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JEE Main 2021 (31 Aug, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2021 (31 Aug, Shift 2)

Q90JEE Main 2021 · 31 AUG · SHIFT 2Easy

Acceleration Due to Gravity

If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is: (Given: r < RE)

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Answer: (d)

  1. gdepth=g(1−x)g_{\text{depth}} = g\left(1 - x\right) and gheight=g(1+x)2g_{\text{height}} = \frac{g}{(1+x)^2}, with x=rREx = \frac{r}{R_E}.
  2. Ratio =(1−x)(1+x)2=(1−x)(1+2x+x2)= (1 - x)(1 + x)^2 = (1 - x)(1 + 2x + x^2)
  3. =1+x−x2−x3= 1 + x - x^2 - x^3
  4. So the answer is (d) 1+rRE−r2RE2−r3RE31 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}.
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