# Tag: physics

### Questions Related to physics

Reflection at a free boundary implies

1. restoring force is zero

2. restoring force is not zero

3. applied force is zero

4. applied force is not zero

Correct Option: A
Explanation:

In a free boundary, the restoring force is always zero

The correct option is (a)

The equation of a progressive wave is given by $y=10 sin (5t-x)$. The wave gets reflected from a open boundary. The equation of the reflected wave is

1. $y=10 sin (5t-x)$

2. $y=-10 sin (5t-x)$

3. $y=10 sin (5t+x)$

4. $y=10 sin (5t+x+\pi)$

Correct Option: C
Explanation:

In reflection of a wave at open boundaries, the phase reversal dosent take place

The correct option is (c)

A wave of frequency $100$ Hz is sent along a string towards a fixed end. When this wave travels back after reflection, a node is formed at a distance of $10$ cm from the fixed end of the string. The speeds of incident(and reflected) waves are?

1. $5$ $cms^{-1}$

2. $10$ $cms^{-1}$

3. $20$ $cms^{-1}$

4. $40$ $cms^{-1}$

Correct Option: A
Explanation:

Given frequency $\nu$=100 H.

Also distance between two successive nodes$=>\dfrac { \lambda }{ 2 } =10\ \lambda =20\quad cm$
And we know $\nu =n\lambda \ n=\frac { \nu }{ \lambda } =\dfrac { 100 }{ 20 } =5\quad cm/s$

A composition string is made up by joining two strings of different masses per unit length $\longrightarrow \mu$ and $4\mu.$ The composite string is under the same tension. A transverse wave pulse : $Y=(6 mm) \sin (5t+40x)$, where '$t$' is in seconds and '$x$' in meters, is sent along the lighter string towards the joint. The joint is at $x=0.$ The equation of the wave pulse reflected from the joint is

1. $(2 mm) \sin(5t-40x)$

2. $(4 mm) \sin(40x-5t)$

3. $- (2 mm) \sin(5t-40x)$

4. $(2 mm) \sin(5t-10x)$

Correct Option: C
Explanation:

Since the wave is travelling from low to more dense medium, thus there will be inversion of the reflected wave . Since it travels back from origin, thus $40x$ will become $-40x.$ Also, amplitude of reflected wave is given by:
${ A } _{ r }=\dfrac { Z _{ 1 }-{ Z } _{ 2 } }{ { Z } _{ 1 }+Z _{ 2 } } A.\quad Z=\mu c=\mu \sqrt { \dfrac { T }{ \mu } } =\sqrt { \mu T } \ Thus\quad { A } _{ r }=\dfrac { \sqrt { \mu T } -\sqrt { 4\mu T } }{ \sqrt { \mu T } +\sqrt { 4\mu T } } A=\dfrac { -1 }{ 3 } A\$
Thus it become $-6/3=-2$.

Which of the following statement is incorrect superposition of waves?
(i) After superposition frequency,wavelength and velocity of resultant wave remains same
(ii) After superposition amplitude of resultant wave is equal to amplitude of either wave
(iii) Mechanical wave cannot superposed with electromagnetic wave
(iv) For superposition two waves should have equal wavelength,frequency and amplitude

1. I & III Only

2. I & II Only

3. I,II, & III Only

4. II & IV Only

Correct Option: A

The wavelength of the first line of Lyman series is $\lambda$. The wavelength of the first line in Paschen series is ________.

1. $108/7$

2. $27/5$

3. $7/108$

4. $5/27$

Correct Option: A
Explanation:
Formula for Lyman series is:
Where, n = 2,3,4,5,.....
Where, R = Rydbergg constant,
$\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)$ and
Paschen series in $\lambda _1$,
$\Rightarrow \dfrac{1}{\lambda _1}=R\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)$

$\Rightarrow \dfrac{\lambda _1}{\lambda}=\dfrac{\dfrac{3}{4}}{\dfrac{7}{16\times 9}}$

$\Rightarrow \lambda _1=\dfrac{3}{4}\times \dfrac{16\times 9}{7}\lambda$

$\Rightarrow \lambda _1=\dfrac{108}{7}\lambda$.

A wave travels on a light string. The equation of the waves is $Y\, = \,A\, sin\,(kx\,-\,\omega\,t+\,30^{\circ})$. It is reflected from a heavy string tied to end of the light string at x = 0 . If 64% of the incident energy is reflected then the equation of the reflected wave is

1. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$

2. $Y\, =\,0.8 \,A\, sin\,(kx\,+\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$

3. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,-\,30^{\circ})$

4. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$

Correct Option: B
Explanation:

There are three things we need to take into account:

• Energy transfer
• Change in velocity
• Change in phase
We know that power delivered is proportional to $A^{2}$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of $180^{\circ}$
Hence the new equation becomes:
$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$
Hence option B.

A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with :

1. a phase change of ${180}^{o}$ with velocity reversed

2. the same phase as the incident pulse with no reversal of velocity

3. a phase change of ${180}^{o}$ with no reversal of velocity

4. the same phase as the incident pulse but with velocity reversed

Correct Option: A
Explanation:

A pulse of wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $\pi$ with the incident wave but wave velocity after reflection reverses its direction.

A wave of length $2m$ is superposed on its reflected wave to form a stationary wave. A node is located at  $x=3m$ The next node will be located at  $x=$

1. $4m$

2. $3.75m$

3. $3.50m$

4. $3.25m$

Correct Option: A
Explanation:

Since wave length is $2m$, half of wavelength is $1m$. Node forms after each half of wave length. So, node will be formed at each $1 m$
So, as node is formed at $3 m$
next node will be formed at $3+1=4\ m.$

A sound wave of frequency $1360 Hz$ falls normally on a perfectly reflecting  wall.  The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ($v = 340 m/s$)

1. $25 cm$

2. $6.25 cm$

3. $62.5 cm$

4. $2.5 cm$

Correct Option: B
Explanation:

$v=f\lambda$
$\lambda=\dfrac{340}{1360}=\dfrac{1}{4}=0.25m$
$=25cm$
As the end is a node, shortest distance at which antinode is formed is $\dfrac{\lambda}{4}=6.25cm$