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An iron sphere of mass 100kg is suspended freely from rigid support means of a rope of length 2m. The horizontal force required to displace it horizontally through 50cm is nearly –A) 980NB) 490NC) 250ND) 112.5N

Last updated date: 18th Jun 2024
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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.

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.