Force $(F)$ and density $(d)$ are related as $F\, = \,\frac{\alpha }{{\beta \, + \,\sqrt d }}$ then dimension of $\alpha $ and $\beta$ are

  • A
    $M^{3/2} L^{-1/2} T^{-2}, M^{1/2} L^{-3/2} T^0$
  • B
    $M^{1/2 }L^{-3/2} T^{-2}, M^{-3/2} L^{-3/2} T^0$
  • C
    $M^{3} L^{-1} T^{-2/3}, M^{2} L^{-3} T^{2}$
  • D
    $M^{2} L^{-1/2} T^{-2}, M^{3/2} L^{-1/2} T^0$

Similar Questions

The time dependence of a physical quantity $P$ is given by $ P = P_0 exp^{(-\alpha t^{2})} $ where $\alpha$ is a constant and $t$ is time. The constant $\alpha$ 

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If the dimensions of length are expressed as ${G^x}{c^y}{h^z}$; where $G,\,c$ and $h$ are the universal gravitational constant, speed of light and Planck's constant respectively, then

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Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as  $E = mc^2$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in $MeV$, where $1\,MeV = 1.6\times 10^{-13}\,J$ ; the masses are measured i unified atomicm mass unit (u) where, $1\,u = 1.67 \times 10^{-27}\, kg$

$(a)$ Show that the energy equivalent of $1\,u$ is $ 931.5\, MeV$.

$(b)$ A student writes the relation as $1\,u = 931.5\, MeV$. The teacher points out that the relation  is dimensionally incorrect. Write the correct relation.

Consider following statements

$(A)$ Any physical quantity have more than one unit

$(B)$ Any physical quantity have only one dimensional formula

$(C)$ More than one physical quantities may have same dimension

$(D)$ We can add and subtract only those expression having same dimension

Number of correct statement is

What is the dimensional formula of $a b^{-1}$ in the equation $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where letters have their usual meaning.

  • [JEE MAIN 2024]