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1.Units, Dimensions and Measurement
hard
Velocity $(v)$ and acceleration $(a)$ in two systems of units $1$ and $2$ are related as $v _{2}=\frac{ n }{ m ^{2}} v _{1}$ and $a_{2}=\frac{a_{1}}{m n}$ respectively. Here $m$ and $n$ are constants. The relations for distance and time in two systems respectively are
A$\frac{ n ^{3}}{ m ^{3}} L _{1}= L _{2}$ and $\frac{ n ^{2}}{ m } T _{1}= T _{2}$
B$L_{1}=\frac{n^{4}}{m^{2}} L_{2}$ and $T_{1}=\frac{n^{2}}{m} T_{2}$
C$L _{1}=\frac{ n ^{2}}{ m } L _{2}$ and $T _{1}=\frac{ n ^{4}}{ m ^{2}} T _{2}$
D$\frac{ n ^{2}}{ m } L _{1}= L _{2}$ and $\frac{ n ^{4}}{ m ^{2}} T _{1}= T _{2}$
(JEE MAIN-2022)
Solution
$\frac{ L _{2}}{ T _{2}}=\frac{ n }{ m ^{2}} \frac{ L _{2}}{ T _{1}}$
$\frac{ L _{2}}{ T _{2}^{2}}=\frac{ L _{1}}{ T _{1}^{2} \times mn }$
$\frac{ n }{ m ^{2}} \times \frac{ T _{2}}{ T _{1}}=\frac{ T _{2}^{2}}{ T _{1}^{2} \times mn }$
$\frac{ n ^{2}}{ m }=\frac{ T _{2}}{ T _{1}}$
$\frac{ L _{2}}{ L _{1}}=\frac{ n ^{4}}{ m ^{2}} \times \frac{1}{ mn }$
$\frac{ L _{2}}{ L _{1}}=\frac{ n ^{5}}{ m ^{3}}$
$\frac{ L _{2}}{ T _{2}^{2}}=\frac{ L _{1}}{ T _{1}^{2} \times mn }$
$\frac{ n }{ m ^{2}} \times \frac{ T _{2}}{ T _{1}}=\frac{ T _{2}^{2}}{ T _{1}^{2} \times mn }$
$\frac{ n ^{2}}{ m }=\frac{ T _{2}}{ T _{1}}$
$\frac{ L _{2}}{ L _{1}}=\frac{ n ^{4}}{ m ^{2}} \times \frac{1}{ mn }$
$\frac{ L _{2}}{ L _{1}}=\frac{ n ^{5}}{ m ^{3}}$
Standard 11
Physics
