# The density of a material in CGS system of units is 4 g / cm^3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be: A) 400 B) 0.04 C) 0.4 D) 40

## 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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