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JEE Main 2015 (11 Apr, Shift 1) · Continuous Bodies: Standard Results & Variable Density

Continuous Bodies: Standard Results & Variable Density question from JEE Main 2015 (11 Apr, Shift 1)

Q8JEE Main 2015 · 11 APR · SHIFT 1Medium

Continuous Bodies: Standard Results & Variable Density

A uniform thin rod AB of length L has linear mass density μ\mu (x) = a + bxL,\frac{bx}{L}, where x is measured from A. If the CM of the rod lies at a distance of (712L)\left(\frac{7}{12}L\right) from A, then a and b are related as:

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Answer: (c)

  1. M=aL+bL2M = aL + \frac{bL}{2}, ∫0Lxμ dx=aL22+bL23\int_0^L x\mu\,dx = \frac{aL^2}{2} + \frac{bL^2}{3}
  2. xcm=L(3a+2b)3(2a+b)=7L12x_{cm} = \frac{L(3a + 2b)}{3(2a + b)} = \frac{7L}{12}
  3. 12(3a+2b)=21(2a+b)⇒3b=6a⇒b=2a12(3a + 2b) = 21(2a + b) \Rightarrow 3b = 6a \Rightarrow b = 2a
  4. So the answer is (c) 2a=b2a = b.
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