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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.
| 38251. |
Two capillary tubes of radius r but of lengths l_(1) and l_(2) are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before |
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Answer» `l_1 + l_2` |
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| 38252. |
What is shear modulus for liquid ? |
| Answer» SOLUTION :ZERO. Because, for LIQUID shear strain `((Delta X)/(THETA)) `is zero. | |
| 38253. |
K is the force constant of a spring. The work done is increasing its extension from I_(1) to I_(1) will be |
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Answer» `K(I_(2)-I_(1))` |
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| 38254. |
The magnitude of gravitational potential energy of a body at a distance .r. from the center of the earth is V. Its weight at a distance .2r. from the center of the earth is |
| Answer» Answer :B | |
| 38255. |
The surface tension of a liquid is 0.5 N/m. If a liquid film is formed on a ring of area 0.02 m^2 then its surface energy will be - |
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Answer» 0.01 Joule |
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| 38256. |
When 8 droplets of water of radius 0.5 mm combine to form a single droplet. The radius of it is .......... |
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Answer» Solution :Volume of 8 droplets of WATER = `8xx 4/3 pi (0.5)^(3)` When each DROPLET combine to form one volume remains conserved `R^(3) = 8 xx (0.5)^(3)` `R^(3) = (8 xx (0.5)^(3))` `R = 2 xx 0.5` = 1 mm |
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| 38257. |
The acceleration displacement graph of a particle moving in a straight line is shown in the figure. Initial velocity of the particle is zero. Find the velocity of the particle when dispalcement of the particle is S=12m |
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Answer» `3sqrt(2)m//s` |
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| 38258. |
Two earth satellites of some mass are revolving round the earth in circular orbits of radii 6400 kg and 19,200 km from the centre of the earth. Ratio of their orbital velocities will be |
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Answer» `1 : SQRT(3)` |
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| 38259. |
For a givenidealgas 6xx 10^(5) J heatenergyissuppliedand the volumeof gas isincreased from 4m^(3) to 6 m^(3) atatmosphericpressure .Calculate(a) theworkdoneby thegas( b) Changein internalenergyof thegas ( c)graphthisprocessinPV andTVdiagram |
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Answer» Solution :Mayer.s relation: Consider u mole of an ideal gas in a CONTAINER with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dt. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. If `C_v` is the molar specific heat capacity at constant volume, from equation. ` C_v= 1/mu( d U )/(dT )` `dU= mu C _vdT` Suppose the gas is heated at constant pressure so that the temperature increases by dT. If .Q. is the heat supplied in this PROCESS and .dV. the change in volume of the gas. `Q=mu C _(p ) dT `...(3) If W is the work done by the gas in this process, then `W= PDV` But from the first law of thermodynamics, `Q=dU + W ` Substituting equations (2), (3) and (4) in (5), we get, `mu C _(p) dT= muC_(v)dT+ p dV ` For mole of ideal gas, the equation of state is given by `PV = mu RTimplies PD V +dp = mu R d T` Since the pressure is constant, dP = 0 `thereforeC_P dT= C_(v)dT+ RdT` ` thereforeC_(P)= C_(V)+R (or )C_(P ) - C_(V )= R ` This relation is called Mayer.s relation It IMPLIES that the molarspecificheatcapacity of an ideal gas at constant pressure is greater than molar specificheat capacityatconstantvolume therealationshowsthat specificheatatconstantpressure`(s_p)`is alwaysthancalculatethe workdoneby thegas (ii) A gas expands from volume `1m^3` to `2m^3` at constant atmospheric pressure. (ii )The pressure P = 1 atm = 101 kPa, ` V_f= 2 m^3 and V_i= 1 m^3` From equation ` W =int_(v_i)^(v_f) pdV = Pint _(vi)^(vi) dv ` Since P is constant. It is taken o’t of the integral. `W=P (V_f -V_f ) = 10 1 xx 10^3xx ( 2-1)= 101 KJ` |
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| 38260. |
State Hooke.s Law? |
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Answer» Solution :A. Hooke.s law STATES that within elastic limits stress is PROPORTIONAL to strain. STRESSE `PROP` strain or `("Stress")/("Strain")`= Constant. |
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| 38261. |
A railway track is banked for a speed v, by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. The radius of curvature of the track is r |
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Answer» `(h)/(d)=(V^(2))/(rg)` |
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| 38262. |
Two particles of masses m_(1) and m_(2) and velocities u_(1) and alpha u_(1)(alpha gt 0), make a elastic head on collision. If the initial kinetic energies of the two particles are equal and m_(1) comes to rest after collision, the |
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Answer» `(u_(1))/(u_(2)) = sqrt(2) + 1` |
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| 38263. |
suggest a situation in which an obiect is accelerated and have constant speed. |
| Answer» SOLUTION :UNIFORM CIRCULAR MOTION. | |
| 38264. |
What is meant by angular harmonic oscillations? Compute the time period of angular harmonic oscillation. Time period and frequency of angular SHM: |
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Answer» Solution :(i) When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. (ii) The point at which the resultant torque acting on the body is taken to be zero is called mean position . If the body is displaced from the mean position , then the resultant torque ACTS such that it is proportional to the angular displacement and this torque has a tendency to bring the body towards the mean position.  (iii) LET `vec(theta)` be the angular displacement of the body and the resultant torque `vec(tau)` acting on the body is `vec(tau)propvec(theta)""...(i)` `vec(tau)=-kvec(theta)""...(2)` k is the RESTORING torsion constant , which is torque per unit angular displacement . If I is the moment of inertia of the body and `vec(alpha)` is the angular acceleration then `vec(tau)=Ivec(alpha)=-kvec(theta)` But, `vec(alpha)=(d^(2)vec(theta))/(dt^(2))` `vec(alpha)=(d^(2)vec(theta))/(dt^(2))=-(K)/(I)vec(theta)""...(3)` `vec(alpha)=omega^(2) theta""...(4)` (iv) This differential equation resembles simple harmonic differential equation . So, comparing equation (3) & (4) we get `omega=sqrt((K)/(I) )"rad"s^(-1)""...(4)` (v) This frequency of the angular harmonic motion is `f=(1)/(2pi)sqrt((K)/(I))HZ""....(5)` (vi) The time period is `T=2pisqrt((I)/(K))" second"....(6)` |
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| 38265. |
The temperatures of the source and the sink of a Carnot engine are 500 K and 300 K respectively. If the temperature of the source diminished to 450 K, what will be the percentage change in efficiency? |
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Answer» Solution :In the first CASE, EFFICIENCY `eta_(1) = 1 - (T_2)/(T_1) = 1 - (300)/(500) = 2/5` In the second case, efficiency `eta_(2) = 1-(T_2)/(T_1) = 1 - (300)/(450) = 1/3` `:.` Fractional change in efficiency `=(eta_(2) - eta_(1))/(eta_1) = (eta_2)/(eta_1)-1 = (1//3)/(2//5)-1 = 5/6-1 = -1/6` `:.` PERCENTAGE change = `-1/6 xx 100 = -16.67%` So, efficiency decreases by 16.67%. |
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| 38266. |
What do you mean by motion in one, two and three dimensions ? |
| Answer» Solution :In a MOTION having three dimensions, all the three COORDINATES specify the position of an object CHANGING with RESPECT to TIME. If a particle moves in three cimensions, then ithas three dimensional motion . Eg: A bird flying in the sky . | |
| 38267. |
As in the diagram three blocks connected together lie on ahorizontal frictionless table and pulled to the right with a force F=50 N if m_(1)=5kg m_(2)=10 kg and m_(3)=15 kg find the tensionsT_(1) and T_(2) |
Answer» Solution : ACCELERATION acquired by all blocks a `= (T_(3))/(m_(1)+m_(2)+m_(3)) = (36)/(36)` 1 `MS^(-2)` `:.` TENSION`T_(2) = (m_(1) + m_(2))` a `= ( 1+8) xx 1 = 9` N |
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| 38268. |
A body falls for 4s from rest. If the acceleration due to gravity of earth ceases to act, then the velocity and distance it travels in the next 3sec is |
| Answer» Answer :A | |
| 38269. |
When n number of particles each of mass m are at distances x_(1)=a, x_(2)=ar, x_(3)=ar^(2)…….x_(n)=ar^(n) units from origin on the x-axis, then the distance of their centre of mass from origin. |
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Answer» Solution :`X_(CM)=(ma+m(ar)+m(ar^(2))+………..+m(ar^(n)))/(m+m+m+……….+m("n terms"))` `X_(cm)=(m(a+ar+ar^(2)+………..+ar^(n)))/(mn)` If `R GT 1` then `X_(cm)=(1)/(n)[(a(r^(n)-1))/(r-1)]=(a(r^(n)-1))/(n(r-1))` If `r lt 1` then `X_(cm)=(1)/(n)[(a(1-r^(n)))/(1-r)]=(a(1-r^(n)))/(n(1-r))` |
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| 38270. |
Two particles are executing SHM in a straightline. Amplitude A and time period T of both the particles are equal. At time t=0, one particle is at displacement x_(1)=+A and the other at x_(2)=(-A)/23 and they are approaching towards each other. After what time they cross each other? |
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Answer» Solution :Equation can be written as : `x_(1)=Acos omega t` and `x_(2)=ASIN(omegat-pi//6)"" "Here" omega=(2PI)/T` Equating `x_(1)=x_(2)""Sinx(omegat-(pi)/6)=COS omegat` `SINOMEGAT"cos"(pi)/6-"sin"(pi)/6cosomegast=cos omegat` `tan omega t=sqrt(3), omegat=(2pi)/T t=(pi)/3`, we get `t=T//6` |
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| 38271. |
A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is 12xx10^(-3) N//m The restoring force constant is 50xx10 N//m. If it is in oscillatory motion, its time period is |
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Answer» `PI` SEC |
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| 38272. |
A point source of light is located 20cm in front of a convex mirror with f=15cm. Determine the positions and character of the image point. |
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Answer» Solution :`1/u+1/v=1/f` Here `u=-20 cm, f=15 cm` `1/v=1/f-1/u= 1/(15 cm)-1/(20 cm)=35/(300 cm)=7/(60 cm)` v=8.6 cm As v is positive, the IMAGE is located behind the MIRROR. |
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| 38273. |
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2m//s^2and the car has acceleration 4m//s^2The car will catch up with the bus after a time of: |
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Answer» `sqrt120 s ` |
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| 38274. |
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. Its range for velocity of projection u is |
| Answer» Answer :B | |
| 38275. |
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train's motion, but the distant objects (hills tops, the Moons, the stars etc) seem to be stationary . (In fact , since you are aware that you are moving , these distant objects seem to move with you. ) |
| Answer» Solution :Near OBJECTS make greater ANGLE than distant (far off) objects at the eye of the observer . When you are moving, the angular change is less for distant objects than nearer objects. So, these distant objects seem to MOVE along with you, but the nearer objects in OPPOSITE direction. | |
| 38276. |
Which of the following expression does not represent SHM |
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Answer» `ACOSWT` |
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| 38277. |
If |vecP+vecQ|=|vecP|+|vecQ|. The angle between the vectors vecPandvecQ is |
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Answer» <P>`0^(@)` `P^(2)+Q^(2)+2PQcostheta=P^(2)+Q^(2)+2PQ` `costheta=1""0-0^(@)` |
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| 38278. |
A sphere of mass M and radius Ris falling in a viscous fluid .The terminal velocity attained by the falling object will be proportional to …… |
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Answer» `R^(2)` |
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| 38279. |
Figure Show plot of PV//T versus P for 1.00xx10^(-3) kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T_(1) gt T_(2) or T_(1) lt T_(2)? (c) What is the value of PV//T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00xx10^(-3) kg of hydrogen, would we get the same value of PV//T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV//T (for low pressure high temperature region of the plot) ? (Molecular mass of H_(2) = 2.02 u, of O_(2) = 32.0 u, R= 8.31Jmol^(-1)K^(-1).) |
| Answer» SOLUTION :(a) The dotted plot corresponds to .ideal gas BEHAVIOUR, (b) `T_(1) gt T_(2)`: (c) `0.26 J K^(-1)`, (d) No, `6.3 XX 10^(-5)` kg of `H_(2)` would yield the same value | |
| 38280. |
Express escape velocity in terms of radius and density of a planet. |
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Answer» SOLUTION :Escape velocity `V_(e) = (2GM)/(R)` `= SQRT(2G xx (4)/(3)(PI R^(3) RHO)/(R)) = sqrt((8)/(3)GR^(2)rho)` |
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| 38281. |
Inquestionnumber 3, the thermalconductivity(in W m^(-1) K^(-1)) if air is |
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Answer» `1/10` `DeltaT_("plate") = ((dQ)/DT)xx (l_("plate"))/(K_("plate") xx A) = (65 xx 4 xx 10^(-3))/(1 xx 1.3) = 0.2 K` `:. DeltaT_("air") = 13 - 2DeltaT_("plate") = 12.6 K` `K_("air") = (((dQ)/(dt))xxl_("air"))/(ADeltaT_("air")) = (65 xx 18 xx 10^(-3))/(1.3 xx 12.6) = 1/14 W m^(-1) K^(-1)` |
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| 38282. |
The gravitational field at some point in space is vec(g) = 3i + 4j N//kg. The force exerted on a 2kg mass placed at that point is |
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Answer» `5N, 53^(@)` with x-axis |
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| 38283. |
What do you call the tendency of a liquid to contract its surface area? |
| Answer» SOLUTION :SURFACE TENSION | |
| 38284. |
The masses of two fixed sheres are M and 2M and the radius of each sphere is R. Their centres are 10R apart. The minimum speed with which a particle of mass M/10 be projected from themid - point of the joining the centres of the two spheres so that it escapes to infinity is .............. |
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Answer» `SQRT((6GM)/(7R))` |
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| 38285. |
What is the acceleration of the blockandtrolley system shown in a Fig.(a), if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m s^(-2) ).Neglect the mass of the string. |
Answer» SOLUTION : Freebodydiagramof theblock ` 30-T = 3a` Freebodydiagramof thetrolley ` T- f_(K)= 20 a` where` f_k= mu _k N = 0.04xx 20 xx 10=- 8 N` solving(1) & (2), a ` = 0.96m//s^2 and T = 27.2N` |
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| 38286. |
A mass M is suspended from a spring of negligiblemass. The spring is pulled a little and then released so that the mass executes SHM of time period T,if the mass is increased by m time period becomes (5T)/3, then the ratio of (m/M) is |
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Answer» `3/2` |
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| 38287. |
Assuming that the frequency gamma of a vibrating string may depend upon (i) applied force (F) (ii) length (l) (iii) mass per unit lengt (m), prove that gamma prop1/l sqrt(F/m) using dimensional analysis. |
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Answer» substitute the dimensional formulae of the above quantities `[M^(0)L^(0)T^(-1)] = [MLT^(-2)]^(x) [L] [ML^(-1)]^(z)` `[M^(0)L^(0)T^(-1)] = [M^(x + z)L^(x+y-z)T^(-2x)]` Comparing the POWERS of M,L,T on both sides, `x+z = 0, x+y-z = 0, -2x = -1` Solving for x,y,z, we get `x = 1/2``y = -1``z = -1/2` Substitute x,y,z VALUES in equ (1) `gamma prop F^(1//2) l^(-1) m^(-1//2) :. gamma prop 1/l sqrt(F/m)` |
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| 38288. |
A pendulum clock loses 12s a day if the temperature is 40°C and gains 4s a day if the temperature is 20°C.The temperature at which the clock will show correct time, and the co-efficient of linear expansion ( alpha) of the metal of the pendulum shaft are respectively: |
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Answer» `55^(@) C, alpha = 1.85 xx 10^(-2) //""^(@) C` |
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| 38290. |
A) For any collision, momentum transfer takes place. B) In any collision, energy transfer takes place. C) Implusive forces are involed in collisions. |
| Answer» Answer :A | |
| 38291. |
A carnot's engine whose source is at 300°C takes in 5000 J of heat in each cycle and gives at 3000 J of heat to the sink. Find the temperature of the sink. |
| Answer» SOLUTION :`T_(1) //T_(2)= Q _(1) //Q_(2) , T _(2) = (Q_(2) //Q _(1)) T_(1) = (3000//5000) 573 = 343.8 K` | |
| 38292. |
If two vectors vecAandvecB form adjacent sides of parallelogram, then the |vecAxxvecB| will give- of parallelogram |
| Answer» Answer :A | |
| 38293. |
Using dimensional analysis check the correctness of the equation T=2pisqrt(l/g), where T is the period of the simple pendulum of length l and g is the acceleration due to gravity. |
| Answer» SOLUTION :`TTO[L].sqrt(l//g)=sqrt([L//LT^(-2)])=sqrt([T^(2)])=T`. Hence correct. | |
| 38294. |
A metal washer has a hole of diameter d_(1) and an external diameter d_(2) such that d_(2) = 3d_(1). If on heating d_(2) increases by 0.3% then d_(1) |
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Answer» DECREASES by 0.1 % |
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| 38295. |
The displacement of a particle executing SHM is given by x = 0.10 sin((2pit)/(T) + phi)where all the quantities are in SI units. The time period of vibration is 3s. At t = 0, the particle is displaced by 0.05 m. Find i) the initial phase,ii) the phase angle corresponding to 0.0866m andiii) the phase difference between any two positions of the particle 2s apart. |
| Answer» SOLUTION :`pi//6` radian or `30^@` , (ii) `pi//3 ` radian or `60^@` , (iii) `4pi//3` radian | |
| 38296. |
A closed bottle containing water at room temperature is taken to the moon and then the lid is opened. The water will ……………… . |
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Answer» freeze |
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| 38297. |
When two organ pipes with fundamental frequencies n_(1)and n_(2) are connected in series, what will be the resultant fundamental frequency? |
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Answer» `(n _(1) + n_(2))` For open pipe `IMPLIES L =(v)/(2f)` (i)For first pipe. `f _(1) = (v )/( 2L _(1)) = L _(1) = (v)/( 2f _(1))` (ii) For second pipe, `f _(2) = (v)/( 2L _(2)) implies L _(2) = (v )/(2f _(2))` Here,` L = L _(1) + L _(2)` `therefore (v )/( 2f)= (v)/( 2f _(1)) + (v)/( 2f _(2))` `therefore (1)/(f) = (1)/(f _(1)) + (1)/(f _(2))` As per the NOTATIONS used in the equation. `n = (n _(1) n _(2))/( n _(1) +n_(2))` |
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| 38298. |
A uniform disk of radius r= 0.6 m and mass M = 2.5 kg is freely suspended from a horizontal pivot located a radial distance d = 0.30 m from its centre . Find the angular frequency of small amplitude oscillations of the disk . |
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Answer» Solution :The MOMENT of inertia of the DISK about a PERPENDICULAR axis PASSING through its centre is `I=1/2mr^(2)` Form the PARALLEL axis theorem , the moment of inertia of the disk about the pivot point is `I'=I+Md^(2)=(2.5xx0.6xx0.6)/(2)+2.5xx0.30xx0.30` `=(0.9)/(2)+0.225` `=0.45+0.225` `=0.675kgm^(2)` The angular frequency of small amplitude oscillations of a compound pendulum is given by `omega =sqrt((Mgd)/(I'))=sqrt((2.5xx9.8xx0.30)/(0.675))` `=sqrt((7.35)/(0.675))=sqrt((0.675)/(10.9))=3.3"rad"/s` |
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| 38299. |
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0^(@)C. What is the change in the diameter of the hole when the sheet is heated to 227^(@)C ? Coefficient of linear expansion of copper =1.70xx10^(-5)K^(-1). |
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Answer» Solution :But area of hole is `A_(1)` at TEMPERATURE `T_(1)=27+273=300K` `:.A_(1)=(pid_(1)^(2))/(4)=(pixx(4.24)^(2))/(4)` `:.A_(1)=4.4944pi" cm"^(2)` Area of hole becomes `A_(2)` at temperature `T_(2)=227+273=500K`. `:.A_(2)=A_(1)(1+beta Delta T)` `(pid_(2)^(2))/(4)=4.4944pi[1+2alpha(T_(2)-T_(1))]" where "beta=2alpha` `:.d_(2)^(2)=4xx4.4944[1+2xx1.7xx10^(-5)XX(500-300)]` `:.d_(2)^(2)=4xx4.944[1+2xx1.7xx10^(-5)xx200]` `:.d_(2)^(2)=4xx4.944[1+0.0068]` `:.d_(2)^(2)=4xx4.944xx1.0068` `d_(2)^(2)=18.0998~~18` `:.d_(2)=sqrt(18.1)=4.2544" cm"` `:.` Change in diameter of hole `Deltad=d_(2)-d_(1)` `=4.2544-4.2400` `=0.0144" cm"` `=1.44xx10^(-2)" cm"` |
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| 38300. |
When a fast running bus is suddenly stopped, the passengers tend to move forward due to |
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Answer» force |
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