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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. |
When .n.number of particles of masses m, 2m, 3m, .... nm are at distances x_(1)=1,x_(2)=4,x_(3)=9,......x_(n)=n^(2) units respectively from origin on the x-axis, then find the distance of their centre of mass from origin |
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Answer» SOLUTION :`x_(cm)=(m(1)+2M(4)+3m(9)+.........+nm(n^(2)))/(m+2m+3m+......+nm)` `=(m(1+2^(3)+3^(3)+......+n^(3)))/(m(1+2+3+.......+n))` `=(((n(n+1))/(2))^(2))/((n(n+1))/(2))=(n(n+1))/(2)` |
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| 2. |
Passage - III : A long horizontal plank of mass 'm' is lying on a smooth horizontal surface. A sphere of same mass 'm' and radius 'r' is spinned about its own axis with angular velocity omega_(0) and gently placed on the plank at t=0. The coefficient of friction between the plank and the sphere is mu. After some time the pure rolling of the sphere on the plank will start. Find the displacement of the plank till the sphere starts pure rolling. |
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Answer» `(omega_(0)^(2)r^(2))/(81mu G)` |
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| 3. |
The molecules of a given mass of a gas have a r.m.s velocity of 200 m/sec at 27^(@)Cand 1.0xx10^(5) N//m^(2) pressure.When the temperature is 127^(@)C and pressure is 0.5xx10^(5) N//m^(2) the r.m.s. velocity in m/sec will be |
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Answer» `100 sqrt(2)` |
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| 4. |
Two bodies having masses 1g and 4g are moving with equal K.E find the ratio of their linear momentum. |
| Answer» SOLUTION :0.043055555555556 | |
| 5. |
The mass of a block of wood is 87.2 g and its volume is 25 cm^3 . Its density with due regard to significant figures is |
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Answer» `3.488g "CM"^(-3)` |
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| 6. |
As a result of addition of detergent to a liquidthe angle of contact . |
| Answer» SOLUTION :Detergents should have SMALL angle of CONTACT so that they have low surface tension and greater ABILITY to WET a surface. Furtheras `h =(2Tcostheta)/("rpg")"i.e", theta`is small cos `theta` will be large so h i.e., penetration will be high. | |
| 7. |
A shell is fired into air at an angle with the horizontal from the ground. On reaching the maximum height |
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Answer» its kinetic ENERGY is not EQUAL to ZERO |
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| 8. |
(i) What is the length of the pendulum of a clock which beats seconds, given that the acceleration due to gravity at a place is equal to 9.8ms^(-2). (ii) Find the time period of another pendulum of length equal to half of it at the same place. (iii) A second pendulum is taken into a mine of depth 640 m and oscillated. What is its time period? (iv) The length of a seconds pendulum at a place is 1.02 m. Find the time period of another pendulum of length0.51 m at the same place. |
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Answer» Solution :The period of simple pendulum is given by `T=2pisqrt(l/g)implies` The length of the pendulum `=1=(gT^(2))/(4pi^(2))implies1=(9.8xx2^(2))/(4xx(3.14)^(2))` `l=(9.8xx2^(2))/(4xx9.86)=0.994m~=1m` (ii) The period of a simple pendulum is given by `T=2pisqrt(l/g)`........1 `"":.g=4pi^(2)l/(T^(2))` i.e.`l/(T^(2))=g/((4p)^(2))=` constant at a given place (2) `(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))implies1.0/(2^(2))=0.50/(T_(2)^(2))`or `T_(2)^(2)=0.50/1.01xx4=2` `:.T_(2)=sqrt(2)=1.414s` (iii) The TIME period of a simple pendulum is `T=2pisqrt(l/g)` When a SECONDS pendulum is taken into a mine g changes (l constant) The ratio of its time periods is `(T_(1))/(T_(2))=sqrt((g_(2))/(g_(1)))` On the SURFACE of the earth `g_(1)=g` and in the mine `g_(2)=g(1-d/R)=(T_(1))/(T_(2))=sqrt((g(1-d/R))/g)"or"(T_(1))/(T_(2))=sqrt(1-d/R)` Here `T_(1)=2s,d=640m=0.640km,` `R=` Radius of the earth `=6400km,T_(2)=?` `2/(T_(2))=sqrt(1-0.640/6400)=sqrt(1-0.0001)` `:.T_(2)=2/(sqrt(1-0.0001))=2.0001s` (iv) The perido of a simple pendulum is given by `T=2pisqrt(l/g)`..........1 i.e `l/(T^(2))=g/(4pi^(2))=` constant at a give place .....2 If `T_(1)` and `T_(2)` are the time periods of the pendulum for length `l_(1)` and `l_(2)` From eq. (2) we can write `(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))`.....3 Here `T_(1)=2s`, when `l_(1)=1.02m` when `l_(2)=0.51m,T_(2)=?` Using eq. 3 `1.02/(2^(2))=0.51/(T_(2)^(2))` or `T_(2)^(2)=0.51/1.02xx4=2` `:.T_(2)=sqrt(2)=1.414s`. When the length increasses the time period of the pendulum increases. So the clock LOSES time or RUNS slow. |
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| 9. |
A uniform rod of mass m and length l is lying on a smooth table. An impulse J acts on the rod momentarily as shown in figure at point R. Just after that: |
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Answer» |
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| 10. |
The radius is 2hati + hatj +hatk, While linear momentum is 2hati +3hatj -hatk. The angular momentum is |
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Answer» `2hati - 4HATK` |
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| 11. |
Calculate the moment of inertia of a cylinder of length 2 m, radius 5 cm and density 8 xx 10^(3) kg//m^3 about (i) the axis of the cylinder and (ii) and axis passing through the centre and perpendicular to its length. |
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| 12. |
Derive an equation for the total pressure at a depth 'h' below the liquid surface. |
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Answer» Solution :Consider a water sample of cross sectional area in the form of cylinder. Let `h_(1)` and `h_(2)` be the depths from th e air -water interface to level I and the 2 of the cylinder, respectively as shown in Figure(a). Let `F_(1)` be the FORCE acting downwards on level 1 and `F_(2)` be the force acting upwards on level 2, such that, `F_(1) = P_(1)` A and `F_(2) = P_(2)` A the mass of the sample to be m and under equilibrium CONDITION, the total upward force `(F_(2))` is balanced by the total downward force `(F_(1) + mg)`, i.e., the gravitational force will act downward which is being exactly balanced by the difference between the force `F_(2) - F_(1)` `F_(2) - F_(1) = mg = F_(G)` where m is the mass of the water available in the sample element. Let `RHO` be the density of the water then, the mass of water available in the sample element is `m = rho V = rhoA(h_(2) - h_(1))` `V = A(h_(2) - h_(1))`  Hence, gravitational force `F_(G) = rho A(h_(2) - h_(1))g` On substituting the value of W in equation `F_(2) = F_(1) + mg rArr P_(2)A = P_(1)A + rho A(h_(2) - h_(1))g` Cancelling out A on both sides, `P_(2) = P_(1) + rho(h_(2) - h_(1))g` The level 1 at the surface of the liquid (i.e., air-water interface) and the level 2 at a depth 'h' below the surface (as shown in Figuer (B)), then the value of `h_(1)` becomes ZERO `(h_(1) = 0)`. and `P_(1) = P_(a)` atmospheric pressure (say `P_(a)`). In the pressure `(P_(2))` at a depth becoes P. Substituting these values in equation. `P = P_(a) + rho gh` which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where `P_(a)` is the atmospheric pressure which is equal to `1.013 xx 10^(5)P_(a)`. If the atmospheric pressure is neglected or ingnored, then `P = rho gh` 
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| 13. |
A particle starts from origin at t= 0 with a velocity 5.0hat(i)m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i+2.0j)"m/s"^(2). (a) What is the y-coordinates of the particle at the instant its x-coordinates is 84m? (b) what is the speed of the particle at this time? |
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Answer» Solution :The position of the PARTICLE is given by `r(t) =v_(0)t+(1)/(2)at^(2)=5.0hat(i)t+"(1/2)"(3.0hat(i)+2.0hat(j))t^(2)` `=(5.0t+1.5t^(2))HAT(i)+1.0t^(2)hat(j)` Therefore, `x(t)=5.0t+1.5^(2), y(t)=+1.0t^(2)` Given `X(t)=84m, t=?` `5.0t+1.5t^(2)=84rArrt=6s` At `t =6s, y=1.0(6)^(2)=36.0m` Now, the velocity `v=(DR)/(DT)=(5.0+3.0t)hat(i)+2.0that(j)` At t = 6s, `v=23.0hat(i)+12.0hat(j)` SPEED `=|v|=sqrt(23^(2)+12^(2))=26ms^(-1)` |
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| 14. |
A body has maximum range R_(1) when projected up the inclined plane. The same body when projected down the inclined plane, it has maximum range R_(2). Find its maximum horizontal inclined plane inte line of the greates slope. |
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Answer» Solution :For upward projection `R_("MAX")=(v_(0)^(2))/(g(1+SIN beta))=R_(1)`………..i For downward projection `R_("max")=(v_(0)^(2))/(g(1-sin beta))=R_(2)`………ii For a projection on horizontal SURFACE substituting `beta=0` Then We have `R_("max")=(v_(0)^(2))/g=R` (say) ...........iii To establish a relation between `R,R_(1)` and `R_(2)`we need to eliminate `sin beta`. Adding `1/(R_(1))` from eq (i) with `1/(R_(2)0` from eq (ii) we have `2/R=1/(R_(1))+1/(R_(2))` Then `R=(2R_(1)R_(2))/(R_(1)+R_(2))` |
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| 15. |
A cylindrical vessel contains a liquid of density rho upto a heighth The liquid is closed by a piston of mass m and area of cross-section A. There is a small hole at the bottom of the vessel. Find the speed with which the liquid comes out of the hole |
Answer» Solution :APPLYING Bernoulli.s THEOREM at 1 and 2, DIFFERENCE in pressure energy between 1 and 2 = difference in KINETIC energy between 1 and 2  or `rhogh+(mg)/(A)=(1)/(2)rhov^(2)` or `v=sqrt(2gh+(2MG)/(rhoA))=sqrt(2(gh+(mg)/(rhoA)))` |
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| 16. |
An ideal gas has a molar heat capacity C_v at constant volume. Find the molar heat capacity C, at constant volume. Find the molar heat capacity of this gas as a function of its volume 'V', if the gas undergoes the following process : P=P_(0)e^(alpha V) |
| Answer» SOLUTION :`C_v +(R )/((1+ ALPHA V))` | |
| 17. |
An ideal gas has a molar heat capacity C_v at constant volume. Find the molar heat capacity C, at constant volume. Find the molar heat capacity of this gas as a function of its volume 'V', if the gas undergoes the following process : T=T_0 e^(alpha V) |
| Answer» SOLUTION :`C_v +(R )/(alphaV)` | |
| 18. |
The image of an extended object placed perpendicular to the principal axis of a mirror, will be erect if (A) the object and the image and both real (B) the object and the image are both virtual (C ) the object is real but the image is virtual (D) the object is virtual but the image is real |
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Answer» (A) (C ) and (D) |
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| 19. |
One end of a uniform rod of mass m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on ti about the clamped end at a uniform angular velocity omega. The force exerted by the clamp on the rod has a horizontal component. |
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Answer» `MOMEGA^(2)L` |
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| 20. |
K is the force constant of a spring. The work done is increasing its extension from l_(1) to l_(2) will be |
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Answer» `K (l_2 - l_1)` |
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| 21. |
Name the physical quantity which has the same meaning asmomentum. |
| Answer» SOLUTION :ANGULAR MOMENTUM | |
| 22. |
Which of the following is the unit of coefficient of visocity a) pascal xx second b) poise c) N//m^(2) d) stoke |
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Answer» a is correct |
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| 23. |
Whatismeantby heating? |
| Answer» Solution :When an object at HIGHER temperature is placed in CONTACT with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the ONE at lower temperature. This energy is CALLED heat. | |
| 24. |
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension inthe string ?What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ? |
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Answer» Solution :`V = 2pi xx 1.5 xx 40/60 = 2pi MS^(-1)` `T = (mv^2)/(R ) = (0.25 xx 4pi^2)/(1.5) = 6.6 N` `200 = (mv_(MAX)^2)/(R ) ` . Which gives`v_(max) = 35 ms^(-1)` |
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| 25. |
How much work to be done in decreasing thevolume of and ideal gas by an amount of 2.4xx10^(-4) m^(3)and constant normal pressure of 1xx10^(5)N//m^(2) |
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Answer» 28 JOULE |
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| 26. |
In determining the temperature of a distant star,one makes use of |
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Answer» KIRCHHOFF's law |
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| 27. |
The momentum of a body is doubled. Its K.E will |
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Answer» be doubled |
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| 28. |
A particle moves in a circular path such that its speed v varies with distance asv = alpha sqrt swhere a is a positive constant. Find the acceleration of particle after transversinga distance S? |
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Answer» `ALPHA^2 SQRT(1/4 - S^2/R^2)` |
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| 29. |
Shape of a string transmitting wave along x-axis some instantis shown. Velocity of point P v - (4pi)cm /s (theta) = tan^(-1) (0.004pi) |
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Answer» Amplitube of wave is 2 mm |
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| 30. |
Parsec is how many light year? |
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Answer» Solution :1 PC= x LIGHT year. `(1 pc)/(1"light year")=(3.08xx10^(16))/(9.46xx10^(15))` `=0.32558xx10^(1)~~3.26` `:.` p1 pc `=3.26` light year |
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| 31. |
s-t graph shown in figure is a parabola. From this graph we find that : |
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Answer» the body is moving with uniform VELOCITY |
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| 32. |
For an ideal gas its adiabatic exponent is gamma . In some process its molar heat capacity varies as C = alpha/T, where alphais a constant. Work performed by one mole of gas during its heating from T_0 to nT_0will be |
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Answer» `alpha log_e n` |
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| 33. |
A man movis 10m in a direction 37^(@) East of North, than 5 sqrt2m in South-East direction and finally 20m in a direction 53^(@) South of West. If the magnitude of resultant displacement of man is (sqrt(1252))/(n)m. Find the value of n |
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| 34. |
A vehicle of mass 500 kg is moving with a velocity of 15ms^(-1). It is brought to rest by a retarding force. Find the distance moved by the vehicle before coming to rest, if the sliding friction between the tyres and the road is 3000N. |
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Answer» 18.75 m |
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| 35. |
If action force acting on a body is gravitational in nature, then reaction force |
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Answer» MAY be a CONTACT FORCE |
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| 36. |
Select the odd man out from the following statement |
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Answer» CENTRIFUGAL force is a pseudo force |
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| 37. |
A siren placed at a railway platformis emitting sound of frequency 5 KHz. A passenger sitting in a moving train A records a frequency of 5.5KHz while the train approaches the siren . During his return journey in a different train B he records a frequency of 6.0 KHz while approaching the same siren . what is the ration of velocity of train B to that of train A ? |
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Answer» Solution :Let , ` u_(A) ` be the velocity of train A . ` therefore ` Frequency of siren relative to the PASSENGER of train A , ` n_(1) = n((V + u_(A))/(V))` ` therefore ` 5((V _ u_(A))/(V)) = 5.5 ` …(1) Similarly , frequency of siren relative to the passenger of train B . ` n_(2) = n((V + u_(B))/(V)) [ u_(B) `= velocity of train B] ` therefore5 ((V + u_(G))/(V))= 6 ` ...(2) From equation (1) , 5.5V -5V = 5u_(A) ` therefore V= (5)/(0.5 )u_(A)` ...(3) ` From equation (2) . 6V - 5V = 5u_(g)` ` therefore V = 5u_(g)` ...(4) COMPARING equation (3) and (4) , ` (5)/(0.5) u_(A) = 5u_(g)` ` (u_(B))/(u_(A)) = (1)/(0.5) = 2/1therefore u_(B) : u_(A) = 2: 1 ` . |
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| 38. |
A person of mass 40kg standing on a smooth horizontal plane of ice throws a body of mass 10kg with a velocity 8ms^(-1). Then the total kinetic energy of person and body. |
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Answer» 100J |
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| 39. |
A sphere of ice at 0^(@)C having initial radius R is placed in an environment having ambient temperature gt0^(@)C. The ice melts uniformly, such that shape remains spherical. After a time .t. the radius of the sphere has reduced to r. Assuming the rate of heat absorpiton is proportional to the surface area of the sphere at any moment, which graph best depicts r (t). |
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Answer»
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| 40. |
A clock with a metallic pendulum gains 5 seconds each day at 22^@C and loses 10 seconds each day at 45^@C . Then, |
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Answer» clock keeps correct time at t`79/3 ""^@C` ` T= 2pi sqrt(L/G)` or`T^2 = 4pi^2 xx L/g `…(i) Partially differentiating (i), we get ` 2T Delta T = 4pi^2 (Delta L)/(g)`….(ii) Dividing (i) by (i) , we get `(Delta T)/(T) = (Delta L)/(2L) = (alpha L Delta t)/(2L ) = 1/2 alpha Delta t` T = 1 DAY = 24 hours = 86400 s `therefore Delta T = 1/2 alpha Delta t xx 86400 ` sec If at this temperature ,clock keeps correct time , then at `22^@C` , gain in time is ` 5 = 1/2 alpha(t - 22) xx 86400 `...(iii) At `45^@C` LOSS in time is ` 10= 1/2 alpha (45- t) xx 86400`...(iv) Dividing (iv) by (iii) , we have `10/5 = (45 - t)/(t - 22)` ` rArr 2(t - 22) = 45 - t` ` 2t - 44 = 45 - t ` ` 3t = 89` ` t = 89/3 ""^@C` substituting this in (iii) , we have ` 5 = 1/2 alpha( 89/3 - 22) xx 86400` `rArr alpha = (5 xx 2 xx 3)/(23 xx 86400) = (30)/(23 xx 0.864) xx 10^(-5)` `= 1.5 xx 10^(-5) ""^@C^(-1)` |
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| 41. |
Consider an incompressible and non-viscous liquid in a container. Density of liquid is p and acceleration due to gravity is g and h represents the vertical separation between two points. All points considered in column I are inside the liquid. Match the statements given in Column 1 with corresponding all possible conditions given in column II |
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Answer» |
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| 42. |
Which of the following examples represent simple harmonic motion and which represent periodic but not simple harmonic motion? |
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Answer» The rotation of earth about its axis. b. It is SHMc. It is SHM d. It is a periodic but not SHM A polyatmic gas molecule has a numbr of natural frequencies and its general motion is the resultant of SHMs of a NUMBER of different frequencies. The resultant motion is periodic but not SHM. |
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| 43. |
Velocity of efflux of a liquid through an orific does not depend on |
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Answer» ACCELERATION DUE to gravity |
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| 44. |
Obtain instantaneous acceleration of a SHM particle with the help of reference circle. |
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Answer» SOLUTION :Acceleration of a particlel executing uniform circular motion with ANGULAR speed `omega` and amplitude A is `(v^2)/(A)" or "A omega^(2)` and direction of acceleration is toward centre so it is negative. The acceleration of projected particle in uniform circular motionl at time t `a(t)= -omega^(2) x(t)` This acceleration is towards the centre :  The CENTRIPETAL force at point P is `omega^(2)A`, its projection on X-axis is `A omega^(2) cos (omega t+phi)` and its direction is opposite to +X-axis hence INSTANTANEOUS acceleration of projected particle at time t is `a(t)= -A omega^(2) cos (omega t+ phi)` `therefore a(t) = -omega^(2) x(t)""[therefore A cos (omega t+ phi)= x(t)]` is the acceleration of SHM particle. |
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| 45. |
A person travelling on a straight line moves with a uniform velocity v_(1) for a distance x and with a uniform velocity v_(2) for the next equal distance. The average velocity v is given by |
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Answer» `V=(v_(1)+v_(2))/(2)` |
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| 46. |
Calculate the terminal velocity with which an air bubble of diameter 0.8 mm rises in a liquid of viscosity 0.250 Ns//m^(2) and density 0.95 x 10^(3)kg//m^(3). Density of air is 1.3 kg//m^(3) |
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Answer» SOLUTION :` ""a= 0.4 xx 10 ^(-3)m, sigma = 0.95 xx 10 ^(3) = 950 kg//m^(3),rho =1.3 kg//m^(3)` `"" eta = 0.250 Ns//m^(2), "" v= ? ` ` "" v= (2(rho -sigma) a^(2)g)/( 9 eta)= ( 2xx (1.3 - 950 )xx(0.4 xx 10 ^(-3))^(2))/( 9xx 0.250 ) ` ` "" = - 0. 135 xx 10^(-3) m//s` |
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| 47. |
An astronaut orbiting in a spaceship round the earth, has centripetal acceleration 2.45m//s^(2). Find the height of the spaceship. (Take R = 6400 km). |
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Answer» Solution :As centripetal acceleration EQUALS to acceleration due to gravity at that height, then `a = g_(h) = (GM)/(r^(2)) = (GM)/((R+h)^(2)) = (gR^(2))/((R+h)^(2))` `rArr a. = (gR^(2))/((R+h)^(2)) = ((R)/(R+h))^(2) = (a)/(G) = (2.45)/(9.8) = (1)/(4)` `rArr (R)/(R+h) = (1)/(2) rArr R+h = 2R ""R = h = 6400 km` `THEREFORE` SPACESHIP is at a height 6400 km. |
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| 48. |
Figures (a)and (b) refer to the steady flow of a (non -viscous) liquid . Which of the two figures is incorrect ? Why ? |
| Answer» Solution :Figure (a)is incorrect . It is because of the fact that at the kink , the velocity of flow of LIQUID is LARGE and HENCE using the Bernoulli.s equation the PRESSURE is less . As a RESULT , the water should not rise higher in the tube where there is a king. This is shown in figure (b) . For this liquid may be incompressible . | |
| 49. |
A body rolled on ice witha velocity of 8 ms^(-1)comes to rest after traveling a distance of 4m for this situation which of the following pair is correct |
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Answer» acceleration =+8 `ms^(-2)` COEFFICIENT of friction =0.8 `v^(2)-u^(2)= 2 as` `0^(2)-8^(2)=2a xx 4` `-64 =8a` `a= (-64)/(8) =-8ms^(-2)` `a= |-8| = 8 ms^(-2)` COEFFICIENTOF friction`mu= (f )/(R )=- (ma)/( mg)` `(a)/(G) = (8)/(9.8)= 0.816` |
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| 50. |
For an object moving with uniform acceleration, travelling 50m in 5^(th) sec, 70m in 7^(th) sec. a) Its initial velocity is 5m//s b) Its acceleration is 20m//s^(2) c) Its travels 100 m in 9th sec d) Its average velocity during 9th sec is 90m//s |
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Answer» a is true |
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