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. |
Viscosity is the property of liqids by virtue of which |
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Answer» Liquid MOLECULES ATTRACT each other |
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| 2. |
The potentail energy of a spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potentail energy stored on it is |
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Answer» `U/4` |
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| 3. |
What is inelastic collision ? In which way it is different from elastic collision. Mention few examples in day to life for inelastic collision. |
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Answer» Solution :Inelastie collision : In collision, the total initial Kinetic energy of the bodies (before collision) is not equal to the total final kinetic energy of the bodies (after collision) then ITIS called s inelastic collision i.e., `"Total kinetic before collision "ne " Total energy after collision"` `({:(,"Total kinetic energy before collision"),(,"Total kinetic energy after collision"):})-({:(,"Total kinetic energy"),(,"after colision"):})=({:(,"loss in energy during collision"),(,"during collision"):})=triangleQ` Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy CONTAINS the kinetic energy term and also a term`triangleQ`, which includes all losses that take place during collision. Note that loss in kinetic energy during collision transformed to another form of energy like sound, thermal, el. Further, if the two colliding bodies stick together after collision such collisions are known as COMPLETELY inelastic collision or perfectly inelastie collision such a collision is found very often. For example when a CLAY putty is thrown on a moving vehicle, the clay putty for ( or Bubblegum) STICKS the moving vehicle and they move together with the same velocity.
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| 4. |
In the given v-t graph. The distance travelled by the body in 5 second will be |
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Answer» 20m |
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| 5. |
What is meant by aan echo ? Explain. |
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Answer» Solution :ECHO: An echo is a repetition of sound produced by the reflection of sound WAVES from a wall, mountain or other obstructing surfaces. Explalation : The speed of sound in air at `20^(@) C `is `344 MS^(-1)` If we shout at a wall which is at 344 m AWAY, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time intervalbetween each sound is `((1)/(10))^(th)` of a second (persistence of hearing ) i.e. , 0.1 s . Then , Velocity = `("Distance travelled")/("Time TAKEN") = (2d)/(t)` `2d = 344 xx 0.1 = 34.4 m , d = 17.2` m The minimum distance from a sound reflecting wall to hear an echo at `20^(@)C ` is 17.2 meter |
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| 6. |
What is linear harmonic oscillator ? And what is non-linear oscillator? |
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Answer» Solution :If a FORCE exerted on any particle is directly proportional to displacement to time t, then oscillating particle is known as linear HARMONIC oscillator. In REAL world, the force may contain SMALL additional terms proportional to `x^(2), x^(3),"….."` etc. these then are called non-linear oscillators. |
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| 7. |
A man is travelling at 10.8 kmph in a topless car on a rainy day. He holds an umbrella at an angle of 37^(@) with the vertical to protect himself from the rain. If cos 37^(@)=0.8 the velocity of the rain drops is |
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Answer» 3 m/s |
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| 8. |
When a plane is taking off on a runway the wind velocities at the top and bottom of the wings are found to be 40 ms^(-1) and 30 ms^(-1) respectively. If the wing span is 60 m^(2) find the dynamics lift on the density of air is 1.27 kg m^(-3). |
| Answer» SOLUTION :`26.7xx10^(3)N` | |
| 9. |
A point light sources lies on the principal axis of concave spherical mirror with radius of curvature 160cm. Its image appears to the back of the mirror at a distance of 70 cm from mirror. Determine the location of the light source. |
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Answer» Solution :`1/u+1/v=2/R` Here v=70 CM `R=-160 cm 1/u=2/R-1/v` `therefore 1/u=2/(-160CM)-1/(70CM) =-15/(560cm)` `therefore u= -560/15 cm=-37cm` The source is at a DISTANCE of 37 cm in front of the mirror. |
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| 10. |
A square plate of 1m side moves parallel to a second plate with velocity 4m/s. A thin layer of water exist between plates. If the viscous force is 2N and the coefficient of viscosity is 0.01 poise then find the distance between the plates in mm. |
| Answer» SOLUTION :`F=(etaAv)/(d)impliesd=(etaAv)/(F)=(0.01xx10^(-1)xx1^(2)xx4)/(0.002)=2m` | |
| 11. |
The displacement of a particle is given by x(t) = (4t ^(2) +8)meter. The instantaneous velocity of a particle at t = 2s is |
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Answer» 12m/s `therefore` Instantaneous velocity `v = (DX )/(dl) = (d)/(DT) (4t ^(2) +8)` `therefore v = 8T` So, at` t = 2s, v = 8(2) = 16 m//s` |
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| 12. |
When two bodies undergo perfectly inelastic collision what is their relative velocity after collision ? |
| Answer» SOLUTION :Zero, SINCE the TWO bodies move with a common VELOCITY. | |
| 13. |
What do you mean by upthrust or buoyancy? |
| Answer» SOLUTION :The upward force EXERTED by a FLUID that opposes the WEIGHT of an immersed object in a fluid is CALLED upthrust or buoyant force. | |
| 14. |
Two indentical springs each of force constant k are connected in paralle and they support a mass m.Calculate the time period of the system |
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Answer» SOLUTION :For parallel combinatin, the effecive force CONSTANT, is `k_(p)=k+k=2K` `:.T_(p)=2pi sqrt((m)/(k_(p)))=2pi sqrt((m)/(k))` Required ratio of the time periods, `(T_(s))/(T_(p))= sqrt((2k)/(k//2))=2` |
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| 15. |
A man of 50 kg is standing at one end on a boat of length 25 m and mass 200kg. If he starts running and when he reaches the other end, he has a velocity 2ms^(-1) with respect to the boat. The final velocity of the boat is (in ms^(-1)) |
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Answer» `(2)/(5)` |
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| 16. |
During an adiabatic process the pressure of gas is found to be proportional to the cube of its absolute temperature. Then ratio C_P/C_V = …….. |
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Answer» Solution :For adiabatic process `P prop T^(GAMMA/(gamma-1))` and it is given that `P prop T^3`, by comparing them, `gamma/(gamma-1)=3` `THEREFORE gamma=3gamma-3` `therefore 3=2gamma` `therefore gamma=3/2` `therefore C_P/C_V=3/2` |
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| 17. |
Two satellite A and B of the same mass are revolving around the Earth in circular orbits such that the distance of B from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on A. |
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Answer» Solution :Thenecessarycentripetalforceis providesby the gravitationalforceof attraction , for circularorbit Force on A ,`(mv^(2))/R = (GMm)/(r^(2))` Force on B , `(mv^(2))/r = (GMm)/(3r)^(2)` The RATIOOF CENTRIPETALFORCE , `(F_(B))/(F_(A)) = (GMm)/(9R^(2)) xx (r^(2))/(GMm)` `(F_(B))/(F_(A)) = 1/9 ` |
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| 18. |
Suppose you have to pile up uniform bricks to erect an inclined wall by offsetting each brick by a constant fraction ,(1)/(n) , of a brick length l over the beneath. How many bricks can be piled by before the wall collapse ? How high a wall can erect if the size of each bricks is 7.5 cm xx 12.5 cm xx 25 cm and the offset is only (1)/(10) of the length of each bricks? What is the total offset you get ? [Hints: The vertical line through CG of the wall must pass through the base. The moment it goes beyond the base brick, the wall will collapse.] |
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| 19. |
A uniform piece of metal sheet is cut in the form as shown in the fig. Locate the centre of mass of the piece. |
Answer» Solution :![]() Let the mass per UNIT AREA be `sigma`. We divide the structure into parts 1, 2 and 3. The mass of part 1 is 300 `sigma`, mass of part 2 is `200 sigma` and that of part 3 in `100 sigma`. The coordinates of centre of mass of part 1 are (5, 15), that of part 2 are (20, 5), and that of part 3 are (15, 25) Using the Eq. `x_(cm)=((300sigma)5+(200 sigma)20+(100 sigma)15)/(600 sigma)~~11.7` `y_(cm)=((300 sigma)15+(200 sigma)5+(100 sigma)25)/(600 sigma)~~13.3` In vector notation `bar(r_(cm))=11.7hat(i)+13.3 HAT(j)` |
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| 20. |
If the rms speed of hydrogen molecules at 300K is 1930ms^(-1). Then what is the rms speed of oxygen molecules at 1200K. |
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Answer» Solution :From RMS speed , `nu_(rms)=SQRT((3RT)/M)` `nu_(O_(2))=sqrt((3RT_(O_(2)))/M_(O_(2))),nu_(H_(2))=sqrt((3RT_(H_(2)))/M_(H_(2)))` `nu_(O_(2))/nu_(H_(2))=sqrt(T_(O_(2))/T_(H_(2)).M_(H_(2))/M_(o_(2)))` `nu_(O_(2))=sqrt((1200/300xx2/32))xx1930=1930/2` `nu_(O_(2))=965ms^(-5)` |
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| 21. |
Two beads A and B of masses m, and m, respectively are threaded on a smooth circular wire of radius .a. fixed in a vertical plane. B is stationary at the lowest point when A is gently dislodged from rest at the highest point. A collided with B at the lowest point. The impulse given to B due to collision is just great enough to carry it to the level of the centre of the circle while A is immediately brought to rest by the impact. a) Find the ratio b) What is the coefficient of restitution between the beads? |
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| 22. |
The fraction of ice that melts by mixing equal masses of ice at -10^(@)C and water at 60^(@)C is |
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Answer» `(6)/(11)` |
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| 23. |
9765 div 14.39= 678.59625. This quotient on rounding off of two significant figures is |
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Answer» 680 |
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| 24. |
In the following table displacement of SHO is in column-I and kinetic energy in column-II is shown. Match them appropriately. . |
| Answer» SOLUTION :`IMPLIES A-iii, B-ii` | |
| 25. |
While deriving the relationship between physical quantities by dimensional analysis, dimensionless constant enters into the relationship. Can you find its magnitude by the method of dimensions? |
| Answer» Solution :No, we cannot find the MAGNITUDE of the dimensionless CONTACT in the method of dimensions because it is a NUMBER. We find its value by EXPERIMENT or through mathematical investigation. | |
| 26. |
The period of geostationary artificial satellite of the earth is |
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Answer» 24 hr |
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| 27. |
Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermal while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is |
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Answer» `2^(GAMMA - 1)` |
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| 28. |
A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere is (G = 6.67 xx 10^(-11) Nm^(2)kg^(-2)) |
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Answer» `6.67 XX 10^(-10)J` |
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| 29. |
A liquid can easily change its shape but a solid can not because |
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Answer» The density of a liquid is SMALLER than that of a solid |
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| 30. |
The translational kinetic energy of diatomic gas for one mole of the gas is equal to |
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Answer» `3/2RT` |
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| 31. |
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K.The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is : |
| Answer» ANSWER :D | |
| 32. |
Two indentical springs each of force constant k are connected in series and support mass m. Calculate time period of the system |
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Answer» SOLUTION :For series combination, the effective FORCE constant is, `k_(s)=(kxxk)/(K+k)=(k^(2))/(2K)=(k)/(2)` `:. T_(s)\= 2PI sqrt((m)/(k_(s)))= 2pi sqrt((m)/(k//2))` |
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| 33. |
When a man is standing, rain drops appears to him falling at 60^(0) from the horizontal from his front side. When he travelling at 5 kmph on a horizontal road they appear to him falling at 30^(0)from the horizontal from his front side. The actual speed of the rain is (in kmph) |
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Answer» 3 |
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| 34. |
The temperature of gas is produced by |
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Answer» the POTENTIAL energy of its molecules |
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| 35. |
A particle is projectedfrom origin at an angle is selectedas X-axis and verticle as Y-axis . It is found that particle passes throughthe points (a, b) and (b, a). Calculate range and angle of projection . |
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Answer» Solution :Equartion of trajectory in terms of range of PROJECTILE can be written as follows : `y=x(1-(x)/(R))TAN theta` Given coodinates must satisfy theabove EQUATION. Hence we can write the following : `b=a(1-(a)/(R)) tan theta` `a=b(1-(b)/(R)) tan theta` Dividing the abovetwo EQUATIONS, we can write the following: `(b^(2))/(a^(2))=(1-(a)/(R))/(1-(b)/(R))` `rArr (b^(2))/(a^(2))=(R-a)/(R-b)rArrb^(2)R-b^(3)=a^(2)R-a^(3)` `rArr R=(a^(3)-b^(3))/(a^(2)-b^(2))=((a-b)(a^(2)_ab+b^(2)))/((a-b)(a+b))` `rArr R=(a^(2)+ab+b^(2))/(a+b)` Now we can substitute R in equation (i) to getthe angle ofprojection . `b=a(1-((a)/(a^(2)+ab+b^(2)))/(a+b))tan theta` `b=a(1-(a^(2)+ab)/(a^(2)+ab+b^(2)))tan theta` `rArr (b)/(c) ((a^(2)+ab+b^(2)0a^(2)-ab)/(a^(2)+ab+b^(2))) tan theta` `rArr (b)/(a)=((b^(2))/(a^(2)+ab+b^(2))) tan theta` `rArr theta =(a^(2)+ab+b^(2))/(ab)` `rArr theta =tan ^(-1)((a^(2)+ab+b^(2))/(ab))`. |
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| 36. |
Assuming Young.s modulus of aluminium to be 7 xx 10 N//m^(2) find energy stored in a rod of cross section 10^(-4)m^(-2) and 0.1 m long when com-pressed by a load of 10 kg. |
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Answer» SOLUTION :ENERGY stored `=(1)/(2)` (stretching force) (elongation) `=(1)/(2)(10xx9.8)` (compression) compression `=Deltal=(Mgl)/(AY)=(10xx9.8xx0.1)/(10^(-4)xx7xx10^(10))=10^(-6)xx1.4` `:.` Energy soted `=(1)/(2)(10xx9.8)(1.4xx10^(-6))=6.86xx10^(-5)` Joule |
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| 37. |
A pendulum bob of mass .m. is held out in the horizontal position and then released from rest. If the string is of length l(I) what is the velocity of the bob and tension in the string when the bob reaches the lowest position (2) If the breaking of the string. |
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Answer» Solution :From law of consevation of mechanical energy. `0 + MGL = (1)/(2) mv^(2) + 0 "" v=sqrt(2GL)` The tension in the string at the lowest position `=(mv^(2))/(r)+mg=(m(2gl))/(L)+mg(thereforer=l)=3MG` (2) `mgl(1-costheta)=(1)/(2)mv^(2)orv^(2)=2gl(1-costheta)` Tension in the string at the lowest position `T=(mv^(2))/(l)+mg` `thereforenmg=(m)/(l)[2gl(1-costheta)]+mg` on simplification `cos theta = (3-n)/(2)` |
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| 38. |
Hail stonces fall from certain height. If only 2% of the mass of the hail stone melt on reacting the ground,, the height from which they fall is ( g = 10 ms^(-2), L = 80 cal/gm and J = 4.2J/cal) |
| Answer» ANSWER :C | |
| 39. |
When a gas expands adiabatically, its internalenergy |
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Answer» INCREASES |
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| 40. |
If the earth suddenly became a thin shell of same radius then length of a day becomes |
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Answer» 24 hr |
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| 41. |
Is torque a vector or a scalar? If it is a vector give its direction. |
| Answer» SOLUTION :TORQUE is a vector. Torque`vectau=barrxxbarF` .The direction of torque is PERPENDICULAR to the PLANE containing `barr` and `barF`. If `barr` and `barF` and in X-Y plane then torque is along Z-axis. Its direction is given by right hand screw rule. | |
| 42. |
Match column - I with column -II {:(,"Column -I",,"Column -II"),("(a)","Gravitational force",,"(p)Central"),("(b)","Viscous force",,"(q) Non-conservative"),("(c)","Electric force",,"(r) Conservative"),("(d)","Friction",,"(s) Mechanical energy is conserved"):} |
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Answer» |
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| 43. |
(A): Young's modulus and shear modulus are relevant only for solids. (R) : Solids have definite shapes and lengths |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 44. |
In Q.118, total distance travelled by the particle from t = 0 to t = t_(0) is: |
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Answer» `(A_(0))/(2 alpha)` |
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| 45. |
The length of minutes hand in a pendulum clock is 10 cm. The speed of the tip of the hand is |
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Answer» `pi//6000 MS^(-1)` |
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| 46. |
A force produces an acceleration of a_(1) in a body and the same force produces an acceleration of A_(2), in another body. If the two bodies are combined and the same force is applied on the combination, the acceleration produced in it is |
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Answer» `a_(1)+a_(2)` |
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| 47. |
A black hole an object whose gravitational field is so strong that even light canot escape from it. To what approximate radius would earth (mass = 5.98 xx 10^(24) kg) have to be compressed to be a black hole ? |
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Answer» `10^(-9)m` |
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| 48. |
Water is filled in tank 3 m height. The base of the tank is at height 1 m above the ground. What should be the height of a hole made in it, so that water can be sprayed uptomaximum horizontal distance on ground? |
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Answer» 3 m from ground |
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| 49. |
A ball of mass 2 kg hanging from a spring oscillates with time period 2pi seconds. Ball is removed when it is equilibrium position, then spring shortens by what length ? |
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Answer» SOLUTION :`T= 2pisqrt((m)/(k))` `:. k= (4pi^(2)m)/(T^(2))= ((4pi^(2))(2))/((2)^(2))` Now `mg= kx_(0)` `:. x_(0)= (mg)/(k)= (2 xx g)/(k)= g` meter or 10M |
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| 50. |
Determine the tension T_(1) and T_(2) in the strings |
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Answer» Solution :As shownin figureresolvethe tension `T_(1)` alonghorizontaland verticaldirection.Asthe BODYIS inequilibrium `T_(1)SIN 60^(@)=4 kgwt = 4 xx 9.8N` From (1) `T_(1) = (4XX 9.8)/(sin 60^(@))` `= (4xx 9.8xx 2)/( sqrt(3)) = 45.26N` from (2) `T_(2) = T_(1) cos 60^(@)` `45.26 xx 0.5 =22.63N` |
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