Question

The density of a material in CGS system of units is $4 \mathrm{~g} / \mathrm{cm}^{3}$.

In a system of units in which unit of length is $10 \mathrm{~cm}$ and unit of mass is $100 \mathrm{~g}$, the value of density of material will be:

A) $400$
B) $0.04$
C) $0.4$
D) $40$

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Answer

To find the density of a material with a different unit system, we can use the following transformation formula:

$$ n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^{-3} $$

Given:

  • The density ( n_1 = 4 , \text{g/cm}^3 )
  • Unit of mass in the new system ( M_2 = 100 , \text{g} )
  • Unit of length in the new system ( L_2 = 10 , \text{cm} )
  • The standard CGS units are ( M_1 = 1 , \text{g} ) and ( L_1 = 1 , \text{cm} )

Now, we substitute these values into the transformation formula:

$$ n_2 = 4 \left[\frac{1 , \text{g}}{100 , \text{g}}\right]^1 \left[\frac{1 , \text{cm}}{10 , \text{cm}}\right]^{-3} $$

Simplifying the terms step by step:

$$ n_2 = 4 \times \frac{1}{100} \times \left(10^{-3}\right)^{-3} $$

Since ((10^{-3})^{-3} = (10^{-3})^{3} = 10^3):

$$ n_2 = 4 \times \frac{1}{100} \times 10^3 $$

$$ n_2 = 4 \times \frac{10^3}{100} $$

$$ n_2 = 4 \times 10 $$

$$ n_2 = 40 $$

Thus, the density of the material in the new unit system is 40. Therefore, the correct answer is:

D) 40


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