Three objectsare at vertices of square PQRS of side 1km speed of these objects are 36km/hr 10m/s 0.8km/min initially they are at rest and start moving in clockwise direction simultaneously Distance between B and C?

To find the distance between points B and C, we first need to determine the location of points B and C in the square PQRS.

Let's label the vertices of the square as follows:
P - top left vertex
Q - top right vertex
R - bottom right vertex
S - bottom left vertex

We are given that the objects start at rest and move simultaneously in a clockwise direction. The velocities of the objects are:
Object A - 36 km/hr
Object B - 10 m/s
Object C - 0.8 km/min

Since they all start at rest, after a certain amount of time, they will reach points on the square based on their velocities.

To determine this, we will convert their velocities into the same units. Let's convert everything to m/s for consistency:
Object A: 36 km/hr = (36 * 1000) / (60 * 60) = 10 m/s
Object B: 10 m/s (no conversion needed)
Object C: 0.8 km/min = (0.8 * 1000) / (60) = 13.333... m/s

Now, let's consider the velocity of Object A, which is 10 m/s. Therefore, after 1 second, Object A will move 10 meters and will reach point Q. Point Q is the vertex closest to Object B.

Now, let's consider the velocity of Object B, which is 10 m/s. Since Object B has the same velocity as the distance covered by Object A in 1 second, we can conclude that after 1 second, Object B will also be at point Q.

Lastly, Object C has a velocity of 13.333... m/s. Since this velocity is greater than the distance covered by Object A and Object B, it will take a shorter time for Object C to reach point Q. In fact, Object C will reach point Q in (1 / 13.333...) seconds.

In summary, based on their velocities and the clock positions on the square:
- After 1 second, both Object A and Object B will be at point Q.
- Object C will also be at point Q but will take a shorter time.

Therefore, the distance between points B and C is 0 meters since they will both reach point Q at the same time.