What is the value of ' x ' in the following equation?
​$$11^2+1.1^2+0.11^2+0.011^2=x$$​​

Correct option is B

Given:
Equation: $$11^2 + 1.1^2 + 0.11^2 + 0.011^2 = x$$​​
Solution:
Calculate the square of each term:
​$$11^2 = 121$$​​
​$$1.1^2 = 1.21$$​​
​$$0.11^2 = 0.0121$$​​
​$$0.011^2 = 0.000121$$​​
Add all the calculated values:
x = 121 + 1.21 + 0.0121 + 0.000121
x = 122.222221
Final Answer
So the correct answer is (b)