First slide

Dimensions of physical quantities

Question

If force $' F',$ acceleration $' A'$ and time $' T'$ are taken as fundamental quantities then the dimensions of energy are.

Moderate

A

$A^{2} T$
B

$F A T^{2}$
C

$F^{2} T$
D

$F A^{- 1} T^{- 1}$

Solution

$E α F^{a} A^{b} T^{c} M L^{2} T^{- 2} = k {(M L T^{- 2})}^{a} . {(L T^{- 2})}^{b} T^{c} M L^{2} T^{- 2} = k M^{a} L^{a + b} T^{- 2 a - 2 b + c} C o m p a r e o n b o t h s i d e M_{,} L_{,} T D i m e n s i o n s M \to a = 1, a + b = 2 L \to b = 2 - 1 = 1 T \to - 2 a - 2 b + c = - 2 s u b s t i t u t e a, b v a l u e s 2 a + 2 b - c = 2 2 + 2 - c = 2 c = 2 E = F^{1} A T^{2} A L T E R N A T E L Y : p o w e r = f o r c e \times v e l o c i t y \frac{e n e r g y}{t i m e} = f o r c e \times a c c e l e r a t i o n \times t i m e s o E = F A T^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App