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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 ring of mass M and radius R is rotating about its axis with angular velocity omega. Two identical bodies each o mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss willbe |
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Answer» `(m(M+2m))/(M)OMEGA^(2)R^(2)` |
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| 2. |
In the x-t plot of a particle in one dimensional motion. Four different equal intervals of time are shown. In which interval, average speed is greatest |
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Answer» 1 |
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| 3. |
Find the position of centre of mass of the system of 3 objects of masses 1 kg, 2kg and 3 kg located at the corner of an equilateral triangle of side 1 m. Take 1 kg mass object at the origin and 2kg along x-axis. |
Answer» Solution : `X_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))` `rArr X_(cm)=(1xx0+2xx1+3xx(1)/(2))/(1+2+3)rArr X_(cm)=(7)/(12)m` `Y_(cm)=(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))` `rArr Y_(cm)=(1xx0+2xx0+3xx(sqrt(3))/(2))/(1+2+3)rArr Y_(cm)=(sqrt(3))/(4)m` `THEREFORE` Co-ordinates of CENTRE of mass `(x_(cm), y_(cm))` `=((7)/(12)m, (sqrt(3))/(4)m)` |
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| 4. |
The process of measurement is basically a …………… |
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Answer» |
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| 5. |
In kinetic theory of gases, which of the following statements regarding elastic collisions of the molecules is wrong |
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Answer» KINETIC energy is LOST in collisions |
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| 6. |
In the system shown in figure A_(0) and 4A_(0) are the area of the pistons of masses m and 2m respectively and the system is in a gravity free space. If one of the piston is displaced slightly by an amount Deltax then the time period of the resulting oscillations is 2pi sqrt((xm)/(yk)). (spring constant are k and 2k) and consider the mass of liquid to be negligible as compared to the masses of the pistons and the piston can slide smoothly without the liquid spilling out.The value of y-x is _________ |
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Answer» `(7Rx)/(4) = (3ma)/(2)` `rArr ((7K)/(6M)) x = a` |
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| 7. |
The velocity of a particle moving in the position direction of the X-axis varies as v=K sqrtS where K is a positive constant. Draw V-t graph. |
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Answer» Solution :`V=K sqrts` `(d S)/(dt)=K sqrtS""therefore underset(0)overset(s)int (d S)/sqrtS =underset(0)overset(t)int K dt` `2sqrtS=Kt and S=1/4 K^(2) t^(2)` `RARR V=(d S)/(dt)=1/4 K^(2)2t+1/2 K^(2)t`  `therefore V ALPHA t` The V-t GRAPH is a straight LINE passing through the origin. |
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| 8. |
Three equal weights of mass m each are hanging on a string passing over a fixed pulley as shown In Fig . What are the tensions in the string connecting weights A to B and B to C ? |
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Answer» |
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| 10. |
A circular racentrack of radius 300 m is banked at an angle of 15^(@). If the coefficient of friction between the wheels of a race - car and the road is 0.2, what is the (a) optimum speed of the race - car to avoid wear and tear on its tures, and (b) maximum. Permissible speed to avoid slipping ? |
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Answer» Solution :On a banked road, the horizontal component of the normal force and the frictional force contribute to PROVIDE centripetal force to keep the car moving on a circular turn without slipping. At the OPTIMUM speed, the normal reaction.s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed `upsilon_(0)` is given by `upsilon_(0)=(Rg tan THETA)^(1//2)` Here `R = 300 m, theta = 15^(@), g = 9.8 ms^(-2)` `upsilon_(0)=(R g tan theta)^(1//2)=(300xx9.8xx tan 15^(@))^(1//2)` On simplification, `upsilon_(0)=28.1 ms^(-1)` The maximum PERMISSIBLE speed `upsilon_(max)` is given by `v_(max)=sqrt((Rg(tan theta + MU))/((1-mu tan theta)))` `= sqrt((300xx9.8xx(tan 15^(@)+0.2))/((1-0.2 tan 15^(@))))` On simplification `upsilon_(max)=38.1 ms^(-1)` |
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| 11. |
A body falling freely from rest from the top of a tower describes 60.1m in the last second of its fall. Find the height of the tower. |
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Answer» |
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| 12. |
Which of the following quantities if zero on an average for the molecules of an ideal gas in equilibrium |
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Answer» KINETIC energy |
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| 13. |
Two waves of same frequency and intensity superimpose on each other in opposite phases.After the superposition the intensity and frequency of waves will. |
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Answer» decreases |
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| 14. |
Give the magnitude and direction of the net force acting on a kite skillfully held stationary in the sky |
| Answer» Solution : As the KITE is held STATIONARY , net force on the kite , in accrodance with Newton,s FIRST LAW,. | |
| 15. |
What are the degrees of freedom in the following cases: |
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Answer» Solution :(i) A particle MOVES along a straight line (i) One, translational. (ii) A particle is projected in space. (ii) Two, translational (iii) MOTION of fan (iii) One, rotational. (iv) Motion of ant. (iv) Two, translational (v) Motionn of mosquito or fly (no ROTATION) (v) THREE, translational. |
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| 16. |
The time for a particle executing SHM with frequency 1Hz is considered from initial phase of pi//4. The time at which potential energy is three times of kinetic energy is (in sec): |
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Answer» `1/12` |
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| 17. |
A uniform gets tube closed at both ends, encloses air columns of equal lengths 5cm,when the tube is placed horizontally.The pressure of the enclosed air is p.When the tube is placed at 60^@ to the vertical the upper and the lower columns of air are of lengths 46 cm and 44.5 cm respectively.Find the value of p.temperature of the system remains constant at 30^@C. |
Answer» SOLUTION :Let the first case length of air column on each side of the mercury thread=l[Fig.6.11] Again length of the tube =44.5+46+5=95.5 cm  `thereforel=(95.5-5)/2=45.25cm` In the secon case `p_2=p_1+5cos60^@=p_1+5/2` `=(p_1+2.5)cmHg` If in this case, `V_1` and `V_2` are the volume of the upper and the lower air columns in the tube, then according to Boyle.s LAW, `p_1V_1=p_2V_2` or, `p_1=46a.=(p_1+2.5) TIMES 44.5a` [where a=cross sectional area of the tube] Solving we GET `p_1=74.17 cm` Again, `pV=p_1V_1` or, `p=(p_1V_1)/V=(74.17 times46a)/(45.25a)=75.4cm` i.e.,`p=75.4cmHg` |
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| 18. |
Two bodies of masses 4kg and 6kg are attached to the ends of a string which passes over a pulley, the 4kg mass is attached to the table top by another string.The tension in this string T_(1) is equal to |
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Answer» 10N |
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| 19. |
(A) : If a physical quantity is a vector, it must have direction. ( R) : Electric Current has a direection therefore it is a vector quantity. |
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Answer» Both (A) and ( R) are TURE and ( R) is the correct explanation of (A) |
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| 20. |
Fundamental unit out of the following is |
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Answer» ampere |
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| 21. |
The mass of moon is 1 % of mass of earth. The ratio of gravitational pull of earth on moon and that of moon on earth will be |
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Answer» `1:1` |
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| 22. |
The orbital speed of Jupiter is ......... |
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Answer» greater than the orbital SPEED of Earth |
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| 23. |
Choose the motion in two dimension from the following. |
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Answer» Motion of a train along a STRAIGHT railway track |
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| 24. |
A uniform circular disc of moment of inertia 7.35 xx 10^(-3)kg m^(2)is suspended by a wire of length 1 m and executes torsional oscillations. The couple per unit twist of the wire is 2 xx 10^(-2)N/m. Calculate the period of oscillation ? |
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Answer» |
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| 25. |
The times of flight of two projectiles aret_1 and t_2. If R be the horizontal range of each of them, then |
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Answer» `t_1t_2propR` |
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| 26. |
If vec(A)=2hat(i)+3hat(j)+4hat(k)and vec(B)=hat(i)+2hat(j)+3hat(k). The value of (2vec(A)-vec(B))*(vec(A)+2vec(B)) is |
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Answer» 30 |
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| 27. |
The natural agular frequency of a particle of mass m attached to an ideal spring of force constant K is |
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Answer» `SQRT(K/m)` |
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| 28. |
A radio signal sent towards the distant planet, returns after "t" s. If "c" is the speed of radio waves then the distance of the planet from the Earth is............... |
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Answer» `C t/2` |
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| 29. |
How can you detect whether the motion of a liquid is streamline or turbulent ? |
| Answer» Solution :Let a liquid flow through a NARROW PIPE of UNIFORM cross SECTION. A coloured solution is injected along the axis of the pipe at the point of entry.If this coloured solution flows like a thread along the axis of the pipe , then it is in streamline motion. The coloured solution spreads all over the liquid in the case of turbulent motion. | |
| 30. |
From the top of a tower two bodies are projected with the same initial speed of 40ms^(-1), first body vertically upwards and second body vertically downwards. A third body is freely released from the top of the tower. If their respective times of times of flights are T_(1),T_(2)andT_(3) identify the correct descending order of the times of flights |
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Answer» `T_(1),T_(2),T_(3)` |
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| 31. |
A satellite is in a circular orbit around a planet. Its period of revolution is T, radius of the orbit is R, orbital velocity V and acceleration 'a', then |
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Answer» V = at and `a = (V^(2))/(R)` |
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| 32. |
The displacement of a body of mass 2 kg varies with time 't' as S= t^(2)+2t, where S is in meters and 't' is in seconds. The work done by all the foces acting on the body during the time interval t= 2s" to "t= 4s is |
| Answer» Answer :B | |
| 33. |
If polar ice caps melt, the length of the day |
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Answer» will increase |
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| 34. |
(A): A man who falls from a height on a cement floor receives more injury than when he falls from the same height on a heap of sand. (R ): The impulse applied by a cement floor is more than the impulse by sand floor. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A's |
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| 35. |
Which of the following pairs is related as in work and force? |
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Answer» ELECTRIC POTENTIAL and Electric intensity |
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| 36. |
Explain with an example that the magnitude of displacement may be zero but the corresponding path length is not zero. |
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Answer» <P> Solution : For EXAMPLE, if the CAR starts from O, GOES to P and then returns to O. The final POSITION coincides with the initial position and the displacement is zero. However, the path length of this journey is OP + PO = 360 m + 360 m = 720 m. Thus, the magnitude of displacement for a course of motion may be zero but the corresponding path length is not zero. |
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| 37. |
The vector product of two vectors overset rarr(A) and overset rarr(B) is another………..whose magnitude is equal to…………..and………….between them. |
| Answer» Solution :vector `overset rarr(C)` , vector of magnitude of `overset rarr(A) and overset rarr(B)`, SINE of smaller angle `theta`. | |
| 38. |
A 1kgstone at the end of a 1 m long string is whirled in a vertical circle at a constant speed of 4ms^(-1). The tension in the string is 16 N, when the stone is at (g = 10 ms^(-2)) |
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Answer» The bottom |
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| 39. |
A ray incident at an angle of incidence 60^@ enters a glass sphere of refractive index mu=sqrt3. This ray is reflected and refracted at the further surface of the sphere. The angle between reflected and refracted rays at this surface is |
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Answer» SOLUTION :`mu= (sin i)/ (sin r)` `therefore sin r= (sin i)/mu= (sin 60^@)/sqrt3=1/2` `therefore r=30^@, PC=QC, therefore angleCPQ=anglePQC=angler=30^@` ANGLE between refected RAY QR and refracted ray QS at the other face `=180^@-r-60^@=180^@-30^@-60^@=90^@` 
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| 40. |
Assertion:A planet is a heavenly body revolving around the sun.Reason:Star is a self-luminous body made up of gaseous material.Select the correct statement from the following statements. |
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Answer» Assertion and REASON are TRUE and reason is the CORRECT explanation of assertion. |
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| 41. |
Two satellites of masses m_(1)andm_(2)(m_(1)gtm_(2)) are moving around the earth in orbits of radii r_(1)andr_(2)(r_(1)gtr_(2)). Which one of the following statements about their velocities is correct? |
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Answer» `v_(1) =v_(2)` |
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| 42. |
A block of mass .m. moving with a speed .upsilon. compresses a spring through a distance .x. before its speed reduces by 25%. The spring constant of the spring is |
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Answer» `(m upsilon^(2))/(4 x^(2))` |
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| 43. |
It is a common observation that rain clouds can be at about a kilometer altitude above the ground. (a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h (g = 10 m//s^(2)). (b) A typical rain drop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground. (c) Estimate the time required to flatten the drop.(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you. (e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through.) |
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Answer» Solution :Given, h = 1 km = 100 m `g = 10 m //s ^(2)` (a) Speed of drop in free FALL, `v ^(2) = 2 gh` `therefore v = sqrt (2gh)` (b) DIAMETER of the drop, `d =4 mm` `therefore ` Radius of the drop, `r = 2 mm = 2 xx 10 ^(-3) m` Mass of a rain drop, `m = rho xx v ` `=4/3 pi r ^(3) rho = 4/3 xx (22)/(7) xx (2 xx 10 ^(-3)) ^(3) xx 10 ^(3) ` `(because rho = 10 ^(3) kg//m ^(2))` `~~ 3.4 xx 10 ^(-5) kg` Momentum of the rain drop, `= p = mv` `= 3.4 xx 10 ^(-5) xx 100 sqrt2` `=4.7 xx 10 ^(-3) kg m//s` `~~5 xx 10 ^(-3) kgm//s` (c) Time REQUIRED to flatten the drop = time taken by the drop to travel the distance equal to the diameter near the ground, `t = (d)/(v)= (4 xx 10 ^(-3))/(1 00 sqrt2) = 0.028 xx 10^(-3) s` `=2.8 xx 10 ^(-5) s ~~ 30 ms` (d) Force exerted by a rain drop, `F = ("Change in momentum")/("Time") = (Delta p )/(t) = (p -0)/(t) = (4.7 xx 10 ^(-3))/(2.8 xx 10 ^(-5)) ~168 N` (e) Radius of the unbrella `=R = 1/2 m` `therefore `AREA of the unbrella, `A = pi R ^(2) = (22)/(7) xx ((1)/(2)) ^(2)= (22)/(28) = (11)/(14) =0.8 m ^(2) ` Number of drops striking the umbrella, `= ("Area of umbrella")/("seperation")= (0.8)/((5 xx 10 ^(-2))^(2)) = 320` `therefore ` Net force `= 320 xx 168 = 53760 N = 54000 N` |
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| 44. |
A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length . Which one of the two objects gets to the bottom of the plane first ? |
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Answer» Both reach at the same time `t = sqrt((2l (1 + (k^2)/(R^2)))/(g sin theta))` SINCE g is constant and l , R and `sin theta` are same for both `therefore (t_(d))/(t_(s)) = (sqrt(1 + (k_(d)^(2))/(R^(2))))/(sqrt(1 + (k_(s)^(2))/(R^(2)))) = sqrt((1 + (R^2)/(2R^2))/(1 + (2R^2)/(5R^2))) ( because k_(d) = (R)/(sqrt2) , k_(s) = sqrt((2)/(5)) R)` `= sqrt((3)/(2) xx (5)/(7)) = sqrt((15)/(14)) implies t_(d) gt t_(s)` Hence , the sphere gets to the bottom first . |
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| 46. |
A body is sliding down an inclined plane having coefficient of friction 1/3. If the normal reaction is three times that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is |
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Answer» `TAN^(-1) (1/2)` |
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| 48. |
On a hot day, a car is left in sunlight with all the windows closed. After sometimes, it is found that air inside the car isconsiderably warmer than the air outside. Explain why? |
| Answer» Solution :Glass TRANSMITS 50% of HEAT RADIATION coming from a HOT source like sun but does not ALLOW the radiation from moderately hot bodies to pass through it. | |
| 49. |
Calculate the efficiency of a Carnot engine operating between 450 K and 300 K. If this engine takes 5,00,000 J from the surce, how much work does it perform ? |
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| 50. |
Liquid is flowing throgh a tube of nonuniform cross section ratio of the radius at entry a and exit end B of the tube is 3 :2. Then the ratio of velocities at entry A and exit B of liquid is …… |
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Answer» `4:9` `(v_(A))/(v_(B))=?` According to EQUATION of CONTINUITY, `A_(A)v_(A)=A_(B)v_(B)` `(v_(A))/(v_(B))=(A_(B))/(A_(A))=(pir_(B)^(2))/(PIRA^(2))` `therefore(v_(A))/(v_(B))=((r_(B))/(r_(A)))^(2)=((2)/(3))^(2)` `(v_(A))/(v_(B))=(4)/(9)` |
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