A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed when it starts pure rolling motion?
A) $\dfrac{{3u}}{5}$
B) $\dfrac{{2u}}{5}$
C) $\dfrac{{5u}}{7}$
D) $\dfrac{{2u}}{7}$

Hint: In the problem, we are supposed to find the speed, when it starts pure rolling motion. We know that the relation of angular momentum with inertia is $mur = {I_c}\omega $ where, r is the radius of sphere, $\omega $ is the angular velocity when sphere starts pure rolling motion, and ${I_c}$ is the moment of inertia and the moment of inertia of a solid sphere about the tangent. By equating the value of $\omega $ we will get an answer.

Complete step by step solution:
Step 1:
Before solving the question, we have to see some definition which we are going to use in question.
The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.
Moment of inertia: Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
We are given that u is the initial velocity of the solid sphere.
Now, By the conservation of angular momentum and relation of the inertia we can write $mur = {I_c}\omega$ …..(1)
Where r is the radius of the sphere, $\omega $ is the angular momentum when the sphere starts pure rolling motion, and ${I_c}$ is the moment of inertia.
Normally, the moment of inertia of a solid sphere about its natural axis passing through the center is –
$I = \dfrac{2}{5}m{r^2}$
Here, we require the moment of inertia of the sphere with respect to the tangent.
To find the moment of inertia about tangent, we have to apply the parallel axis theorem which states that the moment of inertia about any parallel axis is equal to the sum of moment of inertia about the natural axis and the product of mass and square of the distance between the axes.
Applying the parallel axis theorem, we get:
${I_c} = I + m{r^2}$
${I_c} = \dfrac{2}{5}m{r^2} + m{r^2} = \dfrac{{2 + 5}}{5}m{r^2} = \dfrac{7}{5}m{r^2}$
The moment of inertia of a solid sphere about a tangent is $\dfrac{7}{5}m{r^2}$ …… (2)
When a solid sphere starts rolling motion then the speed is, v=$\omega r$ or we can say $\omega = \dfrac{v}{r}$ …… (3)
Putting equation (3) in (1) we get, $mur = \dfrac{7}{5}m{r^2} \times \dfrac{v}{r}$
Cancelling the terms, we get, $u$ = $\dfrac{7}{5}v$
From here we will get its speed when it starts pure rolling motion, $v = \dfrac{{5u}}{7}m{s^{ - 1}}$

Hence option C is correct.

Note: Important step: There are several forms of equation for angular momentum. Still, we need the formula in terms of the moment of inertia since the rolling motion is involved. Also, one should know that there is a different moment of inertia for different bodies and we are required here about the tangent.