Question

A sphere of radius 5 cm weighs 4.4 kg then the weight of the sphere of the same material whose radius is 3 cm is:
(a) 2.64 kg
(b) 1.584 kg
(c) 0.9504 kg
(d) $\dfrac{4}{3}\left( 0.9504 \right)\text{ kg}$

Answer 1

Hint: Find the density of the given sphere by using the formula: $\text{density}=\dfrac{\text{mass}}{\text{volume}}$. Use the formula for the volume of a sphere given by: $V=\dfrac{4}{3}\pi {{r}^{3}}$, where V is the volume of the sphere and r = 5 cm is the radius of the sphere. Once the density of the sphere is determined, find the weight of the second sphere which is made of the same material by using the formula: $\text{mass}=\text{volume}\times \text{density}$. Substitute the value of density obtained in the first step and find the volume of the sphere using the given by: ${{V}^{'}}=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}$, where V’ is the volume of second sphere and r’ = 3 cm is the radius of second sphere.

Complete step-by-step answer:
We know that, density of a substance is the ratio of its mass and volume.
For 1st sphere, we have,
Radius = r = 5 cm, mass = 4.4 kg, therefore, applying the formula: $\text{density}=\dfrac{\text{mass}}{\text{volume}}$ and $\text{Volume}=\dfrac{4}{3}\pi {{r}^{3}}$, we get,

& \text{density}=\dfrac{4.4}{\dfrac{4}{3}\pi \times {{5}^{3}}}\text{ kg/c}{{\text{m}}^{\text{3}}} \\\ & =\dfrac{4.4\times 3}{4\pi \times {{5}^{3}}}\text{ kg/c}{{\text{m}}^{\text{3}}} \\\ & =\dfrac{1.1\times 3}{\pi \times {{5}^{3}}}\text{ kg/c}{{\text{m}}^{\text{3}}} \\\ & =\dfrac{3.3}{\pi \times {{5}^{3}}}\text{ kg/c}{{\text{m}}^{\text{3}}}..................(i) \\\ \end{aligned}$$ Now, it is given that the material of the second sphere is the same as that of the first material. Therefore, the density of the two spheres will remain the same because the same materials have the same density. So, For 2nd sphere, we have, Radius = r’ = 3 cm, therefore, applying the formula: $\text{mass}=\text{volume}\times \text{density}$ and $\text{Volume}=\dfrac{4}{3}\pi {{\left( r' \right)}^{3}}$, we get, $\text{mass}=\dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}\times \text{density}$ Substituting the value of density from equation (i), we get, $\text{mass}=\dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}\text{c}{{\text{m}}^{\text{3}}}\times \dfrac{3.3}{\pi \times {{\left( 5 \right)}^{3}}}\text{ kg}/\text{c}{{\text{m}}^{3}}$ Cancelling the common units and factors, we get, $\begin{aligned} & \text{mass}=4\times {{\left( 3 \right)}^{3}}\times \dfrac{1.1}{{{\left( 5 \right)}^{3}}}\text{ kg} \\\ & =\dfrac{27\times 4\times 1.1}{125}\text{ kg} \\\ & =\dfrac{118.8}{125}\text{ kg} \\\ & =0.9504\text{ kg} \\\ \end{aligned}$ Hence, option (c) is the correct answer. Note: One may see that we have not substituted the value of $\pi $ because in the end it gets cancelled. This reduced our calculation to a great extent. Also, as you can see that we have not changed the unit of volume from cubic centimeter to cubic meter because finally the unit of the volume is getting cancelled. So, these small observations can help us to reduce our calculations. Always remember that, density of two or more objects made up of the same materials have the same densities.