1.
“Heat cannot by itself flow from a body at lower temperature to a body at higher temperature” is a statement of consequence of
(A) second law of thermodynamics
(B) conservation of momentum
(C) conservation of mass
(D) first law of thermodynamics.
Ans Second law of thermodynamics.
2. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp/CV for the gas is
(A) 4/3
(B) 2
(C) 5/3
(D) 3/2
Ans A.
3. Which of the following parameters does not characterize the thermodynamic state of matter?
(A) temperature
(B) pressure
(C) work
(D) volume
Ans C.
4.
One mole of ideal monoatomic gas ( Cp/CV= 5/30) is mixed with one mole of diatomic gas (Cp/CV = 7/5). What is Cp/CV for the mixture?
(A) 3/2
(B) 23/15
(C) 35/23
(D) 4/3
Ans A.
5 If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be
(A) 4
(B) 16
(C) 32
(D) 64.
Ans D.
6. Which of the following statements is correct for any thermodynamic system?
(A) The internal energy changes in all processes.
(B) Internal energy and entropy are state functions.
(C) The change in entropy can never be zero.
(D) The work done in an adiabatic process is always zero.
Ans B.
Thursday, 19 June 2008
Solution for IITJEE test series SHM
Mulitiple choice questions with one answer
1.
Ans c
2.
Ans b
3. Ans (d)
4. ANs (b)
5. Ans (d)
Mulitiple choice questions with more then one answer
6.
Ans b,c
7.
Ans c,d
8.
Ans c,d
9.
Ans a,c
10.
Ans a,c
Assertion and Reason
11.
Ans (d)
12.
Ans (b)
13.
Ans (c)
14.
Ans (b)
MAtrix match type
15.
Ans
A->A
B->B
C->A
D->B
Linked Comphrehension Type
16
Ans (a)
17.
ans (a)
18.
Ans (b)
1.
Ans c
2.
Ans b
3. Ans (d)
4. ANs (b)
5. Ans (d)
Mulitiple choice questions with more then one answer
6.
Ans b,c
7.
Ans c,d
8.
Ans c,d
9.
Ans a,c
10.
Ans a,c
Assertion and Reason
11.
Ans (d)
12.
Ans (b)
13.
Ans (c)
14.
Ans (b)
MAtrix match type
15.
Ans
A->A
B->B
C->A
D->B
Linked Comphrehension Type
16
Ans (a)
17.
ans (a)
18.
Ans (b)
Wednesday, 18 June 2008
IIT JEE Test series (ELectricty) solutions
Multiple choice question with only one answer
1.
Ans (a)
2.
Ans (a)
3.
Ans (A)
4.
Ans (b)
5.
Ans (A)
Multiple choice question with more than one answer
6)
Ans (all)
7)
Ans a,c
8)
Ans a,b,c
9.
Ans a,b
10.
Ans (a)
11.
Ans (c)
12.
Ans (c)
13.
Ans (a)
Matrix Match type
Ans
A-P,R
B-Q,R
C-P,R
D- P,S
Linked Comprehension Type
1)
Ans (a)
2)
Ans (b)
3)
Ans (a)
1.
Ans (a)
2.
Ans (a)
3.
Ans (A)
4.
Ans (b)
5.
Ans (A)
Multiple choice question with more than one answer
6)
Ans (all)
7)
Ans a,c
8)
Ans a,b,c
9.
Ans a,b
10.
Ans (a)
11.
Ans (c)
12.
Ans (c)
13.
Ans (a)
Matrix Match type
Ans
A-P,R
B-Q,R
C-P,R
D- P,S
Linked Comprehension Type
1)
Ans (a)
2)
Ans (b)
3)
Ans (a)
Saturday, 14 June 2008
Capacitors
-Capacitors are important components used in electronics and telecommunication devices for example radio , television recivers , transmitter circuits etc.
-Capacitor is a device used for storing electronic charge.
-All capacitors consists of two metal plates(or conductors) separated by an insulator(air, vacuum or any other dielectric medium).
-Conventional symbol of capacitor is where T is the terminal (positive or negative) of battery joined to the plates.
-Capacitor gets charged when a battery is connected to it or when there is a potential difference between two metal plates of the capacitor.
-Capacitor gets discharged on joining two of it's plates.
-If V is the potential difference between two plates of the capacitor and q is the amount of charge developed on each plate then q/V is constant for the capacitor since q∝V.
-The ratio of charge on either plate to the potential difference between the plates is called capacitance C of the capacitor. Thus,
C=q/V
or q=CV
-Unit of capacitance is Farads(F) or CV-1.
-1F is very large unit of capacitance. Practically capacitors with capacitance of the order of micro farads (μF) are used in circuits of radio recivers , transmitters etc. Thus,
1μF=10-6 (micro)
1nF=10-9 (nano)
1pF=10-12 (pico)
-For any capacitor it's capacitance is constant and depends on shape , size , separation f the two conductors and also on insulating medium being used for making capacitor.
- Capacitance of parallel plate capacitor havinf vacuum or air acting as dielectric or insulating medium is
C=(ε0A)/d
where,
C= capacitance of capacitor
A= area of conducting plate
d= distance between plates of the capacitor
ε0=8.854× 10-12 and is known as electric permitivitty in vacuum.
-If k is the relative permittivity of the dielectric medium then
ε=ε0k
thus capacitance of parallel plate air capacitor in presence of dielectric medium of electric permitivity ε is
C=εA/d
-Capacitance of spherical capacitor having radii a, b (b>a) with
(a) air as dielectric between them
C=(4πε0ab)/(b-a)
(b) dielectric with relative permitivity ε
C=(4πεab)/(b-a)
-Capacitor is a device used for storing electronic charge.
-All capacitors consists of two metal plates(or conductors) separated by an insulator(air, vacuum or any other dielectric medium).
-Conventional symbol of capacitor is where T is the terminal (positive or negative) of battery joined to the plates.
-Capacitor gets charged when a battery is connected to it or when there is a potential difference between two metal plates of the capacitor.
-Capacitor gets discharged on joining two of it's plates.
-If V is the potential difference between two plates of the capacitor and q is the amount of charge developed on each plate then q/V is constant for the capacitor since q∝V.
-The ratio of charge on either plate to the potential difference between the plates is called capacitance C of the capacitor. Thus,
C=q/V
or q=CV
-Unit of capacitance is Farads(F) or CV-1.
-1F is very large unit of capacitance. Practically capacitors with capacitance of the order of micro farads (μF) are used in circuits of radio recivers , transmitters etc. Thus,
1μF=10-6 (micro)
1nF=10-9 (nano)
1pF=10-12 (pico)
-For any capacitor it's capacitance is constant and depends on shape , size , separation f the two conductors and also on insulating medium being used for making capacitor.
- Capacitance of parallel plate capacitor havinf vacuum or air acting as dielectric or insulating medium is
C=(ε0A)/d
where,
C= capacitance of capacitor
A= area of conducting plate
d= distance between plates of the capacitor
ε0=8.854× 10-12 and is known as electric permitivitty in vacuum.
-If k is the relative permittivity of the dielectric medium then
ε=ε0k
thus capacitance of parallel plate air capacitor in presence of dielectric medium of electric permitivity ε is
C=εA/d
-Capacitance of spherical capacitor having radii a, b (b>a) with
(a) air as dielectric between them
C=(4πε0ab)/(b-a)
(b) dielectric with relative permitivity ε
C=(4πεab)/(b-a)
Thursday, 12 June 2008
IITJEE Test series Electric Potential
Multiple choice question with only one answer
1.The elctrical potential energy of an islolated metal sphere of radius R and total Charge Q
a. Q2/4πεR
b. Q2/8πεR
c. Q2/2πεR
d. Q2/16πεR
2. A electric dipole is placed at +q(-a,0) and -q(a,o) in the xy plane.Find the workdone by the electric lines on charge which is moved from point (0,a) to (0,-a).
a. zero
b. p/4πεr2
c. -p/4πεr2
d. none of the above
3.Find the electric field at the centre of the uniformly charged semicircular arc of of Radius R and linear charge density λ
a.λ/2πεa
b.λ/4πεa
c. λ/πεa
d. none of these
4.An electric dipole placed in a uniform electric field experiences
a. A torque but no force
b A force but no torque
c. Both force and torque
d neither a force and torque
5.A charge q is placed at the center of the line joining two equal charges Q.The system of three charges will be in equilibrium if q equal is
a. -Q/2
b. Q/2
c. Q/4
d. -Q/4
Multiple choice questions with one or more answer
6.Choose the correct statement
a.if electric field is zero at the point then electric potential must be also zero at that point
b.Two diffrent equipotential surface can intersect
c.if electric potential is constant in a given region then electric field must be zero in that region
d. Electrons move from higher potential to lower potential
7.A spherical conductor shell has charge Q on it and Radius of the spherical shell is R
a. Electric potential decrease with 0< r < infinity
b Electric field decrease with with 0< r < infinity
c Electric potential is non zero constant for 0< r < =R and decrease for R< r < infinity
d Electric field is zero for 0< r < =R and decrease for R< r < infinity
8.Electric potential V(x,y) of a electrostatic field E=a(yi +xj) where a is constant
a.V(x,y)-V(0,0)=-axy
b.V(x,y)-V(1,1)=-axy+a
c.V(x,y)-V(1,1)=-axy-a
d. None of the these
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
9.
STATEMENT I:Electrix flux through any closed surface around point charge is independent of the size and shape
STATEMENT II.φ=∫ E.da
10.
STATEMENT I:Electric potential inside the spherical conductor shell is nonzero constant
STATEMENT II:Electric field inside the shell is zero
11.
STATEMENT I:Electric Field on the surface of a conductor is less at the sharp corners
STATEMENT II: Surface charge density on conductor surface is inversely proportional to the radius of curvature
1.The elctrical potential energy of an islolated metal sphere of radius R and total Charge Q
a. Q2/4πεR
b. Q2/8πεR
c. Q2/2πεR
d. Q2/16πεR
2. A electric dipole is placed at +q(-a,0) and -q(a,o) in the xy plane.Find the workdone by the electric lines on charge which is moved from point (0,a) to (0,-a).
a. zero
b. p/4πεr2
c. -p/4πεr2
d. none of the above
3.Find the electric field at the centre of the uniformly charged semicircular arc of of Radius R and linear charge density λ
a.λ/2πεa
b.λ/4πεa
c. λ/πεa
d. none of these
4.An electric dipole placed in a uniform electric field experiences
a. A torque but no force
b A force but no torque
c. Both force and torque
d neither a force and torque
5.A charge q is placed at the center of the line joining two equal charges Q.The system of three charges will be in equilibrium if q equal is
a. -Q/2
b. Q/2
c. Q/4
d. -Q/4
Multiple choice questions with one or more answer
6.Choose the correct statement
a.if electric field is zero at the point then electric potential must be also zero at that point
b.Two diffrent equipotential surface can intersect
c.if electric potential is constant in a given region then electric field must be zero in that region
d. Electrons move from higher potential to lower potential
7.A spherical conductor shell has charge Q on it and Radius of the spherical shell is R
a. Electric potential decrease with 0< r < infinity
b Electric field decrease with with 0< r < infinity
c Electric potential is non zero constant for 0< r < =R and decrease for R< r < infinity
d Electric field is zero for 0< r < =R and decrease for R< r < infinity
8.Electric potential V(x,y) of a electrostatic field E=a(yi +xj) where a is constant
a.V(x,y)-V(0,0)=-axy
b.V(x,y)-V(1,1)=-axy+a
c.V(x,y)-V(1,1)=-axy-a
d. None of the these
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
9.
STATEMENT I:Electrix flux through any closed surface around point charge is independent of the size and shape
STATEMENT II.φ=∫ E.da
10.
STATEMENT I:Electric potential inside the spherical conductor shell is nonzero constant
STATEMENT II:Electric field inside the shell is zero
11.
STATEMENT I:Electric Field on the surface of a conductor is less at the sharp corners
STATEMENT II: Surface charge density on conductor surface is inversely proportional to the radius of curvature
Monday, 2 June 2008
Electric Flux and Gauss law
Electric Flux
dφ=E.da
da is the area vector to the surface and it is taken +ve along the outward normal to the surface
dφ=Edacosθ
φ=∫ E.da
For closed surface
φ=∫ E.da
Guass Theorem
Flux in closed surface is equal net charge inside divided by ε
∫ E.da=qin/ε
Some points:
a. E is the electric field present due to all charges in the ssystem not just the charge inside
b.Flux crossing a closed surface does not depend on the shapes and size of gaussian surface
Electric Potential energy of a charge
=qV where V is the potential there
Others important things
1. ∫ E.dl over closed path is zero
2.Electric potential in the spherical charge conductor is Q/4πεR where R is the radius of the shell and the potential is same everywhere in the conductor
3 Conductor surface is a equipotential surface
dφ=E.da
da is the area vector to the surface and it is taken +ve along the outward normal to the surface
dφ=Edacosθ
φ=∫ E.da
For closed surface
φ=∫ E.da
Guass Theorem
Flux in closed surface is equal net charge inside divided by ε
∫ E.da=qin/ε
Some points:
a. E is the electric field present due to all charges in the ssystem not just the charge inside
b.Flux crossing a closed surface does not depend on the shapes and size of gaussian surface
Electric Potential energy of a charge
=qV where V is the potential there
Others important things
1. ∫ E.dl over closed path is zero
2.Electric potential in the spherical charge conductor is Q/4πεR where R is the radius of the shell and the potential is same everywhere in the conductor
3 Conductor surface is a equipotential surface
Sunday, 1 June 2008
Electric Potential
Electric Potential Energy
ΔU=-W
Where ΔU = Change in Potential energy
W= Workdone by the electric lines of forces
For a system of two particles
U(r)=q1q2/4πεr
where r is the seperation between the charges
We assume U to be zero at infinity
Similarliy for a system of n charges
U=Sum of potential energy of all the distinct pairs in the system
For example for three charges
U=(1/4πε)(q1q2/r12+q2q3/r23+q1q3/r13)
Another way to represent
U=1/2ΣqV
where V is the potential at charge q due to all the remaining charges
Electric Potential:
Just liken Electric field intensity is used to define the electric field,we can also use Electric Potential to define the field
Potential at any point P is equal to th workdone per unit test charge by the external agent in moving the test charge from the refrence point(without Change in KE)
Vp=Wext/q
So for a point charge
Vp=Q/4πεr
where r is the distance of the point from charge
Some points about Electric potential1. It is scalar quantity
2.Potential at point due to system of charges will be obtained by the summation of potential of each charge at that point
V=V1+V2+V3+V4
3.Electric forces are conservative force so workdone by the electric force between two point is independent of the path taken
4. V2-V1=-∫ E.dr
5 In cartesion coordinates system
E=Exi+Eyj+Ezk
dr=dxi+dyj+dzk
Now
dV=-E.dr
So dv=-(Exdx+Eydy+Ezdz)
So Ex=∂V/∂x
Similary
Ey=∂V/∂y and Ez=∂V/∂z
Also
E=-[(∂V/∂x)i+(∂V/∂y)j+(∂V/∂z)k]
4
Surface where electric potential is same everywhere is call equipotential surface
Electric field components parallel to equipotential surface is always zero
Electric dipole:
A combination of two charge +q and -q seperated by the distance d
p=qd
Where d is the vector joiing negative to positive charge
Electric potential due to dipole
V=(1/4πε)(pcosθ/r2)
where r is the distance from the center and θ is angle made by the line from the axis of dipole
Electric field
Eθ=(1/4πε)(psinθ/r3)
Er=(1/4πε)(2pcosθ/r3)
Total E=√Eθ2+Er2
=(p/4πεr3)(√(3cos2θ+1))
Torque on dipole=pXE
Potential Energy
U=-p.E
ΔU=-W
Where ΔU = Change in Potential energy
W= Workdone by the electric lines of forces
For a system of two particles
U(r)=q1q2/4πεr
where r is the seperation between the charges
We assume U to be zero at infinity
Similarliy for a system of n charges
U=Sum of potential energy of all the distinct pairs in the system
For example for three charges
U=(1/4πε)(q1q2/r12+q2q3/r23+q1q3/r13)
Another way to represent
U=1/2ΣqV
where V is the potential at charge q due to all the remaining charges
Electric Potential:
Just liken Electric field intensity is used to define the electric field,we can also use Electric Potential to define the field
Potential at any point P is equal to th workdone per unit test charge by the external agent in moving the test charge from the refrence point(without Change in KE)
Vp=Wext/q
So for a point charge
Vp=Q/4πεr
where r is the distance of the point from charge
Some points about Electric potential1. It is scalar quantity
2.Potential at point due to system of charges will be obtained by the summation of potential of each charge at that point
V=V1+V2+V3+V4
3.Electric forces are conservative force so workdone by the electric force between two point is independent of the path taken
4. V2-V1=-∫ E.dr
5 In cartesion coordinates system
E=Exi+Eyj+Ezk
dr=dxi+dyj+dzk
Now
dV=-E.dr
So dv=-(Exdx+Eydy+Ezdz)
So Ex=∂V/∂x
Similary
Ey=∂V/∂y and Ez=∂V/∂z
Also
E=-[(∂V/∂x)i+(∂V/∂y)j+(∂V/∂z)k]
4
Surface where electric potential is same everywhere is call equipotential surface
Electric field components parallel to equipotential surface is always zero
Electric dipole:
A combination of two charge +q and -q seperated by the distance d
p=qd
Where d is the vector joiing negative to positive charge
Electric potential due to dipole
V=(1/4πε)(pcosθ/r2)
where r is the distance from the center and θ is angle made by the line from the axis of dipole
Electric field
Eθ=(1/4πε)(psinθ/r3)
Er=(1/4πε)(2pcosθ/r3)
Total E=√Eθ2+Er2
=(p/4πεr3)(√(3cos2θ+1))
Torque on dipole=pXE
Potential Energy
U=-p.E
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