Energy due to the position of a particle is given by- U - y - y - where - and - are constants- y is distance- The dimensions of - is-A- -M L7 - 2 T -2-B- - M 1 - 2 L 3 - 2 T -2-C- - M 0 L -7 - 2 T 0-D- - M 0 LT 0-

The correct option is A [ M L^7/2T^ 2 ]U=α√(y)y+β nbsp; nbsp; ... (1) Dimension of energy = [ M^1L^2T^ 2 ] nbsp; nbsp; ... (2) Dimension of β=[L^1] ...

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Standard XI

Mathematics

Sum of n Terms in AP

Energy due to...

Question

Energy due to the position of a particle is given by, $U = \frac{α \sqrt{y}}{y + β}$ , where $α$ and $β$ are constants, $y$ is distance. The dimensions of $(α \times β)$ is:

[M^{0} L T^{0}]

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[M^{1 / 2} L^{3 / 2} T^{- 2}]

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[M^{0} L^{- 7 / 2} T^{0}]

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[M L^{7 / 2} T^{- 2}]

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Solution

The correct option is D $[M L^{7 / 2} T^{- 2}]$
$U = \frac{α \sqrt{y}}{y + β}$ ... (1)
Dimension of energy $= [M^{1} L^{2} T^{- 2}]$ ... (2)
Dimension of $β = [L^{1}]$ ... (3)
$∴$ From equation (1)
$[M^{1} L^{2} T^{- 2}] = \frac{α [L]^{1 / 2}}{[L]}$
$α = \frac{[M^{1} L^{2} T^{- 2}] [L^{1}]}{[L]^{1 / 2}}$
$α = [M^{1} L^{5 / 2} T^{- 2}]$
$∴$ Dimension of $[α \times β] = [M^{1} L^{5 / 2} T^{- 2}] [L^{1}]$
$= [M^{1} L^{7 / 2} T^{- 2}]$
$∴$ Option (D) is correct.

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Energy due to the position of a particle is given by, U=α√yy+β, where α and β are constants, y is distance. The dimensions of (α×β) is:

Energy due to the position of a particle is given by, $U = \frac{α \sqrt{y}}{y + β}$ , where $α$ and $β$ are constants, $y$ is distance. The dimensions of $(α \times β)$ is: