A mirror of weight 75N is hung by a cord from a hook on the wall. If the cord makes an angle of 30o with horizontal, the tension in the cord is

When an object is hung by a cord, the cord exerts a force on the object known as tension. In this case, the mirror is hung by a cord from a hook on the wall, and the cord makes an angle of 30 degrees with the horizontal. The weight of the mirror acts downwards, and the tension in the cord acts upwards to balance the weight. To find the tension in the cord, we need to use trigonometry. We can resolve the weight of the mirror into its horizontal and vertical components. The horizontal component is given by Wcosθ, where W is the weight of the mirror and θ is the angle between the cord and the horizontal. The vertical component is given by Wsinθ. In this case, the weight of the mirror is 75N, and the angle θ is 30 degrees. Therefore, the horizontal component is: Wcosθ = 75N x cos 30 = 64.95N The vertical component is: Wsinθ = 75N x sin 30 = 37.5N Since the mirror is in equilibrium, the tension in the cord must be equal to the vertical component of the weight: Tension = Wsinθ = 37.5N Therefore, the tension in the cord is 37.5N. The correct option is: 150N.