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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 ball is projected obliquely with a velocity 49 ms^(-1) strikes the ground at a distance of 245 m from the point of projection. It remained in air for |
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Answer» 10 SEC |
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| 2. |
A railway engin of mass 50 tons is pulling a wagon of mass 40 tons with a force of 4500N. The resistance force acting is1N per ton. The tension is the coupling between the engine and the wagon is |
| Answer» Answer :B | |
| 3. |
A body is projected at an angle 30° with a velocity 42 ms^(-1)Its maximum height is |
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Answer» 22.5 m |
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| 4. |
At a point above the surface of earth, the gravitational potential is -5.12xx10^7Jkg^-1 and accelaration due to gravity, is 6.4ms^-2. Calculate the height of the point above earth's surface. |
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Answer» infinite |
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| 5. |
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? ("mass of the sun " = 2 xx 10^(30) kg). |
| Answer» SOLUTION :GM/`R^(2) = 2.3 xx 10^(12) m s^(-2), omega^(2) R = 1.1 xx 10^(6) MS^(-2)`,here ω is the angular speed of rotation. Thus in the ROTATING frame of the star, the inward force is much greater than the outward centrifugal force at its equator. The object will remain stuck (and not fly off due to centrifugal force). Note, if angular speed of rotation INCREASES say by a factor of 2000, the object will fly off. | |
| 6. |
The moment of inertia of a uniform disc about a normal axis through centre is I. The M.I. about a tangential axis parallel to any of its diameters is |
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Answer» `(7I)/(2)` |
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| 7. |
The bob of a simple pendulum is hollow. Now its time period is T. If it is partially filled with water the time period will be |
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Answer» More than T |
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| 8. |
A girl travels along a straight road due east for the first half distance with speed v_(1)and the second half distance with speedv_(2). What is the average speed of the girl ? |
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Answer» Solution :Given : Total distance = s Let the velocity of the girl for the first HALF distance be `v_(1)` Let the velocity of the girl for the second half distance be ` v_(2)` Formula : Time taken`= ("Distance")/("Velocity")` Time takenforthe first half distance `=((s)/(2))/(V_(1)) = (S)/(2v_(1))` Time taken for the second half distance `= ((s)/(2))/(v_(2))= (S)/(2v_(1))` `{:("Total time"),("taken"):}}= (S)/(2v_(1)) +(S)/(2v_(2))` Average SPEED , `v_(av)= (S)/((s)/(2v_(1))+(s)/(2v_(2)))=(2v_(1)v_(2))/(v_(1)+v_(2))` |
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| 9. |
In S.H.M, there is always a constant ratio between the displacement of the body and its |
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Answer» Velocity |
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| 10. |
A small block of mass M moves with velocity 5m/s towards another block of same mass M placed at a distance of 2m on a rough horizontal surface. Coefficient of friction between the blocks and ground is 0.25. Collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g = 10 m//s^(2)) |
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Answer» Solution :Retardatio due to friction `=MU G = (0.25)(10) = 2.5 m//s^(2)` Collision is elastic, i.e., after collision first block COMES to REST and the second block acquires the velocity of first block. Or we can understand it in this manner that second block is permanetly at rest while only the first block MOVES. Distance travelled by it will be `s=(v^(2))/(2a)=((5)^(2))/((2)(2.5))=5m` `therefore` Final separation will be (5-2) = 3 m |
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| 11. |
Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion. |
Answer» Solution :Let the particle with velocity vbe at point Pat some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.  The angular momentum is `L = r × mv`. Its magnitude is `mvr sin θ` , where `θ` is the angle between r and v as shown in Fig. 7.19. Although the particle changes position with time, the line of DIRECTION of v remains the same and hence `OM = r sin θ` . is a constant. Further, the direction of l is perpendicular to the PLANE of r and v. It is into the page of the figure.This direction does not change with time. Thus, l remains the same in magnitude and direction and is therefore CONSERVED. Is there any external torque on the particle? |
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| 12. |
Explain why two streamline cannot cross each other? |
| Answer» Solution :TANGENT to a STREAMLINE gives the direction of velocity of the fluid particle at that point. If two STREAMLINES cross each other, two tangents can be DRAWN at the intersecting point. This MEANS the fluid has two directions of flow at one point, which is impossible. | |
| 13. |
The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities? |
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Answer» SOLUTION :`(Delta m)/(m) XX 100= 2%, (Delta v)/(v) xx 100= 3%` KINETIC energy `E= (1)/(2) mv^(2)` Maximum error in kinetic energy `(Delta E)/(E )xx 100= (Delta m)/(m) xx 100 + 2 (Delta v)/(v) xx 100` Maximum error in kinetic energy `=2% + 2 xx 3% = 2 % + 6% = 8%` |
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| 14. |
The moment of inertia of a cylinder about its own axis is equal to its moment of inertia about an axis passing through its centre and normal to its length. The ratio of length to radius is |
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Answer» `2:1` |
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| 16. |
A small body of mass m is located on a horizontal plane at the point O. The body acquires a horizontal velocity v_(0). Find the mean power developed by the friction force during the whole time of motion, if the frictional coefficient mu = 0.27 m = 1.0 kg and v_(0) 1.5 m//s. |
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Answer» SOLUTION :Work done against the frictional force W `= - mu mgs` ` therefore (dW)/(dt) = - mu mg((ds)/(dt)) or P = - mu mgvlt P GT= (int_(0)^(v_(0)) Pdv)/(int_(0)^(v_(0))dv) = (int_(0)^(v_(0))-mu mg v dv)/(int_(0)^(v_(0)) dv)` SOLVING, we get ` lt P gt = (-mu m g v_(0))/(2)` Substituting the given values, we get ` lt P gt = - (0.27 xx 1 xx 9.8 xx 1.5)/(2) = -2` Watt. |
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| 17. |
The apparent weight of a man in a lift is less than the real weight when the lift is going |
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Answer» down with UNIFORM SPEED |
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| 18. |
If the equation of state of a gas is expressed as (P + a/(V^2)) (V - b) = RT where P is the pressure, V is the volume and T the absolute temperature and a, b , R are constants, then find the dimensions of 'a' and 'b' ? |
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Answer» Solution :By principle of HOMOGENEITY of dimensions P can added to P only. It means `(a)/(V^(2))` ALSO GIVES PRESSURE. Dimension formulae for pressure `(P)= M^(1) L^(-1) T^(-2)` and Volume (V) `= M^(0) L^(3) T^(0)`. Since `(a)/(V^(2))`= pressure `therefore (a)/((M^(0)L^(3)T^(0))^(2))= M^(1)L^(-1) T^(-2)` `rArr (a)/(M^(0)L^(6)T^(0))= M^(1)L^(-1) T^(-2)` `therefore a= M^(1) L^(5)T^(-2)` Similarly, b will have same dimensions as volume `V-b=` volume `therefore b= M^(0) L^(3) T^(0)` |
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| 19. |
A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P_0. The instantaneous velocity of this car is proportional to |
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Answer» `t^2 P_0` `:. P_0 = mv (dv)/(dt) " or " P_0dt = mvdv` INTEGRATING both SIDES, we get `int_(0)^(t) P_0 dt = m int_0^vvdv , P_0t = (mv^2)/(2) , v = ((2P_0t)/(m))^(1//2)` or `v PROP sqrt(t)`. |
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| 20. |
The length of a smooth inclined plane of inclination 30^(@) is 5m. The work done in moving a 10 kg mass from the bottom of the inclined plane to top is |
| Answer» ANSWER :A | |
| 21. |
(A) : A spring has potential energy, both when it is compressed or stretched.(R ) : In compressing or stretching, work is done on the spring against the restoring foece. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 22. |
Measuring process is essentially ……… Fill in the blanks |
| Answer» SOLUTION :a PROCESS of COMPARISON. | |
| 23. |
Mass of solid floating in liquid is m_(1) and mass of liquid is m_(2). Base area is A. Then pressure at bottom is p_(0) + ((m_(1)+m_(2))g)/(A) Reason : Upward force on liquid from base of vessel is pA, where p is pressure at bottom. |
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Answer» If both Assertion and REASON are correct and Reason is the correct EXPLANATION of Assertion. |
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| 24. |
Give the expression forworkdoneby thegas. |
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Answer» Solution :When a gas expands against pressure, it does WORK on the surroundings. The work done in expansion for volume `V_(1)` to `V_(@)` is GIVEN by `W=int_(V_(1))^(V_(2))PDV` If the pressure remains constant during expansion, `"Then"W=P(V_(2)-V_(1))=PDeltaV` If the volume remains constants, then `W=0`. If there is no external pressure, then no work is done. For example, when a gas expands FREELY in vacuum, no work is done by it. |
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| 25. |
The correct order in which the dimensions of time decreases in the following physical quantities. (A) Power (B) Modulus of elasticity (C ) Moment of inertia (D) Angular momentum |
| Answer» Answer :D | |
| 26. |
A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. If the cyleinder is displced by a small downward displacement a from its equilibrium postion and then released , the restoring force on it will be |
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Answer» `-([K+Apg]a)/(4)` `U. = ((3l)/(4) + a) Arhog ` Force in the spring ` = k(x_(0) + a)` So, the restoring force on the cylinder is `F = -[k(x_(0) + a) - {mg -((3l)/(4) + a)} Arhog] ...(II)` Using (i) and (ii), we get `F = -[k + Arhog]a` |
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| 27. |
Fig. 7. 187 (a) shows a system of two blocks of masses m and M (gt m) placed on a surface , there is no friction between the blocks and the surface . A force F is applied to the system as shown resulting in the motion of the system . Fig. 7.187(b) also shows a system of blocks of masses m and M . Masses m and M have , respectively , the same values as in Fig . 7.187 (a) . There is no friction between the lower block and the surface on which the system is placed but friction is present at the constant of the two blocks . A force F, having the same value as in Fig. 7.187 (a) , is applied to the lower block resulting in the motion of the system such that both blocks move together with a common acceleration . Match Column-I and II. |
Answer»  (a) `"" a = ((F)/(M = m))` Force on M `N = m xx a = (mF)/(M + m)` `a = (F)/(M + m)` (B) Force on m = ` m xx a = ((mF)/(M + m))`  (c) If F is applied on m TOWARDS RIGHT  `a = ((F)/(M + m))` (d) `""a = ((F)/(M + m))` Force on m , f = ma = `((mF)/(M + m))` 
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| 28. |
A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. Time period of vertical S.H.M of the cylinder is |
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Answer» `2pi((m)/(K+APG))^(1//2)` `a. = (F)/(m) = -((k + Arhog)/(m))c` Comparing it with, `a. =-omega^(2)a` `omega^(2) = ((k + Arhog)/(m))` `omega = ((k + Arhog)/(m))^(1//2)` Time period, `T = (2pi)/(omega) = 2pisqrt((m)/(k + Arhog))^(1//2)` |
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| 29. |
Three uniform discs each of mass 'm' and radius 'r' are kept touching each other, such that their centres form a triangle. The M.I. of the system about a median of |
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Answer» `(11)/(4)MR^(2)` |
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| 30. |
Distinguish between liquid body and solid body. |
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Answer» Solution :SHEAR stress can change the shape of a soild keeping its volume fixed. Shape of FLUID changes by APPLICATION of very small shear stress. The shearing streass of FLUIDS is about million times smaller than that of SOLIDS. |
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| 31. |
Three girlsskatting on a circular ice ground ofradius200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig . 4.20 . Whatis the magnitude of the displacement vector for each ? for which girl is this equal to the actual length of path skate ? |
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Answer» |
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| 32. |
In the aboveproblemthe moment of inertia of fourbodies about an axis passing throughany sideof frame is |
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Answer» `4ML^(2)` |
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| 33. |
The acceleration (a) of a particle depends on displacement (s) as a a = 5 + s. Initially s = 0, y = 5, then velocity v corresponding to the displacement is given by |
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Answer» `V = 5+s ` |
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| 34. |
For a particle executing SHM. The KE 'K' is given by K = K_(0) cos^(2) wt. The maximum value of PE is |
| Answer» Answer :A | |
| 35. |
As you have learnt in the text a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth6.0xx 10^(24(kg) , Radius = 6400 km . |
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Answer» SOLUTION :`implies` Mass of the earth `M_(E) 6.0 XX 10^(24) kg,` Radius of earth `R_(E) = 6400 km = 6.4 xx 10^(7)m` DISTANCE of geo stationary satellite from the surface of earth h `= 3600 km = 3.6 xx 10^(7) m` Gravitational potential due to earth.s gravity at height h, `phi= (-GM_E)/(R_(E)+h)` `=-(6.67 xx10^(-11)xx6xx10^(24))/(6.4 xx10^(6)+36 xx 10^(6))` `=-(40.02xx 10^(13))/(4.24xx10^(7))` `=-9.43 xx10^(6)Jkg` |
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| 37. |
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth of below the surface of earth. Then : |
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Answer» d = 2 km |
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| 38. |
The escape velocity for a planet is upsilon_e . A particle is projected from its surface with a speed upsilon .For this particle to move as a satellite around the planet |
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Answer» `upsilon_e/2 LT UPSILON lt upsilon_e` |
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| 39. |
A liquid can easily change its shape but a solid cfan not because |
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Answer» The desity of a liguid is smaller than that of a solid |
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| 40. |
(A) : Centripetal force does no work.(R ) : Force and displacement are perpendicular to each other. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 41. |
Which spring has greater value of spring constant-a hard spring or a delicate springs ? |
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Answer» |
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| 42. |
Guess the effect of temperature on coefficient of friction? |
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Answer» SOLUTION :The coefficient of FRICTION MAY decrease UPON INCREASE in temperature |
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| 43. |
An object of specfic gravity rho is hung from a massless string. The tension in the string is T. The object is immersed in water so that one half of its volume is submerged, the new tension in the string is |
| Answer» SOLUTION :`((2RHO-1)/(2rho))T` | |
| 44. |
Two planet of masses in the ratio 1:2, its radii in the ratio 1:2. Then the ratio of acceleration due to gravity will be .......... |
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Answer» `1:2` `g_1/(g_2)=((M_1)/R_1^2)((R_2^2)/M^2)=(m_1/m_2)(R_2/R_1)^2 =(1/2)(2/1)^2` `g_1/(g_(2))=4/2=2` `g_1 : g_2 = 2:1 ` |
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| 45. |
Derive an expression for the velocity of the body moving in a vertical circle. And also find a tension at the top of the circle. |
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Answer» Solution :(i) A body of mass (m) attached to one end of a massless and inextensible string EXECUTES circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the RADIUS `VEC(r )` of the circular path. (ii) The motion of the body by taking the free body diagram (FBD) at a position where the position vector `vec(r )` makes an angle `theta` with the vertically downwards direction and the instantaneous velocity is as shown in Figure. There are two force acting on the mass. 1. Gravitational force which acts downward 2. Tension along the string.  Applying Newton's second law on the mass, In the tangential direction, `mg sin theta = ma_(t)` `mg sin theta =- m((dv)/(dt))` where, `a_(t)=-((dv)/(dt))` is tangential retardation In the radial direction, `T - mg cos theta = ma_(r ), T - mg cos theta = (mv^(2))/(r )` where, `a_(r )=(v^(2))/(r )` is the CENTRIPETAL acceleration. |
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| 46. |
When some detergent is added to water, the surface tension …………… . |
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Answer» REMAIN UNAFFECTED |
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| 47. |
When a tension 'F' is applied, the elongation produced in uniform wire of length l, radius r is e, when tension 2F is applied, the elongation produced in another uniform wire of length 2l and radius 2r made of same material is |
| Answer» ANSWER :1 | |
| 48. |
Find the stretch in the springs shown in figure . The respective data are given in the figure . The friction and masses in pulleys are negligible . Find the extension in spring? |
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Answer» Solution :(a) As mass M is in equilibrium , the tension in the string with which it is hanging will be Mg. As pulley is massless the tension in the upper string must be such that the restoring force in the string is equal to the tension in it . Thus we have `k_1)x=(Mg)/2 "or" x_(1)=(Mg)/(2k_(1))` (b) Here again mass M is in equilibrium, thus the tension in the string connected to it must be equal to Mg and hence the restoring force in the lower will alxo be Mg . If `x_(1)` be the extension in this spring , we have `k_(1)x_(1)=Mg "or" x_(1)=(Mg)/k_(1)` As pulley is light , the tension in upper string must be twice that of teh lower srting , 2Mg which is equal to restoring force in the upper spring . If `x_(2)` is the extension in the upper spring , we have `k_(2)x_(2)=Mg"or" x_(2) =(2Mg)/k_(2)` Wherever a body is in EQUILIBIUM , due to inertia it is maintained in a state of rest. If any of the supporting force a body is removed , body start from rest with some acceleration in the in the direction opposite to the force applied by teh support , as all other forces acting on body will have a resultant in that direction. We discuss the situation with an EXAMPLE shown in figure. Two masses 5 kg and 10kg are hanging in equilibrium with the strings A and B. If tensions in the strings A and B are `T_(1)"and " T_(2)` respectively , then we have `T_(1)=5g =50N , T_(2) =15g = 150N`  If at some instantwe break string B , tension `T_(1)`becomes zero instantane - ously and block will start falling from rest under gravity and 10 kg block , as it is still in equilibrium, `T_(2)` will instantly change to 10 N to balance its weight. If instead of string B, we break the string A, tension `T_(2)` becomes zero and both the blocks will start falling with acceleration g. Now `t_(1)` is just an internal force for the system of two blocks and as there is no interaction between the two blocks and both are falling under gravity `T_(1)` will ALSO instantly become zero. Now to UNDERSTAND the concept in a better way we slightly modify the situation of previous example. Instead of strings, we use springs as shown in figure. Initially the block are in equilibrium . If spring B breads, its tension will instantly become zero as we are using ideal springs (massless`//`inertialess), but due to inertia initially 10 kg mass will not move from its initial position , so tension in upper spring will not change instantly hence it will accelerate in upward direction with acceleration given as `a=(150-100)/(10)=5 m//s^(2)` If spring a breaks tension in spring A will instantly become zero , but as due to inertia 5kg and 10kg blocks will start moving from rest . As initally just after breaking of spring A two masses are at rest, tension in spring B will not change at this instant which is `T=5g=50N, Thus at this instant (just after spring A is cut ) acceleration of 10 kg mass is `a_(1)(10g+kx)/(10)=(100-:50)/(10)=15 m//s^(2)` and that of 5 kg mass is `a_(2)=(5g-kx)/5=(50-50)/5=0 m // sec^(2)` 
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| 49. |
A volume occupied by a saturated vapour is reducd n times at a constant temperature. What fraction eta of the final volume is occupied by the liquid if the specific volume of saturated vapour be N times greatr than the specific volume of the corresponding liquid? |
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Answer» |
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