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. |
Answer of this question with working |
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Answer» tion:60degreee ,90degreeefollow me FRIENDS |
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| 2. |
A taxi leaves the station X for station Y every 10 minutes simultaneously a taxi leaves station Y the station X every 10 minutes. The taxis move at the same constant speed and go From X to Y or vice a versa in 2 hours . How many taxis coming from the other side will each taxi meet enroute from Y to X? |
| Answer» EXPLANATION:Here's the SOLUTION to your question.Hope it HELPS you...!!!(^_^) | |
| 3. |
(g) A green coloured bangle when crushed becomes white. Where is the green colourgone? |
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Answer» coloured bangle when CRUSHED becomes white because of the PHENOMENON called scattering of LIGHT |
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| 4. |
The answers of the questions |
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Answer» What is your QUESTION bro❓❓ |
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| 5. |
Describe an activity to show that liquid exert pressure on the walls |
| Answer» TION:this MAY HELP you BETA.. read it CAREFULLY | |
| 6. |
Inside a bar magnet, the magnetic field lines(A) are not present(B) are parallel to the cross-sectional area of the magnet(C) are in the direction from N-pole to S-pole(D) are in the direction from S-pole to N-pole |
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Answer» D) INSIDE the BAR magnet the magnetic field LINES are from S- POLE to N- pole. |
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| 7. |
Let r be the distance of a point on the axis of a bar magnet from its centre. The magnetic field at r is always proportional to (A) 1/r² (B) 1/r³ (C) 1/r (D) not necessarily 1/r³ at all points |
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Answer» Magnetic field is inversly proportional to the square of the POSITION vector r.So option (A) is correct.Explanation:since we know the biot savart LAW for a magnetism which sates that magnetic field is (i) directly proportional to current FLOWING in the conductor(ii) directly proportional to length of the conductor(iii)directly proportional to angle between length the conductor and position vector.(IV)inversely proportional to cube of the position vector of the conductor...................(1)so magnetic field is inversly proportional to the square of the position vector r. |
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| 8. |
A bar magnet is oscillating in Earth's magnetic field with periodic time T. If a similar magnet with the same mass and volume has magnetic dipole moment, which is 4 times that of this magnet, then its periodic time will be _____ . (A) T/2 (B) 2T (C) T (D) 4T |
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| 9. |
A magnet of magnetic dipole moment 5.0 A m2 is lying in a uniform magnetic field of 7 x 10⁻⁴ T such that its dipole moment vector makes an angle of 30° with the field. The work done in increasing this angle from 30° to 45° is about _______ J.(A) 5.56 x 10⁻⁴ (B) 24.74 x 10⁻⁴ (C) 30.3 x 10⁻⁴ (D) 5.50 x 10⁻³ |
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Answer» The TOTAL work done in rotating the DIPOLE from to is .Option (A) is correct.Explanation:The work done in rotating from INITIAL ANGLE to a final angle is given by.where, = torque exerted on the magnet due to the MAGNETIC field.We know,where, = dipole moment of the magnet. = magnetic field in which the magnet is placed. = angle between and .Therefore, the work done in rotating from initial angle to a final angle is given byGiven:....Putting these values in above expression of required work done, we get, Thus, the correct option is (A). |
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| 10. |
A very long straight wire carries a current of 5 A. An electron moves with a velocity of 10⁶ m s⁻¹ remaining parallel to the wire at a distance of 10 cm from wire in a direction opposite to that of electric current. Find the force on this electron. (Here the mass of electron is taken as constant) e = -1.6 x 10⁻¹⁹C, μ₀ = 4π x 10⁻⁷SI. |
| Answer» 0 to IL be hdnnsnjviieindhhsjjjfjjdhhd | |
| 11. |
Two parallel very long straight wires carrying currents of 20 A and 30 A respectively are at a separation of 3 m between them. If the currents are in the same direction, find the attractive force between them per unit length. |
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Answer» the ATTRACTIVE FORCE between them is ELECTROMAGNETIC |
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| 12. |
Two rings X and Y are placed in such a way that their axes are along the X and the Y axes respectively and their centres are at the origin. Both the rings X and Y have the same radii of 3.14 cm. If the current through X and Y rings are 0.6 A and 0.8 A respecively, find the value of the resultant magnetic field at the origin. μ₀ = 4π x 10⁻⁷SI. |
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Answer» Two rings X and Y are placed in such a way that their axes are along the X and the Y axes respectively and their centres are at the ORIGIN. Both the rings X and Y have the same RADII of 3.14 cm. If the CURRENT through X and Y rings are 0.6 A and 0.8 A respecively, find the value of the resultant MAGNETIC FIELD at the origin.μ₀ = 4π x 10⁻⁷SI. |
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| 13. |
Q.25 Two blocks are connected by a cord passingover a small frictionless pulley and resting onfrictionless planes as shown in the figure. Theacceleration of the blocks is-100 kgSA350 kg37053°(1) 0.33 m/s2(3) 1 m/s2(2) 0.66 m/s2(4) 1.32 m/s2 |
| Answer» G any ONE of the direction of accand then MAKING the REQUIRED EQUATION | |
| 14. |
(0) 19. Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is [AIPMT-2015] (1) 1.5R (2) 2.5R (3) 4.5R (4) 7.5R |
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Answer» tion:SUPPOSE let the smaller body cover a DISTANCE x before collision then -: Mx =5M(9R - x)X=45R-5x6x= 45RX=45R/6X= 7.5RI HOPE this HELPS you please follow me and mark as brainliest |
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| 15. |
Is here any physics expert who can answer my all asked physics question |
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| 16. |
What are vector Quantities ?? |
| Answer» QUANTITIES are those quantities that have MAGNITUDE as WELL as DIRECTION | |
| 17. |
-18. Find the value of friction forces between the blocks A and B and between B and ground. (Take, g = 10 ms 223.4maximremainu = 0.1-A 5kg|BGround15kgLF = 80 N1 = 0.6(a) 90 N, 5 N(c) 5 N, 75 N(b) 5 N, 90 N(d) 0 N, 80 N |
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Answer» the value between maximum REMAIN u =0.1 a5kg B GROUND 15kg ground L F =80n === 5n , 75 n |
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| 18. |
Find the value of P.E of an object when it is on ground |
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Answer» The value of potential energy when OBJECT on the ground is:P.E = Height of object × weight of objectExplanation:HOPE it's HELP you‼️‼️☑️☑️ |
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| 19. |
a ball of mass 0.20 is thrown vertically upwards with an initial velocity of 20m/s.Calculate the maximum potential energy it gainsas it goes up |
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| 20. |
If we increase velocity 4 Times the Original one's. What will be the Change in Kinetic Energy ? |
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| 21. |
When a galvanometer with a shunt is joined in an electrical circuit 2% of the total current passes through the galvanometer. Resistance of galvanometer is G. Find the value of shunt. |
| Answer» GALVANOMETER with a shunt is joined in an ELECTRICAL circuit 2% of the total current passes through the galvanometer | |
| 22. |
Obtain the formula for the magnetic field produced inside a very long current carrying solenoid wising Ampere's Circuital Law. |
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| 23. |
Derive an expression for the magnetic field at a point on the axis of a current carrying circular ring. |
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Answer» Aim: To study the variation of magnetic field with distance along the axis of a circular coil CARRYING current. Apparatus: Circular coil, compass box, ammeter, rheostat, commutator, cell, key, connection wires, etc. The purpose of the commutator is to allow the current to be reversed only in the coil, while flowing in the same direction in the rest of the circuit. Theory: A current carrying wire generates a magnetic field. According to Biot-Savart’s law, the magnetic field at a point due to an element of a conductor carrying current is, Directly proportional to the strength of the current, iDirectly proportional to the length of the element, dlDirectly proportional to the Sine of the angle θ between the element and the line joining the element to the point andInversely proportional to the square of the distance r between the element and the point. Thus, the magnetic field at O is dB, such that, Then, where, is the proportionality CONSTANT and is called the permeability of free space.Then, In vector form, Consider a circular coil of radius r, carrying a current I. Consider a point P, which is at a distance x from the centre of the coil. We can consider that the loop is made up of a large number of short elements, generating small magnetic fields. So the total field at P will be the sum of the contributions from all these elements. At the centre of the coil, the field will be uniform. As the location of the point increases from the centre of the coil, the field decreases. By Biot- Savart’s law, the field dB due to a small element dl of the circle, centered at A is given by, This can be resolved into two components, one along the axis OP, and other PS, which is perpendicular to OP. PS is EXACTLY CANCELLED by the perpendicular component PS’ of the field due to a current and centered at A’. So, the total magnetic field at a point which is at a distance x away from the axis of a circular coil of radius r is given by, If there are n turns in the coil, then where µ0 is the absolute permeability of free space. Since this field Bx from the coil is acting perpendicular to the horizontal intensity of earth’s magnetic field, B0, and the compass needle align at an angle θ with the vector sum of these two fields, we have from the figure The horizontal component of the earth’s magnetic field varies greatly over the surface of the earth. For the purpose of this simulation, we will assume its magnitude to be B0 = 3.5x10-5 T. The variation of magnetic field along the axis of a circular coil is shown here. HOPE it's helps uuu ✌✌✌ |
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| 24. |
What should be d one to convert a galvanometer into an ammeter. Obtain the formula for the shunt. |
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Answer» A galvanometer can detect only small currents. Thus, to measure large currents it is converted into an ammeter. It can be converted into an ammeter by connecting a low resistance called shunt resistance in parallel to the galvanometer.Let G be the resistance of the galvanometer and Ig be the current for full scale DEFLECTION in the galvanometer, the value of the shunt resistance required to convert the galvanometer into an ammeter of 0 to I AMPERE is,S = I g × G I − I gIg is calculated using the equation, Ig = nk, where n is the number of divisions on the galvanometer and K is the figure of merit of galvanometer.What is figure of merit of a galvanometer?The figure of merit of a galvanometer is defined as the current required in producing a unit deflection in the scale of the galvanometer. It is represented by the SYMBOL k and is given by the equation, Where E is the e.m.f. of the cell and θ is the deflection produced with resistance R.Let ‘l’ be the length of the resistance wire required for a resistance of S ohm, where, r is the radius of the wire and ρ is the resistivity of the material of the wire.hopes it's helps uuu ✌✌✌ |
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| 25. |
Obtain the formula for the Lorentz force on a moving electric charge |
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Answer» Lorentz force, the force exerted on a charged particle q moving with velocity v through an electric E and magnetic field B. The entire ELECTROMAGNETIC force F on the charged particle is called the Lorentz force (after the Dutch PHYSICIST Hendrik A. Lorentz) and is given byF = QE + qv × B. |
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| 26. |
Write the formula for the magnetic field at a point on the axis of a current carrying circular ring and explain with a suitable diagram the right hand rule to find the direction of this magnetic field. |
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Answer» tion:RIGHT hand thumb rulehold your thumb of right hand in the direction of current and curling of fingerstowards the REQUIRES POINT GIVES the direction of magnetic field |
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| 27. |
Can a neutron be accelerated using cyclotron ? Why ? |
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Answer» Why Cyclotron cannot accelerate NEUTRON ? SOLUTION : Cyclotrons work on the principle of Lorentz force. A PARTICLE can EXPERIENCE a Lorentz force only if it is charged. Neutrons being uncharged are not accelerated by a cyclotron. |
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| 28. |
Does the angular frequency of particle depend on its momentum in cyclotron ? Yes or No ? |
| Answer» YES..i THINK so........ | |
| 29. |
Answer the following questions :1. Which law of motion gives the measure of force ? |
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Answer» In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can CAUSE an object with mass to change its velocity (which includes to begin moving from a STATE of rest), i.e., to ACCELERATE. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, MAKING it a vector quantity. It is measured in the SI unit of newtons and represented by the SYMBOL F. |
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| 30. |
If we decrease the velocity 5 times the Original one's. What will be the Effect on it's Kinetic Energy ? |
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Answer» energy will DECREASE by 25 TIMES.Explanation:LET the velocity be v m/s.ke=1/2mv^2.Now, if velocity =v/5m/s, then the new ke=1/2m(v/5)^2=1/2m*(v/25) J.which implies that ke will REDUCE by 25 times of the initial ke.**ke=kinetic energy. |
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| 31. |
4)How much below the surface of the earth the acceleration due to gravity i)reduce to 36 % ii)reduce by 36 % of its value on the surface of the earth Radius of earth=6400km |
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Answer» tion:Let the DEPTH at which acceleration due to gravity ( g) becomes 70% or 7g/10 = d.We know thatg_{d} = g(1 - \frac{d}{r} )g d =g(1− rd )Giveng_{d} = \frac{7g}{10}g d = 107g Also, R = 6400 km [ Radius of earth ]7g/10 = g [ 1 - d / 6400]7/10= 1-d/6400d/6400 = 1 - 7/10d/6400 = 3/10d =3 \times 6400 \div 103×6400÷10d= 1920 km.Therefore,1920 \: \: \: km1920km below the SURFACE of the earth does the acceleration due to gravity BECOME 70% of its value at the surface of earth |
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| 32. |
What is the minimum value of force (in following two cases) required to pull a block of mass m on a horizontal surface having coefficient of friction u? Also find the angle this force makes with the horizontal.(a) If force is parallel to horizontal surface(b) If force is in any direction (Also find the angle this force makes with the horizontal.) |
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| 33. |
Write a report on eye defects project |
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Answer» Explanation:1. MYOPIA or NEARSHIGHTED. Myopia occurs when the eyeball is too long, relative to the focusing POWER of the cornea and LENS of the EYE. 2. HYPEROPIA or FARESIGHTED. This vision problem occurs when LIGHT rays entering the eye focus behind the retina, rather than directly on it. 3. ASTIGMATISM. .4. PRESBYOPIA. |
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| 34. |
Force= a) Mass × velocityb) Mass × accelerationc) Weight × accelerationd) None of the above |
| Answer» FORCE = MASS X ACCELERATION | |
| 35. |
What is physics explain reply fast |
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Answer» y GOOD morning Here is your ANSWER Physics is a BRANCH of science which deals with physical quantities of substance. HOPE it help you Please mark me please. #SIBI....... |
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| 36. |
Aye Mere Gore hone has a radius 4 metres and completes a revolution in 2 seconds then acceleration is |
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Answer» Explanation:2 m/ s^2 |
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| 37. |
How to find the distance of image from the object |
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Answer» tion:You can USE this FORMULA to FIND distance from image USING the formula, m= -v/u = hi/ho. |
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| 38. |
Units & Measurementsall S. I units |
| Answer» TION:Some units of Measurement.the meter (m), the KILOGRAM (kg), the SECOND (s), the KELVIN (K), the ampere (A), the mole (mol), and the candela (cd). | |
| 39. |
What was the Newtons second law of motion |
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Answer» tion:Second LAW of motion --- SAYS that The acceleration of an object as PRODUCED by a NET force is directly proportional to the magnitude of the net force, in the same DIRECTION as the net force, and inversely proportional to the mass of the object. |
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| 40. |
If the speed of a changed particle moving through a magnetic field is increased, then the radius of curvature of its trajectory will (A) decrease (B) increase (C) not change (D) become half |
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Answer» g━━━━✿______________________________✿━━━━@Mg━━━━✿ If the speed of a CHANGED particle moving through a magnetic FIELD is increased, then the radius of curvature of its trajectory will (A) decrease (B) INCREASE (C) not change❤❤ (D) become half |
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| 41. |
Give the formula showing Ampere's Circuital Law. |
| Answer» TION:this is this READ this you all GET your ANSWERS ACCORDINGLY | |
| 42. |
Write the statement of Biot—Savart's Law. |
| Answer» TION:magnetic FIELD along VERTICLE current CARRYING conductor in a closed surface is directly proportional to the current and inversely proportional to the perpendicular distance from the point of conductor.... | |
| 43. |
Vaidhut chetra kise khte h |
| Answer» MAKE your QUESTION CLEAR to UNDERSTAND | |
| 44. |
Two parallel thin wires, each carrying current I are kept at a separation r from each other. Hence the magnitude of force per unit length of one wire due to the other wire is ______ . (A) μ₀I²/r² (B) μ₀I²/2πr (C) μ₀I/2πr (D) μ₀I/2πr² |
| Answer» CEXPLANATION:C is the ANSWER of this QUESTION | |
| 45. |
What is unit vetor? |
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Answer» A UNIT VECTOR is a vector having magnitude of one and is used for defining the direction.For EX- i capHOPE IT HELPS!MARK IT AS BRANLIEST. |
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| 46. |
Define two supplementary SI units ?? |
| Answer» SUPPLEMENTARY Units. The supplementary unitis referred as a DIMENSIONLESS unit which is EMPLOYED with the fundamental units to create the derived units. The supplementary units are utilized in TWO major geometric variables such as PHASE angle and solid angle. | |
| 47. |
Two spheres of uniform density have masses 10kg and 40kg . The distance between the centres of the spheres is 200m.Find the gravitational force between them. |
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Answer» F=Gm1m2/r^2so,G=6.626 *10^-11 Nm^2Kg^-2=6.626*10*40*10^-11/200=6.626*2*10^-11F= 13.26 *10^-11I suppose...HOPE it HELPS |
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| 48. |
Full form of RADAR, and where is used |
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Answer» s an object-detection SYSTEM that uses radio waves to determine the range, ANGLE, or velocity of objects. ... The termRADAR was coined in 1940 by the United STATES Navy as an ACRONYM for RAdio Detection And Ranging or RAdio Direction And Ranging......... |
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| 49. |
Guys help me ..fast its important |
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Answer» hello!!ANSWER:-3)5√2 is the CORRECT answer hello!!answer:-3)5√2 is the correct answer hope U LIKE thishello!!answer:-3)5√2 is the correct answer hope u like thisfollow me.... |
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