Two masses A and B having mass m_a and m_b, respectively, lying in the plane of the figure shown, are rigidly attached to a shaft which revolves about an axis through O perpendicular to the plane of the figure. The radii of rotation of the masses m_a and m_b are r_a and r_b, respectively. The angle between lines OA and OB is 90°. If m_a = 10 kg, m_b = 20 kg, r_a =200 mm and r_b, = 400 mm, then the balance mass to be placed at a radius of 200 mm is----------  kg (round off to two decimal places).

The correct answer is 41.23.

1. Understand the System Configuration:
We are given two masses, $m_a$ and $m_b$, attached to a rotating shaft. The axis of rotation passes through $O$ and is perpendicular to the plane containing the lines of action $OA$ and $OB$ shown in the image. The angle between $OA$ and $OB$ is $90°$.
Let us choose a coordinate system where the line $OA$ lies along the positive x-axis and the line $OB$ lies along the positive y-axis.

2. Given Data:
Mass of A, $m_a=10\text{kg}$
Mass of B, $m_b=20\text{kg}$
Radius of rotation of A, $r_a=200\text{mm}$
Radius of rotation of B, $r_b=400\text{mm}$
Radius of rotation of the balance mass, $r=200\text{mm}$

3. Calculate Centrifugal Force Components:
For a shaft rotating with angular velocity $\omega$, the centrifugal force on any mass $m_i$ at a radius $r_i$ is proportional to the product $m_i\cdot r_i$.
Let us compute the mass-radius products for both masses A and B:
For mass A:
$m_a\cdot r_a=10\text{kg}\cdot 200\text{mm}=2000\text{kg}\cdot \text{mm}$
For mass B:
$m_b\cdot r_b=20\text{kg}\cdot 400\text{mm}=8000\text{kg}\cdot \text{mm}$

4. Determine the Resultant Mass-Radius Product:
Since the two lines of action are perpendicular ($90°$), the resultant mass-radius product ${(m\cdot r)}_{\text{resultant}}$ is found using the Pythagorean theorem:
${(m\cdot r)}_{\text{resultant}}=\sqrt{{(m_a\cdot r_a)}^2+{(m_b\cdot r_b)}^2}$
Substituting the calculated values:
${(m\cdot r)}_{\text{resultant}}=\sqrt{{2000}^2+{8000}^2}$
${(m\cdot r)}_{\text{resultant}}=\sqrt{4000000+64000000}=\sqrt{68000000}\approx 8246.21\text{kg}\cdot \text{mm}$

5. Find the Balance Mass:
To balance the system dynamically, the mass-radius product of the balancing mass $m$ placed at radius $r=200\text{mm}$ must be equal to the resultant mass-radius product:
$m\cdot r={(m\cdot r)}_{\text{resultant}}$
$m\cdot 200=8246.21$
$m=\frac{8246.21}{200}\approx 41.23\text{kg}$

Thus, the required balance mass to be placed at a radius of 200 mm is approximately 41.23 kg.