Answer
427.8k+ views
Hint: We have to analyse the different forces acting on the sphere. The iron sphere experiences a vertical force due to its own mass. The tension on the rope is contributed by this mass. At an angle, we will have to consider the horizontal component of tension on the rope.
Complete answer:
Let us consider the given situation. An iron sphere of mass m is hanging on a rope from a fixed support. Now, as we can see the adjoining figures, the sphere is displaced 50cm from its equilibrium position with a horizontal force. Also, we can see all the components of tension and other forces acting on the spherical body.
From the above diagram, we can conclude the following relations –
\[\begin{align}
& T\cos \theta =mg\text{ --(1)} \\
& T\sin \theta =ma=F\text{ --(2)} \\
\end{align}\]
Where, ma is the horizontal force applied to keep the body at the point C. From the equations (1) and (2), we get,
\[\begin{align}
& \tan \theta =\dfrac{a}{g} \\
& or \\
& a=g\tan \theta \\
\end{align}\]
Now we can consider \[\Delta ABC,\]
\[\begin{align}
& \tan \theta =\dfrac{BC}{AB} \\
& \Rightarrow \tan \theta =\dfrac{0.5}{2} \\
& \Rightarrow \tan \theta =\dfrac{1}{4} \\
\end{align}\]
Also, we have \[a=g\tan \theta \],
\[\begin{align}
& \Rightarrow a=g\tan \theta \\
& \Rightarrow a=\dfrac{g}{4} \\
\end{align}\]
Now. We can compute the required solution for the horizontal force required to balance the iron sphere at the position C by substituting the value of ‘a’ in the equation of ‘F’ as –
\[\begin{align}
& F=ma \\
& given, \\
& m=100kg\text{ and g}=10m{{s}^{-2}} \\
& \therefore F=100kg\times \dfrac{10}{4} \\
& \Rightarrow \text{ }F=250N \\
\end{align}\]
So, the horizontal force required is 250N.
The correct answer is option C.
Additional Information:
The tension on the string is due to the mass hanging on it and the mass of the string itself.
Note:
The horizontal force acting on the system is totally dependent on the angle at which the mass is kept given that the mass doesn’t vary. For a system which has the same ratio of the tangent of the angle will require the same force regardless of the lengths, given that the rope is considered massless.
Complete answer:
Let us consider the given situation. An iron sphere of mass m is hanging on a rope from a fixed support. Now, as we can see the adjoining figures, the sphere is displaced 50cm from its equilibrium position with a horizontal force. Also, we can see all the components of tension and other forces acting on the spherical body.
![seo images](https://www.vedantu.com/question-sets/f8a727fa-83ea-4dc5-bf18-9b390837098c6291006011841504028.png)
From the above diagram, we can conclude the following relations –
\[\begin{align}
& T\cos \theta =mg\text{ --(1)} \\
& T\sin \theta =ma=F\text{ --(2)} \\
\end{align}\]
Where, ma is the horizontal force applied to keep the body at the point C. From the equations (1) and (2), we get,
\[\begin{align}
& \tan \theta =\dfrac{a}{g} \\
& or \\
& a=g\tan \theta \\
\end{align}\]
Now we can consider \[\Delta ABC,\]
\[\begin{align}
& \tan \theta =\dfrac{BC}{AB} \\
& \Rightarrow \tan \theta =\dfrac{0.5}{2} \\
& \Rightarrow \tan \theta =\dfrac{1}{4} \\
\end{align}\]
Also, we have \[a=g\tan \theta \],
\[\begin{align}
& \Rightarrow a=g\tan \theta \\
& \Rightarrow a=\dfrac{g}{4} \\
\end{align}\]
Now. We can compute the required solution for the horizontal force required to balance the iron sphere at the position C by substituting the value of ‘a’ in the equation of ‘F’ as –
\[\begin{align}
& F=ma \\
& given, \\
& m=100kg\text{ and g}=10m{{s}^{-2}} \\
& \therefore F=100kg\times \dfrac{10}{4} \\
& \Rightarrow \text{ }F=250N \\
\end{align}\]
So, the horizontal force required is 250N.
The correct answer is option C.
Additional Information:
The tension on the string is due to the mass hanging on it and the mass of the string itself.
Note:
The horizontal force acting on the system is totally dependent on the angle at which the mass is kept given that the mass doesn’t vary. For a system which has the same ratio of the tangent of the angle will require the same force regardless of the lengths, given that the rope is considered massless.
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