A long capillary tube of mass pi ,g,radius 2, | vedclass

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(A) The forces acting on the capillary tube are its weight $(mg)$ acting downwards and the upward force due to surface tension $(F_s)$ acting along th

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$10.4\,\pi \,mN$

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(A) The forces acting on the capillary tube are its weight $(mg)$ acting downwards and the upward force due to surface tension $(F_s)$ acting along the circumference of the tube at the liquid-air interface. Given: Mass of tube $m = \pi \,g = \pi 	imes 10^{-3} \,kg$. Radius $r = 2\,mm = 2 	imes 10^{-3} \,m$. Surface tension $T = 0.1\,N/m$. Angle of contact $	heta = 0^\circ$. Acceleration due to gravity $g = 10\,m/s^2$. Weight of the tube $W = mg = (\pi 	imes 10^{-3}) 	imes 10 = 10\pi 	imes 10^{-3} \,N = 10\pi \,mN$. The upward force due to surface tension is $F_s = T 	imes (2\pi r) 	imes \cos(	heta)$. Since $	heta = 0^\circ$,$\cos(0^\circ) = 1$. $F_s = 0.1 	imes 2\pi 	imes (2 	imes 10^{-3}) = 0.4\pi 	imes 10^{-3} \,N = 0.4\pi \,mN$. To hold the tube in equilibrium,the external force $F$ applied upwards must satisfy: $F + F_s = W$ $F = W - F_s$ $F = 10\pi \,mN - 0.4\pi \,mN = 9.6\pi \,mN$. Wait,re-evaluating the force balance: The liquid rises in the tube,exerting a downward force on the tube due to surface tension. Thus,the external force $F$ must support both the weight and the downward pull of surface tension. $F = W + F_s = 10\pi \,mN + 0.4\pi \,mN = 10.4\pi \,mN$.

$A$ long capillary tube of mass $\pi \,g$,radius $2\,mm$ and negligible thickness,is partially immersed in a liquid of surface tension $0.1\,N/m$. Take the angle of contact as zero and neglect the buoyant force of the liquid. Find the force required to hold the tube vertically. $(g = 10\,m/s^2)$

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