Messages posted by kishore.subramanian.b

Two small insulating spheres each of mass m = 0.30 milligrams carry charges of equal magnitude (q =

15 pC) but opposite sign are connected by a light, rigid, insulating rod of length d = 5.0 cm, as shown

above. The charges are placed a distance r = 10 cm away from a large conducting sheet carrying a uniform surface charge density σ = 800 nC/m2 (the sheet is oriented perpendicular to the page; the vertical line represents a cross section).

You can assume that the only forces acting in this problem are electrical (ignore gravitational effects). (a) Calculate the electric field due to the sheet of charge at the location of the dipole (you can use the

result we derived in class; no need to rederive using Gauss’s law).

(b) Sketch the electric field generated by the sheet in the figure above.

(c) Calculate the magnitude of the electric field due to the positive charge +q at the location of the negative charge –q.

Your answers to parts (a) and (c) above should demonstrate that the electrical influence of the sheet on the charges will be much larger than the electrical influence of the charges on one another, so for the rest of the problem, you can ignore the electric fields and forces that the point charges create.

In the circuit shown above, the battery maintains a constant potential between points 1 and 2 (that is, the internal resistance of the battery is negligible).

Three identical flashbulbs A, B and C are screwed into their sockets and are lighted when the circuit is closed. After each of the changes suggested in the following lettered questions (a-g), the system is returned to the initial condition shown in the figure before the next change is made.

The question “What happens to…” refers to whether the quantity in question increases, decreases, or remains unchanged.

Indicate your reasoning briefly in answering each question.

(a) How do the brightnesses of bulbs A, B and C compare with each other in the initial condition?

(b) Bulb A is unscrewed and removed from its socket. (Note: Removing a bulb from its socket means that the connection is severed at that point, not that the bulb is replaced by a wire.)

i. What happens to the brightness of each bulb?

ii. What simultaneously happens to the current at points 3, 4, and 5?

(c) Bulb C is unscrewed and removed from its socket i. What happens to the brightness of each bulb?

ii. What simultaneously happens to the current at points 3, 4, and 5?

(d) A wire is connected from the battery terminal at point 1 to point 4. i. What happens to the brightness of each bulb?

ii. What simultaneously happens to the current at point 3?

iii. What simultaneously happens to the potential difference across bulb B?

iv. What simultaneously happens to the potential difference across bulb C?

v. What simultaneously happens to the potential difference between points 1 and 5?

(e) A wire is connected from the battery terminal at point 2 to the socket terminal at point 5. i. What happens to the brightness of each bulb?

ii. What happens to the current at point 2?

(f) A fourth bulb (D) is connected in parallel with bulb B alone, i.e. not in parallel with both B and C. i. What simultaneously happens to the current at point 3?

ii. What simultaneously happens to the potential difference between points 3 and 4?

iii. What simultaneously happens to the potential difference between points 2 and 4?

(g) The connection is broken at point 4, and a resistor is introduced in series with bulbs B and C. What will happen to the brightness of each bulb?

A parallel-plate capacitor has an area of 25 cm2, and a separation distance of 3.50 mm. A charge of 25

µC is placed on the capacitor.

(a) Calculate the capacitance of the capacitor, the voltage across the capacitor, and the electric field between the plates. Show your work below, and record your answers in the boxes provided (don’t forget the units!).

Now, a sheet of plastic 3.50 mm thick is placed between the plates, thereby completely filling the space between them. The dielectric constant of the sheet is κ = 4.0. The charge on the plates remains the same as in part (a).

(b) With the dielectric inserted, calculate the capacitance, the voltage across the capacitor, and the electric field between the plates. Show your work below, and record your final answers in the boxes provided (don’t forget the units!).

Now, the dielectric from part (b) is removed, and a new sheet, also with dielectric constant κ = 4.0 but with half the thickness (1.75 mm) is placed inside the capacitor, thereby filling only half the space between the plates, as shown in the figure. Again, the charge on the plates remains fixed.

An electron traveling horizontally with a speed v0 = 1.5×107 m/s enters a uniform, vertical electric field created by two oppositely charged parallel plates as shown inFigure. The electric field between the plates has a magnitude of E = 2.0 ×104 N/C. The separation between the plates is s = 1.0 cm, and the vertical deflection of theelectron as it passes between the plates is d = 0.60 cm.

(a) What is the potential difference between the plates?

(b) Which plate is at the higher voltage?

(c) Through what potential difference does the electron travel as it passes between the plates?

(d) What is the speed of the electron as it leaves the area between the plates?

u hav to understand n study the topics...

u should solve all the problems from one book fully WRITTEN....for ex. if u solve trignometry from brilliants solve all the problems in the book...from other books just read the problems n solve the different problems...this goes for all three subjects ...

take as many tests as possible n improve u'r level...as u hav joined coaching write their tests fully n review u'r test papers after u get the solutions and check where u hav gone wrong....

u refer to arihant books for reference..there are pleanty of books but brilliants and arihant is sufficient to understand the concepts n problem solving...

The numbers from 1 to 100 are arranged in a 10×10 table so that no
two adjacent numbers have sum larger than S. Find the smallest value of
S for which this is possible

Total number of ways=6!/2!=360

Total number of ways 2 vowels occur together = ³C₂x2!/2!x5! - 4!x3!/2! =360 - 72 = 288

(any 2 together but all three together will also come in this so subracting with all three together)

Hence total number of ways = 360 - 288 =72.

I dont know if this correct...but i think this is the answer...

Total number of pairs is ^2mC₂.one pair must play with all other pairs(^2m-2C₂). i.e

total games=^2mC₂x^2m-2C₂/2 (because same pair will be matched twice)

total games =630=^2mC₂x^2m-2C₂/2..

630=[2m x (2m-1)/2]x[(2m-2) x (2m-3)/2]

hence m=5..

For a function to be convergent to a real number L ..then it has to be bounded .

Here L may be upper bound or lower bound... As the function is bounded it has to be monotonic...i.e (for upper bounded function) or (for lower bound function)...

Hence i think F¹¹(x) has to be zero..

The answer is zero..

Take some function f(x) satisfying the condition and find the ans..

For example in this question we can take ...which is suitable as ..

now find the f''(x)=? which comes out to be zero....

Let P be a complex number z.

A and B be z₁and z₂. PA=|z-z₁| , PB=|z-z₂|

PA/PB=K

i.e |z-z₁|=K|z-z₂| ...

if K=1 then |z-z₁|=|z-z₂| (use z=x+iy if you dont get my explanation)

which is the equation of perpendicular bisector of the line joining the points A and B.

orher wise it is a circle.

Hence ..........

Hence from the three results we can see that

Hence we can say that the value of I depends only on x and not on f(x)....Hence the number of functions of f(x) has to be one..

yes the ans is b....can you pls give me more explanation..i didnt get what you mean...

aieee 2006

inorganic benzene is borazole or borazine...B₃N₃H₆. .

B₂H₆ forms adduct with aqmmonia (e.g.- B₂H₆.2NH₃ ) a white solid , which when heated at 200⁰C forms a stable cyclic compound borazole.

It contains alternate -BH- and -NH- groups with alternate double bonds like in benzene.

how was fiitjee full test 5??

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