IIT JEE Physics Answers to Practice Sets: 2008

Thursday, June 12, 2008

Ch. 29 Electric Field - MQ -1

Prob: Determine the force between two electrons spaced 0.1 nm apart.

Concepts to be used

Coulomb's formula for the electric force between two charges.

(1) F = k*q1*q2/r²

q1,q2 charges
r = separation between charges
k = constant
In SI units k is measured to be 8.98755*10^9 N-m²/C²

The constant k is often written as 1/4πε₀.
The constant ε₀ is called the permittivity of the space and its value is

ε₀ = 8.85419*10^-12 C²/N-m²

q1 = q2 = -1.6*10^-19 C
r = 10^-10 m

F = [9*10^9(1.6*10^-19)²]/(10^-10)²

= 9*2.56(10^(9-38+20))
= 23.04(10^-9)
= 2.304*10^-8

Thursday, May 15, 2008

Question bank - Physics

143 page question bank

http://www.wapa.gov/rm/ScienceBowlRM/questions/97_PHYS.PDF

Questions paper files - Electricity and Magnetism

Tuesday, April 8, 2008

HCV Ch.8 Work and Energy - Problems

Problem 50

From an orkut discussion

Post 1
-------
}
}====>spring
}A
!'''''''!-------------------()-====>pulley
---horizontal surface] !
--------------------------] !
--------------------------] !
--------------------------] !''''''!====>block B

fig. shows 2 blocks A and Beach having a mass of 320g connected by a light string passing over a smooth light pulley.the horizontal surface on which the block A can slide is is smooth.the block A is attached to a springof spring constant 40N/m whose other end is fixed to a support 40cm above the horizontal surface.initially the spring is vertical and unstretched when the system is rreleased to move.find the velocity of the block A at the instant it breaks off the surface below it.g=10m/s^2

Post 2
hey the answer is 1.5m/s^2

i started off the problem by assuming that there will be no extension in the spring until the block A travels a distance L0sinteeta[L=the length of the spring, teeta the angle the spring makes with the vertical]
then i applied conservation of energy[before and after extension]
then i got struck!!!!!!!!!

is this correct what i did?

if not please let me know how to start of with this problem...

Post 3

the exact ans is 1.54m/s..........

when d block has moved a dist x then the vertical component of force of spring balance mg i.e. 3.2N and we calculate extention in spring from there ...... extention comes out to be 0.1m ,,....then apply cons of energy between initial and final pos........

E1= -mgh
E2= 1/2mv^2 -mg(h+x)
E1=E2
h is the length of hanging block which will cancel out x being = Ltan (theeta)

and theeta is calculated when we know extention of spring

........so v=1.54m/s

Saturday, March 22, 2008

JEE Past Objective Questions -Electricity - 1

1983 Screening

1. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 Volts. The potential at centre of the sphere is

a) zero b) 10 V
c) same as at a point 5 cm away from the surface
d) same as at a point 25 cm away from the surface

Answer (b) 10 V

Reason: for a hollow metallic charged sphere the potential inside the sphere is constant and is equal to the potential at the surface.

See HC Verma page 135 point (f)
The electric potential due to a uniformly charged thin spherical shell at an internal point is the same everywhere and is equal to that at the surface.

2. Write the whether the statement is true or false. give brief reasons in support of your answer

Two identical metallic spheres of exactly equal masses are taken. One is given positive charge Q and the other an equal negative charge. Their masses after charging are different.

Answer: True
Reason: giving positive charge to a body involved removal of electrons. giving negative charge involves addition of electrons.

IIT JEE Physics Answers to Practice Sets