Answer

Verified

394.2k+ views

**Hint:**We know that Centre of mass of a body is defined as the point at which the distribution of mass is equal in all directions. We will use the formula of centre of mass symmetric to the y-axis to solve the given problem. Method of integration is used to solve the problem.

**Formula Used:**

We are going to use the following formula to solve the problem:-

${{y}_{cm}}=\dfrac{\int\limits_{0}^{R}{\left[ {{\rho }_{o}}\left( 1-\dfrac{r}{R} \right)2\pi {{r}^{2}}dr \right]\dfrac{r}{2}}}{\int\limits_{0}^{R}{\left[ {{\rho }_{o}}\left( 1-\dfrac{r}{R} \right)2\pi {{r}^{2}}dr \right]}}$

**Complete answer:**

From the figure we cut an element of length $dr$. We know that the centre of mass is calculated by dividing the sum of product of mass moment of inertia and the mean distance by the sum of areas. In this case we have to integrate it to find the whole solid in terms of an element. We have the following parameters with us:-

Mass per unit volume is given as $\rho (r)={{\rho }_{o}}\left( 1-\dfrac{r}{R} \right)$, where $r$ is the radial distance from centre, $R$ is the radius of solid quarter sphere and position of centre of mass of the quarter solid sphere is ${{y}_{cm}}$ .

Area of the cross section is given as $2\pi {{r}^{2}}dr$ and mean distance as $\dfrac{r}{2}$. Now, using following formula we have:-

${{y}_{cm}}=\dfrac{\int\limits_{0}^{R}{\left[ {{\rho }_{o}}\left( 1-\dfrac{r}{R} \right)2\pi {{r}^{2}}dr \right]\dfrac{r}{2}}}{\int\limits_{0}^{R}{\left[ {{\rho }_{o}}\left( 1-\dfrac{r}{R} \right)2\pi {{r}^{2}}dr \right]}}$

$\Rightarrow {{y}_{cm}}=\dfrac{\dfrac{2\pi {{\rho }_{o}}}{2}\int\limits_{0}^{R}{\left[ \left( \dfrac{R-r}{R} \right){{r}^{2}}dr \right]r}}{2\pi {{\rho }_{o}}\int\limits_{0}^{R}{\left[ \left( \dfrac{R-r}{R} \right){{r}^{2}}dr \right]}}$

Integrating further and solving we get,

${{y}_{cm}}=\dfrac{\dfrac{1}{2}\left( \dfrac{{{R}^{4}}}{4}-\dfrac{{{R}^{5}}}{5R} \right)}{\left( \dfrac{{{R}^{3}}}{3}-\dfrac{{{R}^{4}}}{4R} \right)}$

Solving further we get,

${{y}_{cm}}=\dfrac{3}{10}R$

**Hence, option $(A)$ is correct.**

**Note:**

We should be clear about our concept of centre of mass and centre of gravity. Different bodies with different geometric shapes have different points as the centre of mass. Centre of mass does not depend on the gravitational field but the centre of gravity does depend on the gravitational field. Centre of mass and centre of gravity of a body in a uniform gravitational field are always equal but they may be at different points if the gravitational field is not uniform.

Recently Updated Pages

Which of the following materials is likely to have class 8 physics CBSE

What is a quasar What is their importance class 8 physics CBSE

Identify the parts which vibrate to produce sound in class 8 physics CBSE

Write the relation between liter and cm3 class 8 physics CBSE

Define the terms a ray of light and a beam of ligh class 8 physics CBSE

Suppose you are in a car and it is raining heavily class 8 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Which are the Top 10 Largest Countries of the World?

The provincial president of the constituent assembly class 11 social science CBSE

Write the 6 fundamental rights of India and explain in detail