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Units and Measurement question

2015 · Q155
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Units and Measurement question

2015 · Q155

NEETPhysicsUnits and MeasurementMCQ+4 / −1
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
  1. A
    [EV−2T−2]\left[ {E{V^{ - 2}}{T^{ - 2}}} \right][EV−2T−2]
  2. B
    [E−2V−1T−3]\left[ {{E^{ - 2}}{V^{ - 1}}{T^{ - 3}}} \right][E−2V−1T−3]
  3. C
    [EV−2T−1]\left[ {E{V^{ - 2}}{T^{ - 1}}} \right][EV−2T−1]
  4. D
    [EV−1T−2]\left[ {E{V^{ - 1}}{T^{ - 2}}} \right][EV−1T−2]
View written solutionFree

Correct answer: A

Let surface tension

S =kEaaaVbTc
where k is a dimensionless constant

Writing the dimensions on both sides,

[MLT−2L]\left[ {{{ML{T^{ - 2}}} \over L}} \right][LMLT−2​] = {\left[ {M{L^2}{T^{ - 2}}} \right]^a}$$$${\left[ {L{T^{ - 1}}} \right]^b}$$${\left[ T \right]^c}$ <br><br>\left[ {M{L^0}{T^{ - 2}}} \right]===\left[ {{M^a}{L^{2a + b}}{T^{ - 2a - b + c}}} \right]<br><br>Comparing both sides of the equation we get, <br><br>$a$ = 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;....(1) <br>2$a$ + b = 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;....(2) <br>-2$a$ - b + c = - 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;....(3) <br><br>Solving equation (1), (2) and (3), we get <br><br>$a$ = 1, b = - 2, c = - 2 <br><br>$ \therefore $ Dimension of surface tension =\left[ {E{V^{ - 2}}{T^{ - 2}}} \right]$$

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