Question

check the correctness of equation using dimensions analysis of v√2=u+at​

Answer 1

Answer:

according to principle of homogeneity the equation is correct

Explanation:

v=L$T^{-1}$(velocity)

$\sqrt{2\\}$=as it is a constant it does not have dimensional formula

u=L$T^{-1}$(initial velocity)

a=L$T^{-2}$(acceleration)

t=T

substituting the values we get,

L$T^{-1}$=L$T^{-1}$ + L$T^{-2}$ * T

L$T^{-1}$ = L$T^{-1}$ + L$T^{-1}$

therefor the equation is dimensinally correct

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given:- }}}$

$\textsf{ \to A equation is given i.e. \sf v \sqrt2 = u + at }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Check :- }}}$

$\textsf{ \to The correctness of the equation dimensionally.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The equation given to us is v√2 = u + at.Here , a is acceleration , v is final velocity , t is time and u is initial velocity. So ,

$\sf\longrightarrow Dimensional \ formula \ of \ velocity =[v]= \red{ [ M^0 L^1 T^{-1}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ Displacement =[s] =\red{ [ M^0 L^1 T^{0}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ accl^n=[a]=\red{ [ M^0 L^1 T^{-2}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ time=[t]=\red{ [ M^0 L^0 T^{1}]}$

$\underline{\purple{\textsf{Hence here we have in LHS as , }}}$

$\sf:\implies\pink{\boxed{\sf [v]\sqrt 2= [M^0L^1T^{-1}]}}$

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$\underline{\purple{\textsf{Hence here we have in RHS as , }}}$

$\sf:\implies [u] + [at ]= [M^0L^1T^{-1}] + ([M^0L^1T^{-2}])( [M^0 L^0T^1]) \\\\\sf:\implies \sf:\implies[ u] + [at ]= [M^0L^1T^{-1}] + [ M^{(0+0)} L^{(1+0)}T^{(-2+1)}] \\\\\sf:\implies\boxed{\pink{\sf [u]+[at]= [M^0L^1T^{-1}] +[M^0L^1T^{-1}] }}$

$\textsf{ \to Since the dimensions of LHS and RHS is same,}\\\textsf{ i.e.\red{[ \sf M^0L^1T^{-1} ]}.So the equation is dimensionally correct }.$

$\underline{\underline{\textsf{\textbf{ \blue{Hence the equation is dimensionally correct ! }}}}}$