Tension in the string when it makes angle $ \theta $ with the vertical, $ T=\frac{m{{v}^{2}}}{r}+mg\cos \theta $ When the string is horizontal, $ \theta ={{90}^{o}} $ Here, m = 0.6 kg, r = 0.75 m, v = 5 m/ s Hence, $ T=\frac{m{{v}^{2}}}{r}+mg\times 0=\frac{m{{v}^{2}}}{r} $ $ \therefore $ $ T=\frac{0.6\times {{(5)}^{2}}}{0.75}=20\,N $