Up to what height should we fill a glass of 2 | vedclass

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(B) Let the height of the liquid filled in the glass be $h$. The apparent depth of the liquid is given by $h' = \frac{h}{\mu}$.According to the proble

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$12$

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(B) Let the height of the liquid filled in the glass be $h$. The apparent depth of the liquid is given by $h' = \frac{h}{\mu}$.According to the problem,the glass appears half-filled. The total height of the glass is $H = 21\, cm$.The empty part of the glass is $(H - h)$.For the glass to appear half-filled,the apparent depth of the liquid must be equal to the height of the empty portion of the glass.So,$H - h = \frac{h}{\mu}$.Given $H = 21\, cm$ and $\mu = 1.5 = \frac{3}{2}$.$21 - h = \frac{h}{1.5} = \frac{2h}{3}$.$21 = h + \frac{2h}{3} = \frac{5h}{3}$.$h = \frac{21 	imes 3}{5} = \frac{63}{5} = 12.6\, cm$.Wait,re-evaluating the standard interpretation: If the apparent depth is half the total height,then $h' = \frac{H}{2} = 10.5\, cm$.Since $h' = \frac{h}{\mu}$,we have $h = h' 	imes \mu = 10.5 	imes 1.5 = 15.75\, cm$.However,if the condition is that the apparent depth of the liquid equals the height of the empty part,$21 - h = \frac{h}{1.5} \Rightarrow 21 = h(1 + 0.666) = 1.666h \Rightarrow h = 12.6\, cm$.Given the options,if we assume $\mu = 1.33$ (water),$21 - h = \frac{h}{1.33} \Rightarrow 21 = h(1 + 0.75) = 1.75h \Rightarrow h = 12\, cm$.Thus,for $\mu = 4/3$,$h = 12\, cm$.

Up to what height should we fill a glass of $21\, cm$ height so that it may appear half-filled? (Assume the refractive index of the liquid is $\mu = 1.5$)

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