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15.12)  For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

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S Sayak
Answered Jul 07, 2019

(i) (a) All the points vibrate with the same frequency of 60 Hz.

(b) They all have the same phase as it depends upon time.

(c) At different points, the amplitude is different and is equal to A(x) given by

 A(x)=0.06sin(\frac{2\pi }{3}x)

(ii) 

\\A(0.375)=0.06sin(\frac{2\pi }{3}\times 0.375)\\ A(0.375)=0.06\times sin(\frac{\pi }{4})\\ A(0.375)=0.06\times \frac{1}{\sqrt{2}}\\ A(0.375)=0.042m

 

14.22 Show that for a particle in linear SHM the average kinetic energy over a period of
oscillation equals the average potential energy over the same period.

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S Sayak
Answered Jul 07, 2019

Let the equation of oscillation be given by x=Asin(\omega t)

Velocity would be given as 

\\v=\frac{dx}{dt}\\ v=A\omega cost(\omega t)

Kinetic energy at an instant is given by

\\K(t)=\frac{1}{2}m(v(t))^{2}\\ K(t)=\frac{1}{2}m(A\omega cos(\omega t))^{2}\\ K(t)=\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t

Time Period  is given by

T=\frac{2\pi }{\omega }

The Average Kinetic Energy would be given as follows

\\K_{av}=\frac{\int _{0}^{T}K(t)dt}{\int _{0}^{T}dt}\\ K_{av}=\frac{1}{T}\int _{0}^{T}K(t)dt\\ K_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t\ dt\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}cos^{2}\omega t\ dt\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}\left ( \frac{1+cos2\omega t}{2} \right )dt

\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \frac{t}{2} +\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \left ( \frac{T}{2}+\frac{sin2\omega T}{4\omega } \right )-\left ( 0+sin(0) \right ) \right ]\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\times \frac{T}{2}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{4}

The potential energy at an instant T is given by 

\\U(t)=\frac{1}{2}kx^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}(Asin(\omega t))^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t

The Average Potential Energy would be given by

\\U_{av}=\frac{\int_{0}^{T}U(t)dt}{\int_{0}^{T}dt}\\ \\U_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t\ dt\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}sin^{2}\omega t\ dt\\\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}\frac{(1-cos2\omega t)}{2}dt

\\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \frac{t}{2} -\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \left ( \frac{T}{2}-\frac{sin2\omega T}{4\omega } \right )-\left ( 0-sin0 \right ) \right ]\\ U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\times \frac{T}{2}\\ U_{av}=\frac{m\omega ^{2}A^{2}}{4}

We can see Kav = Uav

Q 4. 22  \hat i and \hat j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors \hat i + \hat j, and \hat i - \hat j−? What are the components of a vectorA=2\hat i +3 \hat j along the directions of \hat i + \hat j and \hat i - \hat j? [You may use graphical method]

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D Devendra Khairwa
Answered Jun 06, 2019

Let A be a vector such that:-                 \overrightarrow{A}\ =\ \widehat{i}\ +\ \widehat{j}

Then the magnitude of vector A is given by   :            \left | A \right |\ =\ \sqrt{1^2\ +\ 1^2}\ =\ \sqrt{2}

Now let us assume that the angle made between vector A and x-axis is \Theta.

Then we have:-                 

                                                                \Theta \ =\ \tan^{-1}\left ( \frac{1}{1} \right )\ =\ 45^{\circ}

Similarly, let B be a vector such that:-       \overrightarrow{B}\ =\ \widehat{i}\ -\ \widehat{j}

The magnitude of vector B is     :                       \left | B\right |\ =\ \sqrt{1^2\ +\ (-1)^2}\ =\ \sqrt{2}

Let \alpha be the angle between vector B and x-axis :       

                                                                 \alpha \ =\ \tan^{-1}\left ( \frac{-1}{1} \right )\ =\ -45^{\circ}

 

Now consider   \overrightarrow{C}\ =\ 2\widehat{i}\ +\ 3\widehat{j} :-
Then the required components of a vector C along the directions of (\hat{i}+\hat{j})    is   :-     =\ \frac{2+3}{\sqrt{2}}\ =\ \frac{5}{\sqrt{2}}
and the required components of a vector C along the directions of (\hat{i}-\hat{j})   is    :-    \frac{2-3}{\sqrt{2}}\ =\ \frac{-1}{\sqrt{2}}

What is stopping potential in photoelectric effect?

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P Pankaj Sanodiya
Answered Jul 07, 2019

The Stopping Potential in Photoelectric effect :

The stopping potential is defined as the potential necessary to stop any electron (or, in other words, to stop even the electron with the most kinetic energy) from reaching the other side.

The dimension of \left ( \mu_{\text{o}}\in_{\text{o}} \right )^{-\frac{1}{2}} are :

  • Option 1)

    \left [ \text{L}^{-\frac{1}{2}}\text{T}^{\frac{1}{2}} \right ]

  • Option 2)

    \left [ \text{L}^{\frac{1}{2}}\text{T-}^{\frac{1}{2}}\right ]

  • Option 3)

    \left [ \text{L}^{-1}\text{T}\right ]

  • Option 4)

    \left [ \text{LT}^{-1}\right ]

 

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S solutionqc
Answered Feb 02, 2019


Option 1)

\left [ \text{L}^{-\frac{1}{2}}\text{T}^{\frac{1}{2}} \right ]

Option 2)

\left [ \text{L}^{\frac{1}{2}}\text{T-}^{\frac{1}{2}}\right ]

Option 3)

\left [ \text{L}^{-1}\text{T}\right ]

Option 4)

\left [ \text{LT}^{-1}\right ]

A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is:

  • Option 1)

    \frac{wx}{d}

  • Option 2)

    \frac{wd}{x}

  • Option 3)

    \frac{w(d-x)}{x}

  • Option 4)

    \frac{w(d-x)}{d}

 

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S solutionqc
Answered Feb 02, 2019

 

By \ Equating \ forces . \\N_1+N_2=W \\ Equating \ Torques \ about \ A \\ Wx=N_2d =>\\N_2=Wx/d \\So N_1=W[1- (x/d)] \\.°. \ Normal \ reaction \ on \ A \ is \ W[1- (x/d)]


Option 1)

\frac{wx}{d}

Option 2)

\frac{wd}{x}

Option 3)

\frac{w(d-x)}{x}

Option 4)

\frac{w(d-x)}{d}

In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be:

  • Option 1)

    4P

  • Option 2)

    2P

  • Option 3)

    P

  • Option 4)

    \frac{P}{2}

 

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S solutionqc
Answered Feb 02, 2019

Pressure of a gas

P= \frac{1}{3}mn< v^{2}>

if mass is halved and speed is doubled pressure becomes 2P


Option 1)

4P

Option 2)

2P

Option 3)

P

Option 4)

\frac{P}{2}

 ) 
 

A particle p is projected from a point on the surface of a smooth inclined plane (see fig), simultaneously another particle Q isreleased on the smooth inclined plane from the same position. p and Q collide after t=4sec. The speed of  projection of  P is

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S safeer
Answered Jan 01, 2019

@Arvind gupta 

For Q

U=0,  a=gsin\Theta

So s_q=\frac{1}{2}gsin\Theta\ \ t^2=8gsin\Theta

Now for P

if the angle of projection of P with the incline surface is α 
=> ucosα = speed along the plane, a = gsinθ° = acceleration along the plane 

Similarly usinα = speed perpendicular  to the plane, a = gcosθ° = acceleration perpendicular to the plane 

time of flight (For P) = 2usinα/gcosθ° = 4 s 

usinα = 2gcosθ°

distance travelled down incline = s = ut + ½at² 
Sp= 4ucosα + 8gsinθ° 

So Sp=Sq

= 4ucosα + 8gsinθ° = 8gsinθ° 
=> cosα = 0 or α = 90°

So we get u = 2gcosθ°=

u=2gcos\Theta =2g\frac{1}{2}=g=10m/s^2

 

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

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S safeer
Answered Jan 01, 2019

Velocity=14.7 m/s 
Time taken by the truck to cover 58.8=t=S/v=58.8/14.7 =4sec
Since ball returns back to truck its angle of projection is vertically upwards with respect to truck.
u=speed of projection.
t=4s
h=0
h=4u-1/2 x 9.8x16
0=4u-78.4
4u=78.4
u=78.4/4=19.6 m/s


b) As seen from the road the speed of 
  ball will be resultant of 2 speeds vertical
speed =u=19.6 m/s
Horizontal speed given by truck=14.7m/s
⇒√u²+v²=√19.6²+14.7²=24.5 m/s
Thus angle seen from road:
Tan⁻¹[19.6/14.7] 
=Tan⁻¹1.33
=Tan⁻¹[4/3]
=53°with horizontal.


 

Prevalence of pesticide resistant insects

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S safeer
Answered Jan 01, 2019

  According to Darwin's concept of natural selection, the organisms which are provided with favourable variations would survive because they are the fittest to face their surrounding while the organisms which are unfit for surrounding variations are destroyed. Prevalence of pesticide resistant insects is due to adaptability of these insects for the changes in environment due to use of pesticides.

Capture5.JPG Sir/Madam , can you tell me what is the significance of DNA satellite in DNA fingerprinting? Gene in forms of DNA sequence - (Tip): By defining a cistron as a segment of DNA code for a polypeptide, The structural gene in a transcription unit could be said as monocistronic or polycistronic. Functional unit If inheritance located on the DNA Gene in forms of DNA sequence The monocistronic structural gene Types of RNA Functions of types of RNA Process Of Transcription Working of RNA polymerase Termination Transcription In Eukaryotes Splicing Cappinq Tailing Proposition made by George garow Salient features of genetics code Transcription unit Salient features Of genetics code phenylalanine Convention in deferning two strands of the DNA Convention in deferning two strands of the DNA The Promoter The Terminator t' RNA- the adapter molecule Initiator t-RNA Translation

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S safeer
Answered Jan 01, 2019

A DNA satellite is a region that consists of short DNA sequences repeated many times. The variation between individuals in the lengths of their DNA satellites forms the basis of DNA fingerprinting.
DNA satellites are of two types, i.e. microsatellites and minisatellites. Their characteristic that makes them useful for identification is that they are highly polymorphic. The length of each satellite in DNA is inherited.

The length of satellite regions are highly variable between people. These form small peaks during density gradient centrifugation and thus are invaluable for identification purposes.

In the circuit shown, the current in the 1\Omega resistor is :

 

  • Option 1)

    1.3 A ,from P to Q

  • Option 2)

    0 A

  • Option 3)

    0.13 A ,from Q to P

  • Option 4)

    0.13 A ,from P to Q

 

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V vishal dange
Answered Oct 10, 2018

answer is option 3)0.13A from Q  to P

  •  Let, Current i1 from 6v battery passes through 3 ohm resistance . At Q current breaks and current i2 goes through 3 ohm and ( i1-i2 ) goes through 1 ohm ,.
  • Applying, kirchoffs law in 1st and 2nd loop,we get two eqns as follows
  • 4i1-i2=6
  • 6i2-i1=9
  • solving ,,  i1-i2 =0.13 A
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