Question

16. If density p, accelaration
due to gravity g and frequency
v are the basic quantity, then
dimensions of force is​

Answer 1

SOLUTION :-

• Dimensional formula of density ,

                                  $\sf \rho = [ML^{-3}]$

• Dimensional formula of acceleration due to gravity ,

                                 $\sf g = [LT^{-2}]$

• Dimensional formula of frequency ,

                                 $\sf\nu =[T^{-1}]$

• Dimensional formula of force ,

                                   $\sf F= [MLT^{-2}]$

               

             $\longrightarrow \sf F \propto \rho^x g^y \nu^z \\\\\longrightarrow F= k\rho^x g^y \nu^z \\\\\longrightarrow [MLT^{-2}] = k [ML^{-3}] ^x\ [LT^{-2}]^y \ [T^{-1}]^z \\\\\longrightarrow [MLT^{-2}] = k [M^xL^{-3x}] \ [L^yT^{-2y}] \ [T^{-z}]\\\\\longrightarrow [MLT^{-2}] = k [M^xL^{-3x+y}T^{-2y-z}]$

    On comparing both sides we get ,

    →  x = 1                 ............(1)    

    → -3x + y = 1         ............(2)

    → -2y - z = -2          ............(3)

  Put the value of x in Eq(2) ,

     $\longrightarrow \sf -3 \times 1 + y = 1 \\\\\longrightarrow\bf y = 4$

   

  Put the value of y in Eq(3) ,

   

     $\longrightarrow \sf -2 \times 4 -z = -2\\\\\longrightarrow -z = 6\\\\\longrightarrow \bf z = -6$

  $\underline{\underline{\bf F=k \rho^1 g^4 \nu ^{-6}}}$

                                       

Answer 2

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