A) \[\frac{d-b}{d-c-b+a}\]
B) \[\frac{b-d}{d-c-b+a}\]
C) \[\frac{d-c-b+a}{d-b}\]
D) \[\frac{d-b}{a+b-c-d}\]
Correct Answer: A
Solution :
The real depth \[=\mu \] ( apparent depth) \[\Rightarrow \] In first case, the real depth \[{{h}_{1}}=\mu (b-a)\] Similarly in the second case, the real depth \[{{h}_{2}}=\mu (d-c)\] Since \[{{h}_{2}}>{{h}_{1}},\] the difference of real depths \[={{h}_{2}}-{{h}_{1}}=\mu (d-c-b+a)\] Since the liquid is added in second case, \[{{h}_{2}}-{{h}_{1}}=(d-b)\] \[\Rightarrow \,\,\mu =\frac{(d-b)}{(d-c-b+a)}\]
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