Question:
In a new system of units called star units, $ 1\,kg^* = 10 \,kg $ , $ 1 \,m^* =1 \,km $ and $ 1s^* = 1 $ minute, what will be the value of $ 1J $ in the new system ?
In a new system of units called star units, $ 1\,kg^* = 10 \,kg $ , $ 1 \,m^* =1 \,km $ and $ 1s^* = 1 $ minute, what will be the value of $ 1J $ in the new system ?
Updated On: Jun 14, 2022
- $ 2.4 \times 10^{-8}\, J^* $
- $ 3.6 \times 10^{-4}\, J^* $
- $ 4.2 \times 10^{-3}\, J^* $
- $ 4.2 \times 10^{-2}\, J^* $
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The Correct Option is B
Solution and Explanation
Given that
For star units $1kg^* =10\, kg$
$1m^* = 1 \,km$
and $1s^* = 1 \,m$
The value of $1 \,J$ in star units
$1 \,J = 1 [ML^2 T^{-2}] = 1 [1\, kg\,m^2s^{-2}]$
$= \left[1\times \frac{1}{10} kg + \left(\frac{m^*}{1000}\right)^{2} + \left(\frac{s^*}{60}\right)^{-1}\right] $
$= 3.6 \times 154 J $
For star units $1kg^* =10\, kg$
$1m^* = 1 \,km$
and $1s^* = 1 \,m$
The value of $1 \,J$ in star units
$1 \,J = 1 [ML^2 T^{-2}] = 1 [1\, kg\,m^2s^{-2}]$
$= \left[1\times \frac{1}{10} kg + \left(\frac{m^*}{1000}\right)^{2} + \left(\frac{s^*}{60}\right)^{-1}\right] $
$= 3.6 \times 154 J $
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