If the ratio of fundamental units in two systems is 1 :3, then the ratio of momentum in the two systems is:

1:3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1:9

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

1:27

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

3:1

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

Open in App

Solution

The correct option is A 1:3
$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{1}{3}, \frac{L_{1}}{L_{2}} = \frac{1}{3}, \frac{T_{1}}{T_{2}} = \frac{1}{3}$

$\Rightarrow \frac{{(m o m e n t u m)}_{1}}{{(m o m e n t u m)}_{2}}$ = $\frac{M_{1}^{1} L_{1}^{1} T_{1}^{- 1}}{M_{2}^{1} L_{2}^{1} T_{2}^{- 1}} = \frac{1}{3} \times \frac{1}{3} \times {(\frac{1}{3})}^{- 1} = \frac{1}{3}$

Suggest Corrections

Similar questions

2 : 3

1 : 3

K_{1}

K_{2}

If energy (E), momentum (P) and force (F) are chosen as fundamental units. The dimensions of mass in new system is

v^{'} = \frac{α^{2}}{β} v

a^{'} = (α β) a

F^{'} = (\frac{1}{α β}) F

α

β

Join BYJU'S Learning Program