let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:Racquet force is twice of Player force
Explanation:
Given
ball arrives at a speed of 
ball returned with speed of 
average Force imparted by racquet on the ball is given by

where 
time of contact of ball with racquet


When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet


From 1 and 2 we get

Hence the magnitude of Force by racquet is twice the Force by player
It looks blue as it is only reflecting blue light