363282
A body of weight \(200\,N\) is suspended from a tree branch through a chain of mass \(10\,kg\) . The branch pulls the chain by a force equal to (if \(g = 10\;m/{s^2}\) )
363283
Two identical heavy spheres of equal mass are placed on a smooth cup of radius 3 \(r\) where \(r\) is radius of each sphere. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is
1 \(1\)
2 \(2\)
3 \(3\)
4 \({\rm{None}}\)
Explanation:
Consider any one sphere Let \({N_1}\) is the force between the spheres and \({N_2}\) is the force exerted by the cup on the sphere. As the sphere is in equilibrium \({N_2}\sin 30^\circ = {N_1}\) \(\frac{{{N_2}}}{{{N_1}}} = 2\)
PHXI05:LAWS OF MOTION
363284
A weight \(w\) is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
1 less than \(w\)
2 equal to \(w\)
3 equal to \(2 w\)
4 infinitely large
Explanation:
For equilibrium of body,
\(w = 2T\cos \theta \) \( \Rightarrow \quad T = \frac{w}{{2\cos \theta }}\) For the string to be horizontal, \(\theta = 90^\circ \) \(\therefore \quad T = \frac{w}{{2\cos 90^\circ }} \Rightarrow T = \infty \)
PHXI05:LAWS OF MOTION
363285
A flexible chain of weight \(W\) hangs between two fixed points \(A\) and \(B\) which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is \(\theta \). What is the tension of the chain at the midpoint?
1 \(\frac{W}{2}.{\rm{cosec}}\,{\rm{\theta }}\)
2 \(\frac{W}{2}.\tan \,{\rm{\theta }}\)
3 \(\frac{W}{2}.\cot \,{\rm{\theta }}\)
4 \({\rm{None}}\)
Explanation:
Let \({T_1}\) be the tension at any of the two ends of the chain an \({T_2}\) be the tension at the \(d\) midpoint. \({T_1}\cos \theta = {T_2}\) \({T_1}\sin \theta = \frac{W}{2}\) \( \Rightarrow {T_2} = \frac{W}{2}\cot \theta \)
363282
A body of weight \(200\,N\) is suspended from a tree branch through a chain of mass \(10\,kg\) . The branch pulls the chain by a force equal to (if \(g = 10\;m/{s^2}\) )
363283
Two identical heavy spheres of equal mass are placed on a smooth cup of radius 3 \(r\) where \(r\) is radius of each sphere. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is
1 \(1\)
2 \(2\)
3 \(3\)
4 \({\rm{None}}\)
Explanation:
Consider any one sphere Let \({N_1}\) is the force between the spheres and \({N_2}\) is the force exerted by the cup on the sphere. As the sphere is in equilibrium \({N_2}\sin 30^\circ = {N_1}\) \(\frac{{{N_2}}}{{{N_1}}} = 2\)
PHXI05:LAWS OF MOTION
363284
A weight \(w\) is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
1 less than \(w\)
2 equal to \(w\)
3 equal to \(2 w\)
4 infinitely large
Explanation:
For equilibrium of body,
\(w = 2T\cos \theta \) \( \Rightarrow \quad T = \frac{w}{{2\cos \theta }}\) For the string to be horizontal, \(\theta = 90^\circ \) \(\therefore \quad T = \frac{w}{{2\cos 90^\circ }} \Rightarrow T = \infty \)
PHXI05:LAWS OF MOTION
363285
A flexible chain of weight \(W\) hangs between two fixed points \(A\) and \(B\) which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is \(\theta \). What is the tension of the chain at the midpoint?
1 \(\frac{W}{2}.{\rm{cosec}}\,{\rm{\theta }}\)
2 \(\frac{W}{2}.\tan \,{\rm{\theta }}\)
3 \(\frac{W}{2}.\cot \,{\rm{\theta }}\)
4 \({\rm{None}}\)
Explanation:
Let \({T_1}\) be the tension at any of the two ends of the chain an \({T_2}\) be the tension at the \(d\) midpoint. \({T_1}\cos \theta = {T_2}\) \({T_1}\sin \theta = \frac{W}{2}\) \( \Rightarrow {T_2} = \frac{W}{2}\cot \theta \)
363282
A body of weight \(200\,N\) is suspended from a tree branch through a chain of mass \(10\,kg\) . The branch pulls the chain by a force equal to (if \(g = 10\;m/{s^2}\) )
363283
Two identical heavy spheres of equal mass are placed on a smooth cup of radius 3 \(r\) where \(r\) is radius of each sphere. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is
1 \(1\)
2 \(2\)
3 \(3\)
4 \({\rm{None}}\)
Explanation:
Consider any one sphere Let \({N_1}\) is the force between the spheres and \({N_2}\) is the force exerted by the cup on the sphere. As the sphere is in equilibrium \({N_2}\sin 30^\circ = {N_1}\) \(\frac{{{N_2}}}{{{N_1}}} = 2\)
PHXI05:LAWS OF MOTION
363284
A weight \(w\) is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
1 less than \(w\)
2 equal to \(w\)
3 equal to \(2 w\)
4 infinitely large
Explanation:
For equilibrium of body,
\(w = 2T\cos \theta \) \( \Rightarrow \quad T = \frac{w}{{2\cos \theta }}\) For the string to be horizontal, \(\theta = 90^\circ \) \(\therefore \quad T = \frac{w}{{2\cos 90^\circ }} \Rightarrow T = \infty \)
PHXI05:LAWS OF MOTION
363285
A flexible chain of weight \(W\) hangs between two fixed points \(A\) and \(B\) which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is \(\theta \). What is the tension of the chain at the midpoint?
1 \(\frac{W}{2}.{\rm{cosec}}\,{\rm{\theta }}\)
2 \(\frac{W}{2}.\tan \,{\rm{\theta }}\)
3 \(\frac{W}{2}.\cot \,{\rm{\theta }}\)
4 \({\rm{None}}\)
Explanation:
Let \({T_1}\) be the tension at any of the two ends of the chain an \({T_2}\) be the tension at the \(d\) midpoint. \({T_1}\cos \theta = {T_2}\) \({T_1}\sin \theta = \frac{W}{2}\) \( \Rightarrow {T_2} = \frac{W}{2}\cot \theta \)
363282
A body of weight \(200\,N\) is suspended from a tree branch through a chain of mass \(10\,kg\) . The branch pulls the chain by a force equal to (if \(g = 10\;m/{s^2}\) )
363283
Two identical heavy spheres of equal mass are placed on a smooth cup of radius 3 \(r\) where \(r\) is radius of each sphere. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is
1 \(1\)
2 \(2\)
3 \(3\)
4 \({\rm{None}}\)
Explanation:
Consider any one sphere Let \({N_1}\) is the force between the spheres and \({N_2}\) is the force exerted by the cup on the sphere. As the sphere is in equilibrium \({N_2}\sin 30^\circ = {N_1}\) \(\frac{{{N_2}}}{{{N_1}}} = 2\)
PHXI05:LAWS OF MOTION
363284
A weight \(w\) is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
1 less than \(w\)
2 equal to \(w\)
3 equal to \(2 w\)
4 infinitely large
Explanation:
For equilibrium of body,
\(w = 2T\cos \theta \) \( \Rightarrow \quad T = \frac{w}{{2\cos \theta }}\) For the string to be horizontal, \(\theta = 90^\circ \) \(\therefore \quad T = \frac{w}{{2\cos 90^\circ }} \Rightarrow T = \infty \)
PHXI05:LAWS OF MOTION
363285
A flexible chain of weight \(W\) hangs between two fixed points \(A\) and \(B\) which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is \(\theta \). What is the tension of the chain at the midpoint?
1 \(\frac{W}{2}.{\rm{cosec}}\,{\rm{\theta }}\)
2 \(\frac{W}{2}.\tan \,{\rm{\theta }}\)
3 \(\frac{W}{2}.\cot \,{\rm{\theta }}\)
4 \({\rm{None}}\)
Explanation:
Let \({T_1}\) be the tension at any of the two ends of the chain an \({T_2}\) be the tension at the \(d\) midpoint. \({T_1}\cos \theta = {T_2}\) \({T_1}\sin \theta = \frac{W}{2}\) \( \Rightarrow {T_2} = \frac{W}{2}\cot \theta \)
363282
A body of weight \(200\,N\) is suspended from a tree branch through a chain of mass \(10\,kg\) . The branch pulls the chain by a force equal to (if \(g = 10\;m/{s^2}\) )
363283
Two identical heavy spheres of equal mass are placed on a smooth cup of radius 3 \(r\) where \(r\) is radius of each sphere. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is
1 \(1\)
2 \(2\)
3 \(3\)
4 \({\rm{None}}\)
Explanation:
Consider any one sphere Let \({N_1}\) is the force between the spheres and \({N_2}\) is the force exerted by the cup on the sphere. As the sphere is in equilibrium \({N_2}\sin 30^\circ = {N_1}\) \(\frac{{{N_2}}}{{{N_1}}} = 2\)
PHXI05:LAWS OF MOTION
363284
A weight \(w\) is suspended from the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
1 less than \(w\)
2 equal to \(w\)
3 equal to \(2 w\)
4 infinitely large
Explanation:
For equilibrium of body,
\(w = 2T\cos \theta \) \( \Rightarrow \quad T = \frac{w}{{2\cos \theta }}\) For the string to be horizontal, \(\theta = 90^\circ \) \(\therefore \quad T = \frac{w}{{2\cos 90^\circ }} \Rightarrow T = \infty \)
PHXI05:LAWS OF MOTION
363285
A flexible chain of weight \(W\) hangs between two fixed points \(A\) and \(B\) which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is \(\theta \). What is the tension of the chain at the midpoint?
1 \(\frac{W}{2}.{\rm{cosec}}\,{\rm{\theta }}\)
2 \(\frac{W}{2}.\tan \,{\rm{\theta }}\)
3 \(\frac{W}{2}.\cot \,{\rm{\theta }}\)
4 \({\rm{None}}\)
Explanation:
Let \({T_1}\) be the tension at any of the two ends of the chain an \({T_2}\) be the tension at the \(d\) midpoint. \({T_1}\cos \theta = {T_2}\) \({T_1}\sin \theta = \frac{W}{2}\) \( \Rightarrow {T_2} = \frac{W}{2}\cot \theta \)
