02. Equilibrium of Concurrent Force
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Laws of Motion

146002 The resultant of two forces acting at an angle of \(120^{\circ}\) is \(10 \mathrm{~kg}\)-wt and is perpendicular to one of the forces. That force is:

1 \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
2 \(20 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
3 \(10 \mathrm{~kg}-\mathrm{wt}\)
4 \(\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt}\)
Laws of Motion

146003 A steel wire can withstand a load up to \(2940 \mathrm{~N}\). A load of \(150 \mathrm{~kg}\) is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(80^{\circ}\)
4 \(85^{\circ}\)
Laws of Motion

146005 Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle
between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is

1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are

1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Laws of Motion

146002 The resultant of two forces acting at an angle of \(120^{\circ}\) is \(10 \mathrm{~kg}\)-wt and is perpendicular to one of the forces. That force is:

1 \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
2 \(20 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
3 \(10 \mathrm{~kg}-\mathrm{wt}\)
4 \(\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt}\)
Laws of Motion

146003 A steel wire can withstand a load up to \(2940 \mathrm{~N}\). A load of \(150 \mathrm{~kg}\) is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(80^{\circ}\)
4 \(85^{\circ}\)
Laws of Motion

146005 Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle
between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is

1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are

1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
Laws of Motion

146002 The resultant of two forces acting at an angle of \(120^{\circ}\) is \(10 \mathrm{~kg}\)-wt and is perpendicular to one of the forces. That force is:

1 \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
2 \(20 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
3 \(10 \mathrm{~kg}-\mathrm{wt}\)
4 \(\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt}\)
Laws of Motion

146003 A steel wire can withstand a load up to \(2940 \mathrm{~N}\). A load of \(150 \mathrm{~kg}\) is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(80^{\circ}\)
4 \(85^{\circ}\)
Laws of Motion

146005 Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle
between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is

1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are

1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Laws of Motion

146002 The resultant of two forces acting at an angle of \(120^{\circ}\) is \(10 \mathrm{~kg}\)-wt and is perpendicular to one of the forces. That force is:

1 \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
2 \(20 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\)
3 \(10 \mathrm{~kg}-\mathrm{wt}\)
4 \(\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt}\)
Laws of Motion

146003 A steel wire can withstand a load up to \(2940 \mathrm{~N}\). A load of \(150 \mathrm{~kg}\) is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(80^{\circ}\)
4 \(85^{\circ}\)
Laws of Motion

146005 Forces of \(5 \mathrm{~N}, 12 \mathrm{~N}\) and \(13 \mathrm{~N}\) are in equilibrium. If \(\sin 23^{\circ}=\frac{5}{13}\), then the angle
between \(5 \mathrm{~N}\) and \(13 \mathrm{~N}\) forces is

1 \(23^{\circ}\)
2 \(67^{\circ}\)
3 \(90^{\circ}\)
4 \(113^{\circ}\)
Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is \(16 \mathrm{~N}\). If their resultant is normal to smaller force, and has a magnitude \(8 \mathrm{~N}\), then forces are

1 \(6 \mathrm{~N}, 10 \mathrm{~N}\)
2 \(8 \mathrm{~N}, 8 \mathrm{~N}\)
3 \(4 \mathrm{~N}, 12 \mathrm{~N}\)
4 \(2 \mathrm{~N}, 14 \mathrm{~N}\)