The potential energy of a particle varies with distance x from a fixed origin as V-A-xx-B where A and B are constants- The dimensions of AB are

The correct option is D [M^1L^7/2T^ 2] As B is added to x so, #160; [B]=[x]=[L] Also, #160;[A√(x)]=[Vx] #160;or #160;[A=V√(x ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Dimensional Analysis

The potential...

Question

The potential energy of a particle varies with distance x from a fixed origin as $V = \frac{A \sqrt{x}}{x + B}$ where $A$ and $B$ are constants. The dimensions of $A B$ are

[M^{1} L^{5 / 2} T^{- 2}]

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

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

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

[M^{3 / 2} L^{5 / 2} T^{- 2}]

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

[M^{1} 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 $[M^{1} L^{7 / 2} T^{- 2}]$
As $B$ is added to $x$ so, $[B] = [x] = [L]$
Also, $[A \sqrt{x}] = [V x] o r [A = V \sqrt{x}]$ $= M L^{2} T^{- 2} L^{1 / 2} = M L^{5 / 2} T^{- 2}$
Thus, $[A B] = (M L^{5 / 2} T^{- 2}) (L) = [M^{1} L^{7 / 2} T^{- 2}]$

Suggest Corrections

Similar questions

Q. The potential energy of a particle varies with distance

x

\frac{A x^{1 / 2}}{x^{2} + B}

, where

A

and

B

are constants. The dimensional formula for

A \times B

Potential energy U varies with distance x as $U = \frac{A x^{1 / 2}}{X^{2} + B}$ where A and B are constant. The dimensional formula for $A \times B$ is

Q. The potential energy of a particle varies the distance

x

from a fixed origin as

U = \frac{A \sqrt{x}}{x^{2} + B}

, where A and B are dimensional constants, then find the dimensional formula for AB.

Q. What are the dimensions of

a

and

b

in the relation

F = a \sqrt{x} + b t^{2}

where

F

is force,

x

is distance and

t

is time?

Join BYJU'S Learning Program

The potential energy of a particle varies with distance x from a fixed origin as V=A√xx+B where A and B are constants. The dimensions of AB are

The correct option is D [M1L7/2T−2]As B is added to x so, [B]=[x]=[L] Also, [A√x]=[Vx]or[A=V√x] =ML2T−2L1/2=ML5/2T−2Thus, [AB]=(ML5/2T−2)(L)=[M1L7/2T−2]

The potential energy of a particle varies with distance x from a fixed origin as $V = \frac{A \sqrt{x}}{x + B}$ where $A$ and $B$ are constants. The dimensions of $A B$ are

The correct option is D $[M^{1} L^{7 / 2} T^{- 2}]$
As $B$ is added to $x$ so, $[B] = [x] = [L]$
Also, $[A \sqrt{x}] = [V x] o r [A = V \sqrt{x}]$ $= M L^{2} T^{- 2} L^{1 / 2} = M L^{5 / 2} T^{- 2}$
Thus, $[A B] = (M L^{5 / 2} T^{- 2}) (L) = [M^{1} L^{7 / 2} T^{- 2}]$