The work done on a particle of mass m by a force K[x(x2+y2)3/2^i+y(x2+y2)3/2^j] (K being the constant of appropriate dimensions, when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is:
The work done on a particle of mass m by a force K[x(x2+y2)3/2^i+y(x2+y2)3/2^j] (K being the constant of appropriate dimensions, when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is:
The correct option is D 0
Given that,
Force F=K⎡⎢
⎢
⎢
⎢⎣x(x2+y2)32^i+y(x2+y2)32j⎤⎥
⎥
⎥
⎥⎦
Now, for the small distance is dr travelled by the particle in the direction
d→r=d→x^i+d→y^j
Now, small work done is
dw=F.d→r
The total work done
W=∫dw
W=∫F⋅d→r
W=∫K⎡⎢
⎢
⎢
⎢⎣x(x2+y2)32^i+y(x2+y2)32j⎤⎥
⎥
⎥
⎥⎦⋅[dx^i+dy^j]
W=K0∫a⎡⎢
⎢
⎢
⎢⎣x(x2+y2)32^i⎤⎥
⎥
⎥
⎥⎦+a∫0⎡⎢
⎢
⎢
⎢⎣y(x2+y2)32j⎤⎥
⎥
⎥
⎥⎦
Now, solve the first term of integration
Put,
x2+y2=t
2xdx=dt
Now,
=K20∫a2xdx(x2+y2)32
=K20∫adt(t)32
=K2×2a12
=Ka12
Now, similarly for second term of integration
=−Ka12
Now, work done is
W=Ka12−Ka12
W=0
Hence, the work done on a particle is 0
Given that,
Force F=K⎡⎢ ⎢ ⎢ ⎢⎣x(x2+y2)32^i+y(x2+y2)32j⎤⎥ ⎥ ⎥ ⎥⎦
Now, for the small distance is dr travelled by the particle in the direction
d→r=d→x^i+d→y^j
Now, small work done is
dw=F.d→r
The total work done
W=∫dw
W=∫F⋅d→r
W=∫K⎡⎢ ⎢ ⎢ ⎢⎣x(x2+y2)32^i+y(x2+y2)32j⎤⎥ ⎥ ⎥ ⎥⎦⋅[dx^i+dy^j]
W=K0∫a⎡⎢ ⎢ ⎢ ⎢⎣x(x2+y2)32^i⎤⎥ ⎥ ⎥ ⎥⎦+a∫0⎡⎢ ⎢ ⎢ ⎢⎣y(x2+y2)32j⎤⎥ ⎥ ⎥ ⎥⎦
Now, solve the first term of integration
Put,
x2+y2=t
2xdx=dt
Now,
=K20∫a2xdx(x2+y2)32
=K20∫adt(t)32
=K2×2a12
=Ka12
Now, similarly for second term of integration
=−Ka12
Now, work done is
W=Ka12−Ka12
W=0
Hence, the work done on a particle is 0
