Question

Using screw gauge radius of wire was found to be $$2.50mm$$ . The length of wire found by mm Scale is $$50.0cm$$ . If the mass of the wire was measured as $$25gm$$, the density of the wire is correct S.F. will be $$\left(use\pi =3.14exactly\right)$$ in $$gm/c{m}^3$$ .
A. $$2.5$$
B. $$2.50$$
C. $$25$$
D. $0.25$

Answer 1

Hint: As we know that density is defined as mass per unit volume. It is denoted by$$\rho$$. Density plays an important role in calculating mass and volume.
$$V=m\rho$$
Here, $$V$$ is volume,$$m$$is the mass and$$\rho$$the density.
$$\rho$$= $${\displaystyle \frac{m}{V}}$$ ------- (1)
Volume of the wire can be given by- $$V$$=$$\pi r^{2}l$$
By substituting the volume and mass in equation (1), we can easily determine the value of density of the wire.Density is a scalar quantity because it does not show any direction.

Complete step by step answer:
Given that, radius of wire = $$2.50$$$$mm$$= $$2.50$$$$cm$$
Length of wire =$$50.0$$$$cm$$
Mass of wire = $$25$$$$gm$$
Given, $$\pi$$=$$3.14$$
As we know that
$$V=m\rho$$
$$\Rightarrow \rho$$= $${\displaystyle \frac{m}{V}}$$
$$\Rightarrow \rho$$= $${\displaystyle \frac{25}{\pi r^{2}l}}$$
$$\Rightarrow \rho$$= $${\displaystyle \frac{25}{3.14\times {\left(0.25\right)}^2\times 50.0}}$$
$\therefore \rho =2.5gm/c{m}^3$

Hence, option A is correct.

Additional Information:
Volume is the amount of space occupied by a substance, while mass is the amount of matter.
Mass is physical value while volume is geometrical value. The SI unit of mass is $$kg$$and the SI unit of volume is $$m^3$$.

Note:Volume plays an important role in water conservation, and also when we fill up our vehicles at pumps. Density shows how much a substance occupies volume. The most common example of density is water and oil, when we mix oil and water. Oil is less dense than water so it floats. And a boat also floats on water because the density of the boat is less than water.