Particles of masses 1g, 2g, 3g ….100g are kept at the marks 1cm, 2cm, 3cm …., 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Answer

B, C, D

Explanation

Masses of 1gm, 2gm …………100 gm are kept at themarks 1cm, 2cm…………..100 cm on the xis resectively. A perpendicular axis is passed at the 50th particle.
Therefore on the L.H.S. of the axis there will be 49 particles and on the R.H.S there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment of inertia due to these two particles will be:
`=49xx(1)^2+51xx(1)^2 `
`=100xx1=100gm-cm^2`
similarly if we consider 48th and 52nd term we will get `100xx(2)^2gm-cm^2`.
Threfore we wil get 49 such set and one lone particle at 100 cm. Threfoee totl moment of inertia
`=100(1^2+2^2+3^2+........+49^2)+100(50)^2`
=100(1^2+2^2+3^2+..........+50^2)`
`=100xx(50xx51xx101)/6`
`=100x25xx17xx101`
`=100xx101xx425`
`=4292500gm-cm^2`
`=429kg-m^2=0.43kg-m^2`