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y=x^(2)r+M^(1)L^(1)T^(-2) is dimensionally correct. If r represent displacement, then write dimension of x^(2) |
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :By principle of <a href="https://interviewquestions.tuteehub.com/tag/homogeneity-1028735" style="font-weight:bold;" target="_blank" title="Click to know more about HOMOGENEITY">HOMOGENEITY</a>, <br/> Here `x^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)r=M^(1)L^(1)T^(-2)` <br/> `:.x^(2)=(M^(1)L^(1)T^(-2))/(r)=(M^(1)L^(1)T^(-2))/(L^(1))` <br/> `=M^(1)L^(0)T^(-2)`</body></html> | |