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. |
Assertion(A): Tertiary methyl amine is less soluble in water than methyl amine. Reason (R): Solubility decreases due to the increase in size of the hydrophobic alkyl group. (a) Both A and R are wrong (b) A is correct but R is wrong. (c) Both A and R are correct and R is the correct explanation of A (d) Both A and R are correct but R is not the correct explanation of A |
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Answer» (c) Both A and R are correct and R is the correct explanation of A |
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| 2. |
The reagent used to convert Nitromethane to methyl amine is …….. (a) Zn/NH4Cl (b) Sn/HCI (c) H2SO4 (d) H2S2O8 |
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Answer» The reagent used to convert Nitromethane to methyl amine is Sn/HCI . |
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| 3. |
The order of basic strength for methyl substituted amine solution is …………. (a) N(CH3)3 > N(CH3)2H > N(CH3)H2 > NH3 (b) N(CH3)H2 > N(CH3)2H >N(CH3)3 > NH3 (c) NH3 > N(CH3)H2 > N(CH3)2H > N(CH3)3 (d) N(CH3)2H > N(CH3)2H2 > N(CH3)3 > NH3 |
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Answer» (d) N(CH3)2H > N(CH3)2H2 > N(CH3)3 > NH3 |
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| 4. |
किसी आवेशित गोलीय चालक में विभव (i) गोले के भीतर शून्य होता है तथा गोले के बाहर भी शून्य होता है। (ii) गोले के भीतर अधिकतम होता है तथा गोले के बाहर शून्य होता है। (iii) गोले के भीतर शून्य होता है तथा गोले के बाहर दूरी बढ़ने के साथ कम होता जाता है। (iv) गोले के भीतर अधिकतम होता है तथा गोले के बाहर दूरी बढ़ने के साथ कम होता जाता है। |
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Answer» (iv) गोले के भीतर अधिकतम होता है तथा गोले के बाहर दूरी बढ़ने के साथ कम होता जाता है। |
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| 5. |
निम्न में से कौन-सा वैद्युत-क्षेत्र का मात्रक नहीं है? (i) न्यूटन/कूलॉम (ii) वोल्ट/मीटर (iii) जूल/कूलॉम (iv) जूल/कूलॉम/मीटर |
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Answer» (iii) जूल/कूलॉम |
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| 6. |
वैद्युत-क्षेत्र की तीव्रता का मात्रक है (i) कूलॉम/न्यूटन (ii) जूल/न्यूटन (iii) न्यूटन/कूलॉम (iv) न्यूटन/मी |
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Answer» (iii) न्यूटन/कूलॉम |
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| 7. |
एक R त्रिज्या वाले Q आवेश से आवेशित धातु के खोखले गोले के केन्द्र से r > R दूरी पर वैद्युत विभव का सूत्र लिखिए। |
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Answer» V = 1/4πɛ0 (Q/r) |
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| 8. |
वैद्युत आवेश के क्वाण्टीकरण (quantisation) का मूल कारण क्या है? |
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Answer» इसका मूल कारण यह है कि एक वस्तु से दूसरी वस्तु में इलेक्ट्रॉनों का स्थानान्तरण केवल पूर्णांक संख्याओं में ही हो सकता है। |
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| 9. |
एक बन्द पृष्ठ के भीतर n वैद्युत द्विध्रुव स्थित हैं। बन्द पृष्ठ से निर्गत वैद्युत फ्लक्स होगा। (i) q/ɛ0 (ii) 2q/ɛ0 (iii) -2q/ɛ0 (iv) शून्य |
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Answer» एक बन्द पृष्ठ के भीतर n वैद्युत द्विध्रुव स्थित हैं। बन्द पृष्ठ से निर्गत वैद्युत फ्लक्स शून्य होगा। |
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| 10. |
वैद्युत फ्लक्स को मात्रक है। |
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Answer» (iv) वोल्ट x मीटर |
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| 11. |
A cricket fielder can throw the Cricket ball with a speed v0. If he throws the ball while running with speed u at an angle θ to the horizontal, find(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.(b) what will be time of flight?(c) what is the distance (horizontal range) from the point of projection at which the ball will land?(d) find θ at which he should throw the ball that would maximize the horizontal range as found in (c).(e) how does θ for maximum range change if u > v0, u= v0, u < v0?(f) how does θ in (e) compare with that for u = 0 (i.e., 45°)? |
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Answer» (a) The angle of projection with horizontal seen by spectator will be tan θ = \(\frac{u_y}{u_x}=\frac{v_0sin\,\theta}{u+v_0\,cos\,\theta}\) (b) y = uxt + \(\frac{1}{2}a_yt^2\) , as y = 0, T = \(\frac{2v_0 sin\,\theta}{g},T[v_0sin\,\theta-\frac{g}{2}T]\) = 0 (c) R = ( u+ v0 cos θ)T = ( u+ v0 cos θ) \(\frac{2v_0 sin\,\theta}{g}\) (d) \(\frac{dR}{dv}=0\) \(\frac{v_0}{g}[2ucos\,\theta+v_0cos\,2\theta\times2]\)=0, cos θ = \(\frac{-u\pm \sqrt{u^2+8v^2_0}}{4v_0}\) ⇒ θ = \(tan^{-1}(\frac{v_0}{u})\) (e) If u = v0 cos θ = \(\frac{-v_0\pm \sqrt{v^2_0+8v^2_0}}{4v_0}=\frac{-1+3}{4}=\frac{1}{2}\) ⇒ θ = 60º If u < < v0, then \(8v^2_0+u^2\) ≈\(8v^2_0\) \(\theta_{max}\) = \(cos^{-1}[\frac{-u\pm2\sqrt2 v_0}{4v_0}]\) \(cos^{-1}[\frac{1}{\sqrt2}-\frac{u}{4v_0}]\) If u < < v0 so neglecting \(\frac{u}{4v_0}\) , then \(\theta_{max}\) = \(cos^{-1}(\frac{1}{\sqrt2})\) = 45º If u > v0 and u > > v0 \(\theta_{max}\) = \(cos^{-1}[\frac{v_0}{u}]\) = 0 ∵ [\(\frac{v_0}{u}\) → 0] ⇒ \(\theta_{max}\) = 90º (f) If u= 0, \(\theta_{max}\) = \(cos^{-1}[\frac{0\pm \sqrt{8v^2_0}}{4v_0}]\) = \(\cos^{-1}(\frac{1}{\sqrt2})\) = 45º Hence, when u = 0, θ ≥ 45º |
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| 12. |
एक इलेक्ट्रॉन तथा एक प्रोटॉन जिनकी गतिज ऊर्जाएँ समान हैं, एकसमान चुम्बकीय क्षेत्र के लम्बवत् प्रक्षेपित किए जाते हैं। पथ की त्रिज्या होगी- (i) प्रोटॉन के लिए अधिक (ii) इलेक्ट्रॉन के लिए अधिक (iii) दोनों के पथ समान वक्रीय होंगे। (iv) दोनों पथ सरल रेखीय होंगे |
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Answer» (i) प्रोटॉन के लिए अधिक (∵ त्रिज्या ∝ द्रव्यमान) |
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| 13. |
What are the S.I. and C.G.S units of work? How are they related? Establish the relationship. |
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Answer» S.I unit of work is Joule. C.G.S unit of work is erg. Relation between joule and erg : 1 joule = 1N x 1m But 1N = 105 dyne And 1m=100 cm = 102 cm Hence, 1 joule = 105dyne × 102cm = 107dyne × cm =107erg Thus, 1 Joule = 107 erg |
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| 14. |
A boy of mass 40 kg runs up a flight of 15 steps each 15 cm high in 10 s. Find:(i) the work done and(ii) the power developed by himTake g = 10 N kg-1 |
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Answer» Force of gravity on boy F= mg = 40 × 10 = 400N Total distance covered in 15 steps , S = 15 × 15cm = 225cm = 2.25m Work done by the boy in climbing = Force x distance moved in direction of force Work, W = F × S = 400 × 2.25 = 900J Power developed = work done/time takrn = 900J/10s = 90W |
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| 15. |
State and define the S.I. unit of work. |
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Answer» S.I unit of work is Joule. 1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 meter in its own direction. |
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| 16. |
A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate:(i) the work done and(ii) power spent. |
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Answer» Total distance covered in 30 steps , S= 30 x 20 cm = 600 cm = 6 m Work done by the boy in climbing= Force x distance moved in direction of force Work, W= F x S= 350 x 6 =2100 J Power developed = work done/time taken = 2100J/60s = 35W |
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| 17. |
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig. 6.15).The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calulate(a) work done against gravity(b) work done against force of friction(c) increase in potential energy(d) increase in kinetic energy(e) work done by applied force. |
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Answer» (a) Wg = mg sinθ d = 1×10 × 0.5 ×10 = 50 J. (b) Wf = μmg cosθ d = 0.1 ×10 × 0.866 × 10 = 8.66 J. (c) ΔU = mgh = 1 ×10 × 5 = 50 J (d) a = {F - (mg sinθ + μmg cosθ)} = [10 -5.87] = 4.13 m/s2 v = u + at or v2 = u2 + 2ad ΔK = (1/2)mv2 - (1/2)mu2 = mad = 41.3J (e) W = F d = 100 J |
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| 18. |
A machine raises a load of 750 N through a height of 16 m in 5 s. calculate:(i) energy spent by machine,(ii) power at which the machine works. |
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Answer» (i) Energy spent by machine or work done = F S Work, W = 750 × 16 = 12000J (ii) Power spent = work done/time taken = 12000J/5s = 2400W |
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| 19. |
Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10 m s-2 |
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Answer» Mass = 0.5 kg Energy = 1 J Gravitational potential energy = mgh 1 = 0.5 × 10 × h 1 = 5h Height, h = 0.2 m |
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| 20. |
A man standing at a position X in front of a plane mirror, a distance of y metre from the mirror as shown in figure. When the man moves 5 m away from the mirror, the new distance between the man and his image becomes 20 metre. What is the value of y?(A) 5 m (B)10 m (C) 20 m (D) 40 m |
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Answer» Correct option (A) 5 m Explanation: New distance between the man and the mirror = y+5 ∴ The distance between the man and his image (y+5)x 2 =20 2y +10=20 2y =10 y= 5 m |
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| 21. |
To produce an image by a convex lens, at the position shown (see figure) the object is needed to be placed(A) Between Y and O (B) At Y (C) Between X and Y (D) Ar X |
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Answer» Correct option (D) Ar X Explanation: lf the image is to be produced at 2F, in case of a convex lens, then the object needs to be placed at X (2F). |
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| 22. |
What is the freezing point of water? |
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Answer» 0°C is the freezing point of water. |
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| 23. |
At what temperature condensation of steam takes place? |
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Answer» Condensation of steam takes place at 100°C. |
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| 24. |
What is the boiling point of water? |
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Answer» The boiling point of water is 100°C. |
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| 25. |
Define solubility / What is solubility? |
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Answer» The property of a substance of getting dissolved is called its solubility. |
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| 26. |
Define density / What is density? |
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Answer» The mass of different substances having the same volume can be different. This difference is because of the difference in their densities. Between substances of the same volume, the ones with greater density are heavier than those of lesser density. |
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| 27. |
When does state of substance change? |
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Answer» State of substance changes when it is heated or cooled. |
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| 28. |
Define fluidity / What is fluidity? |
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Answer» Liquids flow downward on a sloping surface. This property is called fluidity. |
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| 29. |
How is fluidity of any liquid determined? |
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Answer» Fluidity of any liquid is determined by how easily it flows. |
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| 30. |
What is the unit of measuring temperature. |
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Answer» Degrees Celsius (°C) is the unit of measuring temperature. |
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| 31. |
How are candles made? |
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Answer» Candles are made by melting paraffin wax. |
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| 32. |
Define brittleness / What is brittleness? |
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Answer» Some substances break into small pieces or particles. Such substances are said to be brittle. This property of substances is called brittleness. |
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| 33. |
Fill in the blanks :(1) The state of a substance changes if it is …………… or …………… .(2) Every substance in our surroundings is found in either the ……………, …………… or gaseous state.(3) On …………… heat, the substance changes from solid to liquid and liquid to gas.(4) When the substance cools, or …………… heat, it changes from gaseous to liquid and liquid to solid state.(5) A …………… amount of heat must be gained or lost before the state of a substance can change.(6) When a substance gets heat, it becomes …………… and then …………… .(7) If the substance is very hot, we could get …………… .(8) A thermometer is used to measure …………… .(9) …………… is the unit of measuring temperature.(10) Nowadays, …………… thermometers are frequently used.(11) Solids have a shape of its …………… .(12) Solids have a …………… volume.(13) Liquids take the shape of the …………… .(14) Liquids have a …………… volume.(15) Air occupies all the available …………… .(16) Evaporation occurs from the …………… of the water.(17) At sea-level, pure water boils at …………… .(18) Condensation of steam takes place at …………… .(19) The temperature of a substance can fall below …………… .(20) Ice melts at …………… .(21) Each substance has a specific boiling point which is also its …………… point.(22) Each substance has a specific melting point which is the same as its …………… point.(23) Candles are made by melting …………… wax.(24) Solid carbon-dioxide is …………… .(25) Liquid …………… is used in animal husbandry.(26) Sand is melted to make …………… .(27) Iron is melted to make …………… .(28) Substances can be identified by studying their …………… .(29) Substances that break into small particles are said to be …………… .(30) The …………… of any liquid is determined by how easily it flows.(31) Between substances of the same volume, the ones with greater density are …………… than those of lesser density.(32) The property of a substance of getting …………… is called its solubility.(33) Minerals from the earth’s crust are …………… to obtain metals.(34) Metals can be converted into …………… by hammering(35) Metals can be stretched and drawn into …………… .(36) All metals are …………… of electricity to a greater or lesser extent.(37) Every metal has a …………… colour by which it can be identified.(38) Metals produce …………… sound.(39) Metals form a …………… group of substances.(40) Heat is the cause of the change of the state of …………… . |
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Answer» (1) heated, cooled (2) solid, liquid (3) gaining (4) loses (5) specific (6) warm, hot (7) scalded (8) temperature (9) Degree Celsius (°C) (10) digital (11) own (12) definite (13) container (14) specific (15) space (16) surface (17) 100°C (18) 100°C (19) 0°C (20) 0°C (21) condensation (22) freezing (23) paraffin (24) dry ice (25) nitrogen (26) glass (27) tools (28) properties (29) brittle (30) fluidity (31) heavier (32) dissolved (33) processed (34) sheets (35) wires (36) conductors (37) specific (38) ringing (39) separate (40) substances |
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| 34. |
What is sublimation? |
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Answer» The change of a solid substance directly into gas or vapour without changing into a liquid is called sublimation. |
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| 35. |
What happens when a substance gains heat? |
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Answer» When a substance gains heat, it changes its state i.e. from solid to liquid and liquid to gas. |
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| 36. |
At what temperature ice melts? |
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Answer» Ice melts at 0°C. |
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| 37. |
Define hardness / What is hardness? |
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Answer» The hardness of a substance is determined by how much resistance it offers to the substances being pushed through it. |
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| 38. |
Define elasticity / What is elasticity? |
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Answer» Some substances change their shape when a force is applied on them but return to their original shape and size when the force is removed. This property is called elasticity. |
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| 39. |
What is the use of solid carbon-dioxide? |
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Answer» Solid carbon-dioxide (dry - ice) is used to make ice cream and to keep it frozen. |
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| 40. |
What is the use of liquid nitrogen? |
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Answer» Liquid nitrogen is used in animal husbandry. |
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| 41. |
What happens when a substance loses heat? |
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Answer» When a substance loses heat, it changes its state from gaseous to liquid and liquid to solid state. |
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| 42. |
How do we tell how hot or cold a substance is? |
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Answer» The temperature on the thermometer will tell us how hot or cold a substance is. |
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| 43. |
Match the columns :Column 'A'Column 'B'1. Boiling watera. > 35°C2. Body temperatureb. 0°C3. Freezing waterc. < 5°C4. Air (summer afternoon)d. < 15°C5. Inside a fridgee. < -18°C6. Air (winter night)f. 100°C7. Inside the freezerg. 37°C |
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| 44. |
When two plane mirrors are placed parallel and facing each other and an object is kept in between them, we get infinite images. But actually only a few images are visible because the intensity of the image is (A) lncreased after each reflection (B) Decreased after each reflection (C) Always constant (D) None of these |
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Answer» Correct option (B) Decreased after each reflection Explanation: After each reflection, some part of light get absorbed by the mirror while most of the part is reflected. Thus, multiple reflections decreases the intensity of the image so only a few images are visible. |
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| 45. |
A rectangular block of glass PQRS has a refractive index 1.6. A pin is placed midway on the face PQ. When observed from the face PS the pin shall (A) Appear to be near P always (B) Appear to be at the centre of PS (C) Appear to be near S (D) None of these |
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Answer» The Correct option is (D) None of these |
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| 46. |
Pressure is defined momentum per unit volume. Is it true? |
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Answer» No, as it is not dimensionally true. Proof : Momentum/Volume = M1L1T-1/L3 = [M1L-2T-1] Pressure = Force/Area = MLT-2/L2 =[M1L-1T-2] |
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| 47. |
Is it true/possible that there is change in temperature of a body without giving/taking heat to/ from it ? |
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Answer» Yes for example in an adiabatic compression, temperature rises and in an adiabatic expansion, temperature falls, although no heat is given or taken from the system in the respective change. |
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| 48. |
Name four physical quantities having the same dimensional formulae. |
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Answer» Length, radius of gyration, displacement and circumference of a circle. |
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| 49. |
If for a liquid in a vessel, the force of cohesion is more than the force of adhesion, (A) the liquid does not wet the solid (B) the liquid wets the solid (C) the surface of the liquid is plane (D) the angle of contact is zero. |
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Answer» (A) the liquid does not wet the solid |
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| 50. |
Define the force of cohension. |
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Answer» It is the force of attraction between the molecules of the same substance. |
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