## Unit – II E |

## Problems on Variation in Acceleration Due to Gravity:

### Problems

- A mass of 5 kg is weighed on a balance at the top of a tower 20 m high. The mass is then suspended from the pan of the balance by a fine wire 20 m long and weighed. Find the change in the weight of a body in mgf assuming the radius of the earth as 6330 km.
- At what height will a man’s weight become half his weight on the surface of the earth. Take radius of the earth as R.
- A meteor is falling. How much gravitational acceleration it will experience when its height from the surface of the earth is equal to three times radius of the earth. Acceleration due to gravity on the surface of the earth is ‘g’.
- Find percentage decrease in the weight of a body when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.
- How much below the surface of the earth does acceleration due to gravity becomes 1 % of the value at the earth’s surface? Assume the radius of the earth as 6380 km.
- Compare the weight of 5 kg body 10 km above and 10 km below the surface of the earth. Given the radius of the earth = 6400 km.
- Compare the weight of body 0.5 km above and 1 km below the surface of the earth. Given the radius of the earth = 6400 km.
- The acceleration due to gravity at a height 1/20 th radius of the earth above the earth’s surface is 9 m/s². Find the value of acceleration due to gravity at an equal distance below the surface of the earth. Given the radius of the earth = 6400 km.
- At what depth below the surface of the earth, a man’s weight becomes half his weight on the surface of the earth. Take radius of the earth as R = 6400 km.
- Find decrease in value of acceleration due to gravity at a point 1600 km below the earth’s surface. R = 6400 km, g = 9.8 m/s²
- What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s².
- Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s².
- Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s²
- Find the difference in weight of a body of mass 100 kg on equator and pole. R = 6400 km, g = 9.8 m/s²
- Determine the linear speed and angular speed at which the earth should rotate so that the weight of a body on the equator would weigh as much as the present. Given R = 6400 km, g = 9.8 m/s²

### Answers

- 31.6 mgf
- 0.414R
- g/16
- 0.25
- 6316 km
- 0.998
- 1
- 9.5 m/s²
- 3200 km
- 2.45 m/s²
- 1 N
- 6.738 m/s²
- 976.6 N, 980 N, 977.5 N
- 3.386 N
- 5 km/s,7.826 10-2 rad/s

Good questions