Solution:

Given, the sphere is rolling on a frictionless surface with a translational velocity v metre per second and it climbs up to a height of h of a smooth inclined plane.

We need to find the value of v.

Using conservation of energy,

The initial energy of the system is equal to the final energy.

Initial energy = Kinetic energy + Potential energy

Final energy = Potential energy (at maximum height)

Initial energy = Final energy

0.5mv^2 = mgh

where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height climbed by the sphere.

Simplifying the above equation, we get

v = √(2gh)

Therefore, the value of v is √(2gh).

Explanation:

When the sphere is rolling on a frictionless surface, its initial energy is purely kinetic energy. As it starts rolling up the inclined plane, some of its kinetic energy gets converted into potential energy. At the maximum height, all the kinetic energy is converted into potential energy. Therefore, the final energy of the system is equal to the potential energy at the maximum height.

Using the law of conservation of energy, we equate the initial energy with the final energy. We get an expression for the velocity of the sphere in terms of the height climbed.

The velocity of the sphere is directly proportional to the square root of the height climbed. This means that the higher the sphere climbs, the greater is its velocity.

NEET Category:

This question belongs to the NEET category, which is the entrance exam for admission to medical and dental courses in India. The question tests the candidate's understanding of the conservation of energy and its application to the motion of a rolling sphere. It also tests their ability to solve mathematical expressions involving square roots.