The consequence of special theory of relativity gives the following relation of an object moving with aspeed ‘v’ approaching to the speed of light ‘c’, relativistic mass ‘m’ and rest mass ‘mo’.2122ocv1mm By using the given dimensions, show that ‘m’ in the given expression approaches infinity (m ∞)

The equation you've provided seems to be a bit garbled, but I believe you're referring to the equation for relativistic mass in Einstein's theory of special relativity. The correct equation is:

m = m0 / sqrt(1 - v^2/c^2)

where:

• m is the relativistic mass,
• m0 is the rest mass,
• v is the velocity of the object, and
• c is the speed of light.

The denominator of this equation, sqrt(1 - v^2/c^2), is known as the Lorentz factor. As the velocity v approaches the speed of light c, the value of v^2/c^2 approaches 1. This means that 1 - v^2/c^2 approaches 0, and therefore the Lorentz factor, being the square root of this quantity, also approaches 0.

Since the relativistic mass m is the rest mass m0 divided by the Lorentz factor, as the Lorentz factor approaches 0, the relativistic mass m approaches infinity. This is a mathematical way of expressing the physical principle that an object's mass increases without bound as its speed approaches the speed of light, which is a key result of special relativity.