A certain amount of energy is measured as 500 units. If the fundamental units of length, mass and time each are doubled then the magnitude of energy in new system will be:

1000 units

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250 units

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500 units

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2000 units

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Solution

The correct option is B 1000 units
The dimensional formula of energy is $E = M^{1} L^{2} T^{- 2}$

Let $E_{1}$ and $E_{2}$ be the old and new energy value in given unit systems
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{M_{1}^{1} L_{1}^{2} T_{1}^{- 2}}{M_{1}^{1} L_{2}^{2} T_{2}^{- 2}}$

Given:
$\Rightarrow M_{2} = 2 M_{1}$

$L_{2} = 2 L_{1}$

$T_{2} = 2 T_{1}$

From equations above:

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{2} \Rightarrow E_{2} = 2 E_{1}$

$\Rightarrow E_{2} = 2 \times 500$

$E_{2} = 1000 u n i t s$

Option A is correct.

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A certain amount of energy is measured as 500 units. If the fundamental units of length, mass and time each are doubled then the magnitude of energy in new system will be:

The correct option is B 1000 unitsThe dimensional formula of energy is E=M1L2T−2Let E1 and E2 be the old and new energy value in given unit systems⇒E1E2=M11L21T−21M11L22T−22 Given:⇒M2=2M1 L2=2L1T2=2T1From equations above:⇒E1E2=12⇒E2=2E1⇒E2=2×500E2=1000 unitsOption A is correct.