Physics - NLM Question with Solution | TestHub

The sphere touches the smooth wall, so the wall's reaction N is horizontal. The string runs from the wall anchor to the sphere's surface along the line through the centre, so the anchor-to-centre distance is

$$\mathrm{AO}=8+5=13\mathrm{cm}$$

The centre is 5 cm from the wall (radius), giving a 5-12-13 triangle. If $\theta$ is the angle the string makes with the wall:

$$\cos \theta =\frac{12}{13}$$

Vertical equilibrium of the sphere:

$$\mathrm{T}\cos \theta =\mathrm{W}\Rightarrow \mathrm{T}=\frac{13}{12}\mathrm{W}$$

Correct option: (1).