Waves - Result Question 47
49. 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 of $1: 3: 5$ ?
[NEET Kar. 2013] (a) $\frac{1500}{23} cm, \frac{2000}{23} cm$
(b) $\frac{1500}{23} cm, \frac{500}{23} cm$
(c) $\frac{1500}{23} cm, \frac{300}{23} cm$
(d) $\frac{300}{23} cm, \frac{1500}{23} cm$
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Answer:
Correct Answer: 49. (a)
Solution:
- (a) From formula, $f=\frac{1}{x} \sqrt{\frac{T}{m}}$
$\Rightarrow f \propto \frac{1}{l}$
$\therefore l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3}$
$=f_2 f_3: f_1 f_3: f_1 f_2$
$=15: 5: 3$
[Given: $f_1: f_2: f_3=1: 3: 5$ ]
Therefore the positions of two bridges below the wire are
$\frac{15 \times 100}{15+5+3} cm$ and $\frac{15 \times 100+5 \times 100}{15+5+3} cm$
i.e., $\frac{1500}{23} cm, \frac{2000}{23} cm$
