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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 block of mass `1kg` is pushed on a movable wedge of mass `2kg` and height `h=30cm` with a velocity `u=6m//sec`. Before striking the wedge it travels `2m` on a rough horizontal portion. Velocity is just sufficient for the block to reach the top of the wedge. Assuming all surfaces are smooth except the given horizontal part and collision of block and wedge is jerkless, the friction coefficition of the rough horizontal part is `:` A. `0.125`B. `0.377`C. `0.675`D. `0.45` |
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Answer» Correct Answer - C Apply work energy theorem `mu mg (2)+mgh=KE_(i)-KE_(f) .....(1)` at the highest point `V_(block )=V _(wed g e)` velocity of the block after passing through the rough surface is `v=sqrt(36-2mu (g))` so applying momentum conservation `1(v)=91+2)v_(f)` `rArr v_(1)=v//3` `rArr mu=0.675` |
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| 2. |
A block of mass `1kg` is pushed on a movable wedge of mass `2kg` and height `h=30 cm` with a velocity `u=6m//sec`. Before striking the wedge it travels `2m` on a rough horizontal portion. Velocity is just sufficient for the block to reach the top of the wedge. Assuming all surface are smooth except the given horizontal part and collision of block and wedge is jerkless, the friction coefficient of the rough horizontal part is `:` An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of `800m` and moving vertically upward at a constant velocity of `5m//s`. The horizontal displacement of balloon from the observer is `1600m`. Shells fired from the gun have an initial velocity of `400 m//s` at a fixed angle `theta ( sin theta =3//5 `and `cos theta =4//5 )`. The observer having gun waits `(` for some time after observing the balloon) and fires so as to destroy the balloon. Assume `g=10m//s^(2)`. Neglect air resistance. A. `0.125`B. `0.377`C. `0.675`D. `0.45` |
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Answer» Correct Answer - C velocity of the block after passing through the rough surface is `v=sqrt(36+2mug(2))=sqrt(36-40mu)` Apply work energy theorem `mu mg (2)+mgh=KE_(i)-KE_(f) ...(1)` at the highest point `V_(block)=V_(wed g e)` `2-mu+3=(1)/(2)1(6)^(2)-(1)/(2)3v^(2)` `mu=(54)/(80)=0.675` |
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| 3. |
A block of mass `m` is pushed towards a movable wedge of mass `nm` and height `h`, with a velocity `u`. All surfaces are smooth. The minimum value of `u` for which the block will reach the top of the wedge is A. Block will reach top of the wedge if `u=sqrt(2gh(1-(1)/(n)))`B. Block will reach top of the wedge if `u=sqrt(2gh(1+(1)/(n)))`C. If the block overshoots `P`, the angle of projectile is `alpha`.D. If the block overshoots `P`, the angle of projectile is less than `alpha`. |
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Answer» When the particle just reaches the top of the wedge `mu=(m+n)v`…..(`1`) `u=sqrt(2gh(1+(1)/(n)))` Angle of projection as observed by the ground will be loss than `alpha` `vecv_(block//ground)=vecv_(block//"wedge")+vecv_(block//ground)` |
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| 4. |
The co-ordinates of a moving particle at any time t are given by x = αt3 and y = βt3. The speed of the particle at time t is given by (1) 3t√(α2 + β2)(2) 3t2√(α2 + β2)(3) t√(α2 + β2)(4) √(α2 + β2) |
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Answer» Answer is (2) 3t2√(α2 + β2) νx = dx/dt = 3 ∝ t2. or νy = dy/dt = 3βt2. ν = √(ν2x + ν2y) = 3t2√(α2 +β2) |
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| 5. |
A painter is applying force himself to raise him and the box with an acceleration of `5(m)/(s^2)` by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If `g=10(m)/(s^2)`, then:A. tension in the rope is 1125 NB. tension in the rope is 2250 NC. force of contact between the painter and the floor is 375 ND. none of these |
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Answer» Correct Answer - A::C For painter : `R = T - mg = ma` `R + T = m(g+a)` ….(1) For the system : `2T -(m _M) g = (m+M)a` `2T = (m+M) (g+a)`…(2) where , `m = 100 kg` `M = 50 kg` `a = 5 m//sec_(2)` `:. T = (150 xx 15)/(2) = 1125 N` and , `R = 375 N`. |
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| 6. |
A painter is applying force himself to raise him and the box with an acceleration of `5(m)/(s^2)` by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If `g=10(m)/(s^2)`, then:A. tension in the rope is `1125 N`B. tension in the rope is `2250 N`C. force of contact between the painter and the floor is `375 N`D. None of these |
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Answer» Correct Answer - A::C For painter `R + T - mg = ma` ltbRgt `r + t = m(g + a)`….(1) For the system , `2T - (m + M)g - (m + M)a` `2T = (m+M)(g+a)….(2)` where, `m = 100 kg` `M = 50 kg` `a = 5 m//sec ^(2)` `:. T = (150 xx 15)/(2) = 1125 N` and , `R = 375 N`. |
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| 7. |
A pulse of light of duration `100ns` is absorbed completely by a small object initially at rest power of the pulse is `30 mW` and the speed of light is `3 xx 10^(8) ms^(-1)` The final momentum of the object isA. `0.3 xx 10^(-17) kg ms^(-1)`B. `1.0 xx 10^(-17) kg ms^(-1)`C. `3.0 xx 10^(-17) kg ms^(-1)`D. `9.0 xx 10^(-17) kg ms^(-1)` |
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Answer» Correct Answer - B Change in momentum `= ("power" xx "total time")/("speed of time") = (P xx t)/(c)` `1.0 xx 10_(-17) kg xx m//s` |
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| 8. |
A thin uniform rod AB of mass m and length l is hinged at one end to the level floor and stands vertically. If it s allowed to fall, with what angular velocity will it strike the floor?A. `sqrt((3g)/(L))`B. `sqrt(3gL)`C. `sqrt((L)/(3g))`D. `sqrt((g)/(3L))` |
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Answer» Correct Answer - B `Mg(1)/(2)=(1)/(2) (mL^(2))/(3) omega^(2) rArr omega =sqrt((3g)/(L))` `V=omega L=sqrt(3gL)` |
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| 9. |
If a man has a velocity varying with time given as `v=3t^(2),v` is in `m//s` and `t` in `sec` then `:` Find out his acceleration after 3 seconds `:`A. `9m//s^(2)`B. `18m//s^(2)`C. `12m//s^(2)`D. `6m//s^(2)` |
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Answer» Correct Answer - B `f=(dv)/(dt)=6t` `t=6xx3=18m//s^(2)` |
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| 10. |
Energy from the sun is recived on earth at the rate of `2` cal per `cm^(2)` per min. If avergae wavelength of solar light be taken at `5500 A` then how many photons are recived on the earth per `cm^(2)` per min ? `(h =6.6 xx 10^(-34)J-s, 1cal = 4.2J)`A. `1.5 xx 10^(13)`B. `2.9 xx 10^(13)`C. `2.3 xx 10^(19)`D. `1.75 xx 10^(19)` |
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Answer» Correct Answer - C Energy received from the sun `= 2 cal cm^(-2) (min)^(-1)` `= 8.4 J cm^(-2) (min)^(-1)` Energy of `1` photon recived from sun, `E = (hc)/(lambda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(5500 xx 10^(-10))` `= 3.6 xx 10^(-19) J` `:.` Number of photos reaching the earth per `cm^(2)` per minute will be `n = ("energy received from sun")/("energy of one photon")` or `n = (8.4)/(3.6 xx 10^(-19)) = 2.3 xx 10^(19)` |
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| 11. |
A thermally insulated vessels contains two liquids with initial temperature `T_(1)` and `T_(2)` and specific heats `C_(0)` and `2C_(0)`, separted by a non conducting partition. The partition is removed, and the difference between the initial temperature of one of the liquids and the temperature `T` established in the vessel turns out to be equal to half the difference between the initial temperatures of the liquids. Determine the ratio of `M_(1)//M_(2)` of the masses of the liquids. |
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Answer» `DeltaQ=0` `C_(0)M_(1)(T-T_(1))+2C_(0)M_(2)(T-T_(2))=0` `2(T_(1)-T)=2(T_(1)-T_(2))` `(M_(1))/(M_(2))=(2C_(0))/(C_(0))=2` |
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| 12. |
The equation of displacement due to a sound wave is s=`s_0`sin^(2)((wt-kx)). if the bulk modulus of the medium is `B`, then the equation of pressure variation due to that sound isA. `B k s_(0) sin(2 omega t-2 k x)`B. `-B k s_(0) sin(2 omega t-2 k x)`C. `B k s_(0)cos^(2) ( omegat-kx)`D. `-B k s_(0)cos^(2) ( omegat-kx)` |
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Answer» Correct Answer - A The equation of pressure variation due to sound is `p=-B(ds)/(dx)=-B(d)/(dx)[s_(0)sin^(2)(omegat-kx)]` `=Bks_(0)sin (2 omegat-2kx)` |
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| 13. |
A machine gun fires 360 bullets per minute, with a velocity of `600 ms^(-1)`.If the power of the gun is 5.4 kW then mass of each bullet isA. 5kgB. 0.5 kgC. 5gD. 0.5g |
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Answer» Correct Answer - B |
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| 14. |
A tennis ball bounces down a flight of stairs, striking each step in turn and rebounding to half to height of the step. The coefficient of restitution isA. `(1)/(2)`B. `(1)/(sqrt2)`C. `(1)/(4)`D. 1 |
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Answer» Correct Answer - B Let h be the height of each step. Then `e=(v)/(u)=(sqrt(2gh))/(sqrt(2g(2h)))=(1)/(sqrt2)` |
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| 15. |
A machine gun fires `10` bullets per second in horizontal direction. Mass of each bullet is `80g` and muzzle velocity is `500m//s`. If the average horizontal force the shooter must apply to keep the machine gun stationary is `Fxx10^(2)N`, find numerical velue of `F`A. `4`B. `5`C. `8`D. `12` |
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Answer» Correct Answer - A Ans. (1) `F_(n et)=(DeltaAp)/(Deltat)=(nmv)/(t)=(10xx80xx500)/(1000xx1)=400N` |
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| 16. |
............ is formed by the union of male and female gametes- (a) Ova (b) Zygote (c) Blastocyst (d) None of these |
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Answer» Zygote is formed by the union of male and female gametes. |
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| 17. |
An artillery shell is fired so as to land in one piece at x=1000m. Instead it explodes at the top of its parabolic trajectory into two pieces of equal mass. The first piece lands at x=750 m . The other piece happens to hit the ground at the same instant. it lands at x=A. 1250 mB. 1750 mC. `-750 m `D. 1000m |
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Answer» Correct Answer - 1 Range =1000 `X_(cm)=(M_(1)x_(1)+M_(2)x_(2))/(M_(1)+M_(2))` `1000=(M/2(750)+M/2(n))/M` 2(1000-375)=x 1250=x |
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| 18. |
The magnetic field at the centre of a current carrying circular loop of radius R, is B1. The magnetic field at a point on its axis at a distance R from the center of the loop is B2. Then the ratio (B1/B2) is :(a) 2√2(b) 1/√2(c) √2(d) 2 |
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Answer» Option : (a) 2√2 |
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| 19. |
The rms current in a circuit connected to a 50 Hz ac source is 15 A. The value of the current in the circuit (1/600)s after the instant the current is zero is -(a) \(\frac{15}{\sqrt2}\)A(b) 15√2A(c) \(\frac{\sqrt 2}{15}\)A(d) 8A |
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Answer» Option : (a) \(\frac{15}{\sqrt2}\)A |
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| 20. |
The self-inductance of a solenoid of 600 turns is 108 mH. The self-inductance of a coil having 500 turns with the same length, the same radius and the same medium will be :(a) 95 mH(b) 90 mH(c) 85 mH(d) 75 mH |
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Answer» Option : (d) 75 mH |
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| 21. |
How the self inductance of a coil depends on number of turns in the coil. |
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Answer» Self inductance of a coil is directly proportional to square of number of turns. L ∝ n2. |
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| 22. |
A particle executes simple harmonic motion between `x = -A and x = + A`. The time taken for it to go from `0 to A//2 is T_1 and to go from A//2 to (A) is (T_2)`. Then.A. `T_1 lt T_2`B. `T_1 gt T_2`C. `T_1 = T_2`D. `T_1 = 2T_2` |
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Answer» Correct Answer - A Method 1 : The velocity of a body executing `SHM` is maximum at its centre and decreases as the body proceeds to the extremes. Therefore, if the time taken for the body to go from `O` to `A//2` is `T_(1)` and to go `A` is `T_(2)`, then obviously `T_(1) lt T_(2)`. Method 2 : Quantitative. Any `SHM` is given by the equation `x = sin omega t`. where `x` is the displacement of the body at any instant `t.a` is the amplitude and `omega` is the angular frequency. When `x =0, omega t_(1) = 0` `:. t_(1) = 0` `{:(When,x=a//2,omegat_(2)=pi//6,t_(2)=pi//6 omega,),(When,x=a,omegat_(3)=pi//2,t_(3)=pi//2omega,):}` Time taken from `O` to `A//2` will be `t_(2) - t_(1) = (pi)/(6 omega) = T_(1)` Time taken from `A//2` to `A` will be `t_(3)-t_(2)=(pi)/(2 omega)-(pi)/(6 omega)=(2pi)/(6 omega)=(pi)/(3 omega) = T_(2)` Hence `T_(2) gt T_(1)`. |
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| 23. |
Give example of a METAL that do not follow Ohm's Law. |
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Answer» Tungsten is an example of a metal that does not follow Ohm's Law. If you were to plot a graph of current against voltage, over a range of voltages, you will find that the result is a curve - showing that current is not directly proportional to voltage, which is the requirement for Ohm's Law. While the ratio of voltage to current will indicate what the resistance happens to be for that particular ratio, you will find that, for tungsten, that ratio continually changes as you increase voltage - proving that tungsten does not obey Ohm's Law. The general rule is that if there is no straight-line relationship between voltage and current, then Ohm's Law doesn't apply. Semiconductors have very low resistance values when passing current in the forward direction, and very high resistance when not passing current in the backward direction, but, at any given value of current and voltage, they still abide by the fundamental laws of nature, including Ohm's Law. |
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| 24. |
Define Equipotential surface |
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Answer» Equipotential surface: Any surface that has same electric potential at every point on it is called an equipotential surface. |
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| 25. |
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?(a) a = 0.7x(b) a = 200x2(c) a = 10x(d) a = 100x2 |
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Answer» Correct answer is (c) ie. a = 10x Represent SHM. This is because acceleration is proportional and opposite to displacement (x). |
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| 26. |
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m. If the position moves with simple harmonic motion with an angular frequency of 200 rev/min, what is its maximum speed? |
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Answer» a = \(\frac{1}{2}\)m, ω = 200 rev/min Umax = aω = \(\frac{1}{2}\) × 200m/min = 100 m/min. |
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| 27. |
The potential of a point B (-20m, 30m) taking the potential of a point A (30m, -20m) to be zero in an electric field `vec E = 10xhati - 20hatj NC^(-1)` isA. 350 VB. `-100 V `C. 300 VD. 3500 V |
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Answer» Correct Answer - D d. `V_(a)-V_(A)=-underset(30)overset(-20)int 10xdx-underset(-20)overset(30)int(-20)dy` or `V_(B)-0=3500V` |
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| 28. |
In wetland farming, the rainfall is in ______ soil moisture requirement of plants during rainy season.1. excess of2. inverse of3. lesser to4. same to |
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Answer» Correct Answer - Option 1 : excess of The correct answer is excess of.
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| 29. |
PCl5 is possible but NCl5 does not exist :(a) in N, d-sub-shell is absent(b) ionization energy of N is very high(c) it does not like Cl(d) None of these |
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Answer» (a) NCl5 in not possible because N does not contain d-orbitals. Only nitrogen has a tendency to form pπ – pπ multiple bonds. Other forms dπ –pπ multiple bonds easily |
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| 30. |
Ammonia is generally manufactured for fertilizers by the reaction(a) 2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3(b) By passing an electric discharge in a mixture of N2 and H2(c) By passing a mixture of N2 and H2 under high pressure and moderate temperature over a catalyst(d) None of these |
| Answer» (c) By Haber’s process | |
| 31. |
Liquid ammonia bottles are opened after cooling them in ice for sometime. It is because liquid NH3(a) Brings tears to the eyes(b) Has a high vapour pressure(c) Is a corrosive liquid(d) Is a mild explosive |
| Answer» (b) Liquid ammonia has high vapour pressure which is lowered down by cooling, otherwise the liquid will bump. | |
| 32. |
NH3 gas is dried over :(a) CaO (b) HNO3 (c) P2O5 (d) CuSO4 |
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Answer» (a) HNO3 and CuSO4 are not drying agents, while P2O5 reacts with NH3 . The moisture present in NH3 is removed by passing it through a tower packed with quicklime (CaO) |
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| 33. |
Which of the following acids possesses oxidising, reducing, and complex forming properties ?A. `NHO_(3)`B. `H_(2)SO_(4)`C. `HCI`D. `HNO_(2)` |
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Answer» Correct Answer - D Intermediate oxidation number. |
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| 34. |
(i) A powdered substance (A) on treatment with fusion mixture gives a green coloured compound (B). (ii) The solution of (B) in boiling water on acidification with dilute `H_(2)SO_(4)` gives a pink coloured compound `(C )`. `(iii)` The aqueous solution of (A) on treatment with NaOH and `Br_(2)-` water gives a compound (D). (iv) A solution of (D) in conc. `HNO_(3)` on treatment with lead peroxide at boiling temperature produced a compound (E) which was of the same coloure at that of (C). (v) A solution of (A) on treatment with a solution of barium chloride gave a white precipitate of compound (F) Which was insoluble in conc. `HNO_(3)` and conc. HCl. The oxidation state of central metal ions of (A), (B) and (C ) compounds are respectively `:`A. `+II, +VI and +VII`B. `+II, +VI and +VI`C. `+II, +VII and +VII`D. `+VI, +VII and +VII` |
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Answer» Correct Answer - A (A)`MnSO_4=+II , (B)MnO_4^(2-)=+VI , (C )MnO_4^(-)=+VI` |
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| 35. |
(i) A powdered substance (A) on treatment with fusion mixture gives a green coloured compound (B). (ii) The solution of (B) in boiling water on acidification with dilute `H_(2)SO_(4)` gives a pink coloured compound `(C )`. `(iii)` The aqueous solution of (A) on treatment with NaOH and `Br_(2)-` water gives a compound (D). (iv) A solution of (D) in conc. `HNO_(3)` on treatment with lead peroxide at boiling temperature produced a compound (E) which was of the same coloure at that of (C). (v) A solution of (A) on treatment with a solution of barium chloride gave a white precipitate of compound (F) Which was insoluble in conc. `HNO_(3)` and conc. HCl. Which of the following is true for compound (C ).A. It oxidises ammonia to nitrogen dioxide in neutral mediumB. It is coloured and colour is due to d-d transitionC. It can be oxidised by ozoneD. It is obtained by alkaline fusion of pyrolusite in presence of air followed by electrolytic oxidation |
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Answer» Correct Answer - D (A)`2MnO_4^(-)+2NH_3 to 2MnO_2+ N_2 + 2OH^(-)+2H_2O` (B)`MnO_4^(-)` is pink coloured and colour is due to charge transfer. (C )In `MnO_4^(-)` , central atom Mn is in highest oxidation state i.e. +VII and hence can not be oxidised. (D)`MnO_2 + OH^(-)+O_2(air)overset("Fusion")toMnO_4^(2-)overset([e^-])toMnO_4^-` |
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| 36. |
Hybridization and structure of XeF4 is(a) sp3d, trigonal bipyramidal(b) sp3, tetrahedral(c) sp3d2, square planar(d) sp3d2, hexagonal |
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Answer» (c) Hybridisation of XeF4 is sp3d2 and structure is square planar |
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| 37. |
Dehydration of alcohols to form an alkene is always carried out with conc. H2SO4 and not with conc. HCl or HNO3. Explain. |
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Answer» In acidic medium alcohols protonated then loses H2O to form a carbo cation. If HCl Cl- strong nucleophile cause nucleophilic substitution, HNO3 causes oxidation. |
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| 38. |
Assertion `:-` Benzaldehyde forms two oximes on reaction with `NH_(2)OH`. Reason`:-` The two oximes arises due to geometrical isomerism around `C-N` bond.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False |
| Answer» Correct Answer - A | |
| 39. |
Which of the following amines yield foul smelling product with haloform and alcoholic KOH? (A) Ethylamine (B) Diethylamine (C) Triethylamine (D) Ethylmethylamine |
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Answer» (A) Ethylamine Primary amines undergo this reaction. |
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| 40. |
Match the proper entries from column-2 to column-1 using the codes given below the columns, if deviation in the Column–II is the magnitude of total deviation (between incident ray and finally refracted or reflected ray) to lie between 0º and 180º. Here n represents refractive index of medium. A. `{:(P,Q,R,S),(4,2,1,3):}`B. `{:(P,Q,R,S),(1,2,4,2):}`C. `{:(P,Q,R,S),(3,1,4,2):}`D. `{:(P,Q,R,S),(2,4,1,3):}` |
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Answer» `C=sin^(-1)((1)/(2//1))=30^(@)` for `i=37` TIR so, `delta=pi-2(37^(@))=104^(@)` `i=25` refraction `deltalt(pi)/(2)-C` `i=45^(@)` TIR so, `delta=pi-2((pi)/(4))=90^(@)` by applying snells law for prism `i=90` `r_(1)=30," "r_(2)=30` `e=45` `delta=90+45-60=75^(@)` |
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| 41. |
An ideal gas consists of a large number of identical molecules. Absolute temperature of the gas is T(in kelvin). Molecular weight of gas is `M` and `R` is gas constant. Mathch the proper entries from column-2 to column-1 using the codes given below the columns. `{:("Column"-1,"Column"-II),((P)"Root mean square speed of molecules is greater than",(1)sqrt((RT)/(M))),((Q)"Most probable speed of molecues is smaller than",(2)1.5sqrt((RT)/(M))),((R)"Average velocity of a molecule is smaller than",(3)2sqrt((RT)/(M))),((S)"Speed of a molecule may be greater than",(4)2.5sqrt((RT)/(M))):}`A. `{:(p,q,r,s),(2,1,3,4):}`B. `{:(p,q,r,s),(3,1,4,2):}`C. `{:(p,q,r,s),(1,3,2,4):}`D. `{:(p,q,r,s),(4,2,3,1):}` |
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Answer» Root mean square speed of molecules `=sqrt((3RT)/(M))=1.732sqrt((RT)/(M))` Most probable speed of molecures `=sqrt((2RT)/(M))=1.44sqrt((RT)/(M))` Average velocity of a molecule is zero Speed of any individual molecule may be anything. |
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| 42. |
Initially two radioactive nucleus have same no of active nucleus their half life are 3 years and 4 years respectively after how many years, number of nucleus of one of the radioactive element is half the number of active nucleus of other radioactive element. |
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Answer» Let `lambda_(1)=(ln2)/3` and `lambda_(2) =(ln2)/4` `(N//2)/N=(N_(0)e^(-lambda_(1)t))/(N_(0)e^(-lambda_(2)t))` t=12 |
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| 43. |
The elements Li(Z = 3), Na (Z = 11)and K (Z = 19) belong to group 1(i) Predict the periods they belong.(ii) Which one of them is least reactive?(iii) Which one of them has the largest atomic radius? Give reason to justify. |
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Answer» As we already know that number of valence electrons is equal to its group number. Here, all of them have only one electron in their valence shell hence, they belong to First group. Lithium is less reactive due to its smallest size and more ionization energy. |
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| 44. |
You want to rent out your newly constructed flat in the heart of the city. Draft an advertisement in more than 50 words to be published in 'The Deccan Herald', Bengaluru under classified column Give all the necessary details. You are Mohan/Mahima of Jayanagar, Bengaluru. |
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Answer» For Rent AVAILABLE for rent a fully furnished, three bedroom flat located in the heart of the city, Dwarika. A drawing hall and balconies on both sides. 24- hour water and electricity. Parking facility also available. Nearest Railway station just 1 km away. For further details contact Mr. Mohan, 1-4-C Jayanagar, Bangaluru, contact number 0xx- xxxxxxxxxx |
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| 45. |
A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground re respectively. (1) g, g(2) g - a, g - a (3) g - a, g (4) a,g |
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Answer» Answer is (3) g - a, g Acceleration observed by the man outside the lift = (g - a) Acceleration observed by the man outside lift = g |
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| 46. |
(a) presbyopia(b) accommodation(c) near-sightedness(d) far-sightedness |
| Answer» (b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens. | |
| 47. |
The international standard thermometer is kept: A. near Washington, D.C. B. near Paris, France C. near the north pole D. near Rome, Italy E. nowhere (there is none) |
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Answer» E. nowhere (there is none) |
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| 48. |
A body of mass 10kg is sliding on a frictionless surface with a velocity 2m/s.What is the force required to keep it moving with a same velocity? |
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Answer» According to the Newton’s law of inertia (Newton’s First Law of Motion) the body can move with a constant speed only if it is acted upon by the balanced forces. |
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| 49. |
A thermometer indicates 98.6◦ C. It may be: A. outdoors on a cold day B. in a comfortable room C. in a cup of hot tea D. in a normal person’s mouth E. in liquid air |
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Answer» C. in a cup of hot tea |
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On a very cold day, a child puts his tongue against a fence post. It is much more likely that his tongue will stick to a steel post than to a wooden post. This is because: A. steel has a higher specific heat B. steel is a better radiator of heat C. steel has a higher specific gravity D. steel is a better heat conductor E. steel is a highly magnetic material |
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Answer» D. steel is a better heat conductor |
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