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] ...

409

You visited us 409 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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.

Suggest Corrections

Similar questions

Q. 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:

Q. Potential energy of a system of particles is

U = \frac{α}{3 r^{3}} - \frac{β}{2 r^{2}}

, where

r

is distance between the particles. Here

α

and

β

are positive constants. Which of the following are correct for the system?

Q. Potential energy of a system of particles is

U = \frac{α}{3 r^{3}} - \frac{β}{2 r^{2}}

, where

r

is distance between the particles. Here

α

and

β

are positive constants. Which of the following are correct for the system?

Q. If

α, β, γ

are three angles given by

α = 2 t a n^{- 1} (\sqrt{2} - 1), β = 3 s i n^{- 1} \frac{1}{\sqrt{2}} + s i n^{- 1} (\frac{- 1}{2}) a n d γ = c o s^{- 1} (\frac{1}{3}),

then

Q. If

α, β

are the solutions of

sin x = - \frac{1}{2}

[0, 2 π]

and

α, γ

are the solutions of

cos x = - \frac{\sqrt{3}}{2}

[0, 2 π],

then

Join BYJU'S Learning Program

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: