Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A magnetic wire of dipole moment `4pi Am^(2)` is bent in the form of semicircle. The new magnetic moment isA. `4 pi Am^(2)`B. `8pi Am^(2)`C. `4 Am^(2)`D. None of these |
|
Answer» Correct Answer - C Here, `M_(i)=4pi` `mL=4pi` `m=(4pi)/(L)` The new distance between poles is `L_(1)=2r=2(L)/(2pi)=(L)/(pi)` `therefore" "M_(f)=mL_(1)=(mL)/(pi)=(4pi)/(pi)` `=4Am^(2)` |
|
| 2. |
A small magnet of dipole moment `M` is kept on the arm of a deflection magentometer set in tan `A` position at a distance of `0.2 m`. If the deflection is `60^(@)`, find the value of `P (B_(H)=0.4xx10^(-4)T)`.A. `2.77 Am^(2)`B. `8 Am^(2)`C. `0.2 Am^(2)`D. None of these |
|
Answer» Correct Answer - A `because" "B=B_(H)tantheta` `"or "(mu_(0))/(4pi)(2M)/(r^(3))=0.4 xx 10^(-4)tan60^(@)` `therefore (mu_(0)M)/(2pi(0.2)^(3))=0.4 xx 10^(-4)tan60^(@)` `M=(0.4xx4pixx(0.2)^(3)xx10^(-4)xxsqrt(3))/(2xxmu_(0))` `=(0.00277xx10^(-4))/(10^(-7))` `therefore " "M=2.77Am^(2)` |
|
| 3. |
A current I flows in a conducting wire of lenth L. If we bent it in a circular form, then calculate its magnetic dipole moment.A. `(ll)/(4pi)`B. `(l^(2)l)/(4pi)`C. `(ll^(2))/(4pi)`D. `(l^(2)l^(2))/(4pi)` |
|
Answer» Correct Answer - B Length of a conducting wire, `2pir=l, r=(l)/(2pi)`, Area of coil, `alpha=pir^(2)=(pi.l^(2))/(4pi^(2))=(l^(2))/(4pi)` `So, " " m=lA=(ll^(2))/(4pi)` |
|
| 4. |
Two similar equal magnetic poles when separated by a distance of `1 m`,A. `10 Am`B. 20 AmC. 50 AmD. 100 Am |
| Answer» Correct Answer - D | |
| 5. |
Calculate the force acting between two small magnets, placed in end on position `0.1 m` apart from their centres. (Given, magnetic moment of each magnet is `5 Am^(2)`).A. 0.6 NB. 0.8 NC. 0.15 ND. 0.2 N |
| Answer» Correct Answer - C | |
| 6. |
Among the following properties describing diamagnetism, identify the property that is wrongly stated.A. Diamagnetic material do not have permanent magnetic momentB. Diamagnetism is explained in terms of electromagnetic inductionC. Diamagnetic materials have a small positive susceptibilityD. The magnetic moment of individual electrons neutralise each other |
|
Answer» Correct Answer - C Diamagnetic materials have negative susceptibility. Thus, (c ) is wrongly stated. |
|
| 7. |
The magnetic induction inside a solenoid is `6.5xx10^(-4)T`. When it is filled with iron medium, then the induction becomes `1.4 T`. The relative permeability of iron will beA. 1578B. 2355C. 1836D. 2154 |
|
Answer» Correct Answer - D `B_("air")=mu_(0)nl, B_("medium")=mu_(0)u_(r )nl=mu_(r )B_(0)` `1.4 = mu_(r )xx 6.5 xx 10^(-4)` `therefore" "mu_(r )=(1.4)/(6.5 xx 10^(-4))=2154` |
|
| 8. |
Inside a long solenoid wounded with 300 turns/m, an iron rod is placed. An iron rod is 0.2 m long, 10 mm in diameter and of permeability `10^(3)`. Calculate the magnetic moment of the rod, if 0.5 A of current is passed through the rod.A. 2.356 Sl unitsB. 1.335 Sl unitsC. 3.664 Sl unitsD. 1.664 Sl units |
| Answer» Correct Answer - A | |
| 9. |
Calculate the angle of dip, if a dip needle oscillating in a vertical plane makes 40 oscillations per minute in a magnetic meridian and 30 oscillations per minute in a vertical plane at right angle to the magnetic meridian.A. `theta=sin^(-1)(0.5625)`B. `theta=sin^(-1)(0.325)`C. `theta=sin^(-1)(0.425)`D. `theta=sin^(-1)(0.235)` |
| Answer» Correct Answer - A | |
| 10. |
In previous problem, the angle of dip isA. `tan^(-1)(0.75)`B. `tan^(-1)(0.5)`C. `tan^(-1)(0.8)`D. None of these |
| Answer» Correct Answer - A | |
| 11. |
The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:A. `tan^(-1)(sqrt(3)/(2))`B. `tan^(-1)(sqrt(3))`C. `tan^(-1)((sqrt(3))/(sqrt(2)))`D. `tan^(-1)((2)/(sqrt(3)))` |
| Answer» Correct Answer - A | |
| 12. |
Magnetic moment of bar magnet is M. The work done to turn the magnet by `90^(@)` of magnet in direction of magnetic field B will beA. zeroB. `(1)/(2)MB`C. MBD. MB |
|
Answer» Correct Answer - D Work done, `W=MB(1-costheta)` `" "theta=90^(@)` `therefore" "W=MB` |
|
| 13. |
A short bar magnet of moment `0*32JT^-1` is placed in a uniform external magnetic field of `0*15T`, if the bar is free to rotate in the plane of the field, which orientations would correspond to its, (i) stable and (ii) unstable equilibrium? What is the potential energy of the magnet in each case?A. when m is anti-parallel to B, U `=4.8 xx 10^(-2)J`B. when m is perpendicular to B, U = 0C. when m is perpendicular to B, U = `4.8 xx 10^(-2)J`D. None of the above |
| Answer» Correct Answer - A | |
| 14. |
A bar magnet has a magnetic moment `2.5JT^(-1)` and is placed in a magnetic field of `0.2 T`. Calculate the work done in turning the magnet from parallel to antiparallel position relative to field direction.A. 1 JB. 2 JC. 3 JD. 4 J |
| Answer» Correct Answer - A | |
| 15. |
With a standard rectangular bar magnet, the time period in the uniform magnetic field is `4 sec`. The bar magnet is cut parallel to its length into `4` equal pieces. The time period in the uniform magnetic field when the piece is used (in sec) (bar magnet breadth is small)A. 16B. 8C. 4D. 2 |
|
Answer» Correct Answer - C Time period of magnet, `T-2pisqrt((l)/(MB))` When magnet is cut parallel to its length into four equal pieces. |
|
| 16. |
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is `T_0` in a uniform magnetic field B. Then, the time period of each part in the same magnetic field isA. `(T_(0))/(2)`B. `(T_(0))/(3)`C. `(T_(0))/(4)`D. None of these |
|
Answer» Correct Answer - B In the case of vertical cutting, `T_(1)=("Original time period")/(n)` `T_(1)=(T_(0))/(3)` (Here, n = number of equal parts) |
|