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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. |
In the tilted position of a floating body, the point at which the vertical line drawn through the centre of buoyancy cuts the central line, name the point of the body. |
| Answer» SOLUTION :METACENTRE | |
| 2. |
The phenomenon of Brownian movement may be taken as evidence of : |
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Answer» EMT of radiation |
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| 3. |
Estimate the volume of a water molecule using the data of one molecule( density of water = 1000kg //m^3)What is its approximate size? |
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Answer» Solution :In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water MOLECULE MAY therefore, be REGARDED as roughly equal to the density of bulk `"water" = 1000 kgm^(-3)`. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 MOLE of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. Since 1 mole contains about `6 xx 10^(23)` molecules (Avogadro.s number), the mass of a molecule of water is `(0.018)//(6 xx 10^(23)) "kg" = 3 xx 10^(-26)` kg. Therefore, a rough estimate of the volume of a water molecule is as follows: `"Volume of a water molecule" = (3 xx 10^(-26) "kg")// (1000 "kg" m^(-3) )` `= 3 xx 10^(-29) m^(3)` ` = (4//3) pi (Radius)^(3)` Hence, `"Radius" ~~ 2 xx 10^(-10) m = 2 Å` |
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| 4. |
A particle of mass 0.1 kg is subjected toa force which varies with distance as shown in figure. It starts its journey from rest at x=0, then its velocity at x=12 m is |
| Answer» Answer :D | |
| 5. |
Two identical spheres are placed at two diametrically opposite points of a smooth circular groove in a horizontal plane. One sphere starts with a speed .u. and collide the other after time .t.. If the coefficient of restitution of colliding bodies is .e.. What is the minimum time after which they again collide ? |
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| 6. |
A hollow sphere of radius R/4 is released from the top of a fixed sphere of radius R as shown. The angle theta made by the radius vector at the instant when smaller sphere leaves contact with the bigger sphere is (assume pure rolling of hollow sphere without slipping) cos^(-1)((10)/(10+x)) where 'x' is |
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Answer» |
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| 7. |
Calculate the safe speed with which a train can negotiate a curve of 50 m radius, where the superelevation of the outer rail above the inner rail is 0.6 m. Given that the distance between the rails is 1 m |
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| 8. |
Two rods A and B of same length and radius are joined together. The termal conductivity of A and B are 2K and K. Under steady state conditions, if temperature difference between the open ends of A and B is 36^(0)C, the temperature difference across .A. is |
| Answer» Answer :C | |
| 9. |
Assertion : In simple harmonic motion, the motion is to and fro and periodic. Reason : Velocity of particle in SHM is v= omega sqrt(a^(2)-x^(2)) Select the correct statement from the following. |
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Answer» Both assertion and REASON are true and reason is the correct explanation of assertion |
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| 10. |
A vertical U-tube of uniform cross-section contains mercury in both its arms. A glycerine (density = 1.3g*cm^(-3)) column of height 10 cm is introduced into one of the arms. Oil of density 0.8g*cm^(-3) is poured into the other arm until the upper surfaces of oil and glycerine are at the same level. Find the height of the oil column. The density of mercury is 13.6g*cm^(-3). |
Answer» Solution :Let the height of the oil column be H  ACCORDING to the figure, AC = ED or, h + BC = 10 or, BC = (10 - h)cm Now, PRESSURE at the point C = pressure at the point D or, `hxx0.8xxg+(10-h)xx13.6xxg=10xx1.3xxg` or, `hxx0.8+136-13.6h=13` or, 12.8h = 123 or, `h=123/12.8=9.61`cm. |
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| 11. |
Check the correctness of the formula S=ut+1/3at^(2) where S is the distance , u is velocity , a is acceleration and t is time . |
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Answer» Solution :DIMENSIONALLY, `LHS=[L]` `RHS=[LT^(-1)][T]+[LT^(-2)][T]^(2)=[L]+[L]` Since, the dimension of each term on both sides are same, the equation is correct dimensionally, However, we cannot say anything about the PHYSICAL correctness of the formula . in fact, we know that the correct formula is `S=ut+1/2at^(2)` Hence, we conslude that dimensional correctness is no guarantee for physical correctness of the formula. however, dimensional incorrectness guarantees the physical incorrectness of the FOMULA. Inspite to the above limitation, the method is STILL helpful to a great extent. |
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| 12. |
A pendulumbob is givenan intialvelocity v at thisbottommostpointand it isfoundthatboblosescirculartrackat a certainpoint and hitsthe pointof suspension. If l is length of the threadthen findv . |
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Answer» Solution :Let VBE the velocityof bobat thepointwhereitloses circulartrackand assumethatstringmakes an angle`theta `with thevertical atthis pointas shown infigure. Let O be thepointwherebobleavesthepointO ison thecircularpathhencewecan applycirculardynamicsfor this point . `T + mg cos theta = ("mu"^(2))/l ` Bob loses circulartrackat this pointhencewe cansustituteT = 0 in aboveequation. ` u = sqrt(gl cos theta ) "" (i)` Let O THEORIGIN and axes are as shownin FIGURE. Bobmovesunder gravity like a projectileafter point O till ithits thepointof suspension `S_(x) = u_(x)t +1/2 a_(x)t^(2) rArr l sin theta= ( u cos theta ) t + 1/2(0)t^(2)` `rArr "" t = (l san theta )/(u cos theta )"" ...(ii)` ` S_(y) - u_(y)t + 1/2 a_(y)T^(2) = l cos theta = (-u sin theta ) t + 1/2 "gt"^(2)` Substituting valueof t from equation (ii) we get ` l COSTHETA = (-u sin theta) ((lsin theta )/(u cos theta )) + 1/2 g ((l sin theta )/(u cos theta ))^(2)` `rArr"" costheta= (- sin theta) ((sin theta )/(cos theta )) + 1/2 gl ((sin theta)/(u costheta))^(2)` `rArr " " cos theta+(sin^(2)theta)/(costheta) =1/2gl((sin^(2)theta)/(u^(2)cos^(2)theta))` ` rArr "" 1/(cos theta)=1/2 gl((sin^(2)theta)/(u^(2)cos^(2)theta))` `rArr "" 1/2 gl((sin^(2)theta)/(u^(2)cos theta))=1 ` `rArr 1/2((glsin^(2)theta)/(gl cos theta xx cos theta))=1 ` `rArr"" tan^(2) theta = 2 rArr tan theta= sqrt(2)"" ....(i)` We can now apply conservationof mechnical ENERGY betweenthe bottommost pointand thepoint where it losesthe circular track . Lossof kinetic energy = gainin gravitational potentialenergy . 
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| 13. |
A uniform pressure P is exerted on all sides of a solid cube at t^(0)C. By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? (K is bulk modulus and alpha is coefficient of linear expansion) |
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Answer» Solution :`K=(P)/((-DELTAV//V)) i.e., (DeltaV)/(V)=(P)/(K)` NUMERICALLY but `(DeltaV)/(V) =gamma Deltat(or) Deltat =(DeltaV)/(V) XX (1)/(gamma) i.e., Deltat=(P)/(K gamma)=(P)/(3K ALPHA)` |
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| 14. |
A jar has mixture of hydrogen and oxygen gases in the ratio 1:5. The ratio of mean kinetic energies of hydrogen and Oxygen molecules is ................ . |
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Answer» `1:5` |
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| 15. |
Derive by the method of dimensions, an expression for the volume of a liquid flowing out per second through a narrow pipe. Asssume that the rate of flow of liwquid depends on (i) the coeffeicient of viscosity eta of the liquid (ii) the radius 'r' of the pipe and (iii) the pressure gradient (P)/(l) along the pipte. Take K=(pi)/(8). |
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Answer» Solution :CONSIDER a liquid flowing steadily through a horizontal capillary tube. Let `v=((V)/(t))` be the volume of the liquid flowing out per second through a capillary tube. It depends on (i) coefficient of viscosity `(eta)` of the liquid, (ii) radius of the tube (r), and (iii) the PRESSURE gradient `((P)/(L))` Then, `vpropeta^(a)r^(b)((P)/(l))^(c)` `v=keta^(a)r^(b)((P)/(l))^(c)""...(1)` Where, k is a dimensionless constant. Therefore, `[v]=("volume")/("time")=[L^(3)T^(-1)]` `[(dP)/(dx)]=("pressure")/("distance")=[ML^(-2)T^(-2)]`, `[eta]=[ML^(-1)T^(-1)]and[r]=[L]` substituting in equation (1) `[L^(3)T^(-1)]=[ML^(-1)T^(-1)]^(a)[L]^(b)[ML^(-2)T^(-2)]^(c)` `M^(0)L^(3)T^(-1)=M^(a+b)L^(-a+b-2c)T^(-a-2c)` So, equating the powers of M, L, and T on both sides, we get `a+c=0,-a+b-2c=3,and-a-2c=-1` On solving three equations, we get `a=-1,b=4andc=1` Therefore, equation (1) becomes, `v=keta^(-1)r^(4)((P)/(l))^(1)` Experimentally, the value of k is SHOWN to be `(pi)/(8)`, we have `v=(pir^(4)P)/(8etal)` The above equation is known as Poiseuille's equation for the flow of liquid through a narrow tube or a capillary tube. |
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| 16. |
During a cyclic process, a heat engine absorbs 500 J of heat from a hot reservoir, does work and ejects an amount of heat 300 J into the surroundings (cold reservoir). Calculate the efficiency of the heat engine? |
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Answer» Solution :The efficiency of heat engine is given by `eta=1-(Q_(L))/(Q_(H))` `eta=1-(300)/(500)=1-(3)/(5)` `eta=1-0.6=0.4` The heat engine has `40%` efficiency, implying that this heat engine converts only `40%` of the INPUT heat into WORK. |
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| 17. |
A satellite is launched in a circular orbit of radius R around earth while a second satellite is launched into a orbit of radius 1.02 R . The Percentage difference in the time period of the two satellite is |
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Answer» 0.7 |
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| 18. |
After falling from rest through a height h a body of mass m begins to raise a body of mass M(Mgtm) connected to it through a pulley. (a) Determine the time it will take for body of mass M to return to its original position. (b) Find the fraction of kinetic energy lost when the body of mass Mis jerked into motion. |
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Answer» Solution :(a) The SPEED of the body `B` just before the string becomes taut is `v=sqrt(2gh)`. When the string is jerked, large impulsive reactions are generated in the string. At this moment effect of gravity is negligible. So momentum of the system is conserved at this INSTANT. Let `v'` be the common speed of the two bodies after they are jerked into motion. From conservation of momentum, we have `MV=(M+m)v'` or `v'=(m)/(M+m)v` The acceleration of the system is `SigmaF=Mg-mg=(M+m)a` or `a=-(M-m)/(M+m)g` The acceleration is negative, (opposite to `v'`) Let the system return to origianl POSITION at time `t`. `0=v't+(1)/(2)at^(2)` or `t=-(2V')/(a)=(2m)/(M-m)sqrt((2h)/(g))` (b) The fractional loss of kinetic energy is `((1)/(2)mv^(2)-(1)/(2)(M+m)v'^(2))/((1)/(2)mv^(2))=(M)/(M+m)` |
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| 19. |
The physical quantity which has dimensional formula as that of ("Energy")/("mass " xx "length") is |
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Answer» Force |
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| 20. |
A car without passengers moving with a constant velocity can be stopped in a distance of 20m. If the passengers add 40% of the weight, its stopping distance for the same breaking force and velocity is (ignore friction) |
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Answer» 14m |
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| 22. |
Why do we go for an atomic standard for defining metre and second? |
| Answer» SOLUTION :Then we can measure LENGTH and time with higher precision and ACCURACY. | |
| 23. |
Eqautions y=2A cos^(2) omegat and y=A (sinomegat+sqrt(3) cosomegat) represent the motion of two particles. |
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Answer» Only one of these is `S.H.M` |
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| 24. |
Two bullets are fired simultaneously, horizontally and with different speed from the same place. Which bullet will hit the ground first ? |
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Answer» the faster one |
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| 25. |
Water flows throw 2 identical tubes A & B, A volume V_(0) of water posses throw the tube A & 2V_(0) throw B in a given time. Which of the following may be correct ? |
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Answer» FLOW in both the tubes are STEADY |
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| 26. |
Which of the ratio of forces represents the Reynolds number Re ? |
| Answer» SOLUTION :`Re=("INITIAL FORCE")/("VISCOUS force")` | |
| 27. |
A blockof mass2 Kgis placedon a smoothinclinedplaneof inclination30 ^(@)with thehorizontalplane. Find(i) accelerationof theblockand (ii )normal reactionsuppliedby theinclinedplaneon theblock . |
Answer» Solution : fromthefigure , (I )accelerationof THEBLOCK`= G sin 30^(@)= 9.8xx(1 )/(2) ` `= 4.9m . S ^(-2) ` (ii )normalreaction, N`= Wcos 30 ^(@)= ( 2 XX 9.8) xx ( SQRT(3) )/( 2) ` `= 9.8 xx 1.732~~ 17 N ` |
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| 28. |
Choose the correct alternatives. |
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Answer» For a general ROTATIONAL motion, ANGULAR momentum L and angular velocity `omega` need not be parallel |
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| 29. |
10 bullets each of mass 10gm are fired in succession into a block of mass 450gm at rest. If velocity of each bullet is 110 ms^(-1) and all the bullets are embeded in the block, the velocity of the block is |
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Answer» `5MS^(-1)` |
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| 30. |
A cyclist comes to a skidding stop in 10 m . During this process , the force on the cycle due to the road is 200 N and is directly opposed to the motion . (a) How much work does the road do on the cycle ? (b) How much work does the cycle do on road ? |
| Answer» SOLUTION :WORK done bythe CYCLE= - 4000 J and work done by FORCE on road = zer | |
| 31. |
A cart loaded with sand moves along a horizontal plane due to a constant force F coinciding in direction with the cart's velocity vector. In the process, sand spills through a hole in the bottom with a constant velocity mu kg //s. Find the acceleration of the cart at the moment t, if at the initial moment t = 0, the cart with the loaded sand has a mass m, and its velocity was equal to zero. The friction is to be neglected. |
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Answer» `(F)/(m_0 - MU t)` |
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| 32. |
When a stationary wave is produced ? |
| Answer» SOLUTION :Stationary waves are PRODUCED when two indentical waves travelling in OPPOSITE DIRECTIONS through a MEDIUM superpose each other. | |
| 33. |
Equal torques act on the disc A and B of the pervious problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are v_(A) and v_(B) respectively. We have |
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Answer» `v_(A)gt v_(B)` |
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| 34. |
Define Ampere : |
| Answer» Solution :One ampere is the constant current, which when maintained in each of the TWO straight PARALLEL conductors of infinite length and negligible cross section, held one metre apart in vacuum shall produce a force per unit length of `2 xx 10^(-7) N//m` between them. | |
| 35. |
A body is moving at constant speed over a frictionaless surface. What is the work done by the weight of the body ? |
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| 36. |
P is the centre of mass of a system of four point masses A,B,C and D, which are coplanar but not collinear. (a) P may or may not coincide with one of the point masses (b) P must lie within or on the edge of atleast one of the triangles formed by taking A,B, C and D three at a time (c ) P must lie on a line joining two of the points A,B,C,D (d ) P lies out side the quadrangle ABCD |
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Answer» a & b are correct |
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| 37. |
State Kepler's three laws. |
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Answer» Solution :1. Law of orbits : Each planet moves around the Sun in an elloptical orbit with the Sun at one of the foci. 2. Law of area: The RADIAL vecotr (LINE joining the Sun to a planet) sweeps EQUAL areas in equal intervals of time. 3. Law of period: 3. Lawo period: The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the SEMI - major axis of the ellipse. |
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| 38. |
Two small blocks of mass m and 4m are connected to two springs as shown in fig. Both springs have stiffness K and they are in their natural length when the blocks are at point O. Both the blocks are pushed so that both the springs get compressed by a distance a. First the block of mass m is released and after it travels through a distance(1-(sqrt(3))/(2)) the second block is also released. (a) At what distance from point O will the two blocks collide? (b) How much time the two blocks need to collide after the block of mass 4m is released? |
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| 39. |
Two bodies of masses 4 kg and 5 kg are moving with equal momentum. Then the ratio of their respective kinetic energies is |
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Answer» `4:5` where p is the momentum and m is the mass of the body. As `p_1 = p_2` (GIVEN) `:. (K_1)/(K_2) = (m_2)/(m_1) = 5/4`. |
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| 40. |
If Planck's constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can in addition be taken to express length, mass and time in terms of the three chosen fundamental quantities? |
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Answer» <P>(A) Mass of electron `(m_(e))` `[c]=[L^(1)T^(-1)], [m_(e)]=M^(1)` `[G]=[M^(-1)L^(3)T^(-2)]` `[e]=[AT],[m_(p)]=[M^(1)]` `[(hc)/(G)]=([ML^(2)-T^(-1)][LT^(-1)])/([M^(1)L^(3)T^(-2)])=[M^(2)]` Mass, `M= sqrt((hc)/(G))` Length, `L=(h)/(CM)=(h)/(C) sqrt((G)/(hC))= (sqrt(GH))/(C(3)/(2))` Now `C=LT^(-1)` `[T]=([L])/([C])=(sqrt(Gh))/(C^((3)/(2)*C))= (sqrt(Gh))/(C^((5)/(2)))` |
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| 41. |
Is radius of gyration a constant quantity? |
| Answer» SOLUTION :No, it CHANGES with the POSITION of AXIS of ROTATION. | |
| 42. |
3 moles of a monoatomic gas requires 45 cal heat for 5^(@)C rise of temperature at constant volume, then heat required for 5 moles of same gas under constant pressure for 10^(@)C rise of temperature is(R=2 cal//"mole"//^(@)K) |
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Answer» 200 cal |
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| 43. |
A 14.5 kg mass fastened to the end of a steel wire of unstretched length 1.0 m is whirled in a vertical circle with an angular velocity of 2 rev.s^-1at the bottom of the circle. The cross sectional area of the wire is 0.065 cm^2. Calculate the elongation of the wire when the mass is at the lowest point of its path. |
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| 44. |
The moment of inertia of same rigid bodies of equal mass 'M' are given along its diameter and other also having equal mass 'M' about given axis are given below. Match the following columns |
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| 45. |
1.00 xx 2.88 is equal to |
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Answer» `2.88` THUS the result when rounded off to three significant digits BECOMES 2.88. |
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| 46. |
Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump DeltaV (=V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P_(1)to P_(2) ? |
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Answer» SOLUTION :`P(V +Delta UPSILON)^(gamma) = (P+DELTAP)V^(gamma)` `P[1+ gamma (Delta upsilon)/(V)] = P(1+(Delta p)/(P))` `gamma (Delta upsilon)/(V) =(Deltap)/(P), (d upsilon)/(dp) = (V)/(gammaP)` `W.D. = int_(P_(1))^(P_(2)) P d upsilon=int_(P_(1))^(P_(2))"P" (V)/(gammaP) dp = ((P_(2)-P_(1)))/(gamma)V` |
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| 47. |
The moment of inertia of same rigid bodies of equal mass 'M' are given along its diameter and other also having equal mass 'M' about given axis are given below. Match the following columns {:("Column-I (Body)","Column-II (Radiusof gyration)"),("(A) Solid cylinder of radius R about its axis",(P)sqrt((2)/(5))R),("(B) Solid sylinder of radius R and length L about its central diameter",(Q)(R)/(sqrt(2))),("(C) Annular cylinder of radius R and outer radius L about cylinder axis",(R)sqrt((R^(2))/(4)+(L^(2))/(12))),("(D) Solid sphere of diameter 2R about its",(S)sqrt((R^(2)+L^(2))/(2))):} |
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| 48. |
In uniform circular motion assuming vecV - Velocity,vec r= radius vector , vecomega= angular velocity relative to centre of the circle, veca= acceleration, which of the following is/are correct? |
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Answer» `|DELTAVECV| NE 0` but `Delta|vecv|=0` |
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| 49. |
A light rod of length l has two masses m_(1) and m_(2) attached to its two ends . The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is |
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Answer» `(m_1 m_2 )/(m_(1) + m_(2)) "" L^(2)` Centre of mass of the system , `I_(1) = (m_(1) xx 0 + m_(2) xx l)/(m_(1) + m_(2)) = (m_(2) l)/(m_(1) + m_(2))` `l_(2) = l - l_(1) = (m_(1) l)/(m_(1) + m_(2))`  REQUIRED moment of inertia of the system , `I = m_(1) l_(1)^(2) + m_(2) l_(2)^(2) = (m_(1) m_(2)^(2) + m_(2)m_(1)^(2)) (l^(2))/((m_(1) + m_(2))^(2))` `= (m_(1) m_(2) (m_(1) + m_(2))l^(2))/((m_(1) + m_(2))^(2)) = (m_(1) m_(2))/(m_(1) + m_(2))""l^(2)` |
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