
Xcm=M1∫0ιx.dM
dM=ρ.dx=(a+b(ıx)2).dx
xcm=∫dm∫xdM=∫ρdx∫xρdx=∫0ι(a+ι2bx2)dx∫0ιx(a+ι2bx2)dx
=a(x)0ι+ι2b(3x3)0ιa(2x2)0ι+ι2b(4x4)0ι
=aι+3bιa2ι2+b4ι2=(3a+b)(2a+b)4ι×3
=43ι(3a+b2a+b)
JEE Main 2020 — Physics Mechanics
A rod of length l has non-uniform linear mass density given by ρ(x)=a+b(lx)2, where a and b are constants and 0≤x≤l The value of x for the centre of mass of the rod is at:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
23(2a+ba+b)L
43(3a+b2a+b)L
34(2a+3ba+b)L
23(3a+b2a+b)L
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