Two unlike parallel forces of 2 N and 16 N act at the ends of a uniform rod of 21 cm length. The point where the resultant of these two forces acts at a distance of .............. from the greater force.

Question

The gravitational force between two masses kept at a certain distance is \( P \) Newton. The same two masses are now kept in water and the distance between them is the same. The gravitational force between these two masses in water is \( Q \) Newton. Then:

Answer 1

Given forces \( F_1 = 2 \, \text{N} \) and \( F_2 = 16 \, \text{N} \), and a rod length of 21 cm, the point where the resultant force acts can be determined using: \[\nx = \frac{F_2 \times d_2}{F_1 + F_2}\n\] Where: - \( F_1 = 2 \, \text{N} \), \( F_2 = 16 \, \text{N} \) - \( d_2 = 21 \, \text{cm} \) Substituting the values: \[\nx = \frac{16 \times 21}{2 + 16} = \frac{336}{18} = 18.67 \, \text{cm}\n\] The distance from the greater force (16 N) is \( 18.67 \, \text{cm} \). The correct answer is 4 cm.

Two unlike parallel forces of 2 N and 16 N act at the ends of a uniform rod of 21 cm length. The point where the resultant of these two forces acts at a distance of .............. from the greater force.

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The Correct Option is A

Solution and Explanation

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