At a given point of time the value of displacement of a simple harmonic oscillator is given as y = A cos (30 ). If amplitude is 40 cm and kinetic
energy at that time is 200 J, the value of force constant is
x  1.0 10^x Nm . The value of x is
                                                                     JEE-2024   
Solution:                
At a given point of time the value of displacement of a simple harmonic oscillator is given as y = A cos (30 ). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is x 1.0 10^x Nm . The value of x is JEE-2024

Bhautik Study is a branch of knowledge that offers comprehensive solutions to a wide range of questions, aiding in preparation for exams such as JEE, NEET, CUET and NDA. 

The magnetic field at the centre of a wire loop formed by two semi circular wires of radii R1 = 2 πm and R2 = 4π m carrying current I = 4A as per figure  given below is  a × 10^–7 T. The value of a is  ______. (Centre O is common for all segments)           

                                                                      JEE-2024