Saved Bookmarks
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. |
Two blocks of equal mass 2 kg are placed on a roughhorizontally surfaces as shown and a force F is applied on the upper block . The systemis initially at rest . Find acceleration of the lower block in m// s^(2) . |
|
Answer» `3m//s^(2)` `a_(c)=(25-5)/4=5m//sec^(2)` [Friction on LOWER block, `f=muN=(1)/10xx50=5N]` |
|
| 2. |
Two bodies of masses 10 kg and 20 kg are located in x-y plane at (0,1) and (1,0). Find the position of their centre of mass. |
|
Answer» |
|
| 3. |
If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t = sqrt(t^(2b) s^c d^(a//2)). It is observed that the time period is directly proportional to sqrt(d/s) . The value of b should therefore be : |
|
Answer» `3//4` |
|
| 4. |
In each situation of Column - I a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface a shown. At t=0 (initially) the angular velocity of the disc is w_(0) and velocity of the centre of mass of the disc is v_(0) (in horizontal direction). The relation between v_(0) and w_(0) is given for each situation and also initial sense of rotation is given for each situation in column I Then match the statments in Column I with the corresponding results in Column II. |
|
Answer» |
|
| 5. |
If the amplitude of body executes SHM becomes doubled, what will be its energy? |
|
Answer» SOLUTION :Four TIMES. Because energy of this BODY. `E propto A^(2)`, HENCE A BECOME doubled, E. will be four times. |
|
| 6. |
If the temperature of the wire is increased, then the Young's modulus will: |
|
Answer» REMAIN the same |
|
| 7. |
The quantity having dimensions only in temperature is |
|
Answer» LATENT heat |
|
| 8. |
1 mole of a gas having gamma = 7/5is mixed with 1 mole of a gas having gamma = 4/3 , what will be the gamma for the mixture? |
|
Answer» `5/11` |
|
| 9. |
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7ms^(-1). If hits the floors of the elevator (length of the elevator 3m) and does not rebound. What is the heat produced by the impact? |
|
Answer» SOLUTION :Mass `m=0.3kg` HEIGHT `h=3m` Accleration due to gravity `g=9.8ms^(-2)`. Potential energy lost by the bolt = mgh `=0.3xx9.8xx3=8.82J` As the bolt does not rebound, the energy is CONVERTED into HEAT. Heat produced = 8.82 J. |
|
| 10. |
A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is |
|
Answer» `(G)/(2(n-1)^(2))` |
|
| 11. |
An object is thrown with initial speed 5ms^(-1) with an angle of projection 30^(@). What is the maximum height and range reached by the particle? |
|
Answer» Range `(R)=(u^(2)sin2theta)/(g)=(25xxsin60^(@))/(9.8)=2.21m` |
|
| 12. |
Derive an expression for escape speed. |
|
Answer» SOLUTION :Laplace.s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the SOUND propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the GAS obeys Poisson.s law (not Boyle.s law as Newton assumed), which is `PV^(GAMMA)` = Constant...(4) where, `gamma = (C_(p))/(C_(v))` which is the ratio between specific heat at constant pressure and specific heat at constant volume. Differentiating equation (4) on both the sides, we get `V^(gamma)dP + P (gammaV^(gamma-1)dV) = 0` or`gammaP = - V (dp)/(dV) = B_(A)`...(5) where, `B_(A)` is the adiabatic bulk modulus of air. Now, substituting equation (5) in equation `V=sqrt(B/rho)` the speed of sound in air is `V_(A) = sqrt((B_(A))/(rho)) = sqrt((gammaP)/(rho)) = sqrt(gammav_(T))` Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take `gamma` = 1.47. Hence, speed of sound in air is `v_(A) = (sqrt(1.4))` (280 m `s^(-1)`) = 331.30 m `s^(-1)`, which is very much closer to experimental data. |
|
| 13. |
In ________ the rotational motion is more than translational motion. |
|
Answer» SLIDING |
|
| 14. |
An idealrefrigeratorkeepsitscontentat 0^(@)Cwhiletheroomtemperatureis 27^(@)C.Calculateitscoefficientof performance. |
|
Answer» Solution :CONTENT placed at TEMPERATURE `T_(L)=0^(@)C=0+273` `T_(L)=273K` REFRIGERATOR placed at temperature `T_(H)=27^(@)C=27+273=300K` Coefficient of performance `(beta)=(T_(L))/(T_(H)-T_(L))=(273)/(300-273)=(272)/(27), beta =10.11` |
|
| 15. |
A man carries a load of 50kg to a height of 40m in 25s. If the power of the man is 1568W, the mass of the man is. |
|
Answer» 20kg |
|
| 16. |
For satellite to appear stationary at a position |
|
Answer» it must rotate with a period EQUAL to that of the EARTH |
|
| 17. |
A black body has maximum omega. L.lambda m at 2000K.Its corresponding omega.L. at 3000 K is |
|
Answer» `3/2lambdam` |
|
| 18. |
A small ball of mass m is suspended from the end A of a L shaped mass lessrigid frame which is fixed to a block of mass m. The block is placed on a smooth table. The ball is given a horizontal impulse so as to impart it a velocity of u. The ball beings to rotate in a circle of radius R about the point A, while the block and the frame slide on the table. Find the tension in the string, to which the ball is attached, at the instant the ball is at the top most position. The rod does not interfere with the string during the motion. |
|
Answer» |
|
| 19. |
Mountain roads rarely go striaght up the slop, but wind up gradually. Why ? |
| Answer» SOLUTION :If roads go to straight up then angle of SLOPE `theta` would be large so frictional FORCE `f=mumgcostheta` would be less and the vehicles may slip. Also greater power would be REQUIRED. | |
| 20. |
A ball is projected inside a room of height 7m with a speed of and at an angle of 37^(@)with horizontal. If its collission with ceiling results in reversal of direction only of y-component of velocity. What is the total time of flight (in sec) till in strikes floor....... |
|
Answer» |
|
| 21. |
The temperatures of equal masses of three different liquids A, B and C are 12^(@)C, 19^(@)C and 28^(0)C respectively. The temperature when A and B are mixed is 16^(@)C and when B and C are mixed, it is 28^(@)C. What will be the temperature when A and C are mixed ? |
|
Answer» |
|
| 22. |
Can a particle in one dimensional motion have zero speed and a non-zero velocity |
| Answer» SOLUTION :No, if the SPEED is ZERO then the VELOCITY will be NECESSARILY zero | |
| 23. |
What is the percentage decrease in the weight of a body at the equatorial region of the earth due to its diurnal motion? Given, radius of the earth = 6400km,pi^(2) = 9.8 |
|
Answer» |
|
| 24. |
The fundamental frequency of a vibrating organ pipe is 200 Hz. |
|
Answer» the FIRST OVERTONE is 400 HZ. |
|
| 25. |
Which of the following pairs is correct |
|
Answer» linear MOMENUTM and angular MOMENTUM `F prop(dp)/(DT)` Newtonthirdlawforceof action =-forceofreaction Torque `prop ` RATE ofchangeof ANGULARMOMENTUM |
|
| 26. |
Among the following find the wrong statement |
|
Answer» Law of GRAVITATION is framed using NEWTON's third law of motion |
|
| 27. |
If a solid sphere of mass 1 kg and radius 0.1 m rolls without slipping at a uniform velocity of 1 m/s along a straight line on a horizontal floor, the kinetic energy is ………. |
|
Answer» `(7)/(5)J` = (translational kinetic energy) + (rotational kinetic energy) `=(1)/(2)mv^(2)+(1)/(2)I.omega^(2)` Moment of inertia `I=(2)/(5)MR^(2)` (for sphere) and `omega=(v)/(R )` `THEREFORE` Total kinetic energy `=(1)/(2)mv^(2)+(1)/(2).(2)/(5)mR^(2).(v^(2))/(R^(2))` `=(1)/(2)mv^(2)+(1)/(5)mv^(2)` `=(7)/(10)mv^(2)` Putting `m=1kg, v=1m//s` Total energy `=(7)/(10)xx1xx(1)^(2)` `=(7)/(10)J` |
|
| 28. |
A cannon is fixed with a smoth massive trolley car at an angle 'theta' as shown in diagram. The trolley car slides from rest down the inclined plane of inclination 'beta', Muzzle velocity of shell fired from cannonis 'U' t_(1), t_(2)are time intants at which shell-1, shell-2 are fired such that they move perpendicular to inclined plane and vertically upwards respectively. {:(" Column-I"," Column-II"),("A) " t_(1)=,"P) "(2u)/(g)),("B) "t_(2)=,"Q) "(2u sin theta)/(g cos beta)),("C) time of flight of shell-1","R) "(u cos theta)/(g sin beta)),("D) time of flight of shell-2","S) "(2ucos (beta+theta))/(g sin 2 beta)),(,"T) " (u)/(g sin beta)),(,):} |
|
Answer» |
|
| 29. |
A lighter body collides head - on and elastically with heavier body at rest then after collision |
|
Answer» Lighter BODY COMES to rest, then heavier body begins to move with initial velocity of lighter body |
|
| 30. |
A longstick rests on the surface a person standing 10 m away the stick with what minimum speed an objectof masss 0.5 kg should he throuwn so that it hits the stick (Assume the coefficientof kinetic friction is 0.7 ) |
|
Answer» Solution :Given: Distance (s) = 10 m Mass of the object (m) = 0.5 kg Coefficient of kinetic friction (`MU`) = 0.7 Work done in moving a body in horizontal SURFACE `omega = mu_R xx s = mu mg xx s ` This work done is EQUAL to INITIAL kinetic energy of the object `1/2 mv^2 = mu mg s ` `|v^2| = 2mug r s = 2 xx 0.7 xx 9.8 xx 10` `v^2 = 14 xx 9.8= 137.2` ` v = 11.71 ms^(-1)` |
|
| 31. |
Assertion :IN extreme position of particle executing SHM, kinetic energy and mechanical energy are same. Reason : At extreme position velocity of a particle executing SHM is zero. |
|
Answer» Both are TURE and the reason is the correct explanation of the ASSERTION. |
|
| 32. |
A planet revolves around sun in an elliptical orbit of eccentricity .e.. If.T. is the time period of the planet then show that the time spent by the planet between the end of the minor axis and close to sun is T(1/4-e/(2pi)) |
Answer» SOLUTION :`(DA)/(dt)`=constant ,  `t_(AB)/T="(Area) SAB"/"(Area) ELLIPSE"=((piab)/4-1/2b(EA))/(piab)` |
|
| 33. |
An inclined track ends in a circular loop ofdiameter .D.. From what heigt on the track a particle should be released so that it completes that loop in the vertical plane ? |
| Answer» Answer :C | |
| 34. |
The idea of calculus was given by |
|
Answer» Newton |
|
| 35. |
The instantaneous angular position of a point ona rotating wheel is given by the equation theta (t) = 2t^(3) - 6t^(2) . The torque on the wheel becomes zero at |
|
Answer» t = 1 s Angular acceleration , `ALPHA = (d^(2) theta)/(dt^(2)) = 12 t -12` When angular acceleration `(alpha)` is zero , then the torque on the WHEEL becomes zero `(becausetau = I alpha)` `implies 12 t - 12 = 0` or `t = 1` s |
|
| 36. |
The density of water is 1000 kg m^(-3). The density of water vapour at 100^(@)C and 1 atm pressure is 0.6 kg m^(-3). The volume of a molecule multiplied by the total number gives ,what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure. |
| Answer» Solution :For a given MASS of water molecules, the density is less if volume is large. So the volume of the vapour is `1000//0.6 = 1//(6 xx10^(- 4))` = times larger. If densities of bulk water and water molecules are same, then the FRACTION of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same AMOUNT, i.e. `6 XX 10^(-4)`. | |
| 37. |
A 60 kg box sledge is travelling horizontally on ice at a speed of 10 ms^(-1). A 90 kg packet is dropped on it vertically. The subsequent velocity of the sledge is |
| Answer» Answer :D | |
| 38. |
Four thin uniform rods each of mass m and length l are arranged to form a square. Find the moment of inertia of the system about an axis (i) Passing through its centre and perpendicular to its plane. (ii) Passing through one of its sides. (iii) Passing through a corner and perpendicular to its plane. (iv) About a disgonal of the system. |
|
Answer» SOLUTION :(i) M.I. about an AXIS PASSING through its centre and perpendicular to its plane `4[(ml^(2))/(12)+m((l)/(2))^(2)]=(4ml^(2))/(12)xx4` `(ii) I_(s)=(ml^(2))/(3)xx2+ml^(2)` `=(5)/(3)ml^(2)`  `(III)` From parallel axes theorem `I=I_(C )+mr^(2)=(4ml^(2))/(3)+4m((l)/(sqrt(2)))^(2)=(10)/(3)ml^(2)` (iv) From perpendicular axes theorem `I_(z )=I_(x)+I_(y)`, `I_(z)=2I_(x), `I_(x)=(I_(z))(2)=(I_(C ))/(2)=(2)/(3)ml^(2)` |
|
| 39. |
An enemy plane is flying horizontally at an altitude of 2km with a speed of 300 ms^(-1). An army man with an anti-aircraft gun on the ground sights hit enemy plane when it is directly overhead and fires a shell with a muzzle speed of 600 ms^(-1). At what angle with the vertical should the gun be fired so as to hit the plane ? |
|
Answer» Solution :Let G be the position of the un and E that of the enemy plane flying horizontally with speed `u = 300 ms^(-1)`, when the shell is fired with a speed `v_(0)` is `v_(x) = v_(0) cos THETA` Let the shell hit the plane a point P and let t be the TIME taken for the shell to hit the plane. It is clear that the shell will hit the plane, if the HORIZONTAL distance EP travelled by the plane in time t= the distance travelled by the shell in the horizontal DIRECTION in the same time, i.e., `u xx v = v_(x)xx t` or `u = v_(x) = v_(0) cos theta` or `cos theta = (u)/(v_(0)) = (300)/(600)`  = 0.5 or `theta = 60^(@)`. Therefore, angle with the vertical `= 90^(@) - theta = 30^(@)`. |
|
| 40. |
Draw the free body diagram that represents the particle accelerating in positive xdirection |
Answer» Solution :The RELATIVE magnitude of forces should be INDICATED when the free body diagram for mass m is drawn.  Case (a): The forces `F_1`and `F_2`have equal LENGTH but opposite direction. So net force along y-direction is ZERO. Since the force is zero, acceleration is also zero along Y-direction (Newton.s second law). Similarly in the x direction, `F_3` and `F_4`have equal length and opposite in direction. So net force is zero in the x direction. So there is no acceleration in x direction. Case (b): The forces `F_1` and `F_2`are not equal in length and act opposite to each other. The figure (b) shows that there are unbalanced forces along the y-direction. So the particle has acceleration in the y-direction. The forces `F_3` and `F_4`are having equal length and act in opposite directions. So there is no net force along the x direction. So the particle has no acceleration in the x direction. Case (c): The forces `F_1` and `F_2`are equal in magnitude and act opposite to each other. The net force is zero in y direction. So in y-direction there is no acceleration. The forces `F_3` and `F_4`are not equal in magnitude and `F_3` is greater than `F_4` . So there is a net acceleration in NEGATIVE x direction Case (d): The forces `F_1`and `F_2`are equal in magnitude and act opposite to each other. The net force is zero in y direction. So there is no acceleration in y-direction. The forces `F_3` and `F_4`are not equal in magnitude. The force `F_4`is greater than the force `F_3`. So there is a net acceleration in the positive x direction. |
|
| 41. |
A bullet moving with speed 150m//s strikes a tree and penetrates 3.5cm before stopping . Find the magnitude of its acceleration and the time taken to stop . |
|
Answer» Solution :`u=150m//s,s=3.5cm=0.035m, V=0, v^(2)-u^(2)=2AS` `0-150^(2)=2 xx a xx 0.035` a=`(-150xx150)/(2xx0.035)=-3.214xx10^(5) m^(-2)` TIME`=(v-u)/a=(0-150)/(3.24xx10^(5))=(-150)/(3.24xx10^(5))=4.67xx10^(-4) sec` |
|
| 42. |
Considera reversibleengineof efficiency1/6 whenthe temperatureof thesinkis reduced by62 ^@C, itsefficiency gets doubled. Thetemperatureof thesourceand sinkrespectivelyare |
|
Answer» 372 K and 310 K |
|
| 43. |
The surface tension of a liquid is 0_(2).5N//m. If a liquid film is formed on a ring of area 0.02m^(2) then its surface energy will be………..Joule |
|
Answer» `0.01` |
|
| 44. |
The gravitational force is a central force. Acceleration due to gravity has different values for different point ont the earth.s surface, g varies with altitude or depth. Gravitational acceleration have different values for different planets and depends on mass and radius of planets. For the given density of planets, gravitational acceleration is directly proportional to radius of planets. The object on equator will have zero weight, when |
|
Answer» EARTH ROTATES 17 times FASTER than the present rotational SPEED |
|
| 45. |
If the distance between the source and the observer is doubled,the intensity of sound becomes |
|
Answer» double |
|
| 46. |
The gravitational force is a central force. Acceleration due to gravity has different values for different point ont the earth.s surface, g varies with altitude or depth. Gravitational acceleration have different values for different planets and depends on mass and radius of planets. For the given density of planets, gravitational acceleration is directly proportional to radius of planets. The values of g at the equator is less than that of pole due to |
|
Answer» shape of EARTH only |
|
| 47. |
If a mass of 10 g is hung from a suspended spring, it elongates by 2mm. If the spring is pulled a little downwards and then released, what will be the frequency of vibration of the spring? |
|
Answer» |
|
| 48. |
The graph between period oscillation T and mass attached m to a spring will be |
|
Answer»
|
|
| 49. |
The average degrees of freedom per moleculefor a gas is 6. The gas performs 25J of work when it expands at constant pressure. The heat absorbed by the gas is |
|
Answer» 75J |
|
