Consider a system of three connected strings, $ S_1, S_2 $ and $ S_3 $ with uniform linear mass densities $ \mu \, \text{kg/m}, 4\mu \, \text{kg/m} $ and $ 16\mu \, \text{kg/m} $, respectively, as shown in the figure. $ S_1 $ and $ S_2 $ are connected at point $ P $, whereas $ S_2 $ and $ S_3 $ are connected at the point $ Q $, and the other end of $ S_3 $ is connected to a wall. A wave generator $ O $ is connected to the free end of $ S_1 $. The wave from the generator is represented by $ y = y_0 \cos(\omega t - kx) $ cm, where $ y_0, \omega $ and $ k $ are constants of appropriate dimensions. Which of the following statements is/are correct:
Understanding the Wave Motion: The wave originates from the leftmost string \( S_1 \), which has a linear mass density \( \mu \), and moves toward the junction \( P \) where it meets a string of higher mass density \( 4\mu \). Then it travels through \( S_2 \) and hits junction \( Q \), again entering a denser string \( S_3 \) with \( 16\mu \), which is fixed at its far right end (a rigid wall).
Let’s analyze the reflection and transmission at each junction.
Step 1: Reflection at Junction \( P \) - A wave moves from a lighter string (\( \mu \)) to a denser string (\( 4\mu \)). - When a wave travels from a lighter medium to a denser medium, partial reflection occurs with a phase change of \( \pi \). - The reflected wave travels in the opposite direction, and the phase shift \( \pi \) changes the cosine to: \[ \cos(\omega t - kx + \pi) = -\cos(\omega t - kx) \] - But since it's moving in the opposite direction, the wave becomes: \[ \cos(\omega t + kx + \pi) \] - Therefore, the reflected wave is: \[ y = \alpha_1 y_0 \cos(\omega t + kx + \pi) \] ✓ So, (A) is correct.
Step 2: Transmission through Junction \( P \) - The wave that transmits into \( S_2 \) continues to travel forward (positive \( x \)-direction), retaining its original phase (no phase change in transmission). - Thus, the transmitted wave would still be of the form: \[ y = \alpha_2 y_0 \cos(\omega t - kx) \] However, the wave number \( k \) could change due to different medium properties. Since \( k \propto 1/v \propto \sqrt{\mu} \), technically it should be adjusted. But the given option doesn't reflect this properly. (B) is partially correct in form but lacks proper adjustment of \( k \). We'll consider it not fully correct.
Step 3: Reflection at Junction \( Q \) - The wave in \( S_2 \) travels toward \( Q \), where \( S_2 \) (with \( 4\mu \)) meets \( S_3 \) (with \( 16\mu \)). Again, this is a reflection at a denser medium, which causes a \( \pi \) phase shift. - The reflected wave moves back to the left, i.e., in the negative \( x \)-direction. - So the correct form of the reflected wave should be: \[ y = \alpha_3 y_0 \cos(\omega t + kx + \pi) \] But in Option (C), it is wrongly written as \( \cos(\omega t - kx + \pi) \), which corresponds to a forward-moving wave. Hence, (C) is incorrect.
Step 4: Transmission through Junction \( Q \) - The wave continues into string \( S_3 \), which has mass density \( 16\mu \). - Since the wave number \( k \propto \sqrt{\mu} \), the new wave number \( k' \) in \( S_3 \) becomes: \[ k' = \sqrt{\frac{16\mu}{\mu}} \cdot k = 4k \] - So the transmitted wave becomes: \[ y = \alpha_4 y_0 \cos(\omega t - 4kx) \] (D) is correct.
A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
In a scattering experiment, a particle of mass $ 2m $ collides with another particle of mass $ m $, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation $ \theta $ of the heavier particle, as shown in the figure, in radians is:
A conducting square loop initially lies in the $ XZ $ plane with its lower edge hinged along the $ X $-axis. Only in the region $ y \geq 0 $, there is a time dependent magnetic field pointing along the $ Z $-direction, $ \vec{B}(t) = B_0 (\cos \omega t) \hat{k} $, where $ B_0 $ is a constant. The magnetic field is zero everywhere else. At time $ t = 0 $, the loop starts rotating with constant angular speed $ \omega $ about the $ X $ axis in the clockwise direction as viewed from the $ +X $ axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ($ V $) in the loop as a function of time:
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
A conducting square loop of side $ L $, mass $ M $, and resistance $ R $ is moving in the $ XY $ plane with its edges parallel to the $ X $ and $ Y $ axes. The region $ y \geq 0 $ has a uniform magnetic field, $ \vec{B} = B_0 \hat{k} $. The magnetic field is zero everywhere else. At time $ t = 0 $, the loop starts to enter the magnetic field with an initial velocity $ v_0 \hat{j} \, \text{m/s} $, as shown in the figure. Considering the quantity $ K = \frac{B_0^2 L^2}{RM} $ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct: