# IIT JEE Physics Answers to Practice Sets

## Wednesday, December 14, 2011

## 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 1/4πε

The constant ε

ε

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

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 1/4πε

_{0}.The constant ε

_{0}is called the permittivity of the space and its value isε

_{0}= 8.85419*10^-12 C²/N-m²q1 = 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

### Questions paper files - Electricity and Magnetism

**Charge and current**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Charge%20%26%20current.pdf

**Resistance and resistivity**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Resistivity.pdf

**Circuits 1**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Circuits%201.pdf

**Circuits 2**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Circuits%202.pdf

**Circuits 3**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Circuits%203.pdf

**Circuits 4**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/AS%20Circuits.pdf

**Internal resistance of batteries**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Internal%20Resistance.pdf

**Electrical energy**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/circuit/Electrical%20Energy.pdf

**Electric field**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/A2gravity/Electric%20fields.pdf

**Magnetic field**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/A2gravity/Magnetic%20Fields.pdf

**electromagnetic induction**

http://www.emanuelschool.org.uk/physics/alevel/aprobs/A2gravity/Electromagetic%20Induction.pdf

## 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

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.

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.

## Thursday, May 17, 2007

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