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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.
| 12801. |
An air bubble of radius r in water is at a depth h below the water surface at some instant if P is atmospheric pressure and d & T are the density and surface tension of water, what is the pressure inside the bubble ? |
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Answer» Solution : `"Excess of PRESSURE INSIDE the AIR BUBBLE in water" =(2T)/(r)` `:.` Total pressure inside the air bubble atmospheric pressure + pressure due to LIQUID column + Excess pressure due to surface tension `=P=hrhog+(2T)/(r)` |
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| 12802. |
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder |
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Answer» INCREASES |
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| 12803. |
The displacement of object is proportional to cube of the time, then acceleration is ..... |
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Answer» constant, but not zero. `therefore y = KT ^(3),` where k is constant `therefore V = (dy)/(DT) = (d)/(dt) (Kt ^(3))` `therefore v = 3 Kt ^(2)` `therefore a = (dv)/(dt) = (d 3Kt ^(2))/(dt) = 6kt` `therefore a prop t[because6k` constant] `therefore` ACCELERATION INCREASE with time . |
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| 12804. |
Why curved road are provided with slope ? |
| Answer» SOLUTION :Whenvehiclemoves oncurvedroadin circularmotioncentripetal FORCEIS producedhencevehicle willslipofffromthe PATH. Tobalancethisforcecentripetalforceisneeded.Toprovidethisforceroadareprovidedwithslope. | |
| 12805. |
A closed vessel is fitted with an evacuating pump which evaluate by catching 1% of the air at each stroke and expelling it out. Find the minimum number of strokes needed to reduce the air pressure in the vessel to 90% (assume slow process) |
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| 12806. |
A spherical ball of steel of radius 2 mm is falling down through glycerine. Determine the terminal velocity of the ball. Specific gravities of steel and glycerine are 8 and 1.3 respectively . Coefficient of viscosity of glycerine, eta=8.3 poise. |
| Answer» SOLUTION :`0.0703m. S^-1` | |
| 12807. |
A constant retarding force of 20 N acts on a body of mass 5 kg moving initially with a speed of 10 ms^(-1). How long does the body take to stop ? |
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Answer» 1.5 s |
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| 12808. |
An electric pump on the ground floor of a building takes 15 minutes to fill a tank of volume 30 m^(3) with water. If the tank is 40 m above the ground and the efficiency of the pump is 30%. Find the electric power consumed by the pump in filling the tank. (Density of water = 1 xx 10^(3) "kg m"^(-3)). |
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| 12809. |
The internal forces acting on an isolated system |
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Answer» Can CHANGE the velocity the SYSTEM |
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| 12810. |
Statement I:In a progressive logitudinal wave, the amplitude of the wave will not be the same at all points of the medium along the direction of motion of the wave. Statement II: there is a continuous change of the phase angle of the wave as it progressive in the direction of motion. |
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Answer» Statement I is true, statement II is true and statement II is the correct explaination for statement I. |
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| 12811. |
A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R.On immersion ina medium of refractive index 1.75 it will behave asa lens |
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Answer» Solution :When glass LENS is immersed in a MEDIUM, its refractive index `^m mu_s. ^m mu_s= ^amu_G/mu_g = 1.50/1.75=6/7` `therefore` By lens MAKER formula `1/f=(m mu_g-1) (1/R_1-1/R_2) or 1/f=(6/7-1)(-1/R-1/R)` or `1/f=(-1/7) (-2/R) or f=(7R)/2=3.5 R` Hence the given lens in medium behaves like convergent lens of FOCAL length 3.5R. |
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| 12812. |
Acceleration displacement graph of a particle executing S.H.M is as shown in given figure. Find the time period of its oscillation (in sec) |
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Answer» Solution :Acceleration `a= -OMEGA^(2)X, i.e., omega^(2)= tan45^(@)= 1` or `(2pi)/(T) =1` or `T= 2pis` |
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| 12813. |
Two rods OA and OB of equal length and mass are lying on xy plane as shown in figure. Let I_(x),I_(y) & I_(z) be the moment of inertias of both the rods abour x,y and z axis respectively. Then: |
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Answer» `I_(X)=I_(y)gtI_(z)` |
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| 12814. |
A rubber cord of radious 2 mm is loaded with a weight of 13 kg . Length of 50 cm is found to be extended to51 cm . Calculate the young's modulus of rubber . |
| Answer» SOLUTION :5.06*10^8Nm^(-2) | |
| 12815. |
Three uniform circular disc, each of mass 0.5 kg and radius 10 cm, are kept in contact with each other as shown in the figure. Find the moment of inertia of the system when it rotates about the axis AB. |
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Answer» Solution :`I=I_(1)+I_(2)+I_(3)` `I=(5MR^(2))/(4)+(5MR^(2))/(4)+(MR^(2))/(4)rArr I=(11 MR^(2))/(4)` `M=0.5 kg, r =10 CM = 10xx10^(-2)m` `THEREFORE` Moment of INERTIA of the SYSTEM `I=(11MR^(2))/(4)=(11xx0.5xx(0.1)^(2))/(4)` `=1.375xx10^(-2)kg m^(2)`. 
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| 12816. |
A person of mass 60 kg jumps from a height of 5 m, onto the ground. If he does not bend his knees on touching the ground, he comes to rest in (1)/(10) s. But if he bends his knees, he takes 1 s to come to rest. Find the force exerted by the ground on him in the two cases. [ g = 10 m cdot s^(2) ] |
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Answer» Solution :Let V = velocity acquired during the free fall through 5 m. `therefore "" v^(2) = 0 + 2 xx 10 xx 5 "" or, v = 10 " m"cdot s^(-1)` This velocity becomes zero due to the impact with the GROUND. Then, change in momentum = `mv - mv = 60 xx 10 - 0= 600 N`s Force applied by the earth for this change, in 1st CASE, `F_(1) = (600)/((1)/(10)) = 6000 N ` in 2nd case, `F_(2)= (600)/(1) = 600 N` Hence, the impact in the 1st case is more severe than in the 2nd case, where the momentum changed at a shower rate. |
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| 12817. |
Sphere of mass M and radius R is surrounded by a spherical shell of mass 2M and radius 2R as shown. A small particle of mass m is released from rest from a height h(lt ltR) above the shell. There is hole in the shell. In what time will it enter the hole at A? |
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Answer» `2sqrt((HR^(2))/(GM))` `E=-(GMm)/(2r)-(GMm)/(2r)-(GMm)/r` Let orbital velocity is `v_(1)` then from MOMENTUM conservation, `mv-mv=2mxxv_(1)rArr v_(1)=0` As velocity of combined MASS just after collision is zero, the combined mass will fall towards earth. at this instant, the total energy of the system only consist of the gravitational potential GIVEN by `U=(GMxx2m)/(2r)` |
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| 12818. |
Sphere of mass M and radius R is surrounded by a spherical shell of mass 2M and radius 2R as shown. A small particle of mass m is released from rest from a height h(ltltR) above the shell. There is hole in the shell. What time will it take to move from A to B? |
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Answer» `=(R^(2))/(sqrt(GMh))`  `r_(1)=a-ae=a(1-e)` `r_(2)=a+ae=a(1+e)....(i)` `mV_(1)r_(1)sin90^(@)=mV_(2)r_(2) sin 90^(@)` `V_(1)r_(1)=V_(2)r_(2)......(ii)` According to conservation of ENERGY at `P` and `A` `1/2mV_(1)^(2)-(GMm)/(r_(1))=1/2mV_(2)^(2)-(GMm)/(r_(2))....(iii)` From `(i),(ii)` and `(iii) V_(1)=sqrt((GM)/a((1+e)/(1-e)))` |
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| 12819. |
Which is more elastic -water or air? |
| Answer» SOLUTION :Water is more elastic than air. Air can be EASILY compressed while WATERIS incompressible. | |
| 12820. |
Moment of inertia of a thin square plate about an axis passing through its diagonal is I. Its moment of inertia about an axis passing through its centre in the plane of the plate and making an angle with the diagonal is |
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Answer» I |
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| 12821. |
The density rho of water of bulk modulus B at a depth y in the ocean is related to the density at surface rho_(0) by the relation |
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Answer» `RHO=rho_(0)(1-(rho_(0)GY)/B)` |
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| 12822. |
Water falling from a 50 m high fall is to be used for generating electric enegy. If 1.8xx10^5kg of water falls per hour and half the gravitational potential energyh can be converted into electric energy, how many 100 W lamps can be it? |
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Answer» `P=100 watt` P.E=mgh` `=1.8xx10^5xx9.8xx50` `=882xx10^5J` Because HALF the potetnial energy is CONVERTED into electricital energy. Electircal energy `=1/2P.E. =441xx10^5J/hr` So, POWER in watt(J/sec)`=(441xx10^5)` `:.` Number of 100 W lamps, that can be lit ` (441xx10^5)/(3600xx100)=122.5~~122` |
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| 12823. |
A particle of mass m is projected with a velocity v making an angle of 45^(0) with the horizontal. The magnitude of the angular momentum of the projectile when the particle is at its maximum height h, are |
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Answer» Zero |
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| 12824. |
A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point directly above the centre of the face, at a height (3a)/(4) above the base. What is the minimum value of F for which the cube begins to topple about an edge? |
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Answer» Solution :In the LIMITING cse normal REACTION will pass through 0. The cube will topple about O if torque of F ecceeds the torque of mg. `impliesF((3a)/(4))gtmg((a)/(2))` `impliesFgt(2)/(3)mg` So, the MINIMUM value of F is `(2)/(3)` mg 
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| 12825. |
Sphere of mass M and radius R is surrounded by a spherical shell of mass 2M and radius 2R as shown. A small particle of mass m is released from rest from a height h(ltltR) above the shell. There is hole in the shell. With what approximate speed will it collide at B? |
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Answer» `sqrt((2GM)/R)`  `r_(1)=a-ae=a(1-e)` `r_(2)=a+ae=a(1+e)....(i)` `V_(2)=(V_(1)r_(1))/(r_(1))RARR V_(2)sqrt((GM)/a((1-e)/(1+e)))` |
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| 12826. |
To go from town A to ton B a plane must fly about 1780 km at an angle of 30^(@) West of North. How far West of A is B ? |
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Answer» 1542 km |
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| 12827. |
1 electron volt is equivalent to |
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Answer» `10^(-7)J` |
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| 12828. |
Some iron beads are embedded in wax ball which is just floating in water. The volume of ball is 18 cm^(3) and relative density of waxis 0.9 .Then mass of the iron trapped in the ball is |
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Answer» 1.8 g According to law of floatation , Weight = upthrust (m + mass of wax)g `=18 xx 1xx g` IGNORING the volume of iron SINCE it is in the FORM of small BEADS `therefore m+18 xx0.9-18xx1` or `m=1.8 g` |
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| 12829. |
At the moment of launching, the mass of a rocket is 150 kg and the mass of fuel in it is 450 kg. Maximum velocity of ejection of the spent fuel is 2 kmcdot s^(-1) . What should be the rate of consumption of fuel so that the initial acceleration is 30 mcdot s^(-2) ? |
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| 12830. |
A star is moving away from the earth at a velocity of10^(5) m s^(-1). What will be the shift in wavelength of the spectral line of length5700 Åas observed on the earth? |
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| 12831. |
A piece of ice is floating in water, taken in a beaker. What will happen to the level of water when all the ice melts? What will happen if the beaker is filled not with water but with liquid (a) denser than water (b) lighter than water? |
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Answer» Solution :If M gram of ice is floating in a liquid of density `rho_(L)` then for its equilibrium Weight of ice = thrust, i.e., `Mg=V_(D)rho_(L)g` So the volume of liquid DISPLACED by the floating ice, `V_(D)=(M//rho_(L))` Now if M gras ice melts completely, water formed will have mass M grams (as mass is conserved). Now if `rho_(w)` is the density of water, the volume of water formed will be `V_(F)=(M//rho_(w))` Here the liquid is water, i.e., `rho_(L)=rho_(w)` , so water displaced by floating ice is equal to water formed by melting of whole ice and HENCE the level of water will remain unchanged. Furthermore: (a) If `rho_(L)gtrho_(w),(M//rho_(L))lt(M//rho_(w)),i.e.,V_(D)ltV_(F)` i.e., water displaced by floating ice will be LESSER than water formed and so the level of liquid in the beaker will rise. (B) If `rho_(L)ltrho_(w),(M//rho_(L))GT(M//rho_(w)),i.e.,V_(D)gtV_(F)` i.e., water displaced by floating ice will be more than more formed and so the level of liquid in the breaker will fall. |
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| 12832. |
The study of the discrete nature of phenomena at the atomic and subatomic levels. |
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Answer» QUANTUM mechanics |
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| 12833. |
A uniform rod of length l is kept vertically on a horizontal smooth at a point O. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end wil remain |
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Answer» at O |
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| 12834. |
A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ? |
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Answer» Solution :When the tube is HELD horizontally, the mercurythread of length of air `= 15 cm`. A lenght of `9 cm` of the tube will be LEFT at teh open end, fig., The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of teh tube be `1 SQ. cm`, `:. P_(1) 76 cm` and `V_(1)=15cm^(3)` When the tube is held vertically, `15 cm` air gets another`9 cm` of air (FILLED in the right handside in the horizontal position) and let `h cm` of mercury flow out tobalance the atmospheric pressure. Then the HEIGHTS of air column and mercury column are `(24+h)cm` and `(76-h) cm` respectively. the pressure of air = `76-(76-h)=h cm` of mercury. `:. V_(2) =(24+h) cm^(3)` and `P_(2) = h cm` If we assume that temperature remains constant, then `P_(1)V_(1) =P_(2)V_(2) or 76 xx 15 = h xx (24+h)`or `h^(2) + 24h - 1140 = 0 or h= -24 +- sqrt((24)^((2)+4 xx 1140))/(2) = 23.8 cm or -47.8 cm`. 
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| 12835. |
Two liquids A and B have their specific heats as S_(A) and S_(B) respectively. If S_(B) lt S_(A) which of the two would you use as a coolant? |
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| 12836. |
A shell fired such that it hits a target at a distance 'b' and having height 'h' Tangent of angle of projection when velocity of projection is minimum is |
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Answer» `((B+H))/(SQRT(b^(2)+h^(2)))` |
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| 12837. |
Two sound waves produce 12 beats in 4 second. By how much do their frequencies differ ? |
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Answer» Solution :`{:("Number of beats"),("produced per second"):}}=(12)/(4)=3s^(-1)` `:. {:("The DIFFERENCE"),("in FREQUENCY"):}}=3Hz` |
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| 12838. |
A massive vertical wall is approaching a man (at rest) with speed 3.5 m/s. The man throws a ball with speed 10 m/s at an angle of 37^(@) with horizontal which after completely elastic collision rebounds and reaches back directly into hands of man. Find the distance between man and wall at the time of projection of ball in metre. (g = 10m//s^(2)) |
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| 12839. |
The change in frequency due to Doppler effect does not depend on ......... |
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Answer» the SPEED of the source |
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| 12840. |
Name the law which helps us to explain the distribution of energy in the spectrum of a black body for (i) short wavelength only(ii)long wavelength only. |
| Answer» SOLUTION :(i) Wien.s LAW(II) RAYLEIGH JEANS law | |
| 12841. |
What is the surface energy ? Find the relation between surface tension and surface energy. Explain why. (i) surface tension ofa liquid is independent of the area of the surface. (ii) detergents should have small angle of contact. |
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Answer» Solution :1st part: N/A 2nd part: (i) Surface tension of liquid is the force acting PER unit length on a line drawn TANGENTIALLY to the liquid surface at rest. Since this force is independent of area of liquid surface. Hence surface tension of a liquid is ALSO independent of the area of the liquid surface. (ii) A CLOTH has narrow spaces in the form capillaries. The rise of liquid in a capillary tube is directly proportional to `cos theta`. If `theta`is small , `cos theta` will be LARGE. Due to this detergent. s penetration in the cloth will be better. Hence, detergents should gave small angle of contact. |
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| 12842. |
A sound absorber attenuates the sound level by 20dB the intensity decreases by a factor of |
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Answer» 100 |
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| 12843. |
During adiabatic expansion a gas cools - explain the reason . |
| Answer» Solution :During adiabatic expamsion no heat is suplied to the gas HENCE work is DONE at the COST of internal ENERGY THUS the temperature of the gas increased . | |
| 12844. |
A ball in thrown up. The magnitude of its momentum first decreases and then increases. Does this violate the conservation of momentum principle. |
| Answer» Solution :The MOMENTUM of a BODY is conserved only when there is no external force ACTING on the body. In the case under DISCUSSION. Force of gravity is acting on the body as an external force. So the change in momentum of the body does not signify any violation of the law of conservation of momentum. | |
| 12845. |
A steel ball initially at a pressure of 10^(5) Pa is heated from 30^(@)C to 130^(@)C keeping its volume constant, Find the final pressure inside the ball. Given that coefficient of linearexpansion of steel is 1.1xx10^(-5)(""^(@)C^(-1)) and Bulk modulus of steel is 1.6xx10^(11)N//m^(2). |
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| 12846. |
Particle A experiences a perfectly elastic collision with a stationary particle B. The particle fly apart symmetrically relative to the initial direction of motion of particle A with angle of divergence theta(0 lt theta lt 90). Ratio between the masses of the two bodies is (1)/(1+2 cos n theta) where 'n' is . |
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| 12847. |
An object moving with uniform acceleration covers 40 m in initial 5 seconds and 65 m in next 5 seconds, then its initial velocity is ....... |
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Answer» `4m//s` ` S _(1) = v _(0) t + 1/2 at ^(2)""…(1)` velocity after 5 seconds, `v=v _(0) + at` `THEREFORE` Displacement in next 5 seconds `S_(2) = vt + 1/2 at ^(2)` `S _(2) = (v _(0) + at) t + 1/2 at ^(2)` `therefore S_(2) = v _(0) t + at ^(2) + 1/2 at ^(2) ""...(2)` `therefore` By substracting equation (1) from equation (2), `S_(2) -S_(1) =at ^(2)` `therefore a = (S_(2) - S_(1))/(t ^(2)) = (65- 40)/((5) ^(2)) = (25)/(25) = 1m s^(-2)` From equation (1), In `S_(1) = v_(0)t + 1/2 at ^(2), S_(1) = 40 m, t = 5S and a =1 m//s ^(2),` `40 = 5v _(0) + 1/2 xx 1 xx 25` `therefore 40-12.5 =5v_(0)` ` therefore 27.5 =5v_(0)` `therefore v _(0) =5.5 m //s` |
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| 12848. |
"It is easier to make a body roll over a surface than to slide". Why ? |
| Answer» SOLUTION :Because the rolling friction between two SURFACES is less than the SLIDING friction between the same two surfaces. | |
| 12849. |
1 MKS unit of power = ....... CGS unit of power. |
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Answer» `10^(2)` `=10^(7)` erg/second |
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| 12850. |
Rate of transfer of heat is maximum in the case of |
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Answer» Conduction |
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