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JEE Main 2019 (12 Jan, Shift 1) · Composite & Cut-Out Bodies

Composite & Cut-Out Bodies question from JEE Main 2019 (12 Jan, Shift 1)

Q10JEE Main 2019 · 12 JAN · SHIFT 1Medium

Composite & Cut-Out Bodies

The position vector of the centre of massr⃗cm\vec{r}_{cm} of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

Figure

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Answer: (c)

2L: (L, L)L: (2L, L/2)L: (2.5L, 0)LO
  1. The bar has three straight parts (the left vertical line is the y-axis, not the bar); mass ∝\propto length.
  2. Top part: length 2L2L, centre (L, L)(L,\ L). Vertical part: length LL, centre (2L, L/2)(2L,\ L/2). Bottom part: length LL, centre (2.5L, 0)(2.5L,\ 0).
  3. Total length 4L4L: xcm=2L(L)+L(2L)+L(2.5L)4L=138Lx_{cm} = \frac{2L(L)+L(2L)+L(2.5L)}{4L} = \frac{13}{8}L
  4. ycm=2L(L)+L(L/2)+L(0)4L=58Ly_{cm} = \frac{2L(L)+L(L/2)+L(0)}{4L} = \frac{5}{8}L
  5. So the answer is (c) r⃗cm=138L x^+58L y^\vec{r}_{cm} = \frac{13}{8}L\,\hat{x} + \frac{5}{8}L\,\hat{y}.
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