Glass has density of $2.5g/c{m^3}$ and a volume of $2c{m^3}.$ Mass of glass will be:
$
A.{\text{ 2g}} \\
B.{\text{ 2}}{\text{.5g}} \\
{\text{C}}{\text{. 1}}{\text{.5g}} \\
{\text{D}}{\text{. 5g}} \\
$
Answer
603k+ views
Hint: In physics density is a very basic concept and it can be defined as the mass per unit volume. It is denoted by row ($\rho $). It is commonly measured in gram per centimetre cube. Here we will use the formula - $\rho = \dfrac{M}{V}$ where M is the mass and V is the volume.
Complete step by step answer:
Given that: Density, $\rho = 2.5g/c{m^3}$
Volume, $V = 2c{m^3}$
Now, according to formula the density is –
$\rho = \dfrac{M}{V}$
Now, place the known values in the above equation
$2.5 = \dfrac{M}{2}$
Do cross multiplication and make the unknown Mass “M” the subject-
$
M = 2.5 \times 2 \\
\therefore M = 5g \\
$
Therefore, the required answer is - The Mass of glass will be $5g.$
Hence from the given multiple choices- the option D is the correct answer.
Note: Remember the basic standard formulas to solve these types of questions. Always check the given units and the units asked in the solution. All the units should be in the same format. Since here all the units were given in the CGS (Centimetre Gram Second) system of units it was just the substitution of values in the formula.
Complete step by step answer:
Given that: Density, $\rho = 2.5g/c{m^3}$
Volume, $V = 2c{m^3}$
Now, according to formula the density is –
$\rho = \dfrac{M}{V}$
Now, place the known values in the above equation
$2.5 = \dfrac{M}{2}$
Do cross multiplication and make the unknown Mass “M” the subject-
$
M = 2.5 \times 2 \\
\therefore M = 5g \\
$
Therefore, the required answer is - The Mass of glass will be $5g.$
Hence from the given multiple choices- the option D is the correct answer.
Note: Remember the basic standard formulas to solve these types of questions. Always check the given units and the units asked in the solution. All the units should be in the same format. Since here all the units were given in the CGS (Centimetre Gram Second) system of units it was just the substitution of values in the formula.
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