SATHEE: Waves - Result Question 47

Waves - Result Question 47

49. The length of the wire between two ends of a sonometer is $100 c m$ . 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 of $1 : 3 : 5$ ?

[NEET Kar. 2013] (a) $\frac{1500}{23} c m, \frac{2000}{23} c m$

(b) $\frac{1500}{23} c m, \frac{500}{23} c m$

(c) $\frac{1500}{23} c m, \frac{300}{23} c m$

(d) $\frac{300}{23} c m, \frac{1500}{23} c m$

Show Answer

Answer:

Correct Answer: 49. (a)

Solution:

(a) From formula, $f = \frac{1}{x} \sqrt{\frac{T}{m}}$

$\Rightarrow f \propto \frac{1}{l}$

$∴ l_{1} : l_{2} : l_{3} = \frac{1}{f_{1}} : \frac{1}{f_{2}} : \frac{1}{f_{3}}$

$= f_{2} f_{3} : f_{1} f_{3} : f_{1} f_{2}$

$= 15 : 5 : 3$

[Given: $f_{1} : f_{2} : f_{3} = 1 : 3 : 5$ ]

Therefore the positions of two bridges below the wire are

$\frac{15 \times 100}{15 + 5 + 3} c m$ and $\frac{15 \times 100 + 5 \times 100}{15 + 5 + 3} c m$

i.e., $\frac{1500}{23} c m, \frac{2000}{23} c m$

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