A) 4 : 1
B) 1 : 2
C) 1 : 1
D) 1 : 3
Correct Answer: B
Solution :
\[{{n}_{1}}={{n}_{2}}\] \[T\to \]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.
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