A block of mass 1 kg connected to a spring of stiffness 10 N m^-1 is operating in a viscous medium such that the damping ratio (or damping factor) is equal to the ratio of the damped frequency to the natural frequency. The magnitude of the damping ratio for this system is ________ (rounded off to 2 decimal places).

The correct answer is 0.71.

Let us analyze the system to determine the damping ratio (denoted by ζ).

For a damped spring-mass system, let:
- m be the mass of the block, where m = 1 kg.
- k be the stiffness of the spring, where k = 10 N m^-1.
- ω_n be the natural frequency of the system.
- ω_d be the damped natural frequency.
- ζ be the damping ratio (or damping factor).

The natural frequency is given by the relation:
$${\omega }_n=\sqrt{\frac{k}{m}}$$

The damped frequency ω_d is related to the natural frequency ω_n by the formula:
$${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$$

According to the problem statement, the damping ratio (ζ) is equal to the ratio of the damped frequency to the natural frequency:
$$\zeta =\frac{{\omega }_d}{{\omega }_n}$$

Substituting the expression for ω_d into this condition:
$$\zeta =\frac{{\omega }_n\sqrt{1-{\zeta }^2}}{{\omega }_n}$$

This simplifies directly to:
$$\zeta =\sqrt{1-{\zeta }^2}$$

To solve for ζ, we square both sides of the equation:
$${\zeta }^2=1-{\zeta }^2$$

Rearranging the terms, we get:
$$2{\zeta }^2=1$$

Thus:
$${\zeta }^2=\frac{1}{2}=0.5$$

Taking the positive square root since the magnitude of the damping ratio is a positive quantity:
$$\zeta =\sqrt{0.5}\approx 0.7071$$

Rounding off to two decimal places, we get:
$$\zeta \approx 0.71$$

Therefore, the magnitude of the damping ratio for the system is 0.71.