Two wires  and  have the same radius r and respective densities  and  such that  They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary waves is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in to is |
 [JEE Online 08-04-2017] |
A) 4 : 1
B) 1 : 2
C) 1 : 1
D) 1 : 3
Correct Answer: B
Solution :
| [b] \[{{n}_{1}}={{n}_{2}}\] |
| \[T\to \]same |
| \[r\to \]same |
| \[l\to \]same |
| \[n=\frac{p}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}d}}\] |
| \[{{n}_{1}}={{n}_{2}}\] |
| \[\frac{{{p}_{1}}}{\sqrt{{{d}_{1}}}}=\frac{{{p}_{2}}}{\sqrt{{{d}_{2}}}}\] |
| \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{1}{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec
