, respectively. The dimension of $ ab $ is equivalent to that of:
From the given equation \( \left( P + \frac{a}{V^2} \right)(V - b) = RT \), we have the following dimensions for each variable: \[ [a] = \left[ P \right] \left[ V \right]^2 = ML^{-1}T^{-2}L^2 = M L T^{-2} \] \[ [b] = [V] = L^3 \] Now, \( [ab] = (M L T^{-2})(L^3) = M L^4 T^{-2} \).
Thus, the dimensions of \( ab \) correspond to the dimension of compressibility.
Match List-I with List-II.
| List-I (A) Coefficient of viscosity (B) Intensity of wave (C) Pressure gradient (D) Compressibility | List-II (I) [ML-1T-1] (II) [MT-3] (III) [ML-2T-2] (IV) [M-1LT2] |
The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
