InterviewSolution
Saved Bookmarks
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.
| 1501. |
The electric field in a region is given by `E = 3/5 E_0hati +4/5E_0j` with `E_0 = 2.0 x 10^3 N//C`. Find the flux of this field through a rectangular surface of area `0.2 m^2` parallel to the y-z plane. |
| Answer» `phi_(E)=vec(E).vec(S)=(3/5 E_(0) hat(i)+4/5 E_(0) hat(j)).(0.2 hat(i))=240 (N-m^(2))/C` | |
| 1502. |
An infinite number of charges each equal to q, are placed along the X-axis at `x = 1, x = 2, x = 4, x = 8`,…….. and so on. (i) find the electric field at a point `x = 0` due to this set up of charges. (ii) What will be the elctric field if the above setup, the consecutive charges have opposite signs. |
|
Answer» Correct Answer - `(q)/(3pi in_(0)) , (q)/(5pi in_(0))` (i) Electric field at `x = 0` `E = (1)/(4pi in_(0)) [(q)/(1^(2)) + (q)/(2^(2)) + (q)/(4^(2)) + ….]` `= (q)/(4pi in_(0)) [(1)/(1) + (1)/(4) + (1)/(16) + …..]^(**)` `E = (q)/(4pi in_(0)) [(1)/(1- 1//4)] = (q)/(5pi in_(0))` |
|
| 1503. |
A point charge Q is placed at the corner of a square of side a, then find the flux through the square. |
| Answer» The electric field due to Q at any point of the square will be along the plane of square and the electric field lines are perpendicular to square, so `phi=0`. In other words we can say that no line is crossing the square so, flux = 0. | |
| 1504. |
Two charges of `-4 muC and +4muC` are placed at the points `A (1,0,4) and B (2,-1,5)` location in an electric field `vec(E) = 0.20 hat(i) V//cm`. Calculate the torque acting on the dipole.A. `2.31 xx 10^(-4) N//m`B. `7.98 xx 10^(-4) C//m`C. `7.11 xx 10^(-4) C//m`D. `7.04 xx 10^(-4) C//m` |
|
Answer» Correct Answer - B |
|
| 1505. |
Two charges of `-4 muC and +4muC` are placed at the points `A (1,0,4) and B (2,-1,5)` location in an electric field `vec(E) = 0.20 hat(i) V//cm`. Calculate the torque acting on the dipole. |
|
Answer» Correct Answer - `1.13xx10^(-4) N-m` As A `(1, 0,4), B (2,-1,5)` `:. 2 vec(a) = vec(AB) = [(2-1) hat(i) + (-1 - 0) hat(j) + (5-4) hat(k)]` `= [hat(i) - hat(j) + hat(k)]` `q = +- 4xx10^(-6) C, E = 0.20 hat(i) V//cm = 20 hat(i) V//m` As `vec(tau) = vec(p) xx vec(E) = q (2 vec(a)) xx vec(E)` `:. vec(tau) = 4xx10^(-6) (hat(i) - hat(j) + hat(k)) xx20 hat(i)` `vec(tau) = 8xx10^(-5) (hat(k) + hat(j))` Magnitude of torque, `tau = 8xx10^(-5) sqrt(1^(2) + l^(2)) = 1.13xx10^(-4) N-m` |
|
| 1506. |
A charged cylinder of radius 3 mm has surface density of charge `4muC//m^(2)`. It is placed in a medium of dielectric constant 6.28. The electric intensity at a point at a distance of 1.5 M from its axis isA. `1.44V//m`B. `2.44V//m`C. `3V//m`D. `0.5V//m` |
|
Answer» Correct Answer - A `E=(sigmaR)/(epsi_(0)kr)=(4xx10^(-6)xx3xx10^(-3))/(8.85xx10^(-12)xx6.28xx1.5)` `=1.44V//m.` |
|
| 1507. |
A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius `b(bgt a`). The solid sphere is given a charge Q. A student measures the potential at the surface of the solid sphere as V and the potential at the surface of spherical shell as `V_b`. After taking these readings, he decides . to put charge of `-4Q` on the shell. He then noted the readings of the potential of solid sphere and the shell and found that the potential difference is `/_V` .He then connected the outer spherical shell to the earth by a conducting wire and found that the charge on the outer surface of the shell as He then decides to remove the earthing connection from the shell and earthed the inner solid sphere. Connecting the inner sphere with the earth he observes the charge on the solid sphere as `q_2`. He then wanted to check what happens if the two are connected by the conducting wire. So he removed the earthing connection and connected a conducting wire between the solid sphere and the spherical shelll. After the connections were made he found the charge on the outer shell as `q_3`.Potential difference `(/_CV)` measured by the student between the inner solid shere and outer shell after putting a charge `-4Q` isA. `V_a-3V_b`B. `3(V_a-V_b)`C. `V_a`D. `V_a-V_b` |
|
Answer» Correct Answer - D Potential difference i such situation depens on inner charge only. So, potential difference will remain unchanged. Hence `/_V=V_a-V_b` |
|
| 1508. |
Find out flux through the given Gaussian surface. |
| Answer» `phi=Q_("in")/epsi_(0)=(2muC-3muC+4muC)/epsi_(0)=(3xx10^(-6))/epsi_(0) Nm^(2)//C` | |
| 1509. |
A charge `q` is located at the centre of a cube. The electric flux through any face isA. `(piq)/(6(4pi epsilon_(0)))`B. `(q)/(6(4pi epsilon_(0)))`C. `(2pi q)/(6(4pi epsilon_(0))`D. `(4pi q)/(6(4pi epsilon_(0)))` |
|
Answer» Correct Answer - D |
|
| 1510. |
A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius `b(bgt a)`. The solid sphere is given a charge Q. A student measures the potential at the surface of the solid sphere as V and the potential at the surface of spherical shell as `V_b`. After taking these readings, he decides . to put charge of `-4Q` on the shell. He then noted the readings of the potential of solid sphere and the shell and found that the potential difference is `/_V` .He then connected the outer spherical shell to the earth by a conducting wire and found that the charge on the outer surface of the shell as He then decides to remove the earthing connection from the shell and earthed the inner solid sphere. Connecting the inner sphere with the earth he observes the charge on the solid sphere as `q_2`. He then wanted to check what happens if the two are connected by the conducting wire. So he removed the earthing connection and connected a conducting wire between the solid sphere and the spherical shelll. After the connections were made he found the charge on the outer shell as `q_3`. `q_3` isA. `(Q(a+b))/(a-b)`B. `(Qa^2)/b`C. `(Q(a-b))/b`D. `-(Qb)/a` |
|
Answer» Correct Answer - C Whole inner charge transfers to shell. `:.` Total charge on shell `=q_2-Q` `=Q(a/b-1)` |
|
| 1511. |
A point charge is kept at the centre of a metallic insulated spherical shell. ThenA. Electric field out side the sphere is zeroB. electric field inside the sphere is zeroC. Net induced charge on the sphere is zeroD. Electric potential inside the sphere is zero |
| Answer» Correct Answer - C | |
| 1512. |
If a point charge q is placed at the centre of a cube, then find out flux through any one face of cube. |
|
Answer» Flux through all 6 faces `=q/epsi_(0)`. Since, all the surface are symmetrical So, flux through one face `=1/6 q/epsi_(0)` |
|
| 1513. |
Mark the wrong statement-A. Equipotential surface never cross and otherB. for a uniformly charged nonconducting sphere, the electric potential at the centre of the sphere is 1.5 times that at the surfaceC. if potential in a certain region in non zero constant, then the electric field in that region will also be non zero constantD. Inside a spherical uniformly charged shell, the electric field is zero but the electric potential is the same as that at the surface. |
| Answer» Correct Answer - C | |
| 1514. |
A and B are two concentric spherical shells. If A is given a charge `+q` while B is earthed as shown in figure then A. A and B both will have the same charge densitiesB. the potential inside A and outside B will zeroC. the electric field between A and B is non zeroD. the electric field inside A and outside B is non zero |
| Answer» Correct Answer - C | |
| 1515. |
A and B are two concentric spherical shells. If A is given a charge `+q` while B is earthed as shown in figure then A. the charge appearing on inner surface of B is `-Q`B. the field inside and outside A is zeroC. the fieldm between A and B is not zeroD. the charge appearing on outer surface of B is zero |
|
Answer» Correct Answer - A::C::D |
|
| 1516. |
A and B are two concentric spherical shells. If A is given a charge `+q` while B is earthed as shown in figure then A. The charge appearing on inner surface of B is -QB. The field inside and outside a is zero.C. The field between a and B is not zero.D. The charge appearing on outer surface of B is zero. |
| Answer» Correct Answer - A::C::D | |
| 1517. |
An electric dipole is kept in the electric field produced by a point chargeA. dipole wil experience a force.B. dipole will experience a torque.C. it is possible to find a path (not closed) in the field on which work required to move the dipole is zero.D. dipole can be in stable equilibrium. |
| Answer» Correct Answer - A::C | |
| 1518. |
There is a hemisphere of radius R having a uniform volume charge density `rho`. Find the electric potential and field at the centre |
|
Answer» Correct Answer - `E=(rhoR)/(4 in_(0));V=(rhoR^(2))/(4 in_(0))` |
|
| 1519. |
Figure shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged? A. AB. BC. CD. D |
| Answer» Correct Answer - A::D | |
| 1520. |
An imaginary closed surface P is constructed around a neutral conducting wire connected to a battery and a switch as shown in figure. As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface P is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface : A. remains unchangedB. remains zeroC. is increasedD. is decreased |
| Answer» Correct Answer - A::B | |
| 1521. |
A small, electrically charged bead can slide on a circular, frictionless, thin, insulating ring. Charge on the bead is Q and its mass is m . A small electric dipole, having dipole moment P is fixed at the centre of the circle with the dipole’s axis lying in the plane of the circle. Initially, the bead is held on the perpendicular bisector of the dipole (see fig.) Ignore gravity and answer the following questions. (a) Write the speed of the bead when it reaches the position `theta` shown in the figure. (b) Find the normal force exerted by the ring on the bead at position `theta`. (c) How does the bead move after it is released? Where will the bead first stop after being released? (d) How would the bead move in the absence of the ring? |
|
Answer» Correct Answer - (a). `v=sqrt(2-(QKPcostheta)/(mr^(2)))` (b). Zero (c). The bead oscillates. It stops at a point dametrically opposite to its starting point. (d). Exactly same as it would move in the presence of the ring |
|
| 1522. |
A long cylindrical volume (of radius R) contains a uniformly distributed charge of density `rho`. Consider a point P inside the cylindrical volume at a distance x from its axis as shown in the figure. Here x can be more than or less than R. Electric field at point P is : A. `(rho x)/(2 epsi_(0))` if `x lt R`B. `(rho x)/epsi_(0)` if `x lt R`C. `(rho R^(2))/(4 epsi_(0) x)` if `x gt R`D. `(rho R^(2))/(2 epsi_(0) x)` if `x gt R` |
| Answer» Correct Answer - A::D | |
| 1523. |
Consider two spheres of the same radius R having uniformly distributed volume charge density of same magnitude but opposite sign `(+rho` and `-rho)` the spheres overlap such that the vector joining the centre of the negative sphere to that of the positive sphere is `vecd`. `(d lt lt R)`. Find magnitude of electric field at a point outside the spheres at a distance r in a direction making an angle `theta` with `vecd`. Distance r is measure with respect to the mid point of the line joining the centers of the two spheres. |
|
Answer» Correct Answer - `(rhoR^(3)d)/(3epsi_(0)r^(3))sqrt(3cos^(2)theta+1)` |
|
| 1524. |
A thin conducting ring of radius `R` is given a charge `+Q`, Fig. The electric field at the center `O` of the ring due to the charge on the part `AKB` of the ring is `E`. The electric field at the center due to the charge on part `ACDB` of the ring is A. `E` along `KO`B. `3E` along `OK`C. `3E` along `KO`D. `E` along `OK` |
|
Answer» Correct Answer - 4 Electric field at `O` is zero. Electric field at `O` due to `ACDB` is equal and opposite to electric field at `O` due to `A K B` ,i.e. `E` along `OK`. |
|
| 1525. |
(a) There is a long uniformly charged cylinder having a volume charge density of `rho C//m^3`. Radius of the cylinder is R. Find the electric field at a point at a distance x from the axis of the cylinder for following cases (i) x lt R (ii) x gt R What is the maximum field produced by the charge distribution at any point? (b) The cylinder described in (a) has a long cylindrical cavity. The axis of cylindrical cavity is at a distance a from the axis of the charged cylinder (see figure). Find electric field inside the cavity. |
|
Answer» Correct Answer - (a). (i) `(rhox)/(2epsilon_(0)x)` (ii) `(rhoR^(2))/(2epsilon_(0)x)` field at the surface is maximum `E_(max)=(rhoR)/(2epsilon_(0))` (b). `vecE=(rhoveca)/(2epsilon_(0))` |
|
| 1526. |
An infinitely long time charge has linear charge density `lamdaC//m` the line charge is along the line `x=0,z=2m` find the electric field at point (1,1,1) m. |
|
Answer» Correct Answer - `vecE=(lamda)/(4pi epsilon_(0))(hati-hatk)` |
|
| 1527. |
A simple pendulum has a length `l` & mass of bob m. The bob is given a charge q coulomb. The pendulum is suspended in a uniform horizontal electric field of strength E as shown in figure, then calculate the time period of oscillation when bob is slightly displaced from its mean position. A. `2pi sqrt(l/g)`B. `2pi sqrt({l/(g+(qE)/m)})`C. `2pi sqrt({l/(g-(qE)/m)})`D. `2pi sqrt(l/sqrt(g^(2)+((qE)/m)^(2)))` |
| Answer» Correct Answer - D | |
| 1528. |
A pendulum has a bob of mass m carrying a positive charge q. Length of the pendulum string is L. Beneath the pendulum there is a large horizontal dielectric sheet of charge having uniform surface charge density of `sigma C//m^2`. [figure (i)] (a) Find the time period of small oscillations for the pendulum (b) Now the dielectric sheet of charge is tilted so as to make an angle `beta` with horizontal. Find the angle (a) that the thread makes with vertical in equilibrium position. Find time period of small oscillations in this case. [figure (ii)] |
|
Answer» Correct Answer - (a). `T=2pisqrt((L)/(g-(qsigma)/(2mepsilon_(0))))` (b). `T=2pisqrt((L)/(sqrt(g^(2)+((qE)/(m))^(2)-2g(qE)/(m)cosbeta)))` |
|
| 1529. |
A pendulum bob of mass `30.7 xx 10^(-6)kg` and carrying a chargee `2 xx 10^(-8)C` is at rest in a horizontal uniform electric field of `20000V//m`. The tension in the thread of the pendulum is `(g = 9.8 m//s^(2))`A. `3 xx 10^(4)N`B. `4 xx 10^(-4)N`C. `5 xx 10^(-4)N`D. `6 xx 10^(-4)N` |
|
Answer» Correct Answer - C |
|
| 1530. |
A `4mu F` capacitor is charged to 400 volts and then its plates are joined through a resistor of resistance `1K Omega`. The heat produced in the resistor isA. `0.16 J`B. `1.28 J`C. `0.64 J`D. `0.32 J` |
|
Answer» Correct Answer - D |
|
| 1531. |
Charges can neither be created nor be destroyed is the statement of law of conservation of ………… |
|
Answer» Charges can neither be created nor be destroyed is the statement of law of conservation of charge |
|
| 1532. |
The value of s, for air or vacuum is ……… |
|
Answer» The value of s, for air or vacuum is 1 |
|
| 1533. |
Write down the value of obsolute permittivity of free space.A. `9 xx 10^(9) NC^(2)//m^(2)`B. `8.85 xx 10^(-12) N-m^(2)//C^(2)s`C. `8.85 xx 10^(-12) C^(2)//N-m^(2)`D. `9 xx 10^(9)C^(2)//N - m^(2)` |
|
Answer» Correct Answer - C |
|
| 1534. |
The unit of permittivity in free space (s0) is …………… |
|
Answer» The unit of permittivity in free space (s0) is C2 N-1 m-2 |
|
| 1535. |
The unit of electric charge is ………… |
|
Answer» The unit of electric charge is coulomb |
|
| 1536. |
Bodies which allow the charge to pass through them are called ……… |
|
Answer» Bodies which allow the charge to pass through them are called conductor. |
|
| 1537. |
Three charged particles are collinear and are in equilibrium, thenA. all the charged particles have the same polarityB. the equilibrium in unstableC. all the charged particles cannot have the same polarityD. both (a) and (c ) are correct |
|
Answer» Correct Answer - B::C When three charged particles are collinear and are in equilibrium, then a small displacement of one charged particle from its position will make the system of charges unstable. Thus, the option (b) is true. Since all charges are in equilibrium, so net force on any of charges in equilibrium is zero. It will be so if two equal and opposite forces act on every charge . It will be so if the three charged particles do not have the same plartity. Thus, option (c) is true. |
|
| 1538. |
Bodies which do not allow the charge to pass through them are called ……… |
|
Answer» Bodies which do not allow the charge to pass through them are called insulators |
|
| 1539. |
A proton and an electron are placed in a uniform electric field.A. The magnitude of the electric forces acting on them will be equalB. The electric forces acting on them will be unequalC. The magnitude of their accelerations will be equalD. Their accelrations will be equal |
|
Answer» Correct Answer - A As `F = qE, E` is uniform and also , `q_("proton") = q_("electron")` therefore,magnitude of electric forces will be equal. This is the only correct choice. |
|
| 1540. |
No current flows between two charged bodies when connected (a) if they have the same capacitance (b) if they have same quantity of charge (c) if they have the same potential (d) if they have the same charge density |
|
Answer» (c) if they have the same potential |
|
| 1541. |
A point charge is brought in an electric field. The electric field at a nearby point (i) will increase if the charge is `+ve` (ii) will decrease if the charge is `-ve` (iii) may increase if the charge is `+ve` (iv) may decrease if the charge is `-ve`A. will increase if charge is positiveB. may increase if charge is positiveC. will increase if charge is negativeD. may increase if charge is negative |
|
Answer» Correct Answer - B::D Electric field may increase for positive charge and may decrease for negative charge. |
|
| 1542. |
Electric field lines about a negative point charge are (a) circular, anticlockwise (b) circular, clockwise (c) radial, inwards (d) radial, outwards |
|
Answer» (c) radial, inwards |
|
| 1543. |
The number of electric lines of force crossing through the given area is ………… |
|
Answer» The number of electric lines of force crossing through the given area is electric flux |
|
| 1544. |
The lines of force are far apart, when electric field E is ………… |
|
Answer» The lines of force are far apart, when electric field E is small |
|
| 1545. |
The lines of force are close together, when electric field E is ………… |
|
Answer» The lines of force are close together, when electric field E is large |
|
| 1546. |
Assertion: A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge. Reason: The electric field is independent of the nature of charge.(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.(c) If assertion is true but reason is false.(d) If the assertion and reason both are false.(e) If assertion is false but reason is true. |
|
Answer» (d) If the assertion and reason both are false. Explanation: Electric field at the nearby-point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought. |
|
| 1547. |
Define ‘Electric field. |
|
Answer» The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by \(\vec E\) = \(\frac{\vec F}{q_0}\) The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC-1 ). |
|
| 1548. |
Write a short note on superposition principle. |
|
Answer» According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges. \(\vec F\)tot = \(\vec F\)12 \(\vec F\)13 \(\vec F\)14 \(\vec F\)1n |
|
| 1549. |
The electrostatic potential inside a charged spherical ball is given by `phi=ar^2+b` where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:A. ` - 24 pi a varepsilon_0`B. `-6 a varepsilon_0`C. `-24 pi a varepsilon_0^r`D. `-6 a varepsilon_0r` |
| Answer» Correct Answer - B | |
| 1550. |
The electrostatic potential inside a charged spherical ball is given by `phi=ar^2+b` where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:A. `-24 pi a epsi_(0) r`B. `-6pi a epsi_(0) r`C. `-24 pi a epsi_(0)`D. `-6 a epsi_(0)` |
| Answer» Correct Answer - D | |