InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Eight charges, 1muC, -7muC , -4muC, 10muC, 2muC, -5muC, -3muC, and 6muC`, are situated at the eight cornersof a cube of side 20 cm. A spherical surface of radius 80 cm encloses this cube. The center of the sphere coincides with the center of the cube. Then, the total outgoing flux from the spherical surface (in units of Vm) isA. `36 pi xx 10^3`B. `684 pi xx 10^3`C. zeroD. none of these |
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Answer» Correct Answer - C c. `1-7-4+10 +2 - 5 - 3 +6 = 0` Sum of all the charges is zero, so net flux is zero. |
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| 2. |
A charge q is placed at the centre of the open end of a cylindrical vessel . The flux of the electric field through the surface of the vessel is A. ZeroB. `q/(epsilon_(0))`C. `q/(2epsilon_(0))`D. `(2q)/(epsilon_(0))` |
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Answer» Correct Answer - c |
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| 3. |
The electric field in a region is radially outward with magnitude `E=Ar`. Find the charge contained in a sphere of radius a centred at the origin. Take `A=100 V m^(-2)` and `a=20.0 cm`. |
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Answer» Correct Answer - `3 xx 10^(-10) C` Consider a spherical shell of radius x. The electric flux through this surface is `phi = int vecE * dvecS = E_r4pir^2` Therefore, the electric flux through spherical suface of radius R will be `phi = int vecE * d vecS = E_r 4pir^2` When `r = R, E_R = alphaR`, we have `phi = alphaR4piR^2` By Gauss theorem, net electric flux is `1/epsilon_0 xx` change enclosed :. `alpha R4piR^2 = 1/epsilon_0 Q_(encl osed)` or `Q_(enclosed) = (4piepsilon_0)alphaR^3` Given `R = 0.30m, alpha = 100 Vm^(-2)` `Q_(en clo sed) = 1/(9 xx 10^(9)) xx 100 xx (0.30)^3 = 3 xx 10^(-10)C`. |
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| 4. |
Figure, shown above, shows three situations involving a charged particle and a uniformly charged spherical shell. The charges and radii of the shells are indicated in the figure. If `F_(1),F_(2)` and `F_(3)` are the magnitudes of the force on the particle due to the shell in situations (I),(II) and (III) then A. `F_(3) gtF_(2) gtF_(1)`B. `F_(1) gtF_(2)=F_(3) `C. `F_(3)= F_(2)gtF_(1) `D. `F_(1)gt F_(2)gtF_(3) ` |
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Answer» Correct Answer - c |
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| 5. |
A linder charge having linear charge density `lambda` pentrates a cube diagonally and then it pentrates a spere diametrically as shown. What will br the ratio of flux coming out of cube and sphere ? A. `1/2`B. `2/ (sqrt3)`C. `(sqrt3)/2`D. `1/1` |
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Answer» Correct Answer - c |
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| 6. |
A cylinder of radius R and length L is placed in a uniform electric field E parallel to the axis. The total flux for the surface of the cylinder is given byA. `2piR^(2)E`B. `piR^(2)//E`C. `(piR^(2)- piR) E `D. zero |
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Answer» Correct Answer - d |
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| 7. |
In a certain region of space, there exists a uniform electric field of value ` 2xx 10^2hatkVm^(-1)`. A rectangular coil of dimension `10 cm xx 20 cm ` is placed in the xy plane. The electric flux through the coil isA. zeroB. 30 V mC. 40 V mD. 50 V m |
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Answer» Correct Answer - C c. `vecE = 2000hatk, vecA = 10 xx 20 xx 10^(-4)hatk` `phi = vecE * vecA = 40 Vm`. |
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| 8. |
A spherical conductor `A` contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge `q_(1)` at the center of one cavity and `q_(2)` at the center of the other cavity . Another charge `q_(3)` is placed at a large distance `r` from the center of the spherical conductor. The force acting on conductor `A` will beA. zeroB. `(q_(3) (q_(1) + q_(2)))/(4 pi epsilon_(0) r^(2))`C. `( - q_(3) (q_(1) + q_(2)))/(4 pi epsilon_(0) r^(2))`D. `(q_(3) q_(1) + q_(2) q_(3) + q_(1) q_(2))/( 4 pi epsilon_(0) r^(2))` |
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Answer» Correct Answer - B We see that that surface of two spherial cavities carries charges `q_(1), - q_(2)` respectively, and since the sphere A was not charged orginally, its sphereical surface must carry induced charges `(q_(1) + q_(2))`. Therefore, force acting on conductor A is `q_(3) (q_(1) + q_(2))//4 pi epsilon_(0) r^(2)` |
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| 9. |
A spherical conductor `A` contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge `q_(1)` at the center of one cavity and `q_(2)` at the center of the other cavity . Another charge `q_(3)` is placed at a large distance `r` from the center of the spherical conductor. If `q_(1)` is displaced from its center slightly (being always inside the same cavity ), then the correct representation of field lines inside the same cavity isA. B. C. There will be no field lines inside the cavity.D. |
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Answer» Correct Answer - B Thereis charge inside cavity , so field lines will be present , but the potential at surface of conductor is same everywhere . So field lines should be perpendicular . |
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| 10. |
A spherical conductor `A` contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge `q_(1)` at the center of one cavity and `q_(2)` at the center of the other cavity . Another charge `q_(3)` is placed at a large distance `r` from the center of the spherical conductor. Which of the following statements are true ?A. Charge `q_(3)` applies a larger force on the charge `q_(2)` than on charge `q_(1)`.B. Charge `q_(3)` applies a smaller force on the charge `q_(2)` than on charge `q_(1)`.C. Charge `q_(3)` applies equal force on both the charges.D. Charge `q_(3)` applies no force on any of the charges. |
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Answer» Correct Answer - D Charges outside cavity of a conductor have no influence on the field inside cavity. |
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| 11. |
The cube in figure has edge length `sqrt2m` and is oriented as shown in a region of uniform electric field , in newtons per coulomb, is given by `(a)6.00hati , (b) -2.00 hatj, and (c) -3.00 hatj+ 4.00 hatk` . (d) what is the total flux through the cube for each field ? |
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Answer» Correct Answer - (a) zero `(b) -4.0N.m^(2)//C` © zero (d) zero |
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| 12. |
In a region of space, the electric field is given by `vecE = 8hati + 4hatj + 3hatk`. The electric flux through a surface of area 100 units in the xy plane isA. 800 unitsB. 300 unitsC. 400 unitsD. 1500 units |
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Answer» Correct Answer - B b. `vecE = 8hati + 4hatj + 3hatk, , vecA = 100hatk` So `phi = vecE * vecA = 300 units`. |
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| 13. |
A suface has the area vector `vecA(2hati + 3hatj)` what is the flux erof a uniform electric though the area if the field is a`vecE=4hatiN//c and vecE = 4 hatkN//C` ? |
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Answer» Correct Answer - `8N - m^(2)//C; (b)0` |
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| 14. |
Three charges `q_1 = 1 xx 10^(-6) C, q^2 = 2 xx 10^(-6)C, and q_3 = -3 xx 10^(-6) C ` have been placed as shown in figure. Then the net electric flux will be maximum for the surface A. `S_1`B. `S_2`C. `S_3`D. same of all three |
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Answer» Correct Answer - A a. Charge inside `S_1` = `q_1 +q_2 = 3 xx 10^(-6)C` Charge inside `S_2` = `q_2 + q_3 = -1 xx 10^(-6)C` Charge inside `S_3` = 0 Charge inside `S_1` is greatest. So, flux through `S_1` is maximum. |
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| 15. |
Figure shows four charges `q_1,q_2,q_3,and q_4` fixed is space. Then the total flux of the electric field through a closed surface S, due to all the charges, is A. not equal to the total flux through S due to `q_3 and q_4`B. equal to the total flux through S due to `q_3 and q_4`C. zero if `q_1 + q_2 = q_3 +q_4`D. twice the total flux through S due to `q_3 and q_4 if q_1+q_2 = q_3+q_4`. |
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Answer» Correct Answer - B b. Net flux is due to charges inside S only. |
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| 16. |
In the figure a hemispherical bowl of bowl of radius R is shown Electric field of intensity E is present perpendicular to the circular cross section of the hemisohere. A. The magnitude of electric flux through the hemispherical surface in sittation I will be `EpiR^(2)`B. The magniude of electric flux through the hemisoherical surface in surface II will be `2EpiR^(2)`C. the magnitude of electric flux thorugh the hemisperical surface in situation II will be `(EpiR^(2))/2`D. the magnitude of electric flux through the hermispherical surface in situation II will be zero. |
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Answer» Correct Answer - b,d |
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| 17. |
Figure showns four solid spheres, each with charge Q uniformly distributed through its volume. The figure also shows a point P for each sphere. All at the same distance from the centre of the sphere. Rank the spheres according to the magnitude of the electric field they produce at point P. A. ` I and II" tie " to III to IV`B. `II to III to IV`C. `III to II to I to IV`D. all tie |
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Answer» Correct Answer - a |
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| 18. |
A ring of radius R is placed in the plane its centre at origin and its axis along the x-asis and having uniformly distributed positive charge. A ring of radius `r(ltltR)` and coaxial with the larger ring is moving along the axis with constant velcity , then the variation of electrical flux `(phi)` passing through the smaller ring with position will be best represent by : A. B. C. D. |
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Answer» Correct Answer - c |
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| 19. |
Electric charge is uniformly distributed along a along straight wire of radius `1 mm`. The charge per centimeter length of the wire is `Q` coulomb. Another cyclindrical surface of radius `50 cm` and length `1 m` symmetrically enclose the wire ask shown in figure. The total electric flux passing through the cyclindrical surface is A. `Q/epsilon_(0)`B. `(100Q)/epsilon_(0)`C. `(10Q)/((piepsilon_(0)))`D. `(100Q)/((piepsilon_(0)))` |
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Answer» Correct Answer - b |
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| 20. |
A Gaussian surface S encloses two charges `q_(1)= q and q_(2) = -q`the field at p is where `vecE_(1), vecE_(2)and vecE_(3)` are the field contributed by `q_(1),q_(2)and q_(3)at` P respectively. A. `vecE_(1)+vecE_(2)`B. `vecE_(1)+vecE_(2)+ vecE_(3)`C. ` vecE_(3)`D. `vecE_(1)+vecE_(2)- vecE_(3)` |
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Answer» Correct Answer - b |
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| 21. |
An electric field given `vecE= 4.0hati -3.0(y^(2) + 2.0)j` pierces a Gaussian cube of edge length 2.0 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position z is in meters.) what is the electric flux through the (a) top face, (b) bottom face, (c) left face , (b) bottom facem (c) left face, and (d) back face ? (e) what is the net electric flux through the cube ? |
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Answer» Correct Answer - `72N.m^(2)//C; (b)+24Nm^(2)//C (c)-16N.m^(2)//c (d) zero (e) -48N.m^(2)//C.` |
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| 22. |
figure shows, in cross section, two Gaussian spheres and two Gaussian cubes that are centered on a positively charged particle on a positively charged partical . Rank greatest first, and indicate whether the manitudes are unifrom or variable along each surface. A. net flux through all the four Gaussion surface will be equalB. the magniude of the electric . Fields on the surface (i) and (ii) will be constantC. the magnitude of the electric fields on the sufaces (ii) and (iv) will be variableD. The magnitude of the electric fields on all the surfaces will be constant |
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Answer» Correct Answer - a,b,c |
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| 23. |
In figure , a solid sphere of radius a = 2.00cm is concentric with a spherical conductiosn shell of inner raedius b =2.00 a and outer radius c=2.40 a. the sphere has a net unifrom charge `q_(1)=+5.00C`. The shell has a net charge `q_(2) =-q_(1)`. Distance (a) r=0 (b) r= a/2.00 (c) r=a (d) r = 1.50 a (e) r= 2.30a and (f) r= 3.50 a ? what si the net charge on the (g) inner and (h) outer surface of the shell ? |
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Answer» Correct Answer - (a)zero `(b) 5.6 xx 10^(13) N//C (c)1.1xx10^(14) N//C(d) 5 xx 10 ^(13)N//C` € zero (f) zero (g) -5.00 C (h)zero |
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| 24. |
Consider two concentric spherical surfaces `S_1` with radius a and `S_2` with radius 2a, both centered at the origin. There is a charge +q at the origin and there are no other charges. Compare the flux `phi_1` through `S_1` with the flux `phi_2` through `S_2`.A. `phi_1 = 4phi_2`B. `phi_1 = 2phi_2`C. `phi_1 = phi_2`D. `phi_1 = phi_2//2` |
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Answer» Correct Answer - C c. Flux through both will be same as the net charge enclosed by both in same. |
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| 25. |
Positive and negative charges of equal magnitude lie along the symmetry axis of a cylinder. The distance from the positive charge to the left - end cap of the cylinder is the same as the distance from the negative charge to the right - end cap. What is the flux of the electric field through the closed cylinder ?A. zeroB. `+ Q//epsilon_(0)`C. `+ 2Q//epsilon_(0)`D. `- Q//epsilon_(0)` |
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Answer» Correct Answer - A Net charge enclosed by the cylinder is zero , so flux through the cylinder is zero. |
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