WRONG PROBLEM OF HC VERMA

·         WRONG PROBLEM OF H C VERMA

H C Verma  (Part 2)  Gauss’s Law   Page # 141        Problem # 4

Q. The electric field in a region is given by E = (E_ox/L) i . Find the charge contained inside a cubical volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a. Take E_o = 5 ×10³ N/C, L = 2 cm and a = 1 cm.  

SOLUTION:

NOTE: This question has conceptual mistake.

•  As electric field is given by the eqn. E = (E_ox/L) i , it means electric field is towards positive x-axis right of y-z axis and it is towards –ve x axis  left of y-z axis. 
• It means field lines are parallel.
• Parallel field lines mean that field is uniform.
• But as field depends on value of x, it can’t be uniform.
• Hence, there is a contradiction.

• Figure explanation.
• Plane x = 0 corresponds to y-z plane , shown by left face in the figure.
• Plane x = a corresponds to y-z plane , shown by right face in the figure.
• Plane y = 0 corresponds to x-z plane , shown by bottom face in the figure.
• Plane y = a corresponds to x-z plane , shown by top face in the figure.
• Plane z = 0 corresponds to x-y plane , shown by back face in the figure.
• Plane z = a corresponds to x-y plane , shown by front face in the figure.

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Rotational motion 6-10

Rotational motion 1-5

52

Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.

this question is similar to last example of hcv. please write down that example in comment box.

Consider any particle say A which moves towards B with a speed of v. But the particle B is itself moving with a speed v along BC. So movement of particle B along the AB direction will be v.cos∠CBG = v.cos60° =v/2.  

Now the relative speed of Particle A with respect to particle B 

= v-v/2 = v/2,     This is the effective speed with which they come closer. To travel a distance of 'a' the time taken 

= Distance/speed 

= a/(v/2) =2a/v.

It's simple, just find the component of velocity of any particle towards the center, and it won't change throughout the motion (imagine this, it's the "emotions" of physics .... Dinesh will understand the statement :P)

So, the component of the velocity towards center, you can find it easily, and then the distance between center and any particle, then just get the time= . not tough at all.....

You may want to see the solution I wrote to your problem (square one)

Note that this will happen if and only if the polygon is regular polygon and the speed of each of the particles is the same.

The time it takes is exactly 2 times the time it takes for any particle to travel the length of one side of the hexagon. If there were 4 particles in a square, the time is the same as travelling a side. Given any regular polygon, since at any time after pursuit begins, the particles are at the corners of the same regular polygon but a shrinking one, the time it takes is the same as finding the length of logarithmic spirals, which have the property of self-similarity.

Okay, if you want to prove they will, first draw a hexagon slightly smaller the first, but inside it, such that after each particle has traveled a very short distance, they're all still headed towards each other. By property of self-similarity, this can be repeated until the innermost hexagon has shrunk to a point. The paths traced out by the particles is a logarithmic spiral, which has a finite arc length from any point on it down to the center.

A shrinking hexagon indicates that the particles are getting closer to each other, because the sides are shrinking.

As a counterexample, if each particle was constrained to move only along a circular path common to all of them, then they'll never get any closer.

Hey friends i just tried to deduce a formula for such type of problems  where  is side length and  denotes the total sides of the regular polygon

Ess Kp 
11 months, 4 weeks ago

ANSWER: 
2a/v

Plz admire this image for understanding my answer... and plz consider it as a regular hexagon

**now, Initial separation between two particles = Side of hexagon = a

Final separation = 0

Therefore, Relative displacement between two particles = a

Particle B has a component v cos60 along particle BC (a side)

Therefore, relative velocity with which B and C approaches each other = v - vcos60 = v/2

Since, v is constant, thus time taken by these two balls to meet each other is given by

=(Relative displacement) / Relative velocity = a/(v/2) = 2a/v

So, the time taken by the particles to meet each other = 2a/v

In reality each particle will follow a curved path and eventually meet at the center of the hexagon.*

o

51

Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time lag B finds between seeing and hearing the drum beating by A ?

As seen in above diagram the resultant velocity of sound will be along AB and its magnitude AE can be calculated from the right angled triangle ACE.  

AE²=AC²-CE²  = v²-u²  

AE = √(v²-u²)   

So the time lag B finds between seeing and hearing the drum beating by A 

= Distance AB /velocity AE 

= x/√(v²-u²)

50

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. 
A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.


 The beats are a unit of time.
t = d/v

In the first experiment, the wind is assisting the speed of sound. The total speed is v + u. Therefore the time = x / (v + u)
11 = x / (v + u)
11v + 11u = x

In the second experiment, the wind is going against the travel of direction. v - u
The equation you get is
12v - 12u = x

The two x's are an equal distance So you can begin by equating the two distances.
12v - 12u = 11v + 11u
v = 23 u

Now you can go to one of the original equations.
12v - 12u = x
12*23u - 12u = x
12u (23 - 1) = x
12u * 22 = x
u = x / 264

v = 23*u
v = 23 * x/264

Answer: In the first case resultant velocity = u+v  

So the time taken t₁ = x/(u+v)  

→ u+v = x/t₁  ..................(A)   

In the second case resultant velocity = v-u   

Now the time taken = t₂ = x/(v-u)   

→ v-u = x/t₂    ................(B)  

Adding the equations (A)  and  (B)  

2 v = x/t₁ + x/t₂ 

→ v = x/2  * (1/t₁+1/t₂)   It is the velocity of sound in still air.

 Subtracting (B) and (A) we get 

2u = x/t₁ - x/t₂   

→ u = x/2  * (1/t₁ - 1/t₂)      It is the velocity of wind . 

the velocity of sound and wind are added vectorially.
therefore, in the first case,

v + u = x/t1 ........................ (1)

v - u = x/t2 ...........................(2)
on adding, we get
v = x (1/t1 + 1/t2) /  2
on subtracting,
u = x (1/t1 - 1/t2) /  2

in the second case,

the resultant velocity and u are at 900 to each other
therefore,if u make a rt. triangle, 
                                   
v acts as hypotenuse and u as a leg and resultant vel. towards B makes the other leg. therefore,
Resultant vel. =   v2 - u2 
time taken =     x      
                            v2 - u2 

49

Q 49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

I will form the diagram for the given problem.

Let us first assign the unknown variables.
t = Total time of flight (unknown)
A = Angle of destination vector from north = 30∘
B = Angle of airplane's propulsion, measured from East (to find)
V = Final Velocity (Unknown)
u = Air Velocity = 20ms−1
s = Airplane Velocity = 150ms−1

The air velocity only provides a northward displacement, ie the Eastward displacement should come from the East component of the plane's velocity.

The eastward displacement is 500×cos(90∘−A)km
The East component of the airplane's velocity is s×cos(B)

So the equation is 
500×cos(90∘−A)=s×cos(B)×t

You will obtain an equation in B and t.

Now, the northward displacement is due to both the airplane's propulsion and the wind velocity.
The net northward component of velocity is
s×sin(B)+u

The distance is 
500×sin(90∘−A)

Thus the equation is
500×sin(90∘−A)=(s×sin(B)+u)×t

You have another equation in B and t.

Solve for B