SATHEE: Physics And Measurement Question 110

Physics And Measurement Question 110

Question: If velocity $v$ , acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v, A$ and $F$ would be

Options:

A) $F A^{- 1} v$

B) $F v^{3} A^{- 2}$

C) $F v^{2} A^{- 1}$

D) $F^{2} v^{2} A^{- 1}$

Show Answer

Answer:

Correct Answer: B

Solution:

$L \propto v^{x} A^{y} F^{z}$

$\Rightarrow$ $L = k v^{x} A^{y} F^{z}$

Putting the dimensions in the above relation $[M L^{2} T^{- 1}] = k {[L T^{- 1}]}^{x} {[L T^{- 2}]}^{y} {[M L T^{- 2}]}^{z}$

therefore $[M L^{2} T^{- 1}] = k [M^{z} L^{x + y + z} T^{- x - 2 y - 2 z}]$

Comparing the powers of $M, L$ and $T$

$z = 1$ -(i)

$x + y + z = 2$ -(ii)

$- x - 2 y - 2 z = - 1$ -(iii) On solving (i), (ii) and (iii) $x = 3, y = - 2, z = 1$

So dimension of $L$ in terms of $v, A$ and $f$

$[L] = [F v^{3} A^{- 2}]$

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Physics And Measurement Question 110

Question: If velocity v , acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v,A and F would be

Options:

Answer:

Solution:

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Question: If velocity $v$ , acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v, A$ and $F$ would be