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The potential energy of a particle varie...

The potential energy of a particle varies with distance x from a fixed origin as `V= (Asqrt(X))/(X + B)` where A and B are constants . The dimension of AB are

A

`[M^(1)L^(5//2)T^(2)]`

B

`[M^(1)L^(2)T^(2)]`

C

`[M^(3//2)L^(5//2)T^(2)]`

D

`[M^(1)L^(7//2)T^(2)]`

Text Solution

AI Generated Solution

The correct Answer is:
To find the dimensions of the product \( AB \) where the potential energy \( V \) is given by the equation \[ V = \frac{A \sqrt{x}}{x + B} \] we will follow these steps: ### Step 1: Identify the dimensions of \( V \) Potential energy \( V \) has the dimensions of energy, which can be expressed as: \[ [V] = [M L^2 T^{-2}] \] ### Step 2: Identify the dimensions of \( x \) The variable \( x \) represents distance, so its dimensions are: \[ [x] = [L] \] ### Step 3: Determine the dimensions of \( B \) In the expression \( x + B \), both terms must have the same dimensions for the addition to be valid. Since \( x \) has dimensions of length \( [L] \), we conclude that: \[ [B] = [L] \] ### Step 4: Analyze the expression for \( V \) Rearranging the equation \( V = \frac{A \sqrt{x}}{x + B} \), we can express it as: \[ V (x + B) = A \sqrt{x} \] ### Step 5: Substitute the dimensions into the equation The dimensions of \( \sqrt{x} \) are: \[ [\sqrt{x}] = [L^{1/2}] \] Thus, the left-hand side becomes: \[ [V] \cdot [L] = [M L^2 T^{-2}] \cdot [L] = [M L^3 T^{-2}] \] And the right-hand side becomes: \[ [A] \cdot [L^{1/2}] = [A] \cdot [L^{1/2}] \] ### Step 6: Equate the dimensions Since both sides of the equation must have the same dimensions, we can write: \[ [M L^3 T^{-2}] = [A] \cdot [L^{1/2}] \] ### Step 7: Solve for the dimensions of \( A \) To isolate \( [A] \), we rearrange the equation: \[ [A] = \frac{[M L^3 T^{-2}]}{[L^{1/2}]} = [M L^{3 - 1/2} T^{-2}] = [M L^{5/2} T^{-2}] \] ### Step 8: Calculate the dimensions of \( AB \) Now we can find the dimensions of the product \( AB \): \[ [AB] = [A] \cdot [B] = [M L^{5/2} T^{-2}] \cdot [L] = [M L^{5/2 + 1} T^{-2}] = [M L^{7/2} T^{-2}] \] ### Final Answer Thus, the dimensions of \( AB \) are: \[ [AB] = [M L^{7/2} T^{-2}] \]
To find the dimensions of the product \( AB \) where the potential energy \( V \) is given by the equation \[ V = \frac{A \sqrt{x}}{x + B} \] we will follow these steps: ...
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