The work done by a force of  (2xi^+3j^)N in displacing the particle from (1,0,2) m to (2,y,2) m is zero. The value of y is

Work done=∫F→.dr →   here F=force=iFx+jFy+kFz; dr=displacement=idx+jdy+kdzW=∫Fx  dx+∫Fydyby question work done=0, substitute the given equation0=∫122x  dx  + ∫0y3  dy0=2x2212+3(y−0)0=22[22-12]+3y0=3+3y⇒y=−1