A proton and an α-particle are accelerated from rest by 2V and 4V potentials, respectively. The ratio of their de-Broglie wavelength is
- 2:1
- 4:1
- 8:1
- 16:1
The Correct Option is B
Solution and Explanation
Top Questions on Waves
- The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is \( \frac{5}{x} \). The value of \( x \) is ________.
- In an open organ pipe \( \nu_3 \) and \( \nu_6 \) are 3rd and 6th harmonic frequencies, respectively. If \( \nu_6 - \nu_3 = 2200\ \text{Hz} \), then the length of the pipe is _________ mm. (Take velocity of sound in air as \(330\ \text{m s}^{-1}\).)
Two loudspeakers (\(L_1\) and \(L_2\)) are placed with a separation of \(10 \, \text{m}\), as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. A voice recorder, initially at point \(A\), at equidistance to both loudspeakers, is moved by \(25 \, \text{m}\) along the line \(AB\) while monitoring the audio signal. The measured signal was found to undergo \(10\) cycles of minima and maxima during the movement. The frequency of the input signal is _____________ Hz.
(Speed of sound in air is \(324 \, \text{m/s}\) and \( \sqrt{5} = 2.23 \))
- The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be _________ cm/s.
- 5th harmonic of a closed organ pipe matches with 1st harmonic of an open organ pipe. Find ratio of their lengths.
Questions Asked in JEE Main exam
The given circuit works as:
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Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
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