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

y=x^(2)r+M^(1)L^(1)T^(-2) is dimensionally correct. If r represent dis

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