SATHEE: Physics And Measurement Question 196

Physics And Measurement Question 196

Question: The potential energy of a particle varies with distance $χ$ as $U = \frac{A x_{}^{1 / 2}}{x_{}^{2} + B}$ where A and B are constants. The dimensional formula for A x B is

Options:

A) $M_{}^{1} L_{}^{7 / 2} T_{}^{- 2}$

B) $M_{}^{1} L_{}^{1 / 2} T_{}^{- 2}$

C) $M_{}^{1} L_{}^{5 / 2} T_{}^{- 2}$

D) $M_{}^{1} L_{}^{9 / 2} T_{}^{- 2}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Here $x^{2}$ has the dimensions of $L^{2}, B = [L^{2}]$ .

Also $M L^{2} T^{- 2} = \frac{A L^{1 / 2}}{L^{2}}$ or $A = M L^{7 / 2} T^{- 2}$

$∴ A \times B = M L^{11 / 2} T^{- 2}$

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Physics And Measurement Question 196

Question: The potential energy of a particle varies with distance χ as U=Ax1/2x2+B where A and B are constants. The dimensional formula for A x B is

Options:

Answer:

Solution:

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Question: The potential energy of a particle varies with distance $χ$ as $U = \frac{A x_{}^{1 / 2}}{x_{}^{2} + B}$ where A and B are constants. The dimensional formula for A x B is