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. |
Let us look at a charged particle which is moving in a circle with a constant speed. This is uniform circular motion that you have studied earlier. Thus, there must be a net force acting on the particle, directed towards the centre of the circle. As the speed is constant, the force also must be constant, always perpendicular to the velocity of the particle at any given instant of time. Such a force is provided by the uniform magnetic field \(\overrightarrow{B}\) perpendicular to the plane of the circle along which the charged particle moves. |
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Answer» When a charged particle moves in uniform circular motion inside a uniform magnetic field \(\overrightarrow{B}\) in a plane perpendicular to \(\overrightarrow{B}\) the centripetal force is the magnetic force on the particle. As in any UCM, this magnetic force is constant in magnitude and perpendicular to the velocity of the particle. |
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| 2. |
Magnetic Resonance Imaging (MRI) technique used for medical imaging requires a magnetic field with a strength of 1.5 T and even upto 7 T. Nuclear Magnetic Resonance experiments require a magnetic field upto 14 T. Such high magnetic fields can be produced using superconducting coil electromagnet. On the other hand, Earth’s magnetic field on the surface of the Earth is about 3.6 × 10-5 T = 0.36 gauss. |
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Answer» Magnetic Resonance Imaging (MRI) is a noninvasive imaging technology that produces three dimensional detailed anatomical images. Although MRI does not emit the ionizing radiation that is found in X-ray imaging, it does employ a strong magnetic field, e.g., medical MRIs usually have strengths between 1.5 T and 3 T. The 21.1 T superconducting magnet at Maglab (Florida, US) is the world’s strongest MRI scanner used for Nuclear Magnetic Resonance (NMR) research. Since its inception in 2004, it has been continually conducting electric current of 284 A by itself. Because it is superconducting, the current runs through some 152 km of wire without resistance, so no outside energy source is needed. However, 2400 litres of liquid helium is cycled to keep the magnet at a superconducting temperature of 1.7 K. Even when not in use this magnet is kept cold; if it warms up to room temperature, it takes at least six weeks to cool it back down to operating temperature. The 45 T Hybrid Magnet of the Lab (which combines a superconducting magnet of 11.5 T with a resistive magnet of 33.5 T) is kept at 1.8 K using 2800 L of liquid helium and 15142 L of cold water. |
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| 3. |
Give 3 examples of two dimensional shapes. |
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Answer» Triangle, rectangle circle etc. |
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| 4. |
Plane figures are called (a) multi-dimensional (b) three-dimensional (c) two-dimensional (d) none of the above |
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Answer» (c) two-dimensional |
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| 5. |
Number of faces in a cuboid are(a) 6(b) 8(c) 12(d) 10 |
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Answer» Number of faces in a cuboid are 6. |
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| 6. |
Number of vertices in a cuboid are(a) 12(b) 8(c) 6(d) 4 |
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Answer» Number of vertices in a cuboid are 8. |
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| 7. |
An example of three dimensional figure is(a) cuboid(b) circle(c) square(d) rectangle |
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Answer» An example of three dimensional figure is cuboid. |
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| 8. |
What are two dimensional shapes? |
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Answer» The plane figures having two measurements, length and breadth, are called two dimensional shapes. |
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| 9. |
Recognize polyhedron shapes in the following organized shapes also tell that which shapes are used to organized these shapes. |
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Answer» Figure (i) and (ii) and Polyhedron. Figure (i) is made by combination of Cuboid and Pyramid. Figure (ii) is made by combination of Prism and Pyramid. |
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| 10. |
Write number of edges and faces in a triangular prism. |
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Answer» Number of edges in a triangular prism = 9 and number of faces in a triangular prism = 5. |
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| 11. |
Give 3 examples of three dimensional shapes. |
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Answer» Cuboid, sphere, cylinder etc. |
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| 12. |
Number of faces in a cube(a) 12(b) 8(c) 6(d) 10 |
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Answer» Number of faces in a cube 6. |
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| 13. |
What is the definition of Prism? |
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Answer» A prism is a polyhedron whose base and top are congruent polygons and whose other faces, i.e., lateral faces are parallelogram in shapes. |
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| 14. |
Is a square, prism same as a cube? Explain. |
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Answer» Yes, It can be a cube. But it can be a cuboid also. |
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| 15. |
If a polyhedron has 8 faces and 8 vertices, find the number of edges in it. |
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Answer» Faces = 8 Vertices = 8 using Eulers formula, F + V – E = 2 8 + 8 – E = 2 -E = 2 – 16 E= 14 |
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| 16. |
Can a polyhedron have: (i) 3 triangles only ? (ii) 4 triangles only ? (iii) a square and four triangles ? |
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Answer» (i) No. (ii) Yes. (iii) Yes. |
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| 17. |
What do you mean by pyramid? |
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Answer» A polyhedron having a polygonal base and triangular sides with a common vertex, is called a pyramid. |
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| 18. |
Is a square prism same as a cube? |
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Answer» Yes, a square prism is same as a cube. |
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| 19. |
In a polyhedron, number of faces is 5 and number of edges is 9. Find the number of vertices. |
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Answer» Number of faces (F) = 5 We know by Euler’s formula |
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| 20. |
Find number of edges in a polyhedron which have 9 vertices and 9 faces. |
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Answer» Number of vertices (V) = 9 Euler’s formula : V + F = E + 2 |
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| 21. |
Can a polyhedron have 10 faces, 20 edges and 15 vertices? |
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Answer» Since, F + V = E + 2 As 10 + 15 ≠ 20 + 2 ∴ A polyhedron cannot have 10 faces, 20 edges and 15 vertices. |
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| 22. |
Define a regular polyhedrons. |
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Answer» A polyhedron is said to be regular if its faces are made up of regular polygons and the same number of faces meet at each vertex. |
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| 23. |
Can a polyhedron have 8 faces, 26 edges and 16 vertices? |
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Answer» Number of faces = 8 Number of vertices = 16 Number of edges = 26 Using Euler’s formula F + V – E ⇒ 8 + 16 – 26 ≠ -2 ⇒ -2 ≠ 2 No, a polyhedron cannot have 8 faces, 26 edges and 16 vertices. |
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| 24. |
Find the number of faces in polyhedron having vertices 10 and edges 16. |
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Answer» Number of vertices (V) = 10, Number of edges (E) =16, Number of faces (F) = ? Euler formula V + F = E + 2 ⇒ 10 + F = 16 + 2 or F = 16 + 2 – 10 = 8 |
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| 25. |
The base of a triangular field is three times its altitude. If the cost of sowing the field at ₹ 58 per hectare is ₹ 783, find its base and height. |
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Answer» Take x and the height and 3x as the base of the triangular field We know that Area of triangle = ½ × b × h By substituting the values Area of triangle = ½ × x × 3x So we get Area of triangle = 3/2 x2 1 hectare = 1000 sq. metre It is given that Cost of sowing the field per hectare = ₹ 58 Total rate of sowing the field = ₹ 783 So we can find the total cost by Total cost = Area of the field × ₹ 58 By substituting the values (3/2) x2 × (58/10000) = 783 By cross multiplication x2 = (783/58) × (2/3) × 10000 On further calculation x2 = 90000 By taking the square root x = √90000 So we get x = 300 m Base = 3 × 300 = 900 m Therefore, base = 900 m and height = 300 m. |
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| 26. |
If the area of a trapezium is 28 cm2 and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm. |
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Answer» Given that, Let the length of other parallel side of trapezium = x cm Length of one parallel side of trapezium = 6 cm Area of trapezium = 28 cm2 Length of altitude of trapezium = 4 cm We know that, Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel sides i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides 28 = 1/2 (6 + x) × 4 28 = (6 + x) × 2 (6 + x) = 28/2 (6 + x) = 14 x = 14 – 6 x = 8 ∴ Length of the other parallel side of trapezium = 8 cm |
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| 27. |
The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height. |
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Answer» Let altitude of the triangular field be h m Then base of the triangular field is 3h m. We know that area of triangle = ½ x b x h Area of the triangular field = ½ (h x 3h) = 3h2/2 m2 ……. (i) The rate of cultivating the field is Rs 24.60 per hectare. Therefore, Area of the triangular field = 332.10 /24.60 = 13.5 hectare = 135000 m2 [Since 1 hectare = 10000 m2] …… (ii) From equation (i) and (ii) we have, 3h2/2 = 135000 m2 3h2 = 135000 x 2 = 270000 m2 h2 = 270000/3 = 90000 m2 = (300)2 h = 300 m Hence, Height of the triangular field = 300 m and Base of the triangular field = 3 x 300 m = 900 m |
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| 28. |
Find the area of the quadrilateral ABCD given in Fig.. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other. |
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Answer» Given diagonal AC = 48 cm and diagonal BD = 32 m Area of a quadrilateral = ½ (Product of diagonals) = ½ (AC x BD) = ½ (48 x 32) m2 = 768 m2 |
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| 29. |
Give an example of a metal which (a) is liquid at room temperature. (b) is the best conductor of heat |
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Answer» (a) Mercury (b) Silver, Copper |
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| 30. |
True of False. If False, give the correct statement1. Two different elements may have similar atoms.2. Compounds and elements are pure substance.3. Atoms cannot exist alone; they can only exist as groups called molecules4. NaCl represents one molecule of sodium chloride5. Argon is mono atomic gas. |
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Answer» 1. True 2. True 3. True 4. True 5. True |
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| 31. |
Define the following terms with an example of each: a. Element b. Compound c. Metal d. Non-metal e. Metalloid |
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Answer» (a) Element: It is a substance that cannot be broken down into simpler substance by chemical means Ex. : Oxygen, Hydrogen, Gold & Helium. (b) Compound A compound is a pure substance that is formed when the atoms of two or more elements combine chemically in definite proportions. Ex : H?0, NaCl. (c) Metal A chemical element that is an effective conductor of electricity and heat can be defined as a metal. Ex.: Copper, Iron, Silver, etc (d) Non-Metal Non-metal is an element that doesn’t have the characteristics of metal including, (i.e.) ability to conduct heat or electricity luster or flexibility. Ex. Carbon Iodine, Sulphur. (e) Metalloid : Metalloid is a chemical element that exhibits some properties of metals and some of non-metals. Metalloids are generally semi- conductors. Ex. : Silicon. Arsenic, Antimony and Boron. |
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| 32. |
Fill in the blanks :1. Mercury: liquid at room temperature:: Oxygen: _________ 2. Non metal conducting electricity: _______ :: Metal conducting electricity: Copper3. Elements: combine to form compounds:: Compounds: ________ 4. Atoms: fundamental particle of an element:: ________ fundamental particles of a compound |
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Answer» 1. Gas at room temperature 2. Graphite 3. can be split into elements 4. elements |
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| 33. |
An element which is always lustrous, malleable and ductile (a) non-metal (b) metal (c) Metalloid (d) gas |
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Answer» Correct answer is (b) metal |
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| 34. |
Fill in the blanks :1. The smallest particle of matter that can exist by itself ______ 2. A compound containing one atom of carbon and two atoms of oxygen is ______ 3. ______ is the only non-metal conducts electricity. 4. Elements are made up of ______ kinds of atoms. 5. ______of some elements are derived from Latin or Greek names of the elements. 6. There are ______ number of known elements. 7. Elements are the ______ form of pure substances. 8. The first letter of an element always written in ______ letter. 9. Molecule containing more than three atoms are known as ______10. ______ is the most abundant gas in the atmosphere. |
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Answer» 1. atom 2. CO2 3. Graphite 4. same 5. Symbol 6. 118 7. simplest 8. capital 9. polyatomic molecule 10. Nitrogen |
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| 35. |
What is a chemical formula? |
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Answer» A chemical formula is a symbolic representation of one molecule of an element or a compound. |
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| 36. |
The smallest unit of an element is __________ (a) atom (b) molecule (c) compound (d) none |
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Answer» Correct answer is (a) atom |
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| 37. |
______ can be formed by the same or different kinds of atoms. (a) Atom (b) Molecule (c) Gases (d) None |
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Answer» Correct answer is (b) Molecule |
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| 38. |
What is the fundamental particle of an element? |
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Answer» Atom is the fundamental particle of an element. |
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| 39. |
Matter is its simplest form is called _______ (a) molecule (b) Metals (c) element (d) none |
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Answer» Correct answer is (c) element |
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| 40. |
________ is the first scientist who used the term element. (a) New ton (b) Einstein (c) Robertr boyle (d) Robert hook |
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Answer» Correct answer is (c) Robertr boyle |
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| 41. |
What is arbor vitae? |
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Answer» Core of white matter in cerebellum is called as arbor vitae. |
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| 42. |
How many spinal nerves occur in man? |
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Answer» There are 31 pairs of spinal nerves in man. |
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| 43. |
Name the parts of the brain which functions as endocrine glands. |
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Answer» The parts of brain which functions as endocrine glands are hypothalamus, pituitary body, pineal body. |
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| 44. |
What is meant by cyclotron frequency? |
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Answer» The cyclotron frequency, or the magnetic resonance frequency, is the frequency of revolution of a charged particle of charge per unit mass \(\cfrac {q}{m}\) in a magnetic field of induction B inside a cyclotron. The cyclotron frequency, f = \(\cfrac{qB}{2\pi m}\). The frequency of the alternating voltage applied to the dees of the cyclotron should be equal to the cyclotron frequency. |
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| 45. |
What is resonance condition in a cyclotron ? OR What should be the frequency of the alternating voltage applied between the dees of a cyclotron? |
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Answer» The frequency of the alternating voltage between the dees of a cyclotron should be equal to the cyclotron frequency so that a positive ion exiting a dee always sees an accelerating potential difference to the other dee. This equality of the frequencies is called the resonance condition. |
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| 46. |
If R is the radius of the dees and B the magnitude of the magnetic field induction in which positive charges (q) of mass m escape from the cyclotron, then their maximum speed vmax is(A) \(\cfrac{qR}{Bm}\)(B) \(\cfrac{qm}{BR}\)(C) \(\cfrac{qBR}{m}\)(D) \(\cfrac m{qBR}\) |
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Answer» Correct option is (C) \(\cfrac{qBR}{m}\) |
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| 47. |
The mass of neutron is nearly equal to that of ……. A) proton B) electron C) α – particle D) β – particle |
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Answer» Correct option is A) proton |
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| 48. |
What is molecular mass? |
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Answer» Molecular mass: Molecular mass or molecular weight refers to the mass of a molecule. It is calculated as the sum of the mass of each constituent atom multiplied by the number of atoms of that element in the molecular formula. |
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| 49. |
In the figure ……………. has more energy. A) K – shell B) L – shell C) M – shell D) All are equal |
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Answer» C) M – shell |
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| 50. |
According to Thomson, the total …………… of the atom is uniformly distributed throughout the atom. A) volume B) density C) pressure D) mass |
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Answer» Correct option is D) mass |
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