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y=x^(2)r+M^(1)L^(1)T^(-2) is dimensional...

`y=x^(2)r+M^(1)L^(1)T^(-2)` is dimensionally correct. If r represent displacement, then write dimension of `x^(2)`

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By principle of homogeneity,
Here `x^(2)r=M^(1)L^(1)T^(-2)`
`:.x^(2)=(M^(1)L^(1)T^(-2))/(r)=(M^(1)L^(1)T^(-2))/(L^(1))`
`=M^(1)L^(0)T^(-2)`
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