A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 1 kg-m2 and its rate of rotation 2 rev/sec. The angular momentum of the body in kg-m2/sec is A. 1.257 B. 12.57 C. 13.57 D. 1.357

1.257
12.57
13.57
1.357

The correct answer is $\boxed{\text{A}}$.

The angular momentum of a body is given by the equation:

$$L = I\omega$$

where $I$ is the moment of inertia of the body and $\omega$ is the angular velocity of the body.

In this case, we are given that the moment of inertia of the body is $I = 1\text{ kg m}^2$ and the angular velocity of the body is $\omega = 2\text{ rev/s}$. We can convert the angular velocity to radians per second by multiplying it by $2\pi$, so $\omega = 2\pi\cdot 2\text{ rev/s} = 4\pi\text{ rad/s}$.

Substituting these values into the equation for angular momentum, we get:

$$L = I\omega = (1\text{ kg m}^2)(4\pi\text{ rad/s}) = 4\pi\text{ kg m}^2\text{ rad/s}$$

This is equal to $\boxed{1.257\text{ kg m}^2\text{ rad/s}}$ in SI units.

Each of the other options is incorrect because it does not represent the correct value of the angular momentum of the body.

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