M.I of a solid sphere about its diameter is $64\,kg\,{m^2}$. If that sphere is recast into $8$identical small spheres, then M.I. of such small sphere about its diameter is
$\left( A \right)\,\,8\,Kg\,{m^2}$
$\left( B \right)\,\,4\,kg\,{m^2}$
$\left( C \right)\,\,3\,kg\,{m^2}$
$\left( D \right)\,\,2\,kg\,{m^2}$
Answer
594.9k+ views
Hint:
In the question, the diameter of the solid sphere is given. But the sphere is divided into eight equal parts. By substituting the known values in the equation of moment, we get the value of the moment of inertia of such a small sphere of the diameter.
Moment of Inertia of the sphere is $I = \dfrac{2}{5}m{r^2}$
Where, $m$be the mass of the given sphere and $r$be the radius of the sphere.
Complete step by step solution:
Given that Diameter of the solid sphere$d = 64\,kg\,{m^2}$
Let the mass of given sphere is $m$and the radius of the sphere is $r.$
Moment of Inertia of the sphere is $I = \dfrac{2}{5}m{r^2}$.
Here, now we divide the sphere into eight equal small spheres
So, the equation of the sphere is,
$\Rightarrow 8\left( {\dfrac{4}{3}\pi {R^3}} \right) = \dfrac{4}{3}\pi {r^3}$
Simplify the above equation, we get the value of $r$,
$\Rightarrow r = \dfrac{R}{2}$
Now we have to find the moment of inertia of the each small spheres, we get
$\Rightarrow I = \dfrac{2}{5}m{r^2}$
Substitute the value of $r$in the equation, we get
$\Rightarrow I = \dfrac{2}{5}m{\left( {\dfrac{R}{2}} \right)^2}$
There are eight small spheres, so we written the equation of the moment of the inertia as,
$\Rightarrow I = \dfrac{2}{5} \times \dfrac{m}{8}{\left( {\dfrac{R}{2}} \right)^2}$
Simplify the above equation, we get
$\Rightarrow I = \dfrac{{64}}{{32}}$
Perform the divide operations in the above equation, we get
$\Rightarrow I = 2\,Kg\,{m^2}$
Therefore, the moment of the inertia of the small diameter is $2\,Kg\,{m^2}$.
Hence from the above options, option (D) is correct.
Note:
In the question, the sphere is used to find the moment of inertia. Spheres are divided into identical spheres which means it splits into equal spheres. But, So, diameter of the sphere is about the recast of the spheres. Consider those values, we get the moment of inertia of the sphere.
In the question, the diameter of the solid sphere is given. But the sphere is divided into eight equal parts. By substituting the known values in the equation of moment, we get the value of the moment of inertia of such a small sphere of the diameter.
Moment of Inertia of the sphere is $I = \dfrac{2}{5}m{r^2}$
Where, $m$be the mass of the given sphere and $r$be the radius of the sphere.
Complete step by step solution:
Given that Diameter of the solid sphere$d = 64\,kg\,{m^2}$
Let the mass of given sphere is $m$and the radius of the sphere is $r.$
Moment of Inertia of the sphere is $I = \dfrac{2}{5}m{r^2}$.
Here, now we divide the sphere into eight equal small spheres
So, the equation of the sphere is,
$\Rightarrow 8\left( {\dfrac{4}{3}\pi {R^3}} \right) = \dfrac{4}{3}\pi {r^3}$
Simplify the above equation, we get the value of $r$,
$\Rightarrow r = \dfrac{R}{2}$
Now we have to find the moment of inertia of the each small spheres, we get
$\Rightarrow I = \dfrac{2}{5}m{r^2}$
Substitute the value of $r$in the equation, we get
$\Rightarrow I = \dfrac{2}{5}m{\left( {\dfrac{R}{2}} \right)^2}$
There are eight small spheres, so we written the equation of the moment of the inertia as,
$\Rightarrow I = \dfrac{2}{5} \times \dfrac{m}{8}{\left( {\dfrac{R}{2}} \right)^2}$
Simplify the above equation, we get
$\Rightarrow I = \dfrac{{64}}{{32}}$
Perform the divide operations in the above equation, we get
$\Rightarrow I = 2\,Kg\,{m^2}$
Therefore, the moment of the inertia of the small diameter is $2\,Kg\,{m^2}$.
Hence from the above options, option (D) is correct.
Note:
In the question, the sphere is used to find the moment of inertia. Spheres are divided into identical spheres which means it splits into equal spheres. But, So, diameter of the sphere is about the recast of the spheres. Consider those values, we get the moment of inertia of the sphere.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Which gland is known as mixed gland class 11 biology CBSE

