Question

potential energy = m^x g^y h^z find value of x y z . m =mass of particle . g = gravitational acceleration . h= height of particle​

Answer 1

Dear student,

It is given that potential energy = m^x g^y h^z. And we also know by formula that potential energy = mgh ( for gravitational potential ).

We need to find the values of x, y and z.

Method 1

Since,      mass = m

                         = M¹ ____(1)

acceleration due to gravity = g

                          = LT⁻² ____(2)

             height =  h

                         = L¹ _____(3)

From (1), (2) and (3) and equating with m^x g^y h^z, we obtain;

       

       m^x g^y h^z = M¹L¹T⁻² × L¹

       m^x g^y h^z = M¹L²T⁻²

                 x, y, z = 1, 2, - 2

Hence, the values of x, y and z are 1, 2 and - 2 respectively.

Method 2

We know that any form of energy has a fixed dimension. And we know by dimensional analysis that the dimensions of energy are M¹L²T⁻².

Equating  m^x g^y h^z with the dimensions of energy, we get;

      m^x g^y h^z = M¹L¹T⁻² × L¹

      m^x g^y h^z = M¹L²T⁻²

                x, y, z = 1, 2, - 2

Therefore, the x = 1, y = 2 and z = - 2.

Answer 2

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Given :             

E = G phqcr

∴   [ML 2T−2]=[M−1L3T−2]p× [ML 2T−1]q ×[M0LT−1]r

⟹ [ML 2T−2] =[M(−p+q)L(3p+2q+r)T(−2p−q−r)] 

Equating both sides, we get:            

−p+q=1                  . . . . . . . .(1)

3p+2q+r=2          . . . . . . . . (2)

−2p−q−r=−2      . . . . . . . . . .(3)

Adding (2) and (3),  we get:    p+q=0   . . . . (4)

Solving (1) and (4),        

⟹p=−21    and   q=21

Now from (2),              

3×(−21)+2×21+r=2

⟹r=25