IIT JEE , AIEEE Forum - Physics , Electricity

sulekha_hi

A BOB OF A SIMPLE PENDULUM HAS A MASS OF 40G AND A POSITIVE CHARGE OF 4.0*10^6C.IT MAKES 20 OSCILLATIONS IN 45s.A VERTICAL ELECTRIC FIELD POINTING UPWARDAND OF MAGNITUDE2.5*10^4 N/C IS SWITCHED ON HOW MUCH TIME WILL ITNOW TAKE TO COMPLETE 20 OSCILLATIONS?

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aditya_arora04

Hi,

Look, the question is a bit tricky and lengthy. Though it is a good one !!!

We, know that,

Time period of pendulum = 2(pie)( L / g )^1/2

Here, L = length of pendulum.

From here you can calculate the length of the pendulum.

Now, when we start the electric field, effective acceleration due to gravity changes.

In mean position, mg_eff = mg + qE

g_eff = g + (qE / m)

Put it in place of g in the formula given in the starting.

So, formula becomes , T = 2(pie)( L / ( g + qE/m))^1/2

So, you get the answer.

Hope you have got it !!!!

partha

net accn of the bob in the upward dirn=10-((4*10^-6)*(2.5*10^4))
=9.9 ms^-2
as t is inversely prop to g
required t=45*(10/9.9)^1/2
=45.23 s

elessar_iitkgp

Let the time taken for n( = 20) when the electric field is off be T and T' when its on.
Then,
T/n = 2

(L/g)
(T/n)² = 4

² (L/g) ----------(1)
When the electric field is switched on,
T'/n = 2

(L/(g-qE/m))
(T'/n)² = 4

² (L/(g-qE/m)) ---------------(2)
Dividing (2) by (1)
(T'/T)² = (g/(g-qE/m))
T' = T

(g/(g-qE/m)) =

Substitute the values and get the answer.

phyana

elessar has done it correctly,ofcourse the others also.

entrance method

just calculate the effective acc=g-qE/m

time period is inversely proportional to square of acclaration.
find by how much times the acc increases.
square that value
multiply it with previuos time ths get the answer

Mr.IITIAN007

Elessar is right.That is the correct answer.

krishna.gopal

Sulekha, i think u made some mistake in righting charge on bob. Any wat elessar is right

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