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1.Units, Dimensions and Measurement
hard
If energy $(E),$ velocity $(V)$ and time $(T)$ are chosen as the fundamental quantities, the dimensional formula of surface tension will be
A
$[EV^{-2}T^{-1}]$
B
$[EV^{-1}T^{-2}]$
C
$[EV^{-2}T^{-2}]$
D
$[E^{-2} V^{-1}T^{-3}]$
(AIPMT-2015) (AIEEE-2012)
Solution
$S = E ^{ x } V ^{ y } T ^{ z }$
$S = \frac FL$ $\Rightarrow[ T ]=\left[ MT ^{-2}\right]$
$[ E ]=\left[ ML ^2 T ^{-2}\right]$
${[ V ]=\left[ LT ^{-1}\right] }$
${[ T ]=[ T ] }$
${\left[ MT { }^{-2}\right]=\left[ E ^{ x } V ^{ y } t ^{ z }\right]=\left[ ML ^2 T ^{-2}\right]^{ x }\left[ LT ^{-1}\right]^{ y }[ T ]^{ z } }$
Comparing coefficients
$x=1$
$2 x+y=0$
$-2 x-y+z=-2$
Solving above equations $x =1, y =-2, z =-2$
Substituting in expression of surface tension:
$S = EV ^{-2} T ^{-2}$
Standard 11
Physics
Similar Questions
Match the following two coloumns
| Column $-I$ | Column $-II$ |
| $(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ – 3}}{A^{ – 2}}$ |
| $(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ – 3}}{A^{ – 2}}$ |
| $(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ – 3}}{A^{ – 1}}$ |
| $(D)$ Specific conductance | $(s)$ None of these |
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