Dear Prachi

Gravitational field at a distance r from the infinitely long rod with liner mass density d = 2Gd/r

here r=L and force will be field x mass

force= 2GMd/L

All the best.                                                           

AKASH GOYAL

AskiitiansExpert-IIT Delhi

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively :Click here to download the toolbar..

Gravitational Force is -Gm1m2 / r^2

since r is the radius and it is the prependicular distance between the rod and the point of mass M

the mass of the rod is d*V , since a rod is cylindrical the volume is pi*r*r*h and h = infinity

so mass of rod = d*infinity = infinity

Gravitational Force =  - G * infinity * M / L^2

                           =  -  infinity / l^2

since infinity divided by any positive number is also infinity

                           =  - infinity !

therefore gravitational force is - infinity

please approve if the answer is correct