Waves Ques 44
44. 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} $ $cm$
Show Answer
Answer:
Correct Answer: 44.(a)
Solution:
- (a) From formula, $f=\frac{1}{x} \sqrt{\frac{T}{m}}$
$\Rightarrow f \propto \frac{1}{l}$
$\therefore l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3}$
[Given: $f_1: f_2: f_3=1: 3: 5$ ]
$=f_2 f_3: f_1 f_3: f_1 f_2$
$=15: 5: 3$
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$
BETA
