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# Solve the following :

Question:

Particles of masses $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}, \ldots . ., 100 \mathrm{~g}$ are kept at the marks $1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, \ldots \ldots$, $100 \mathrm{~cm}$ respectively on a metrescale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Solution:

The perpendicular bisector passes from $50 \mathrm{~cm}$ mark. So, there will be 49 particles on LHS and 50 particles on RHS.

$\mathrm{I}=\left[49(1)^{2}+51(1)^{2}\right]+\left[48(2)^{2}+52(2)^{2}\right] \ldots \ldots \ldots+\left[1(49)^{2}+99(49)^{2}\right]+100(50)^{2}$

$=(100)\left[1^{2}+2^{2}+\ldots \ldots \ldots+50^{2}\right]$

$=(100)\left[\frac{50 \times(50+1)(2 \times 50+1)}{6}\right] \mathrm{gm}-\mathrm{cm}^{2}$

$\mid \approx 0.43 \mathrm{~kg} / \mathrm{cm}^{2}$

$[\because$ Sum of square of $n$ natural numbers,

$\left.\Delta=\frac{n(n+1)(2 n+1)}{6}\right]$