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Waves question

2013 · Q163
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Waves question

2013 · Q163

NEETPhysicsWavesMCQ+4 / −1
The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1 : 3 : 5.
  1. A
    150023cm,50023cm{{1500} \over {23}}cm,{{500} \over {23}}cm231500​cm,23500​cm
  2. B
    150023cm,{{1500} \over {23}}cm,231500​cm, 30023cm{{300} \over {23}}cm23300​cm
  3. C
    30023cm,150023cm{{300} \over {23}}cm,{{1500} \over {23}}cm23300​cm,231500​cm
  4. D
    150023cm,200023cm{{1500} \over {23}}cm,{{2000} \over {23}}cm231500​cm,232000​cm
View written solutionFree

Correct answer: D

From formula, f=1xTmf = {1 \over x}\sqrt {{T \over m}} f=x1​mT​​

⇒1f∝l \Rightarrow {1 \over f} \propto l⇒f1​∝l

∴l1:l2:l3=1f1:1f2:1f3 \therefore {l_1}:{l_2}:{l_3} = {1 \over {{f_1}}}:{1 \over {{f_2}}}:{1 \over {{f_3}}}∴l1​:l2​:l3​=f1​1​:f2​1​:f3​1​

= f2f3 : f1f3 : f1f2
[Given: f1 : f2 : f3 = 1 : 3 : 5]

= 15 : 5 : 3

Therefore the positions of two bridges below the wire are

15×10015+5+3cm{{15 \times 100} \over {15 + 5 + 3}}cm15+5+315×100​cm and 15×100+5×10015+5+3cm{{15 \times 100 + 5 \times 100} \over {15 + 5 + 3}}cm15+5+315×100+5×100​cm

⇒150023cm,  200023cm \Rightarrow {{1500} \over {23}}cm,\,\,{{2000} \over {23}}cm⇒231500​cm,232000​cm

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