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ICSE Class 10 Physics 2013 Paper

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Question : 44 of 70
Marks: +1, -0
A coin placed at the bottom of a beaker appears to be raised by 4.0cm4.0 \mathrm{cm} . If the refractive index of water is 43\frac{4}{3} , find the depth of the water in the beaker.
Solution:  
Given : Shift =4cm,μ=43=4 \mathrm{cm}, \mu = \frac{4}{3},
Real depth == ?
We know that Shift
  =   Real depth   (1−  1aμm)\;=\; \text{ Real depth } \; \left(1-\; \frac{1}{{}_a\mu_m}\right)
4  =   Real depth   (1−  34)4\;=\; \text{ Real depth } \; \left(1-\; \frac{3}{4}\right)
4  =   Real depth   ×  144\;=\; \text{ Real depth } \; \times \; \frac{1}{4}
   Real depth     =  4×41\; \text{ Real depth } \;\;=\; \frac{4 \times 4}{1}
  =16cm.\;=16 \mathrm{cm}.
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