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NEET PHYSICSEasy

A body weighing 100 N100 \text{ N} on the surface of the Earth weighs x kg m s−2x \text{ kg m s}^{-2} at a height 19RE\frac{1}{9}R_E above the surface of Earth. The value of xx is: (take g=10 m s−2g=10 \text{ m s}^{-2} at the surface of Earth and RER_E is the radius of Earth)

A

72

B

54

C

81

D

62

Step-by-Step Solution

  1. Formula: The acceleration due to gravity at a height hh above the surface of the Earth is given by g(h)=g(1+h/RE)2g(h) = \frac{g}{(1 + h/R_E)^2} .
  2. Weight Relationship: The weight of a body is W=mgW = mg. Therefore, the weight at height hh is Wh=mg(h)=mg(1+h/RE)2=Wsurface(1+h/RE)2W_h = m g(h) = \frac{mg}{(1 + h/R_E)^2} = \frac{W_{surface}}{(1 + h/R_E)^2}.
  3. Substitution: Given the weight on the surface Wsurface=100 NW_{surface} = 100 \text{ N} and height h=19REh = \frac{1}{9}R_E. Wh=100(1+1/9RERE)2=100(1+19)2W_h = \frac{100}{\left(1 + \frac{1/9 R_E}{R_E}\right)^2} = \frac{100}{\left(1 + \frac{1}{9}\right)^2}
  4. Calculation: Wh=100(109)2=10010081=100×81100=81W_h = \frac{100}{\left(\frac{10}{9}\right)^2} = \frac{100}{\frac{100}{81}} = 100 \times \frac{81}{100} = 81 The weight is 81 N81 \text{ N} (or 81 kg m s−281 \text{ kg m s}^{-2}). Thus, x=81x = 81.
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