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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. |
A particle is oscillating according to the equation x=5cos(0.5pit) where t is in seconds. The particle moves from the position of equilibrium to the position of maximum displacement in time……………… |
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Answer» 1S |
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| 2. |
(A): Collision between two fundamental particles is elastic (R ): Fundamental particles lose their shape in collision |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A's |
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| 3. |
A vector has x component of -25.0 units and y component of 40.0 units find the magnitude and direction of the vector. |
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Answer» SOLUTION :Consider a vector `vecA = A_("x")HAT(i)+A_("y")hat(j)` `:.vecA = -25hat(i)+40hat(j)` `|vecA|=sqrt(A_("x")^(2)+A_("y")^(2))=sqrt((-25.0)^(2)+(40.0)^(2))` `=47.16~~47.2` UNITS. and `tanalpha=(A_(y))/(A_(x)) = (40.0)/(-25.0)=1.6` `:. alpha = TAN^(-1)(-1.6)=-58.0^(0)` with -ve x-axis. (This is in clock WISE direction). This is equivalent to `(122^(@))` in anticlockwise direction with the x-axis. |
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| 4. |
One petametre is equal to |
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Answer» `10^15 m` |
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| 5. |
The net external torque on a system of particles about an axis is zero. Which of the following are compatile with it? (a) The forces may be acting radially from a point on the axis (b) The forces may be acting on the axis of rotation (c )The force may be acting parallel to the axis of rotation (d) The torque caused by some forces may be equal and opposite to that caused by other forces. |
| Answer» Answer :D | |
| 6. |
The measured mass and volume of a body are 53.63 g and 5.8 cm^3 respectively, with possible errors of 0.01 g and 0.1 cm^3. The maximum percentage error in density is about |
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Answer» 0.002 |
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| 7. |
सिद्ध कीजिए कि |vecA xx vecB|^(2) + |vecA * vecB|^(2) = (AB)^(2). |
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Answer» zero `:.|vec(A)xxvec(B)|=ABsinthetaand|vec(A).vec(B)|=ABc ostheta` `:.|vec(A)xxvec(B)|^(2)+|vec(A).vec(B)|^(2)=(ABsintheta)^(2)+(ABcostheta)^(2)` `=A^(2)B^(2)sin^(2)cos^(2)theta=A^(2)B^(2)` |
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| 8. |
The coolant in a chemical or nuclear plant (i.e. the liquid used to prevent different parts of a plant from getting too hot) should have hight specific heat, comment? |
| Answer» Solution :So, that it ABSORBS more heat with comparatively small CHANGE in temperature and EXTRACTS large amount of heat. | |
| 9. |
A particle if fired vertically upward with a speed of 15 km/s from the surface of earth. The speed with which it moves in intersteller space is - (Take escape speed from surface of earth as 11.2 km/s) |
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Answer» ZERO |
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| 10. |
The angular momentum of a rotating body is increased by 20 %. What will be the increase in its rotational kinetic energy? |
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Answer» Solution :KINETIC ENERGY K.E=`(L^(2))/(2L),E ALPHA L^(2),(E_(1))/(E_(2))=(L^(2))/(L_(2)^(2)),L_(1)=L,L(2)=L+(20L)/(100)=(120L)/(100)=(120L)/(100)=(12L)/(10)` `therefore (E_(1))/(E_(2))=L^(2)/((12L)/(10))=(L^(2)xx100)/(144L^(2)),(E_(1))/(E_(2))=(100)/(144)therefore(E_(2))/(E_(1))=(144)/(100)` `(E_(2)-E_(1))/(E-(1))=(144-100)/(100)=(44)/(100)therefore `% increase in rotaional kinetic energy =44 % |
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| 11. |
(A) For any collision, momentum transfer takes place (B) In any collision, energy transfer takes place. (C ) Impulsive forces are involved in collisions. |
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Answer» A and C are true |
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| 12. |
A thermodynamics process is carried out from an original state D to an intermediate state E by the linear process shown in figure, the total work is done by the gas from D to E to F is |
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Answer» 100 J |
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| 13. |
A vessel of volume 3V contains gas at pressure 4 P, and another vessel of volume 2V contains the same gas at pressure 1.5 P_0 , Both vessels have same temperature. When both vessels are connected by a tube of negligible volume, the equilibrium pressure is nP_0//2, where n is an integer, Find the value of n. |
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Answer» |
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| 14. |
Gulab Jamuns' (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twise big (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following: |
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Answer» Both size gulab jamuns will GET HEATED in the same time. |
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| 15. |
A steel wire of length 1.5m and diameter 0.25cm is loaded with a force of 98N The increase in length of the wire is 1.5xx10^-4m.Calculate the tensile stress and the fractional change in length of the wire. |
| Answer» SOLUTION :L=1.5m, r=d/2 =`0.125xx10^-2 m,F=98N, deltaL=1.5xx10^-4m`STRESS `=F/A = F/pi r^2 =98//3.14xx(0.125xx10^-2)^2=1.99xx10^7Nm^-2 `FRACTIONAL CHANGE in LENGTH, `deltaL//L= 1.5xx10^-4//1.5 =10^-4` | |
| 16. |
The error in the measurement of the length of the simple pendulum is 0.2% and the error in time period 4%. The maximum possible error in measurement of L//T^(2)is |
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Answer» `4/2%` |
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| 17. |
Are AU and A are equal unit of length ? |
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Answer» SOLUTION :No. `1AU=1.496xx10^(11) m and ` `1Å=10^(-10) m` |
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| 18. |
What is (i) dry friction and (ii) fluid friction ? |
| Answer» SOLUTION :FORCE of friction between two solids is CALLED DRY friction and that between a liquid and any body is called FLUID friction. | |
| 19. |
A rope of length 10m and linear density 0.5kg/m is lying length wise on a smooth horizontal floor. It is pulled by a force of 25 N. The tension in the rope at a point 6m away from the point of application is |
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Answer» Solution :`"Force on unit length" =(F)/(L),` `"Force on l UNITS of length "=((F)/(L))l` `"TENSION at "(L-l)" length "=(F)/(L)(L-l)` |
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| 20. |
In a thermodynamics process which one of the following statement is not true ? |
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Answer» In an ISOCHORIC PROCESS pressure remains CONSTANT. |
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| 21. |
The distance between two particles of masses m_(1)andm_(2) is r. If the distance of these particles from the centre of mass of the system are r_(1)andr_(2) respectively, then show that r_(1)=r((m_(2))/(m_(1)+m_(2)))andr_(2)=r((m_(1))/(m_(1)+m_(2))) |
Answer» Solution : Position vector of c.m. `VEC(r)_(cm)=(m_(1)vec(r_(1))+m_(2)vec(r_(2)))/(m_(1)+m_(2))""....(1)` Taking c.m. at the origin `vec(r_(cm))=vec0` `therefore 0=m_(1)vec(r_(1))+m_(2)vec(r_(2))[because" From eq. " (1)]` Now c.m. is on the LINE joining `m_(1)andm_(2)` and the directions of `vec(r_(1))andvec(r_(2))` are mutually opposite. `therefore 0=-m_(1)r_(1)+m_(2)r_(2)` `therefore m_(1)r_(1)=m_(2)r_(2)""......(2)` `therefore m_(1)(r_(1))=m_(2)(r-r_(1)) [because r=r_(1)+r_(2)]` `therefore m_(1)r_(1)=m_(2)r-m_(2)r_(1)` `therefore (m_(1)+m_(2))r_(1)=m_(2)r` `therefore r_(1)=r[(m_(2))/(m_(1)+m_(2))]` Again taking `r_(1)=r-r_(2)" in "m_(1)r_(1)=m_(2)r_(2)` `therefore m_(1)(r-r_(2))=m_(2)r_(2A)` `therefore m_(1)r-m_(1)r_(2)=m_(2)r_(2)` `therefore m_(1)r=(m_(1)+m_(2))r_(2)` `therefore r_(2)=r[(m_(1))/(m_(1)+m_(2))]` |
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| 22. |
No physicist has ever "seen" an electron, yet, all physicists believe in existence of electrons . An intelligent but superstitious man advancesthis analog to argure that 'ghosts' exist even though no one has "seen"one . How will you refute his argument ? |
| Answer» SOLUTION :Many phenomena which depend UPON the existence of ELECTRONS have been PREDICTED and actually observed in everyday life. There is no phenomenon which can be explained on the basis that ghosts exist though they are not seen. So, obviously, the COMPARISON between two situations does not make any sense. | |
| 23. |
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 pi rad/s. Which of the two will start to roll earlier ? The coefficient of kinetic friction is mu_(k)=0.2 |
| Answer» Solution :The DISC would begin to ROLL earlier than the ring | |
| 24. |
A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion. 1. y= asin""(2pit)/(T) 2. y= asinvt 3. y= (a)/(T)sin""(t)/(a) 4. y= (asqrt(2))(sin((2pit)/(T))+ cos((2pit)/(T))) where a is the maximum displacement of the particle, v is the speed of the particle, T is the time period of motion. Then dimensionally |
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Answer» 1 and 2 are WRONG |
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| 25. |
Energy required to move a body of mass m from an orbit of radius 2R to 3R is.......... |
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Answer» `(GMm)/(12R)` `U_2 =-(GM_em)/(3R)` `:.` Change in potential energy `=U_2 -U_1` `(GMm)/(3R)-(-(GMm)/(2R))` `=(GMm)/R [((-2)+3)/6]` `=(GMm)/(6R)` |
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| 26. |
An organ pipe, open at both ends produces 5 beats per second. When vibrated with a source of frequency 200 Hz. The second harmonic of the same pipe produces 10 beats per secondwith a source of frequency 420 Hz. The frequency of source is : |
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Answer» 205 Hz `f=\200+-135 Hz (or) 205 Hz` second harmonic of open pipe. `2f=420+-10=410 Hz (or) 430 Hz` (or) `f=205 Hz (or) 215 Hz` The COMMON frequency is 205 Hz. |
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| 27. |
Moment of inertia of a disc of radius R about a diametric axis is 35 "kg m"^(-2). The MI of the disc about a parallel axis at a distance R//2 from the centre is |
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Answer» `31.25 "kg-m"^(2)` `I=I_(V)+md^(2)=(ma^(2))/(2)+md^(2)""(because a=R,d=R)` `rArr` Moment of inertia `= (3)/(2) ma^(2)=(3)/(2)(25)` `:.` Moment of inertia `=(75)/(2)=37.5 "kg m"^(-2)`. |
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| 28. |
The spectral line emitted by a star, known to have a wavelength of 6500 Å.What is the speed of the star in the line of sight relative to the earth ? Is the star approaching or receding ? Speed of light in air=3xx10^(8) ms^(-1). |
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Answer» |
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| 30. |
Obtain the relation between angular momentum of a particle and torque acting on it. |
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Answer» Solution :The TOTAL angular moment of a system of particles is the angular moment of individual particles. For a systm of n particles, `vecL=vec(l_(1))+vec(l_(2))+vec(l_(3))+....vec(l_(n))` `=underset(i=1)overset(n)sumvec(l_(i))` where `i=1,2,3,….n` and `vec(l_(i))=vec(r_(i))xxvec(p_(i))` where `vec(r_(i))` is the position vector of the `i^(th)` particle with RESPECT to a given origin and `vec(p_(i))` is the linear momentum of the particle. The total angular momentum of the system of particle as `therefore vecL=underset(i=1)overset(n)sumvec(l_(i))=underset(i=1)overset(n)sumvec(r_(i))xxvec(p_(i))` Differentiating w.r.t. time `therefore (dvecL)/(dt)=underset(i=1)overset(n)sumvec(tau_(i))....(1) [because vec(r_(i))xxvec(p_(i))=vec(tau_(i))]` `therefore vectau=underset(i=1)overset(n)sumvec(r_(i))xxvec(F_(i))[because vectau=vecrxxvecF]` Here, `vec(F_(i))` is the force on the `i^(th)` particle is the vector sum of external forces `F_("iext")` acting on the particle and the INTERNAL forces `F_("i(int)")` exerted on it by the other particles of the system. `therefore vec(F_(i))=vecF_(i("ext"))+vecF_(i("int"))` `therefore vectau=vectau_(("ext"))+vectau_(("int"))` where `vectau_((ext))=sumvec(r_(i))xxvecF_(i(ext))` and `vec(tau)_("(int)")=sumvec(r_(i))xxvecF_(i("int"))` but the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joiningthe two particles. The contribution of the internal forces to the total torque on the system is zero. So `vectau_("(int)")=0` and therefore `vectau=vectau_("(ext)")` From eqn. (1), `(dvecL)/(dt)=vectau_(("ext"))` Hence, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques EQUATION `(dvecL)/(dt)=vectau_(("ext"))` is the rotational ANALOGUE of `(dvecp)/(dt)=vec(F_(ext))`. |
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| 31. |
A particle of mass 'm' is attached to three identical springs A, B and C each of force constant 'k' as shown in figure. If the particle of mass 'm' is pushed sightly against the spring 'A' and released. Find the period of oscillation |
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Answer» Solution :When the particle of mass .m. at .0. is pushed by .y. in the direction of A, SPRING .A. will be compressed by `y. = y cos 45^@ ` , so the TOTAL restoring force on the mass .m. along .A0. is `RF = F_A + F_B cos 45 + F_c cos 45 = ky + 2(ky^1) cos 45` `=ky + 2k (ycos 45) cos 45` `F=-k^1 y ` with `k^1 = 2k impliesT = 2pi sqrt((m)/(k^1)) = 2pi sqrt((m)/(2k))` |
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| 32. |
What is the angular velocity of (a) a second hand and (b) minute hand of a clock ? |
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Answer» |
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| 33. |
A uniform rod of length l is kept vertically on a horizontal smooth surface at a point O. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain |
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Answer» at `O` |
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| 34. |
Two walls of thickness in the ratio 1:3 and thermal conductivities in the ratio3:2 from a composite wall of a builiding . If thefree surfaceof the wall be at temperatureof 30^(@)C and 20^(@)Crespectively , whatis thetemperature of theineterface ? |
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Answer» Solution :`R_(1)=(l_(1))/(K_(1)A)= (3l)/(KA) and R_(2) = (l_(2))/(K_(2)A) = (2L)/(3KA)` Effective thermal resistance`R = R_(1) +R_(2) = (11)/(3) (l)/(KA)` Now , thermalcurrent `i=(Delta T)/(R) = (30-20)/((11l)/(3KA)) = (30KA)/(11l)` ALSO , themal currentthrough the first slal will be ` i = ((30 -T))/(R_(1))` (where T = temperature of theinterface) i.e., `(30KA)/(11l) = ((30-T)KA)/(3l) rArr 90-330T rArr T = 21.82^(@)C` |
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| 35. |
Which two physical quantities remains constant in law of conservation of momentum? Which quantity remain zero? |
| Answer» SOLUTION :Linear MOMENTUM and centre of MASS REMAINS constant and ACCELERATION of centre of mass remains zero. | |
| 36. |
Explain how the interference of waves is formed. |
Answer» Solution :Interference is a phenomenon in which two waves superimpose to form a resultant wave of GREATER,lowe or the sameamplitude. Let us consider two harmonic waves having identical frequencies constant phase difference `varphi` and same dwave form (can be trated as coherent source), but having amplitudes `A_(1) and A_(2)` then `y_(1)=A_(1)s in(kx-OMEGAT) ""....(1)` `y_(1)=A_(1)s in(kx-omegat+varphi) ""....(2)` Suppose they move simultaneously in a particular direction, then interference occur (i.e., overlap of these two waves). Mathematically `y=y_(1)+y_(2) ""...(3)` Hennce by substituting equation (1) and equation (2) in equation (3) we get `y={{:(A_(1)sin(kx-omegat)),(+A_(2)sin(kx- omegat+varphi)):}` USING trigonometric IDENTITY `sin (alpha+beta)=(sin alpha COS beta+ cos alpha sin beta)`, we get `y={{:(A_(1)sin(kx-omegat)),(+A_(2)sin(kx- omegat+varphi)),(+cos(kx-omega t) sin varphi"]"):}` `{{:(sin (kx- omega t)(A_(1)+A_(2)cos varphi)),(+A_(2)sin varphi cos(kx-omegat)""....(4)):}` Let us re-define `Acos theta=(A_(1)+A_(2)cos varphi) ""...(5)` and `A sin theta=A_(2) sin varphi ""....(6)` Then equation (4) can be rewritten as `{{:(A sin (kx-omegat)cos theta),(+A cos (kx- omegat) sin theta):}` `{{:(A "("sin (kx-omegat)cos theta),(+sin theta cos (kx- omegat) sin theta):}` `A=A sin (kx- omegat+theta) ""...(7)` By squaring and adding equation (5) and (6), we get `A^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos varphi...(8)` Since intensity is square of the amplitude `(I=A^(2))`, we get `I=I_(1)+I_(2)+ 2sqrt(I_(1)I_(2)) cos varphi "" ...(9)` This means the resultant intensity at any point depends on the phase difference at that point. |
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| 37. |
A rain drop of radius 2 mm falls from a height of 500 m above the ground .It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height , it attains its maximum (terminal) speed, and moves with uniform speed thereafter .What is the work done by the gravitational force on the drop in the first and second half of its journey ?What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 1 ms^(-1) ? |
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Answer» Solution :Work DONE by gravitational force during first half height , W=mgh [ Where , `m =(rho) (4/3pir^(3))(g)(h)` `10^(3) xx4/3 xx3.14 xx(2xx10^(-3))^(3)xx9.8 xx250` `= 82058.66 xx10^(-6)` ` approx 0.082 J` The change in kinetic energy of drop during first half of motion, `DeltaK= 1/2mv^(2) -1/2 mv_(0)^(2) -1/2mv_(0)^(2) " " v=10ms^(-1) , v = 0 ` ` m = rho xx4/3 pir^(3)=3.35 xx10^(-5)` `=1/2 xx3.35 xx10^(-5) xx(10)^(2)` `= 0.00167 J` Work done by gravitational force during whole motion , = mgh ` = 3.35 xx10^(-3) xx9.8 xx500` `=0.16415 J` Work done by the resistive forces during whole motion = change in kinetic motion - work done by gravitatonal force ` = 0.001675 -0.1645` `= - 0.163 J` . |
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| 38. |
The number of signifigures in the numbers 4.8000 xx 10^(4) and 48000.50 are respectivelys |
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Answer» 5 and 6 |
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| 39. |
Doppler effect in sound is called asymmetric, Explain. |
| Answer» Solution :Observed frequency of the source moving towareds STATIONARY observer and vice versa are not EQUAL. For this reason it is CALLED ASYMMETRIC. | |
| 40. |
A person runs along a circular path of radius 5m. Find the magnitude of the displacement if he runs all the way around the circle ? |
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Answer» 10 m |
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| 41. |
The diameter of a solid solid disc is 0.5 m and its mass is 16 kg. What torque will increase its angular velocity from zero to 120 rotations/ minute in 8 seconds? |
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Answer» Solution :Given : diameter =0.5m, Mass `=16 kg` Formula: MOMENT of inertia of solid disc `I=1/2MR^(2)` `I=1/2 MR^(2)` `=1/2xx16xx[0.5/2]^(2)` `=1/2` ANGULAR VELOCITY, `OMEGA=(120xx2pi)/60=4pi` And change in angular momentum `DeltaL=I omega-0=I omega` Angular impulse `tauxxt=DeltaL` `impliestau=(DeltaL0/l=1/8xx1/2xx4pi` `=(pi)/4Nm` |
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| 42. |
Two circular loops of radii R and nR are made of same wire. If their M.I about their normal axis through centre are in the ratio 1 : 8, the value of n is |
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Answer» 6 |
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| 43. |
An object falling through a fluid is observed to have acceleration given by a = g - bv where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed ? |
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Answer» Solution :When speed becomes constant acceleration `a = (dv)/(dt) =0` `a = g - bv `(GIVEN) It is CLEAR from above equation as speed increases acceleration will DECREASE. At a certain speed say `v _(0),` acceleration will be zero and speed will remain constant. `implies a = g - B v _(0)` `therefore 0 =g - bv _(0)` `therefore v _(0) = (g)/(b)` |
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| 44. |
Assertion : A particle performing SHM at certain instant is having velocity v. It again acquire a velocity v for the first time after a time interval of T second. Then the time period of oscillation is T second. Reason : A particle performingSHM can have same velocity at two instants in one cycle. |
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Answer» If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. But `t_(PQ)` is not the time PERIOD of oscillation. 
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| 45. |
The minimum number of unequal forces in a plane that can keep a particle in equilibrium is |
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Answer» 4 |
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| 46. |
If energy E, velocity v and time T are chosen as th e fundamental quantities. Find dimensional formula of surface tension? |
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Answer» Solution :SURFACE Tension , `S =MT^(-2)` Energy , `E=ML^(2)T^(2)`, Velocity, `v=LT^(-1)`, Time ,`t=T^(1)` `:. S. alpha (E)^(a) (v)^(b)(t)^(c)`and `MT^(-2)=(ML^(2)T^(-2))^(a)(LT^(-1))(T)^(c)` Comparing POWER of M,L and T on either SIDE `a=1, 2a+b=0rArr b=-2` `-2a-b+c=-2 rArr c=-2` `:. S.=E^(1) (V)^(-2)(T)^(-2)` |
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| 47. |
A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure-6.30. A particle of mass m' is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this - particle due to the sphere and the shell if (a) r lt x lt 2r, (b) 2rlt x lt 2R "and"(c) xgt2R. |
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Answer» |
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| 48. |
If linear momentum of a bodyis increased by 50 % Its kinetic energy increasesby ……….. |
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Answer» 0.5 ` :. K prop p^(2)` ` :. (K_(2))/(K_(1)) = (P_(2)^(2))/(P_(1)^(2))` ` :. (K_(2)-K_(1))/(K_(1)) xx100 %= (P_(2)^(2)-P_(1)^(2))/(P_(1)^(2)) xx100 % ` ` = 5/4 xx100 % = 125 % ` |
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| 49. |
As astronaut orbiting the earth in a circular orbit 120 km above the surface of Earth, gently drops a spoon out of space-ship. The spoon will…………. |
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Answer» fall vertically down to the Earth |
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| 50. |
A car moving with constant acceleration covers the distance between two points 180 m apart in 6 sec. Its speed as it passes the second point is 45 m/s. What is its acceleration and its speed at the first point |
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Answer» `-5m//s^(2), 15m//s` |
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