361323
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is \({0.07 {Nm}^{-1}}\), then the excess force required to take it away from the surface is
1 \(19.8\,mN\)
2 \(198\,N\)
3 \(1.98\,mN\)
4 \(99\,N\)
Explanation:
Surface tension force acts along the circumference of the disc. Excess force required to balance the surface tension force is \({F_{ext}} = T \times 2\pi R\) \({F_{ext}} = \frac{7}{{100}} \times 2 \times 3.14 \times \frac{{4.5}}{{100}}\) \( = 197.82 \times {10^{ - 4}} = 19.8\,mN\)
NEET - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361324
Maximum possible mass of a greased needle of length \(l\) floating on water surface of tension \(T\) is \(CTl.\) Find value of \(C.\) (Take \(g = 10\;m{\rm{/}}{s^2}\))
1 0.2
2 0.6
3 0.9
4 0.7
Explanation:
Let the mass of the needle be \(m\). As the liquid surface is distorted, the surface tension forces acing on both sides of the needle make an angle \(\theta\), say, with vertical as shown in the figure. Since the forces acting on the needle are \(F,F\) and \(mg\), resolving the forces vertically for its equilibrium, \(\sum {{F_y}} = F\cos \theta + F\cos \theta - mg = 0\) \( \Rightarrow m = \frac{{2\;F\cos \theta }}{g}\) where \(F = Tl\) \(\therefore m = \frac{{2\;Tl\cos \theta }}{g}\) For \(m\) to be maximum, \(\cos \theta=1\) \(\therefore {m_{\max }} = \frac{{2\;Tl}}{{\;g}} = \frac{{2Tl}}{{10}} = 0.2\;Tl\) \( \Rightarrow C = 0.2\)
361323
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is \({0.07 {Nm}^{-1}}\), then the excess force required to take it away from the surface is
1 \(19.8\,mN\)
2 \(198\,N\)
3 \(1.98\,mN\)
4 \(99\,N\)
Explanation:
Surface tension force acts along the circumference of the disc. Excess force required to balance the surface tension force is \({F_{ext}} = T \times 2\pi R\) \({F_{ext}} = \frac{7}{{100}} \times 2 \times 3.14 \times \frac{{4.5}}{{100}}\) \( = 197.82 \times {10^{ - 4}} = 19.8\,mN\)
NEET - 2024
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361324
Maximum possible mass of a greased needle of length \(l\) floating on water surface of tension \(T\) is \(CTl.\) Find value of \(C.\) (Take \(g = 10\;m{\rm{/}}{s^2}\))
1 0.2
2 0.6
3 0.9
4 0.7
Explanation:
Let the mass of the needle be \(m\). As the liquid surface is distorted, the surface tension forces acing on both sides of the needle make an angle \(\theta\), say, with vertical as shown in the figure. Since the forces acting on the needle are \(F,F\) and \(mg\), resolving the forces vertically for its equilibrium, \(\sum {{F_y}} = F\cos \theta + F\cos \theta - mg = 0\) \( \Rightarrow m = \frac{{2\;F\cos \theta }}{g}\) where \(F = Tl\) \(\therefore m = \frac{{2\;Tl\cos \theta }}{g}\) For \(m\) to be maximum, \(\cos \theta=1\) \(\therefore {m_{\max }} = \frac{{2\;Tl}}{{\;g}} = \frac{{2Tl}}{{10}} = 0.2\;Tl\) \( \Rightarrow C = 0.2\)
