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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 5501. |
During the fire he ________ the people into groups which carried sand and water to throw on to the flames. A) called B) dispersed C) organized D) made E) planned |
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Answer» Correct option is C) organized |
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| 5502. |
Explain the Dilution Factor calculation method |
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Answer» Dilution Factor = final volume/initial volume. DF = Vf /Vi For Example : How would you make 500 mL of a 1:250 dilution? Solution: DF = Vf /Vi Vi = Vf /DF = 500mL/250 = 2.00 mL Pipet 2.00 mL of your stock solution into a 500 mL volumetric flask. Add diluent to the mark on the flask (you will have added about 498 mL of water). You now have a 1:250 dilution of your original solution. |
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| 5503. |
Assertion:- The colour of old lead paintings can be retored by washing with dilute solution of `H_(2)O_(2)` Reason:- Black lead sulphide is oxidised by `H_(2)O_(2)` to white lead sulphate.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 Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False. |
| Answer» Correct Answer - A | |
| 5504. |
Hello, Tom. What's been happening lately? A) What's in? B) What's next? C) What's up? |
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Answer» Correct option is C) What's up? |
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| 5505. |
Could you tell me what the amount of the bill is? A) what's the dirt B) what's the doc C) what's the damage |
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Answer» Correct option is C) what's the damage |
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| 5506. |
Is demineralised or distilled water useful for drinking purpose? If not, how can it be made useful? |
| Answer» Demineralised or distilled water does not contain any caton or anion, and hence even the useful ions or minerals are missing. Hence, it is not useful for drinking purposes. To make it useful for drinking purposes, addition of useful minerals in adequate amounts should be done. | |
| 5507. |
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? |
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Answer» Demineralised or distilled water is not useful for drinking purposes because it does not contain even useful minerals. Therefore, to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised or distilled water. |
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| 5508. |
1. Explain why hydrogen peroxide is stored in coloured plastic bottles. 2. Write any two uses of H2O2 . |
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Answer» 1. In the presence of metal surfaces or traces of alkali (present in glass conatiners), the decomposition of H2O2 (2H2O2 → 2H2O + O2) is catalysed. It is therefore stored in wax-lined glasses or plastic vessels in dark. 2. 1. As a hair bleach and as a mild disinfectant. 2. In the syntheseis of hydroquinone, tartaric acid and certain food products and pharmaceuticals. |
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| 5509. |
Why H2O2 is used as mild antiseptic? |
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Answer» The oxidising property of hydrogen peroxide and harmless nature of its products such as water and oxygen, leads to oxidation of pollutants in water and act as a mild antiseptic. |
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| 5510. |
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true? (A) The temperature distribution over the filament is uniform (B) The resistance over small sections of the filament decreases with time (C) The filament emits more light at higher band of frequencies before it breaks up (D) The filament consumes less electrical power towards the end of the life of the bulb |
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Answer» (C) The filament emits more light at higher band of frequencies before it breaks up (D) The filament consumes less electrical power towards the end of the life of the bulb When filament breaks up, the temperature of filament will be higher so according to wein’s law \((\lambda _m ∝\frac{1}{T},v_ m∝T)\) , the filament emits more light at higher band of frequencies. As voltage is constant, so consumed electrical power is P = v2 / R As R increases with increase in temperature so the filament consumes less electrical power towards the end of the life of the bulb. |
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| 5511. |
Four types of oscillatory system a simple pendulum a physical pendulum a torsional pendulum and a spring mass system each taken on to the moon of same time period are taken to the mass if line period will have it unchanged?A. only spring-mass systemB. spring-mass system and torsional pendulumC. spring-mass system and physical pendulumD. none of these |
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Answer» Correct Answer - B Both the spring-mass system & torsional pendulum have no dependence on gravitational acceleration for their times periods. |
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| 5512. |
Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is `25` minutes and for smaller one is `16` minutes, the ratio of thermal conductivities of the materials of larger sphere to that of smaller sphere is:A. `4:5`B. `5:4`C. `25:8`D. `8:25` |
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Answer» Correct Answer - 4 `H=Q/t=(KA)/(l)(theta_(1)-theta_(2))` `=(mL)/(t)=(K4piR^(2))/(l)(theta_(1)-theta_(2))` `(VrhoL)/(t)=(K4piR^(2))/(l)(theta_(1)-theta_(2))` `(4)/(3)(pir^(3)rhoL)/(t)=(K4pir^(2))/(l) (theta_(1)-theta_(2))K propto (rl)/(t)` `(K_(1))/(K_(2))=(r_(1))/(r_(2)).(l_(1))/(l_(2))xx(t_(2))/(t_(1))` `(K_(1))/(K_(2))=(2r)/(r).(l)/(4l).(16)/(25)=(8)/(25)` |
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| 5513. |
If `S_(n)` denotes sum of first `n` tem of a series and if `S_(n+2)-S_(n)=n^(2)` then `S_(20)` is `("given that" T_(1)+T_(2)=0)`A. `1540`B. `1140`C. `770`D. `1120` |
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Answer» Correct Answer - B If `S_(n)` denotes ………… `S_(n+2)-S_(n)=n^(2)` `T_(n+2)+T_(n+1)=n^(2)` `T_(1)+T_(2)=0` `T_(3)+T_(2)=1` `T_(4)+T_(3)=4` `T_(5)+T_(4)=9` and so on Now sequence is `T_(1), -T_(1), 1+T_(1),3-T_(1),6+T_(1),10-T_(1)`............. `i.e. S_(20) = T_(1)-T_(2)+1+3+6+10+` ....... up to 20 terms `S_(20 = 1+3+6+10+` ......... up 18 terms `S_(20) = sum_(r=1)^(18)(r^(2))/(2)+(r )/(2)` `S_(20) = 1140`. |
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| 5514. |
A wall is made up of two layers, \( A \) and \( B \). The thickness of the two layers is the same, but the materials are different. The thermal conductivity of \( A \) is double that of \( B \). If in thermal equilibrium, the temperature difference between the two ends is \( 36^{\circ} C \), then the difference in temperature between the two surfaces of \( A \) will be:1. \( 6^{\circ} C \)2. \( 12^{\circ} C \)3. \( 18^{\circ} C \)4. \( 24^{\circ} C \) |
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Answer» Correct option is (2) 12∘C Consider two walls 'A' and 'B' of thickness 't' each, and thermal conductivity of '2K' and 'K' respectively. The temperatures at the left of A = Ta The temperatures at the right of B = Tb The temperature at the junction = T Heat flow is a constant in steady state: \(Q = KA \frac{dT}{dx}\) = constant Equating heat flow for both walls we get: \(2KA \frac{T-T_a}{t - 0} = KA \frac{T_b - T}{2t - t}\) Which simplifies to: 3T = Tb + 2Ta Also given that the temperature difference between the walls is 36∘C Tb − Ta = 36 Combining the two equation in T, Ta, Tb and eliminating Tb We get T − Ta = 12∘C |
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| 5515. |
Two circular disc `A` and `B` with equal radii are blackened. They are heated to same temperature and are cooled under identical conditions. What inference do your draw from their cooling curves? A. A and B have same specific beatsB. specifice heat of A is lessC. specific heat of B is lessD. nothing came be said. |
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Answer» Correct Answer - B `-(dt)/(dt)=(4sigmaAT_(0)^(3))/(ms)(T-T_(0))` `-(dT)/(dt)prop(1)/(3)` |
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| 5516. |
One end of a light string of length `L` is connected to a ball and the other end is connected to a fixed point `O`. The ball is released from a rest at `t=0` with string horizontal and just taut. The ball then moves invertical circular path as shown. The time taken by ball to go from position `A` to `B` is `t_(1)` and from `B` to lowest position `B` to lowest position `C` is `t_(2)`. Let the velocity of ball at `B` is `vec(v)_(B)` and `C` is `vec(v)_(C)` respectively. If `|vec(v)_(C)|=|vec(v)_(B)|` then the value of `theta ` as shown isA. `cos^(-1) (1)/(4)`B. `sin^(-1) (1)/(4)`C. `cos^(-1) (1)/(2)`D. `sin^(-1) (1)/(2)` |
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Answer» Correct Answer - B `V_(B)=sqrt(2gLsintheta)` and `V_(C)=sqrt(2gL)` `V_(C)=2V_(B)` Then `2gL=4(2gL sin theta )` or `sin theta =(1)/(4) ` or `theta=sin ^(-1)((1)/(4))` |
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| 5517. |
A lift starts from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is A. `20m`B. `64m`C. `32m`D. `128m` |
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Answer» Correct Answer - D At `t=4 sec. V=0+(4)(4)=16m//sec.` At `t=8sec. V=16m//sec`. At `t=12 sec, V=16-4(12-8)=0` For `0` to `4 sec,s_(1)=(1)/(2)at^(2)=(1)/(2)(4)(4)^(2)=32m` For 4 to `8 sec,s_(2)=16(8-4)=64m` For `8` to `12 sec,s_(3)=16(4)-(1)/(2)(4)(4)^(2)=32m` So `s_(1)+s_(2)+s_(3)=32+64+32=128m` Alter`:` Draw `v-t` graph Area of `v-t` graph `=` displacement. |
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| 5518. |
Two boats start from mid point of river in case (i) and in mid point of pond of still water in case (ii) If all velocity shown in figure are relative to water then mark the correct option about time taken to reach the nearest bank. A. Boat A takes least timeB. Boat B takes maximum timeC. All the four boats A,B,C and D take same timeD. Time of reaching bank can not be compared. |
| Answer» `:=(d//2)/(v sin theta)` | |
| 5519. |
A lift starts going up from rest, it acceleration is shown in fig, plotted against time. Time when lift comes to instantaneous rest is .. A. 4 secB. 8 secC. 10 secD. 12 sec |
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Answer» Area a-t curve give `DeltaV` `1xx4=(t-8)xx2` `4=2t-16` `t=10` sec. |
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| 5520. |
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.) |
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Answer» (a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential. (b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground. (c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral. (d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere. |
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| 5521. |
The force of attractions between two charges `8muC` and `-4muC` is 0.2 N. Find the distance of separation.A. 1.2 mB. 12 mC. 120 mD. 0.12 m |
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Answer» Correct Answer - A Given, that, F=0.2 N, `Q_(1) = 8muC`, `Q_(2)=4muC` `therefore` Force between two charges, F=`K(Q_(1)Q_(2))/r^(2))` or 0.2 = `9 xx 10^(9) xx (8 xx 10^(-6) xx 4 xx 10^(-6))/0.2` or r=12cm |
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| 5522. |
Two small spheres each of mass 'm' kg and having equal charge are suspended from a point by insulating threads each of length 'I' metre but negligible mass. If '0' is the angle which each string makes with vertical when equilibrium has been reached, find charge on one sphere. |
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Answer» T cos θ = mg T sin θ = Fe tan θ = \(\frac{Fe}{mg}\) \(Fe = \frac{kq^2}{r^2}\) \(Fe = \frac{kq^2}{(2d)^2}\) \(Fe = \frac{kq^2}{4l^2 sin^2\theta}\) \(tan\theta= \frac{kq^2}{4l^2sin^{2} \theta\times mg}\) \(q^2 = (4mgl^2\,sin^2\theta\, tan\theta)4\pi\varepsilon\) |
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| 5523. |
A conducting sphere of radius `10 cm` is charged `10 muC`. Another uncharged sphere of radius `20 cm` is allowed to touch it for some tome. After that if the sphere are separted, then surface density of chsrges, on the spheres will be in the ratio ofA. `1 : 1`B. `4 : 1`C. `1 : 3`D. `1 : 4 |
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Answer» Correct Answer - B (b) When one charged and other uncharged conducting spheres are in contact then charge on each sphere is half the charge on first charged sphere. When both the spheres are placed in contact with each other, the charge on each, `q_(1) = q_(2) = q = (10 + 0)/(2) muC = 5 mu C` Surface charge density is given by, `sigma = (q)/(4 pi r)` Hence, `sigma_(1) = (q_(1))/(4 pi r_(1)^(2))` and `sigma_(2) = (q_(2))/(4 pi r_(2)^(2))` `implies (sigma_(1))/(sigma_(2)) = (q_(1))/(4 pi r_(1)^(2)) (4 pi_(2)^(2))/(q_(2)) = (5 xx (20)^(2))/((10)^(2) xx 5) = (4)/(1)` |
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| 5524. |
A long straight non-conducting string carriers a charge density of `40 muC//m`. It is pulled along its length at a speed of `300 m//sec`. What is the magnetic field at a normal distance of `5 mm` from the moving string?A. `B=4.8 xx 10^(-7)T`B. `B=3.2 xx 10^(-7)T`C. `B=2.5 xx 10^(-7)T`D. `B=5 xx 10^(-7)T` |
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Answer» Correct Answer - A The moving string behaves like a current, `i=(40 xx10^(-6)C//m)(300 m//sec)` `=1.2 xx 10^(-2)` ampare Then `B=(mu_(0)i)/(2 pi r)=(2 xx 10^(-7)xxx1.2 xx 10^(-2))/(0.005)` `=4.8xx10^(-7)` tesla |
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| 5525. |
A dog with mass M has its string attached to one end of a spring which runs without friction along a horizontal overhead rod. The other end of the springs is fixed to a wall The spring constant is K. The string is massless and inextensible and it maintains a constant angle `theta` with the overhead rod, even when the dog moves. There is friction with coefficient mu between the dog and the gound What is the maximum distance (in cm) that the dog moving slowly can stretch the spring beyond its natural length? Use M=30 kg,`theta=400N//m and mu=1/3` |
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Answer» Correct Answer - `0020` For the dog, `N+ Tsintheta-Mg=0` (vertical) `F-Tcostheta=0` (horizontal) For maximum extension `f=muN` For spring `Tcostheta -kx=0` (horizontal) We have four unknows (N,T,F,x) and four equations Solve for T `Tcostheta=kx Rightarrow T=kx//costheta` Substitute for F and solve for N `muN-Tcostheta=0 Rightarrow N=Tcostheta//mu=kx//mu` |
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| 5526. |
A monkey of mass 20 kg rides on a 40 kg trolley moving at a constant speed of 8 m/s along a horizontal track. Frictional force can be neglected. If the monkey jumps vertically, with respect to the moving trolley, the catch the overhanging branch of tree, the speed of the trolley after the monkey has jumped off is :-A. 16 m/sB. 12 m/sC. 8 m/sD. 6 m/s |
| Answer» Correct Answer - A | |
| 5527. |
A man of mass `80 kg` stands on a plank of mass `40 kg`. The plank is lying on a smooth horizontal floor. Initianlly both are at rest. The man starts walking on the plank towards north and stops after moving a distance of `6 m` on the plank. ThenA. The center of mas os plank man system remains stationary.B. The plank will slide to the north by a distance 6mC. The plank will slide to the south by a distance 6mD. The plank will slids to the south by a distance 12m |
| Answer» Correct Answer - A | |
| 5528. |
If two waves are in phase and have same amplitude then resultant wave hasA. half of amplitude of single waveB. same amplitude as single waveC. twice of amplitude of single waveD. thrice of amplitude of single wave |
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Answer» C. twice of amplitude of single wave |
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| 5529. |
When two waves meet, their displacementsA. add upB. cancel outC. destruct each otherD. subtract down |
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Answer» The Correct option is A. add up |
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| 5530. |
For listening radio in cars, external radio aerials are used becauseA. radio waves have shorter wavelengthB. radio waves have longer wavelengthC. radio waves cannot pass through windowD. radio waves require a medium to propagate |
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Answer» B. radio waves have longer wavelength |
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| 5531. |
Principle of superposition can be applied toA. EM wavesB. sound wavesC. radio wavesD. all of above |
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Answer» D. all of above |
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| 5532. |
“The aerial distance between two towers is 4km. But speedometer of car shows 5.6km when travel from one tower to another” 1. By reading this statement explain the concept of distance and displacement. 2. What is the numerical ratio of displacement of object to distance? Explain. 3. A particle is moving along a circular trace of radius ‘R’. What is the distance travelled and displacement of the particle in half revolution? |
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Answer» 1. Distance and the length of the path covered by the object. It is a scalar quantity. Displacement is the length between initial point and final point. 2. \(\frac{dispalcement}{distance}≤1\) For the straight-line path, displacement is equal to the distance travelled. But for the curved path displacement is less than the distance travelled. 3. distance = πR displacement = R + R = 2R. |
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| 5533. |
A car travels half the distance with constant velocity of 40 kmph and the remaining half with a constant velocity 60 kmph. The average velocity of the car in kmph is(a) 40 (b) 45 (c) 48 (d) 50 |
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Answer» Correct answer is (C) 48 Average velocity = Vav = \(\cfrac{s}{\frac{s}{80}+{\frac{s}{120}}}\) = 48 kmph. |
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| 5534. |
Which of the following pairs have same dimensional formula for both the quantities?1. Kinetic energy and torque2. Resistance and Inductance3. Young’s modulus and pressure(a) (1)only(b) (2) only(c) (1) and (3) only(d) All of three |
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Answer» Correct answer is (c) (1) and (3) only |
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| 5535. |
How many seconds are there in a light fermi?(a) 10-15(b) 3.0 × 108(c) 3.33 × 10-24(d) 3.3 × 10-7 |
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Answer» Correct answer is (c) 3.33 × 10-24 One light fermi is time taken by light to travel a distance of 1 fermi ie. 10-15m 1 light fermi = \(\frac{10^{-15}}{3×10^8}\) = 3.33 × 10-24s. |
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| 5536. |
Two objects A and B travel from P to Q through two different paths as shown in figure. If both A and B takes the same time interval to travel from P to Q, then which of the following statements are true?(a) A and B have same speed. (b) A and B have same velocity (c) A and B have same average velocity. (d) The speed of A is greater than that of B(e) The speed of B is greater than that of A |
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Answer» (d) The speed of A is greater than that of B |
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| 5537. |
A closed organ pipe of length 28 cm closed at one end is found to be at resonance when a tuning fork of frequency 850Hz is sounded near the open end. If velocity of sound in air is 340 m/s, then theA. air in the pipe is vibrating in fundamental modeB. air in the pipe is vibrating in first overtoneC. end correction of the pipe is 1cmD. end correction of the pipe is 2cm. |
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Answer» BD `lambda=(v)/(u)=40 cm` `(3lambda)/(4)=28+eimplies"first overtone"` |
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| 5538. |
The area under velocity-time graph for a particle in a given interval of time represents (a) velocity (b) acceleration (c) work done (d) displacement |
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Answer» (d) displacement: Area under velocity-time graph represents displacement of a particle in a given interval of time. |
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| 5539. |
Two straight lines drawn on the same displacement time graph make angles 30° and 60° with time axis respectively in the figure.1. Which line represents greater velocity? 2. What is the ratio of the velocity of line A to the velocity of line B? |
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Answer» 1. B 2. \(\frac{V_A}{V_B}= \frac{tanθ_A}{tanθ_B} \) \(\frac{tan 30°}{tan 60°} = \frac{\frac{1}{\sqrt{3}}}{\sqrt3} = 1:3\) |
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| 5540. |
The x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph. |
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Answer» No, wrong, x-t plot does not show the trajectory of a particle. Context: A body is dropped from a tower (x = 0) at t = 0. |
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| 5541. |
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh-1. What is the magnitude of average velocity, and average speed of the man over the following intervals of time:1. 0 to 30 min,2. 0 to 50 min,3. 0 to 40 min? |
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Answer» Average speed over the interval of time from 0 to 30min = \(\frac{20km}{30min} = \frac{2.5 km}{\frac{1}{2}h}\) = 5 kmh-1 Magnitude of average velocity over the interval of time from 0 to 30 min is 5kmh-1. This is because the “distance travelled” and the ‘magnitude of displacement’ over the interval of time from 0 to 30 min are equal. Distance covered from 30 to 50 minutes = 7.5kmh-1 × \(\frac{20}{60}\) h = 2.5km Total distance covered from 0 to 50 minute = 2.5 km+ 2.5 km = 5km Total time = 50min = \(\frac{50}{60}\) h = \(\frac{5}{6}\) h Average speed over the interval of time from 0 to 50min = \(\frac{5km}{5/6h}\) = 6kmh-1 The displacement over the interval of time from 0 to 50 min is zero. So the magnitude of average velocity is zero. Distance covered from 30 to 40 min = 7.5kmh-1 × \(\frac{1}{6}\) h = 1.25km Total distance covered from 0 to 40 minute = 2.5 km + 1.25 km = 3.75km Average speed over the interval of time from 0 to 40min = \(\cfrac{3.75km}{\frac{40}{60}h}\) = 5.625kmh-1 The “magnitude of displacement” is (2.5 -1.25) km, i.e., 1.25 km. Time interval = \(\frac{2}{3}\)h The ‘magnitude of average velocity’ = \(\frac{1.25 km}{2/3h}\), = 1.875 kmh-1. |
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| 5542. |
Look at the graphs (a) to (d) in the following figure carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of the particle. |
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Answer» None of the four graphs can represent one-dimensional motion of the particle. In fact, all the four graphs are impossible. 1. A particle cannot have two different positions at the same time. 2. A particle cannot have velocity in opposite directions at the same time. 3. Speed is always positive (non-negative). 4. Total path length of a particle can never decrease with time. Note: The arrows on the graphs are meaningless. |
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| 5543. |
Martha is from the United States. She’s ______. A) American B) United Statian C) United Statianese |
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Answer» Correct option is A) American |
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| 5544. |
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7kms-1? The operating frequency of the scanner is 4.2MHz. |
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Answer» λ = ? u = 1.7kms-1 = 1700ms-1 v = 4.2 × 106 Hz, u = vλ = or λ = \(\frac{u}{v}\) or λ =\(\frac{1700}{4.2×10^6}\) m = 4.05 × 10-4 m = 0.405mm. |
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| 5545. |
A steel rod 100cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53kHz. What is the speed ‘ of sound in steel? |
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Answer» l = 1m, v = 2.53 × 103HZ, \(\frac{λ}{2}\) = I or λ = 2m u = v λ = 2.53 × 103 × 2ms3-1 = 5.06 × 103 ms-1 = 5.06kms-1 |
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| 5546. |
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of the wire is 3.5 × 10-2kg and its linear mass density is 4.0 × 10-2kgm-1. What is1. the speed of a transverse wave on the string and2. the tension in the string? |
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Answer» mass of wire M = 3.5 × 10-2kg Linear density µ = mass / length = M/l = 4.0 × 10-2kg ∴ Length of wire l = \(\frac{M}{μ}\)=\(\frac{3.5×10^{−2}} {4 \times 10^{-2}}\)m = 0.875m In the fundamental mode, λ =2l = 2 × 0.875m = 1.75m 1. Speed of transverse waves u = v λ = 45 × 1.75ms-1 = 78.75ms-1 2. u= \(\sqrt{\frac{T}{μ}}\) or T = µu2 = 4 × 10-2(78.75)2N = 278.06N. |
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| 5547. |
The woman ________ for her husband’s life when he was found guilty of murder. A) bid B) disputed C) pleaded D) debated E) sued |
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Answer» Correct option is C) pleaded |
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| 5548. |
The wet forest of the Amazon basin are known as which one of the following ? (a) Campos (b) Lianos (c) Pampas (d) Selves |
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Answer» The Right option is : (d) Selves |
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| 5549. |
I know you’ve got it - so come on, ______! A) hand it on B) hand it out C) hand it over D) hand it in |
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Answer» Correct option is C) hand it over |
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| 5550. |
A `10kg` weight is suspended with a wire `AB` of linear mass density `0.01kg//mj`. Length of `AP` segment is `1m` and it is vibrated with a tuning form in its second overtone. Now instead of `10kg`. A `40kg` weight is suspended. By what minimum distance should we displace the pulley forward to achieves the resonance with same tuning fork `(g=10m//sec^(2))` A. `0.33m`B. `0.67m`C. `1m`D. `1.33m` |
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Answer» Correct Answer - A Initially `f_(0)=3/(2l)sqrt(T/(mu))` `t_(0)=3/2 3/(2xx1)sqrt(100/0.01)=150Hz` In the second case suppose the tunning fork is in resonance with `n^(th)` harmonic `f_(0)=n/(3l)sqrt(T/(mu))` `150=n/(2l)sqrt(400/0.01)` `l=n/1.5` `n=1 l_(1)=1/1.5=0.67` `n=2 l_(2)2/1.5=1.33` `n=3 l_(3)=3/1.5=2m` |
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