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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. |
If R is radius of the earth, the height above the surface of the earth where the weight of a body is 36% less than its weight on the surface of the earth is |
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Answer» 4R/5 |
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| 2. |
(i) Three blocks A, B and C are placed in an ideal Atwood machine as shown in the figure. When the system is allowed to move freely it was found that tension in the string connecting A to C was more than thrice the tension in the string connecting A and B. The masses of the three blocks A, B and C are m1, m2 and m3, respectively. State whether the following statements are true or false [All masses have finite non zero values and the system has a non zero acceleration]. (i) m_(3) can have any finite value (ii) m_(1) gt 2m_(2) (ii) In an Atwood machine the sum of two masses is a constant. If the string can sustain atension equal to ((24)/(30))of the weight of thesum of two masses, find the least acceleration of the masses. The string and pulley are light. (iii) A load of w newton is to be raised vertically through a height h using a light rope. The greatest tension that the rope can bear isetaw (eta gt 1). Calculate the least time of ascent if it is required that the load starts from rest and must come to rest when it reaches a height h. |
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Answer» (b) True (II) `(G)/(5)` (iii)`sqrt((2etah)/((eta-1)g))` |
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| 3. |
In the measurement of volume of solid 5sphere using the formula V = 4/3 pi r^3if the error committed in the measurement of the radius r is 1.5%, the percentage error in the volume measurement is |
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Answer» 0.03 |
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| 4. |
Explain why: (a) A body with large reflectivity is a poor emitter (b) A brass tumbler feels much colder than a wooden tray on a chilly day. (c) An optical pyrometer (for measuring high temperature) calibrated for an ideal black body radiation gives too low value for the temperature of a red-hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace. (d) The earth without its atmosphere would be inhospitably cold. (e) Heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water. |
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Answer» Solution :(a) A body with large reflectivity is a poor ABSORBER of heat and poor absorbers of heat are poor emitters. (b) Because brass tumbler is a good conductor of heat, when we touch a brass tumbler on a chilly day, heat quickly flows from our body to the tumbler and it appears colder. On the other hand, wood is a bad conductor of heat so on touching, heat does not FLOW from our body to the wood and it appears comparatively hotter, (C) LET T be temperature of red-hot iron piece. When red-hot iron piece is in open having temperature of surroundings `T_0`, Heat RADIATED per second per unit area `E_1 = sigma (T^4 - T_0^4) `....(i) When body is in the furnace having temperature T, then heat radiated per second per unit area `E_2 = sigma T^4`.....(ii) From (i) and (ii),we have `E_1 lt E_2`So the optical pyrometer gives low value of the temperature of a body placed in open. (d) The heat radiated by the earth is reflected back by the atmosphere. If atmosphere is not present then all the heat will escape from the earth surface and the earth will be too cold to live. (e) Because steam at `100^@C` contains more heat than boiling water at `100^@C` . Secondly 1 gm of steam possess 540 calories more heat than 1 gm of water at `100^@C` . Thus heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water. |
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| 5. |
The pulley of Atwood.smachine has a moment of inertia .I. about its axis and its radius is R. Find the magnitude of acceleration of the two blocks assuming the string is light and does not slip on he pulley. |
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Answer» Solution :SUPPOSE the block of mass .M. goes down with an ACCELERATION .a.. The ANGULAR acceleration of the pulley is, therefore `alpha = (a)/(R )` `Mg-T_(1)=Ma` `T_(2)-mg=ma` `T_(1)R-T_(2)R=I alpha = L(a)/(R )` Solving the equations `a=((M-m)gR^(2))/(I+(M+m)R^(2))` 
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| 6. |
The radius of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be : |
| Answer» ANSWER :B | |
| 7. |
A 1kg ball moving at 12 ms^(-1) collides head on with a 2kg ball moving in the opposite direction at 24 m/s. The velocity of each ball after the impact, if the coefficient of restitution is 2/3, is |
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Answer» `-28 m//s , -4 m//s` |
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| 8. |
While rolling, the path of center of mass of an object is |
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Answer» straight line |
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| 9. |
Galileo’s law of odd numbers : “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. |
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Answer» Solution :Let us divide the time interval of motion of an object under free fall into MANY equal intervals `TAU`and find out the distancestraversed during successive intervals of time. Since initial velocity is zero, we have `y=-(1)/(2)g t^(2)` Using this equation, we can calculate the position of the object after different time intervals, `0, tau, 2tau, 3tau`... which are given in second column of Table 3.2. If we take `(-1//2)g t^(2)` as`y_(0)` - the position coordinate after FIRST time interval `tau`, then third column gives the positions in the UNIT of `y_(0)`. The fourth column gives the distances traversed in successive `tau s`. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. |
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| 10. |
White light is incident on the interface of glass and air as shown in the figure . If green light is just totally internally reflected then the emerging ray in air contains 1) yellow, orange, red 2) violet,indigo,blue 3) all colours 4) all colours except green |
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Answer» SOLUTION :Critical angle `theta_c= sin^-1(1/mu)` Wavelenght increases in the sequence of VIBGYOR. According to Caunchy.s formula refractive index `(mu)` decreases as the wavelength increases. HENCE, the refractive index will increase in the sequence ROYGBIV. THe critical angle `theta_c` will thus increase in the same order VIBGYOR . For green light the incidence angle is just EQUAL to the critical angle. For yellow, orange and RED the critical angle will be greater than the incidence angle. So, these colours will emerge from the glass air interface |
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| 11. |
A vector perpendicular to the vector (i + 2j) and having magnitude 3 sqrt5 units is |
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Answer» `3HAT(i) + 6 hat(i)` |
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| 12. |
A particle of mass m is projected with speed u at an angle theta with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point. |
| Answer» Solution :`tau =rF = R XX mg = (v_(0)^(2) sin 2theta//2g) xx mg = mv_(0)^(2) sin theta//2` | |
| 13. |
. A curve is drawn expressing the kinetic energy of a particle as a function of the distance traversed (on X-axis). The slope of this curve represents the instantaneous |
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Answer» VELOCITY |
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| 14. |
Subtraction of vectors obeys |
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Answer» COMMUTATIVE LAW |
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| 15. |
If two bodies of masses 3 gm and 5 gmare moving with linear momenta in the ratio 2 : 3, the ratio of their kinetic energies is |
| Answer» Answer :D | |
| 16. |
The masss of an athlete is 40 kg. He can cover a distance of 150m is maximum of 15 sec and minimum of 10 sec then the range of his kinetic energy if |
| Answer» Answer :A | |
| 17. |
A steel wire of 1mm diameter and of length 1m is stretched by applying force of 10N. If the increase in length is 0.064mm, find (i) the stress, (ii) the strain and (iii) the Young's modulus of the wire. |
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Answer» Solution :`R = 0.5 xx 10^(-3) m, L = 1m` `F = 10N, e = 0.064 xx 10^(-3) m` i) stress ` = F/A = (F)/(pir^2) = (10)/(pi (0.5 xx 10^(-3))^2) = 1.273 xx 10^7 NM^(-2)` ii) stress ` = e/L = (0.064 xx 10^(-3))/(1) = 0.064 xx 10^(-3)` iii) Y = stress/strain ` = (1.273 xx 10^7)/(0.064 xx 10^(-3)) = 1.989 xx 10^11 Nm^(-2)` |
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| 18. |
Two black bodies at temperautre 400K and 500K are placed in an evacuated enclosure whose wall are at 300K . Find the ratio of their rates of cooling. |
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Answer» Solution :HEAT LOST per SECOND by radiation E = `bar(SIGMA)(T^(4)-T_(0)^(4))` Ratio of the rates of cooling= `E_(1)/E_(2)` `E_(1)/E_(2)=((T_(1)^(4)-T_(0)^(4)))/((T_(2)^(4)-T_(0)^(4))` `T_(1)=400K, T_(2) = 500K, T_(0)= 300K` `E_(1)/E_(2)=(400^(4)-300^(4))/(500^(4)-300^(4))` `E_(1):E_(2)=175:544` |
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| 19. |
A charge particle moves towards another charged particle. Under what conditions the total momentum and the total energy of the system conserved ? |
| Answer» Solution :Coulomb force is ACTING in between the charged particles.Internalforce is a conservation force. If no external force of the work done by external forces were then the mechanical energy of the system and ALSO a lite mementum also REMAINS CONSTANT. | |
| 20. |
The area under the force, time graph is |
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Answer» momentum |
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| 21. |
(Figure 5.196) shows an elevator cabin, which is moving downwards with constant acceleration a . A particle is projected from corner A, directly towards diagonally opposite corner C. Then prove that (a) particle will hit C only when a = g. (b) Particle will hit the wall CD if a lt g. ( c) Particle will hit the roof BC if a gt g. . |
| Answer» Solution :Superimpose an upward acceleration `a` on the SYSTEM. The box becomes STATIONARY. The particle has an upward acceleration `a` and a downward acceleration `G. If a = g`, the particle has no acceleration and will hit `C`. If `a GT g`, the particle has a net upward acceleration, and if `a lt g`, the particle has a net downward ACCLERATION. | |
| 23. |
The stress-strain graphs for two materials are shown in figure. (Assume same scale). |
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Answer» Material (ii) is more elastic than material (i) and HENCE material (ii) is more brittle. |
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| 24. |
A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one rod. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed V_(0) strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth of the mass of the rod. The final angular velocity of the rod is omega and the ratio of the kinetic energy of the system after collicion to the kinetic energy of the bullet before collision is 1/x where x is |
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Answer» |
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| 25. |
Inside a uniform spherical shell (a)the gravitational potential is zero (b)the gravitational field is zero (c)the gravitational potential is same everywhere (d)the gravitational field is same everywhere |
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Answer» only a and B are TRUE |
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| 26. |
A big size balloon of mass M is held stationary in air with the help of a small block of mass M//2 tied to it by a light string such that both float in mid air. Describe the motion of the balloon and the block when the string is cut. Support your answer with calcutations |
Answer» SOLUTION :Free block DIAGRAM of balloon and block shown below :  When the balloon is held stationary in air, the forces acting on it get BALANCE Up thrust = Wt. of Balloon + Tension in string U = Mg + T M for the small block of MASS`(M)/(2)` floating stationary in air `T=(M)/(2)` g `:. "" U=Mg+(M)/(2)g=(3)/(2)Mg` When the string is cut T = 0, the small block begins to fall freely, the balloon rises up with an acceleration ‘a’ such that U-Mg=Ma `(3)/(2)Mg-Mg=Ma` ` a=(g)/(2)` in the UPWARD direction. |
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| 27. |
Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is V. if the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become |
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Answer» V/4 |
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| 28. |
4 kg of a liquid contained in a vessel at 47^(@)C is placed in a surrounding of 22^(@)C. The temperature falls at the rate of 1.600^(@)C//s. When the temperature of the liquid drops to 37^(@)C, the liquid looses heat at the rate of 4608 J/s. Find the specific heat capacity of the liquid. |
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Answer» |
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| 29. |
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 ms^(-1)to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? |
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Answer» `sin^(-1) (1/3) , 16KM ` |
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| 30. |
Obtain the equation rate of emission of heat for perfect black body by using Stefan Boltzmaan law. |
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Answer» SOLUTION :If Q is the amount of RADIANT energy from substance in time .t. and IRON cross-sectional area .A., then, `[P=W/t]` `W=Q/(At)""`[Power H, radiant energy Q] `:.Q=AWt` `(dQ)/(dt)AW` `=AesigmaT^(4)" "[because" From equation (1)"]` Now `T=T_(1)` at temp. `Q=Q_(1)`, `(dQ_(1))/(dt)=AsigmaeT_(1)^(4)` and `T=T_(2)` at temp. `Q=Q_(2)`, `(dQ_(2))/(dt)=AsigmaeT_(2)^(4)` Here, `T_(1)gtT_(2)` by taking `(dQ_(1))/(dt)gt(dQ_(2))/(dt)` `:.` By keeping substance of temp. `T_(1)` in surroundings to temp. `T_(2)`, rate of emission of energy, `(dQ)/(dt)=(dQ_(1))/(dt)-(dQ_(2))/(dt)` `=AsigmaeT_(1)^(4)-AsigmaeT_(2)^(4)` `=Asigmae(T_(1)^(4)-T_(2)^(2))` If `T_(1)=T` and `T_(2)=T_(S)`, `(dQ)/(dt)=Asigmae(T^(4)-T_(S)^(4))" where "(TgtT_(S))` |
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| 31. |
Heat and work are equivalent. This means, |
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Answer» when we SUPPLY heat to a body we do work on it |
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| 32. |
In a diatomic molecule, the rotational energy at a given temperature |
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Answer» does not OBEYS Maxwell.s distribution |
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| 33. |
A block of mass'm' is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is mu . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction is |
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Answer» `MU MG ` |
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| 34. |
A rigid bar of mass M is supported symmetrically by three wires each of length 1. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to ...... |
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Answer» `(Y _("Copper"))/(Y _("IRON "))` `Y= (FL)/(A Delta L ) = (FL )/(.pi ((D)/( 2 )) ^(2). Delta L )` `therefore Y= (4 FL)/(pi D ^(2) Delta L)` `therefore D ^(2) = (4 FL)/( pi Delta L Y)` `therefore D = sqrt (( 4 FL )/( pi Delta L Y) ) , pi 4, F , (L)/(Delta L )` are same. `therefore D PROP (1)/( sqrtY)` `therefore (D _("Copper"))/( D _("Iron "))= sqrt (( Y _("Iron "))/(Y _("Copper")))` |
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| 35. |
A 100cm long glass capillary tube closed at both ends, has a mercury thread of length 10 cm.When the tube is horizontal the mercury thread stays at the middle of the tube with air columns of equal length on either side,at 76 cmHg pressure and 27^@C. Now the temperature of one side is changed to 0^@C,and of the other side to 127^@C. Find the length and the pressure of the air column kept at 0^@C. Neglect expansions of glass and mercury. |
Answer» Solution :Let the area of cross section of the capillary be `a cm^2` [Fig.6.8].  In the FIRST case,volume of air on either side of the mercury thread in the tube,`V_1=45a cm^3`,temperature `T_1=273+27=300K`,pressure of air `p_1=76cmHg`. In the second case let the left side of the Hg thread inside the tube be at `0^@C` or 273 K and the other side be at `127^@C` or 400 K.Let the pressure of the confined air on each side =PCMHG. Let L=LENGTH ofthe air column at `0^@C`,(90-l)=length of the other air column at `127^@C`. Using equation of state, `(p_1V_1)/T_1=(p_2V_2)/T_2.(76times45a)/300=(pla)/273`........(1) and `(76times45a)/300=(p(90-l)a)/400`........(2) From (1) and (2) we get, `(ptimes l times a)/273=(p(90-l)a)/400` or,`400l=273(90-l)or,673l=273times90` `thereforel=(90times273)/673=36.5cm` From equation (1) `(76times45a)/300=(ptimes36.5a)/273or,p=85.3cmHg` `therefore` The length of the air column at `0^@C` is 36.5 cn=m and its pressure is 85.3 cmHg. |
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| 36. |
A body of mass 8kg is moved by a force F =3x N, where x is the distance covered . Initial position is x= 2m and final position is x = 10m . If initialy the body is at rest , find the final speed . |
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Answer» Solution :`F=ma, F= m. (dv)/(dt)` ` 3x = m. (dv)/(dx).(dx)/(dt) , 3x=8. (dv)/(dt).v , 3x dx = 8V dv ` ` 3int_(2)^(10)x dx = 8 int_0^(v) v dv , 3[(x^(2))/(2)]^(10)=8[(v^(2))/(2)]^(v) , 3[100-4]=8v^(2) , v^(2) = (3 XX 96)/(8) =36 , v=6 m//s` |
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| 37. |
A car is moving in a circular horizontal track of radius 10m with a constant speed of 10ms^(-1). A plumb bob is suspended from the roof car by a light rigid rod of length 1m. The angle made by rod with the vertical is |
| Answer» ANSWER :C | |
| 38. |
A hollow vertical cylinder of radius R and height h has smooth internal suface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v_(0) tangential to rim.It leaves the lower rim point Q vertically below P. The number of revolutions made by the particle will: |
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Answer» `h/(2piR)` |
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| 39. |
A cannon ball has a range Ron a horizontal plane, such that the corresponding possible maximum heights reached are H_1 and H_2 . Then, the correct expression for R is |
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Answer» `((H_(1)) +H_(2))/2` `H_1=(u^2 sin^(2) theta)/(2g) and H_2 =(u^2 sin^(2) (90^@ - theta))/(2g)=(u^2 cos ^(2) theta)/(2g)` `:. H_1H_2 =1/4 ((u^2 sin thetacos theta)/(g))^2=1/4xxR^2/4 ( :. R=(u^2 sin 2 theta)/g)` `R=4sqrt(H_1H_2)` |
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| 40. |
A particle performing SHM with a frequency of 5Hz and amplitude 2 cm is initially in the positive extreme position. What is the equation for its displacement ? |
| Answer» SOLUTION :x=0.02 `COS PI TM` | |
| 41. |
N divisions on the main scale of a vernier callipers coincide with (N + 1) divisions on the vernier scale. If each division on the main scale is 'a' units, then the least count of the instrument is |
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Answer» `(a)/(N+1)` |
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| 42. |
The time period of simple pendulum at the surface of earth is T. If it is taken to a height equal to the radius of earth. Find the period of new oscillations. |
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Answer» SOLUTION :At a height R. new `G^(1)= (GM)/(r^(2))= (GM)/(2R)^(2)=(g)/(4):. T PROP (1)/(sqrt(g))` `(T^(1))/(T_(0))= sqrt((g)/(g_(1)))= sqrt(((g)/(g))/(4))= 2`or `T^(1)= 2T_(0)` |
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| 43. |
A cubical block of mass 'm' rests on rough horizontal surface. mu is coefficient of static friction between block and the surface. A force "mg" acting on cube at an angle " theta" with vertical side of cube pulls the block. If the block is to be pulled along the surface then the value of cot (theta//2) is |
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Answer» LESS than `MU` |
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| 44. |
A block of mass m is suspended by different springs of force constant shown in figure. |
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Answer» `T_(1)=T_(2)=T_(4)` |
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| 45. |
The displacement of a particle executing SHM is given by x = 0.10 sin ((2pit)/(T)+phi)where all the quantities are in SI units. The time period of vibration is 3s. At t = 0, the particle is displaced by 0.05m. Findthe initial phase, |
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Answer» Solution :For the particle in SHM, the DISPLACEMENT , `x= 0.10sin((2pit)/(T)+PHI)` `= 0.10sin((2PI)/(3)t+phi)( :.T=3s)` At t=0, x= 0.05 `0.05 = 0.10 SIN((2pi)/(3) xx0+phi)= 0.10sinphi` `sinphi= (1)/(2) IMPLIES phi= (pi)/(6) "radian"` or `30^(@)` |
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| 46. |
The displacement of a particle executing SHM is given by x = 0.10 sin ((2pit)/(T)+phi)where all the quantities are in SI units. The time period of vibration is 3s. At t = 0, the particle is displaced by 0.05m. Findthe phase angle corresponding to 0.0866 m |
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Answer» Solution :For the particle in SHM, the displacement , `x= 0.10sin((2pit)/(T)+phi)` `= 0.10sin((2pi)/(3)t+phi)( :.T=3s)` substituting x= 0.0866m in `x= 0.10sin((2pi)/(3)t+phi)` `0.0866= 0.10 sin ((2pi)/(3)t+phi)` `sin((2pi)/(3)t+phi)=(0.0866)/(0.10)= 0.866` `:.`PHASE angle `((2pi)/(3)t+phi)= (PI)/(3)` radian or `60^(@)` |
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| 47. |
The displacement of a particle executing SHM is given by x = 0.10 sin ((2pit)/(T)+phi)where all the quantities are in SI units. The time period of vibration is 3s. At t = 0, the particle is displaced by 0.05m. Findthe phase difference between any two positions of the particle 2 s apart. |
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Answer» Solution :For the particle in SHM, the displacement , `x= 0.10sin((2pit)/(T)+phi)` `= 0.10sin((2PI)/(3)t+phi)( :.T=3s)` Phase angle at `t= t_(1)` is `phi_(1)= ((2pi)/(3)xxt_(1)+phi)`and `t= t_(2)` is `phi_(2)= ((2pi)/(3)xxt_(2)+phi)` Phase difference `= phi_(2)-phi_(1)` `=((2pi)/(3)t_(2)+phi)- ((2pi)/(3)t_(1)+phi)= (2pi)/(3)(t_(2)-t_(1))` `= (2pi)/(3) xx 2= (4PI)/(3) "RADIAN" ( :. t_(2)-t_(1)=2)` |
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| 48. |
Calculate the force required to double the length of a wire of diameter 2mm. (Y=1.2xx10^(11)N//m^(2)) |
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Answer» Solution :Strain `=(Deltal)/(l)=(l)/(l)1, :. Deltal=2l-1=l` `Y+("Stress")/("Strain")=(F)/(pir^(2)x1)`, where `tau= 1 mm =10^(-3)m` `rArr F= gamma X pi R^(2)=1.2xx10^(11) XX pi xx10^(-6)` `F=1.2xx pi xx10^(5)N=3.768xx10^(5)N=3.810^(5)N` `:.` Force `=3.8xx10^(5)N` |
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| 49. |
A cylindrical pipe of uniform cross section and of length 120 cmis closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water-column increases by 0.7 cm. Calculate the Poisson's ratio of the material of the pipe. |
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Answer» Solution :Poisson.s ratio , `sigma=(LATERAL strai n)/(longitudi nal stra i n)` Here, longtitudinal STRAIN=`l/L=1/120` Let the initial difference of the pipe be D, the DECREASE in its diameter due to elongation be d. Initial volume of water =`pi/4 D^2 TIMES 120` and final volume of water `=pi/4(D-d)^2times(120+0.7)` `=pi/4(D-d)^2times120.7` Since the volume of water inside the cylinder remains unchanged. `pi/4D^2times120=pi/4times(D-d)^2times120.7` or,`(D-d)/D=sqrt(120/120.7)or,1-d/D=sqrt0.9942 =0.997` or,`d/D=1-0.997=0.003` `thereforesigma=(d//D)/(l//L)=0.003/(1/120)=0.003times120=0.36` |
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