Q

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?

S Sayak

(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.

S Sayak

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 vector$A=2\hat i +3 \hat j$ along the directions of $\hat i + \hat j$ and $\hat i - \hat j$? [You may use graphical method]

D Devendra Khairwa

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     is   :-     $=\ \frac{2+3}{\sqrt{2}}\ =\ \frac{5}{\sqrt{2}}$
and the required components of a vector C along the directions of    is    :-    $\frac{2-3}{\sqrt{2}}\ =\ \frac{-1}{\sqrt{2}}$

What is stopping potential in photoelectric effect?

P Pankaj Sanodiya

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 are :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S solutionqc

Option 1)

Option 2)

Option 3)

Option 4)

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)

• Option 2)

• Option 3)

• Option 4)

S solutionqc

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

Option 1)

Option 2)

Option 3)

Option 4)

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)

S solutionqc

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)

)

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

S safeer

@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.

S safeer

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
Tanâ»¹[19.6/14.7]
=Tanâ»¹1.33
=Tanâ»¹[4/3]
=53°with horizontal.

Prevalence of pesticide resistant insects

S safeer

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

S safeer

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 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

V vishal dange