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The weight of a body on the suface of th...

The weight of a body on the suface of the earth is 100 N. The same at a height equal to a radius of the earth will be

A

25 N

B

40 N

C

75 N

D

100 N

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The correct Answer is:
To solve the problem step by step, we will use the concept of gravitational force and how it changes with height. ### Step 1: Understand the weight of the body on the surface of the Earth The weight \( W \) of a body on the surface of the Earth is given as 100 N. The weight is defined as: \[ W = m \cdot g \] where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface of the Earth, which is approximately \( 9.8 \, \text{m/s}^2 \). ### Step 2: Calculate the mass of the body From the equation \( W = m \cdot g \), we can rearrange it to find the mass \( m \): \[ m = \frac{W}{g} = \frac{100 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 10.20 \, \text{kg} \] ### Step 3: Determine the gravitational acceleration at height equal to the radius of the Earth When the body is taken to a height \( h \) equal to the radius of the Earth \( R \), the new gravitational acceleration \( g_h \) can be calculated using the formula: \[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \] Since \( h = R \), we have: \[ g_h = \frac{g}{(1 + 1)^2} = \frac{g}{4} = \frac{9.8 \, \text{m/s}^2}{4} \approx 2.45 \, \text{m/s}^2 \] ### Step 4: Calculate the weight of the body at the new height Now, we can find the weight \( W_h \) of the body at this height: \[ W_h = m \cdot g_h = 10.20 \, \text{kg} \cdot 2.45 \, \text{m/s}^2 \approx 25 \, \text{N} \] ### Conclusion The weight of the body at a height equal to the radius of the Earth is approximately **25 N**. ---

To solve the problem step by step, we will use the concept of gravitational force and how it changes with height. ### Step 1: Understand the weight of the body on the surface of the Earth The weight \( W \) of a body on the surface of the Earth is given as 100 N. The weight is defined as: \[ W = m \cdot g \] where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface of the Earth, which is approximately \( 9.8 \, \text{m/s}^2 \). ...
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