In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well.
But for big addition problems, where the limits could reach to even infinity, integration methods are used. Integration and differentiation both are important parts of calculus. The concept level of these topics is very high. Hence, it is introduced to us at higher secondary classes and then in engineering or higher education.
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We can say, integration is the calculation of an integral. Integrals in mathematics are used to find many useful quantities such as areas, volumes, displacement, etc. When we speak about integrals, it is related to usually definite integrals. The indefinite integrals are used for anti-derivatives. Integration is one of the two major calculus topics in Mathematics, apart from differentiation(which measure the rate of change of any function with respect to its variables). It’s a vast topic which is discussed at higher level classes like in Class 11 and 12
To download the important questions of indefinite integration click below.
Answers of all the questions are given at the end of all questions of the assignment.
1. for which if the following depends of drift velocity v on electric field E is Ohm’s law obeyed ? (a) v α E (b) v= constant (c) v α E ½ (d) v α E 2 .
2. A potential difference is applied across the end of the metallic wire . If the potential difference is doubled , the drift velocity ; (a) will be doubled, (b) will be halved, (c) will be quadrupled, (d) no change .
3. The maximum current that flows in the fuse wire , before it blows out , varies with the radius r as :(a) r3/2 (b) r (c) r2/3 (d) r ½
4. The mass of three wires of copper are in the ratio of 1:3:5 and their length are in the ratio 5:3:1 . T he ratio of their electric resistance is ; (a) 1:3:5 (b) 5:3:1 (c) 1:15:125 (d) 125:15:1
5. The temperature (T) dependence of resistivity of a semiconductor represented by :
6. A wire of resistance R b= is stretched till its radius is half of its original value . Then the radius of the stretched wire is ; (a) 2R (b) 4R (c) 8R (d) 16R :
7. A wire till its length is doubled of its original length , then the resistance of the stretched wire is; (a) 2R (b) 4R (c) 8R (d) 16R :
8. If a copper wire is stretched to make it 0.1% longer, then the percentage change in its resistance is approximately (a) 0.1% (b) 0.2 % (c) 0.4% (d) 0.8% .
9. If a copper wire is stretched to make its radius decrease by 0.1% , then the % change in its resistance is approximately ; (a) 0.1% (b) 0.2 % (c) 0.4% (d) 0.8% .
10. A current of 5 amp exist in the 10 ohm resistance, for 4 minutes . How many coulombs pass through any cross-section of the resistor in this time ? (a) 12 (b) 120 (c) 1200 (d) 12000.
11. The temperature coefficient of resistance of a wire is 0.00125/0C at 300K is one Ohm. The resistance of the wire will be 2 ohm at (a)1154K (b) 1100k (c) 1400k (d) 1127k .
12. Drift speed of electrons is of the order of ; (a) zero (b) 10-2 (c) 105 (d) 100m/s ;
13. Which represents the negative dynamic resistance?
(a) AB (b) BC (c)CD (d) DA
14. When a current flows in a wire , there exist in electric field in the direction of ; (a) flow of current, (b) opposite to the flow of current (c) perpendicular to the flow of current (d)at the angle of 450 to the flow of current.
15. The drift velocity of free electron in a conductor is v when a current I is flow in it . If both the radius and current are doubled , then drift velocity will be; (a) v (b) v/2 (c) v/4 (d) v/8 .
16. The sides of rectangular block are 2,3 and 4cm . the ratio of maximum to minimum resistance between its parallel face is; (a) 4 (b)3 (c) 2 (d) 1 ;
17. The electric resistance of certain wire is R, if its length and radius both are doubled , then ; (a) resistance will be doubled and the specific resistance will be halved (b) resistance will be halved and the specific resistance will be remain same (c) resistance will be halved and the specific resistance will be doubled (d) resistance will be unchanged and also the specific resistance will be unchanged.
18. Total no. of electrons present in 4 amp current flowing for 1 sec is ; (a) 2.5 x 109 (b)2.5x 1020 (c) 1.25×1013 (d) 6x 108 .
19. If the electric current through a bulb is 3.2 A, the no of electrons flowing through it in 1 sec is; (a) 2 x 109 (b) 2 x 1019 (c) 3.2 x 109 (d) 1.6 x 1018 .
20. The resistance of a wire is 5 ohm at 500c and 6 ohm at 1000c , the resistance of the wire at 00c will be; (a) 3 ohm (b) 2 ohm (c) 1 ohm (d) 4 ohm .
21. Two wires have length ,diameter and resistivity all are in the ratio 1:2 the resistance of the first wire is 10 ohm, the resistance of the second wire is ; (a) 5 (b) 10 (c) 20 (d) infinity.
22. Two copper wire have their masses in the ratio 2:3 and the length in the ratio 3:4 , the ratio of their resistance is ; (a) 4:9 (b) 27:32 (c) 16:9 (d) 27:128 (e) 1:2 ,
23.On increasing the temperature of the conductor its resistance increases because the ; (a) relaxation time increases (b) mass of electron increases (c) electron density decreases (d) relaxation time decreases .
24 .The resistance of a wire is R ohm . It is melted and stretched to n times its original length , the new resistance will be ; (a) R/n (b) n2R (c)R/ n2 (d) nR .
25. The current flowing through a wire depends on time as I = 3t2 +2t +5 . The charge flowing through the cross-section of the wire in time from t=0 to t=2 sec is; (a) 22c (b) 20c (c) 18c (d) 5c.
ans- Ans- 1-a ;2- a ; 3-a ;4- d ;5-b ;6-d ; 7-b ; 8-b ; 9-c ;10-c ; 11-d; 12- b ; 13 -c; 14-a ;15-b ;16- a; 17-b ; 18-a ; 19-b ; 20-d ; 21-b ; 22-b ; 23-d ;24-b ;25-a
In this topic we will get some, Very important questions of ray optics. Students preparing for board examination must solve all the questions given below.
Very important questions of ray optics
1. Derive the mathematical relation between refractive indices μ1 and μ2 of two radii and radius of curvature R for refraction at a convex spherical surfaces. Consider the object to be a point source on the principle axis in rarer medium of refractive index μ1 and a real image formed in the denser medium .
Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm . The distance of light source from the glass surface is 100cm . At what position the image will be formed?
2. A point object is placed in front of a double convex lens with its spherical faces of radii of curvature R1 and R2 . Show the path of rays due to refraction at first and subsequently at the second surface to obtain the formation of the real image of the object. Hence obtain the lens maker’s formula for a thin lens.
(i) A double convex lens having both faces of the same radius of curvature has refractive index 1.5. Find out the radius of curvature of the lens required to get the focal length of 20 cm . (ii) How the focal length of the above lens changes when it is placed in liquid of refractive index (a) 1.3 (b) 1.5 and (c)2.5 .
To solve this questions students must watch this video
3. Draw the ray diagram to show the refraction of a ray of monochromatic light passing through a glass prism. Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.
A ray of light incident on face AB of an equilateral glass prism ABC , shows minimum deviation of 300 calculate the speed of light through the prism. Also find the angle of incidence at face AB so that the emergent ray along the face AC .
4.What do you mean by the term total internal reflection also write the conditions to get the total internal reflection, define the term critical angle for a pair of media , how the critical angle is related to the refractive index of the medium. A point source is kept at the bottom of the pond 2m deep ( μ= 4/3 )draw the ray diagram and calculate the area of water surface through which light emerges out of the pond.
5. Draw a ray diagram showing the image formation of compound microscope, hence obtain the expression for magnifying power when (I) image is formed at infinity (II) image is formed at least distance of distinct vision. (a)Why must both the objective and eyepiece of a compound microscope have short wavelength? (b) while viewing through a compound microscope, why should our eyes be placed not on the eye piece but at short distance away from it for best viewing?
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6. Draw a labelled ray diagram of astronomical telescope, for(I) normal adjustment, (II) at least distance of distinct of vision. Also derive the expression for magnifying power .You are given the three lance L1 , L2 , L3 having power 3D , 6D and 10D respectively with aperture 8cm,1cm and 1cm respectively, which two lances will you use as an eyepiece and objective to construct (a) compound microscope and (b) Astronomical telescope. Give reasons.
7.Explain reflecting type telescope draw the ray diagram for (I) Newtonian and (II)Cassegrainian type telescope. Write advantages of reflecting type telescope over refracting type telescope.
8.Derive lens formula in a convex lens using suitable ray diagram 1/v -1/u = 1/f . (a) A convex les is used to obtain the image of an object on the screen 10 cm from the lens . if the magnification is 19, find the focal length of the lens. (b) An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen , twice the size of the object.
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In this assignment Complete assignment of magnetic effect of current all kind of questions , objective , short answer type questions, long answer type questions and all kind of numericals with hots questions.
After solving these questions students will be able to solve the all kind of questions related to this chapter. Specially objective type question will help the students to solve the questions for all competitive examinations like JEE and NEET along with board examinations.
physics classes by Nayan jha already provided assignment of electric current . But this is the Complete assignment of electric current . In students can solve all kind of questions like objective,fill in the blanks,very short, short , long answer type , numericals and hots questions . This assignment will help the students to prepare for 12th board and all kind of competitive examination like JEE and NEET .Specially the objective questions are very helpful for board and competitive examinations.
Before to solve this assignment students must learn complete assignment of electric current .
For which of the following dependence of drift velocity ʋ on electric field E is ohm’s law obeyed? (a)ʋ ꭀ E (b) ʋ = constant (c) ʋ ꭀ E1/2 (d) ʋ ꭀ E2
Ans-a
in metals the time of relaxation of electrons:
(a)increases with increasing temperature
(b)decreases with increasing temperature
(c)does not depend on temperature
(d)changes suddenly at 400 K
Ans-b
Constantan wire is used for making standard resistance because it has:
(a)low specific resistance
(b)high specific resistance
(c)negligible temperature coefficient of resistance
(d)high melting point
Ans-c
A potential difference is applied across the ends of a metallic wire. If the potential difference is doubled, the drift velocity:
(a)will be doubled (c) will be quadrupled
(b)will be halved (d) will remain unchanged
Ans-a
The maximum current that flows in the fuse wire, before it blow out, varies with the radius r as:
(a)r3/2 (b) r (c) r2/3 (d) r1/2
Ans-a
A wire of resistance R is cut into n equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be:
(a)nR (b) R/n (c) n/R (d) R/n2
Ans-d
The masses of three wires of copper are in the ratio of 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of their electrical resistances is:
Kirchhoff’s first law is based on the law of conservation of:
(a) charge (b) energy (c) momentum (d) sum of mass and energy
Ans-a
Kirchhoff’s second law is based on the law of conservation of:
(a)charge (b) energy (c) momentum (d) sum of mass and energy
Ans-b
A call of emf E is connected across a resistance r. The potential difference between the terminal of the cell is found to be V. The internal resistance of the cell must be:
(a) 2(E – V)V/r (b) 2(E – V)r/E (c) (E – V)r/V (d) (E – V)r
Ans-c
PDF of the Complete assignment of electric current is given below-
Students as you know , physics classes by Nayan jha already provided many kind of assignment of the chapter Electrostatics . But this is the Complete assignment of electrostatics . It contains objective type,fill in the blanks,1 marks,2 marks,3 marks and 5 marks questions with numerical and HOTS .
It will help the students to revise complete chapter .which will help the students preparing for 12th board and all kind of competitive examinations .
Before to solve the questions of these assignment student must learn complete notes of these two chapters , electric charges and field and electrostatic potential and capacitance .
After revise the entire chapters of electric charges and field and electrostatic potential and electrostatics.Students may solve the questions of related chapters. Here is the Complete assignment of electrostatics –
In this topic we will discuss about the Energy stored in a capacitor and energy density, when a capacitor is charged from zero to Q .
ENERGY STORED IN A CAPACITOR –
Suppose the plates of the capacitor is initially uncharged , and the charge is to be transferred from plate 2 to plate 1 , at a small time charge dQ transferred and at the final time it reaches from 0 to Q .
Let V is the potential difference across the plates then work done for small instant is
dW = V . dQ = (Q/C)dQ .
integrating both sides,
∫ dW =∫(Q/C)dQ from limit 0 to Q , then we get ,
net work done W = Q2/2C ,
also as we know Q=CV, so work done may be written as ,
Numerical based on electric charge ,electric field .
Numerical based on electric dipole ( electric field at axial and equatorial point, torque and potential energy stored in dipole )-
Two charges + 30μc and -30μc are placed 1cm apart . calculate electric field at a point on the the axial and equatorial line at a distance of 20 cm from the centre of the dipole .
Ans- 6.25 x 105 N/C and 3,25 x 105‑ N/C .
An electric dipole of dipole moment 4 x 10-5 C-m is placed in uniform electric field of 103 N/C making an angle of 300 with the direction of field . Determine the torque exerted by the electric field on the dipole. Ans- 2 x 10-8Nm .
An electric dipole is placed at an angle of 600 with an electric field of magnitude 4 x 105 N /C . It experience a torque of 8√3 Nm . If the length of the dipole is 2cm , determine the magnitude of either charge of the dipole . Ans – 2 x 10-3 C .
An electric dipole of length 10cm having charges of magnitude 6 x 10-3 C , placed at 300 with respect to a uniform electric field experience a torque of magnitude 6√3 N-m . Calculate (i) magnitude of the field and (ii) potential energy of the dipole . Ans- 2√3x 104; -18J
An electric dipole consist of two opposite charges of magnitude q=1×10-6C separated by 2.0 cm . The dipole is placed in an external electric field of 1×105 N/C . What maximum torque does the field exert on the dipole ? How much work must an external agent to do turn the dipole end for end , starting from position of alignment ( ϴ=00 ) . Ans- 2×10-3 N-m ; 4×10-3J .
(i)Two point charges 4Q and Q are separated by 1m in air at what point on the line joining the two charges the electric field intensity is zero. (ii) Two identical metallic sphere A and B having charges +4Q and -10Q are kept at a certain distance apart . A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B . Spheres A and B are then brought in contact and then separated . Find the charges on the sphere A and B .
7.Calculate the amount of work done in turning an electric dipole of dipole moment 3×108 C-m from its position of unstable equilibrium to stable equilibrium, in a uniform electric field of intensity 103N/c.
8.The sum of two point charges is 7µc they repel each other with a force of 1N when kept 30 cm apart in free space , calculate the value of each charge.
This topic contain numerical for the practice of the chapter electric field
Before to solve the numerical based on electric field students has to study ( learn ) , and all the notes based on electric field , electric dipole ( axial and equatorial point)
Two point charges +16μc and -9μc are placed 8cm apart in air . Determine the position of the point at which the resultant electric field is zero . [ Ans- 24 cm]
A particle of mass m and charge q is thrown at speed u against a uniform electric field E . How much distance will it travel before coming to momentary rest . ? [Ans- mu2/2qE]
A particle of mass m and charge q is released from rest in uniform electric field of intensity E . Calculate the kinetic energy it attains after moving a distance a distance x between the plates . [Ans- qEx]
Eight identical point charges of q coulomb each are placed at the corners of a cube of each side 0.1 m . calculate the electric field at the centre of the cube . calculate the field at the centre when one of the corner charge is removed . [Ans – 0 ; 1.2 x 1012 .]
Two charges + 30μc and -30μc are placed 1cm apart . calculate electric field at a point on the the axial and equatorial line at a distance of 20 cm from the centre of the dipole . [ Ans- 6.25 x 105 N/C and 3,25 x 105‑ N/C .]
Four particles each of charge q are be placed on the vertices A ,B, C, D of a regular pentagon ABCDE . the distance of each corner from the centre is ‘a’ . Calculate the electric field at the centre of the pentagon . [ Ans- q/4∏ϵ0 a2 along OE .]
A small sphere of mass 1g carries a charge of +6μc . The sphere is suspended by a string in an electric field of 400N/C acting downwards. Calculate the tension in the string . What will be the tension if the charge on the sphere be -6μc . [Ans- 1.22x 10-2 N . ; 74 x 10-4 N .]
An electron falls through a distance of 1.5 cm in uniform electric field of value 2 x104N/C . when the direction of electric field is reversed , a proton falls through the same distance . compare the time of fall in each case .
A particle of mass ‘m’ and charge -q enters the region between two charged plates initially moving along x-axis with speed vx ( as shown in the fig) . The length of the plate is L and an uniform electric field E is maintained between the plates . Show that the vertical deflection of the particle at the far edge of the plate is QelL2 / (2m vx2) .
As shown in the figure three charged particles in a uniform electrostatic field . Give the sign of three charges . Which particle has the highest charge to mass ratio .
this assignment will provide the important Objective questions of ray optics for students preparing for class-12th board examination and for the students preparing for JEE/NEET and other competitive examinations.
Reflection and Refraction of Light at Plane Surface ( Objective questions of ray optics)
When light travels from glass to air, the incident angle is and the refraction angle is .the true relation is :(a)θ 1 =θ 2 (b) θ 1 >θ2 (c)θ1 <θ2 (d)Not predictable
Two mirrors are kept at 60 to each other and a body is placed in the middle.The total number of images formed is:(a)Three(b)Four (c)Five (d)Six
If an object is placed unsymmetrically between two plane mirrors inclined at an angle 72 ,then the total number of images formed is:(a)5 (b)4 (c)2 (d)infinite
If an object is placed symmetrically between two plane mirrors, inclined at an angle of 72 .than the total total number of images formed is:(a)5 (b)4 (c)2 (d)Infinite
5) A ray of light is incident on a plane mirror at an angle of incidence of 30⁰ ,the deviation
Produced by the mirror is :
(a) 30 ⁰ ( b) 60⁰
(c) 90⁰ (d) 120
6) when a ray of light enters a glass slab from air:
(a) its wavelength decreases
(b) its wavelength decreases
(c) its frequency increases
(d) neither wavelength nor frequency changes
7) When a light enters from air to water, then its:
(a) frequency increases and speed decreases
(b) frequency is the same but the wavelength is smaller in water then in air:
(c) frequency is same but wavelength in water is greater than in air
