Exams › NEET › Physics

The horizontal component of a force of 10 N inclined at \(30^\circ\) to the vertical is

1. 3 N
2. \(5\sqrt{3}\) N
3. 5 N
4. \(10\sqrt{3}\) N

Correct answer: 5 N

Solution

If a force makes \(30^\circ\) with the vertical, it makes \(60^\circ\) with the horizontal. The horizontal component is \(10\cos 60^\circ = 5\text{ N}\).

Related NEET Physics questions

• Set the angles made by following vectors with x -axis in increasing order:
• Consider the following two statements about a physical scenario. Statement 1 and Statement 2 are both given. Which option correctly describes their relationship?
• Two forces of magnitudes 5 N and 4 N act on a particle. Which of the following could be a possible magnitude of their resultant?
• Two forces act at a point such that the sum of their magnitudes is 16 N. Their resultant is perpendicular to the smaller force and has magnitude 8 N. What is the magnitude of the smaller force?
• Forces of magnitude \(3P\) and \(2P\) act at a point and their resultant is \(R\). If the force \(3P\) is doubled to \(6P\) while \(2P\) remains unchanged, the new resultant becomes \(2R\). What is the angle between the original two forces?
• A force of 6 kg-wt and another of 8 kg-wt are applied together at a point. Which of the following could be the magnitude of their combined resultant?

⚔️ Practice NEET Physics free + battle 1v1 →