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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. |
Three identical rings, each of mass M and radius R are placed in the same touching each other such that their centers form the vertices of an equilateral triangle. The M.I. of the system about an axis passing through centre of one the rings and perpendicular to plane is |
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Answer» `("MR"^(2))/(2)` |
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| 2. |
At what points is the energy entirely kinetic |
| Answer» SOLUTION :ENERGY is KINETIC at the MEAN POSITION. | |
| 3. |
A stone thrown vertically up with velocity v reaches three points A, B and C with velocities v,(v)/(2)and(v)/(4) respectively. Then AB:BC is |
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Answer» `1:1` |
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| 4. |
For a body rolling along a level surface, the translational and rotational K.E. are equal. The body is |
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Answer» SOLID CYLINDER |
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| 5. |
Angular momentum is |
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Answer» moment of momentum |
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| 6. |
Water can be heated in a cup of paper because : |
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Answer» Paper is GOOD CONDUCTOR of heat |
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| 7. |
Is a single isolated force possible in nature. |
| Answer» Solution :No. According to Newton.s third law, for EVERY ACTION, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is ALWAYS a reaction. So it is IMPOSSIBLE. | |
| 8. |
A player takes 0.1 s in catching a ball of mass 150 g moving with velocity of 20m/s. the force imparted by the ball on the hands of the player is |
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Answer» 3N `=(m(v-u))/(t)= (150xx10^(-3)XX(0-20))/(10^(-1))` `= -20xx150xx10^(-2)= -30N` This is the force exertedby the hand on the BALL. `therefore` The ball EXERTS a force of 30 N on the hands of the player. |
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| 9. |
A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia. |
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Answer» increases |
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| 10. |
Temperature of a steel block changes from 80^(@)C to 75^(@)C in time t_(1), from 75^(@)C to 70^(@)C in time t_(2) and from 70^(@)C to 65^(@)C in time t_(3) then. . . . . . . . . |
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Answer» `t_(1)=t_(2)=t_(3)` `(dQ)/(dt)` is PROPORTIONAL to the temperature differnce. It decreases as temperature decreases. So at high temperature, time is LESS for `5^(@)C` DECREASE in temperature and at LOWER temperature, time is more for `5^(@)C` decrease in temperature. `:.t_(1)lt t_(2) lt t_(3)` |
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| 11. |
The angel subtended by the vector A = 6hat(i) + 3hat(j) + 4hat(k) with the y-axis is |
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Answer» `sin^(-1)((3)/(61))` |
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| 12. |
A wire has a mass (0.3+0.003) gm, radius (0.5+0.005) mm and length (0.6+0.006) cm. The maximum percentage error in the measurement of density is |
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Answer» 1 |
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| 13. |
In the friction cone experiment the hanging mass, which was 250g remained stationary when the wheel was rotated at the rate of 1//2 revolutioin per second. Calculate Joule's mechanical equivalent of heat if the rise in temperature was 4^(@)C in 15 minutes.The total water equivalent of cones and water was 100g, diameter of the wooden disc =50cm sp. heat capacity of water =1000cal per kg per kelvin. |
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Answer» |
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| 14. |
Two masses connected with a string. When a force F is applied on mass m_1 . The force acting on m_1is |
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Answer» `(m_1 F)/(m_1 + m_2)` |
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| 15. |
An observer moves towards a stationary source of sound. The percentage change in the apparent frequency is |
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Answer» zero |
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| 16. |
The dimensional formula for Magnetic Moment of a magnet is |
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Answer» `M^(@)L^(2)T^(@)A^(1)` |
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| 17. |
I a uniformly positively charged sphere a very thin tunnel has been made along the diameter as shown in the figure below. A carged particle -q having mass m is released from rest at one end of tunnel. For the situation described, mark out the correct statement (s) : [Neglect gravity] |
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Answer» Charged particle will perform SHM about centre of the sphere as MEAN position |
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| 18. |
What is the natuer of the graph in the following : i) volume and temperature on Kelvin scale, keeping pressure constant. ii) pressure and temperature on Kelvin scale, keeping volume constant. |
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Answer» Solution :i) SHAPE of GRAPH with volume (V) and temperature (T) in Kelvin SCALE of constant pressure is a straight LINE with some slope. It intersects X-axis at zero Kelvin. II) Shape of P - T graph with volume as constant is a straight line with some slope. It intersects X-axis at zero Kelvin. |
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| 19. |
A monkey climbs up a slippery pole for 3 and subsequently slips for 3. Its velocity at time t is given by v (t) = 2t (3 - t), 0 ltt lt 3 and v(t) = -(t - 3) (6-t) for 3 lt tlt 6 s in m/s. It repeats this cycle till it reaches the height of 20 m. (a) At what time is its velocity maximum ?(b) At what time is its average velocity maximum ? (c) At what time is its acceleration maximum in magnitude ? (d) How many cycles (counting fractions) are required to reach the top ? |
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Answer» Solution :For maximum velocity, `(dv)/(dt) = 0` Given velocity, `v = 2t (3-t) = 6t - 2t ^(2)` (a) From maximum velocity, `(dv)/(dt) =0` `THEREFORE (d)/(dt) (6t - 2t ^(2)) =0` `therefore 6- 4t =0` `therefore t = (6)/(4) = 3/2 s = 1.5 s` (b) From equ. (i) `v = 6t - 2t ^(2)` `therefore (dx)/(dt) = 6t - 2t ^(2)` `therefore (dx)/(dt) = 6t - 2t ^(2)` `therefore dx = (6t - 2t ^(2)) dt` `therefore` Distance travelled in time interval 0 to 3 s. `x _(1) = int _(0) ^(3) ( 6t - 2t ^(2)) dt` `= [ ( 6t ^(2))/( 2 ) - ( 2t ^(3))/( 3) ] _(0) ^(3) = [ 3t ^(2) - (2)/(3) t ^(3) ] _(0) ^(3)` `= 3 XX 9 - 2/3 xx 3 xx 3 xx 3=27-18 =9m` Average velocity `= ("Displacement")/("Time")=9/3=3m//s` Given, `x = 6t - 2t ^(2)` `3 = 6t - 2t ^(2)` `therefore 2t ^(2) - 6t - 3=0` `therefore t = (6pm sqrt (6 ^(2) - 4 xx 2 xx 3 ))/(2 xx 2 ) = ( 6pm sqrt ( 36 - 24))/( 4)` `= (6pm sqrt12)/(4) = (3pm 2 sqrt3)/(2)` CONSIDERING positive sign only. `therefore t = (3 + 2 sqrt3)/( 2) = (3 + 2 xx 1.732)/(2) = 9/4s` (c) In a periodic motion when velocity is zero acceleration will be maximum PUTTING v =0 in equ. (i), `0= 6t - 2t ^(2)` ` therefore 0= t ( 6- 2t)` ` = t xx 2 (3-t) =0` `therefore t = 0 or t =3 s` (d) Distance covered in 0 to `3s = 9m` Distance covered in 3 to `6s int _(3) ^(5) (18-9t +t ^(2)) dt` `x _(2) = (18 t - (9t^(2))/( 2) + (t ^(3))/( 2)) _(3) ^(6)` `= 18 xx 6- 9/2xx 6^(2) + (6 ^(3))/(3)` `- (18 xx 3 - (9 xx 3 ^(2))/(2) + ( 3 ^(3))/( 3))` `= 108 -9 xx 18 + (6 ^(3))/( 3) ` `- 18 xx 3 + 9/2 xx 9- (27)/(3) = 108 - 18 xx 9+ (216)/(3) ` `- 54 + 4.5 xx 9-9 =- 4.5 m` `therefore` Total distance travelled in oen cycle, `= x _(1) + x _(2) = 9 -4.5 = 4.5 m` Number of cycles covered, `= ("total distance ")/("distance covered in each cycle")= (20)/(4.5) ~~4.44 ~~5` |
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| 20. |
State Kirchoff's law of heat radiation and mention its mathematicalform. |
| Answer» Solution :The RATIO of the emission to the absorptive powers of any body at a GIVEN temperature in a given wavelenght range is a constant and is equal to the emission POWER of a perfectly BLACK body at the same temperature and for the same wavelenght range. The mathematicalfrom of Kirchhoff.s law is `(e_(lambda))/(a_(lambda)) = E_(lambda)` | |
| 21. |
A solid body starts rotating about a stationary axis with an angular acceleration beta = at where a = 2xx10^(-2) rad//sec^(2) How soon from beginning of rotation will be total acceleration vector of an arbitrary point of the body form an angle alpha = 60^(@) with its velocity vector |
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Answer» |
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| 22. |
A heavy wheel of radius 20 cm and weight 10kg is to be dragged over a step of height 10cm, by a horizontal force F applied at the centre of the wheel. The minimum value of F is |
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Answer» 20 KGWT |
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| 23. |
If the unit of length is doubled, unit of time is halved and unit of force is quadrupled, the unit of power would change by the factor |
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Answer» `1//4` |
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| 24. |
A 1.5 kg hammer moving with a velocity of 10 m/s strikes a nail for 0.005 s. The average force exerted on the nail is |
| Answer» ANSWER :C | |
| 25. |
A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown. |
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Answer» VERTICALLY downward |
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| 26. |
A solid sphere of mass 2.0 kg and diameter 0.30 m is suspended on a wire Find the period of angular oscillation for small displacement if the torque constant of the wire is 6.0 xx 10^(-3)N -m/radian. |
| Answer» SOLUTION :`T= 2PI sqrt(I/C) = 2pi sqrt(2/5 (MR^2)/(C)) = 2 xx 3.14 xxsqrt((2xx2xx(0.15)^2)/(5xx6xx10^(-3)) = 10.9 s` | |
| 27. |
Use the formula v= sqrt((gamma p)/(rho)) to explain why the speed of sound in air (a) is independent of pressure. (b) increases with temperature. (c ) increases with humidity |
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Answer» Solution :Assume ideal gas law `p = (rho RT)/(M)`, where `rho` is the density, M is the molecular mass, and T is the temperature of the gas. This gives `v= sqrt((gamma RT)/(M))`. This shows that v is : (a) Independent of pressure (b) INCREASES as `sqrtT` (c ) The molecular mass of WATER (18) is less than that of `N_(2) (28) and O_(2) (32)` Therefore as humidity increases, the effective molecular mass of air DECREASES and hence v increases. |
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| 29. |
Two identical blocks A and B each of mass 2 kg hanging stationsary by a light inextensible flexible string passing over a light and frictionless pulley. A shot C, of mass 1 kg moving vertically with velocity 9 m/s collides with block B and get stuck to it. Calculate : Time after which block B starts moving downward Maximum height reached by B Loss of mechanical energy upto that instant |
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Answer» |
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| 30. |
A projectile has a range of 50 m and reaches a maximum height of 10m . Calculate the angle at which the projectile is fired . |
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Answer» Solution :Here R = 50 m, H = 10M , `theta` = ? ` R = (u^(2) sin 2 theta)/(g)` `= (2u^(2) sin theta cos theta)/(g) "" . . . (1)` MAXIMUM height, `H = (u^(2) sin^(2) theta)/( 2G) "". . . (2)` Dividing (2) by (1) , we get `(H)/(R) = (u^(2) sin^(2) theta)/( 2g) XX (g) / (2 u^(2) sin theta cos theta)` `= (1)/(4) tan theta` (or)`tan theta = (4H)/(R) = (4 xx 10)/(50) = 0 . 8` (or) `theta = tan^(-1) (0.8) = 38 . 66^(@)` |
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| 31. |
If vecA + vecB = vecRand2vecA + vecB s perpendicular to vecB then |
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Answer» A = R |
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| 32. |
The position x of a particle varies with time t as x= at^(2)-bt^(3). The acceleration of the particle will be zero at time t equal to |
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Answer» `a/b` |
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| 33. |
A marble floor appears cold than a cemented floor, because marble has |
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Answer» LOW conductivity |
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| 34. |
A truck blowing horn of frequency 500Hz travels towards a vertical mountain and driver hears echo of frequency 600Hz. If velocity of sound in air is 340m//s then speed of truck is |
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Answer» `31m//s`  The sound REFLECTED from mountain appears to be coming from mirror of SOURCE. Thus image appears to be moving TOWARDS the vehicle (DRIVER) `v_(0)=v_(s)=x :.n.=n((V+V_(0))/(V-V_(s)))` `:.(l_("air"))/(l_("water"))=sqrt((T_("air"))/(T_("water")))` |
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| 35. |
The radius vector is 2hati+hatj+hatk while linear momentum is 2hati+3hatj+hatk. Then the angular momentum is |
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Answer» `-2hati+4hatk` |
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| 36. |
Figures (i) and (ii) below show the displacement - time graphs of two particles moving along the x -axis. We can say that |
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Answer» both the PARTICLES are having UNIFORMLY accelerated motion. |
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| 37. |
A particle moving along straight line has velocity v=mus^(2) where s is the displacement. If s= s_0 then which of the following graph best represent s versus t? |
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Answer»
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| 38. |
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s^(-1) . If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball.(Assume linear motion of the ball) |
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Answer» Solution :Mass ofball m= 0.15kg andinitialvelocityof ball `v_(0)= -12ms^(-1)` Finalvelocityafterhittingv= 12 `ms^(-1)` impulseof force`=m DELTA v ` `=0.15(v-v_(0))` =0.15[1212] =3.6 Ns DIRECTIONOF impulseis from BATSMAN to bowler. |
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| 39. |
Why aeroplanes and cars have streamline shape? |
| Answer» Solution :AEROPLANES and cars have streamline shape in order to reduce the friction of air. During MOTION, all particles coming in FRONT of them will be directed tangentially. So they can MOVE with greater SPEED easily. | |
| 40. |
An open pipe resonates to a frequency f_(1) and a closed pipe resonates to a frequency f_(2). If they are joined together to form a longer tube, then it will resonate to a frequency of (neglect end corrections) |
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Answer» `(f_(1)f_(2))/(2f_(2)+f_(1))` |
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| 41. |
(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? |
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Answer» SOLUTION :(a)100 rev/min (use angular momentum CONSERVATION). (B)The new kinetic energy is 2.5 times the initial kinetic energy of rotation.The CHILD uses his internal energy to INCREASE his rotational kinetic energy. |
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| 42. |
A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A. Its potential energy is given as U(x) = ax^4 where alpha is a positive constant. The x-coordinate of particle where its potential energy is one third the corresponding kinetic energy of particle is |
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Answer» `+- A/2` |
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| 43. |
A mass of 6kg is suspended by a rope oflength 2m from a celling a force of 50N in the horizontaldirection is applied at themidpoint of the rope as shown in the figure what is the angle the rope makes with the vertical is equalibrium take g=10 ms^(-2)neglect mass of the rope |
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Answer» Solution :Givenm= 6KG length = 2 m force=50 N g=10`MS^(-2)` Asshownin thediagramthere are threeforcesactingon the midpointp of the ropeSupposethe ropemakesan angle`theta` with theverticel inequilibriumresolvingthe forceshorizontally and VERTICALLY we get `T_(1) SIN theta=T_(3) = 50 N` `T_(1) cos thetaT_(2)= 6kgwt = 60N` Dividing (1) by (2) we get tan `theta ` `=(5)/(6) ` (or) `thetatan ^(-1) ((5)/(6))` `=tan^(-1)(0.833)` `=theta= 39 .8^(@)` |
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| 44. |
A mixture of two non-reactive ideal gases isenclosed in a vessel consisting of one mole of a monatomic gas .A. and .n. moles of diatomic gas .B. at a temperature .T.. If the adiabaticconstant of the gaseous mixture is(13)/(9)then the value of .n. is |
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Answer» 5 |
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| 45. |
State in the following cases, whether the motion is one, two or three dimensional: Motion of a butterfly in search of honey. |
| Answer» SOLUTION :THREE DIMENSIONAL | |
| 46. |
The bob of a pendulum of length 980 cm is released from rest with it string making an angle 60^(@) with the vertical. It collides with a ball 20g resting on a smooth horizontal surface at the position of rest of pendulum. If the collision is perfectly elastic, with what velocity will the bob move immediately after the collision ? Mass of the bob is 10g |
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Answer» `490 CM s^(-1)` |
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| 47. |
Onahorizontalsurface. 2 kgblockis placedeastwardand 4Kgblocksare nowconnectedwitha horizontalstring. If12Nhorizontal force is appliedon 4 Kgblockalongwestdirection, find(i)commonaccelerationof twoblocks(ii)tension in thestring |
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Answer» Solution :The12Nforcealongwestdirectionactson( 4+2 )Kg .Or 6Kg mass. (I )Common acclerationoftwo BLOCKS`=(12)/( 6 ) = 2 m.s ^(-2)` (ii )theforceappliedon 2kgmass alongwestdirectionby thestring`=2xx2=4N ` ` therefore ` TENSIONIN thestring= 4N |
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| 48. |
(A) : The planets appear to move slower when they are farther from the sun than when they are nearer. (R) : Law of conservation of angular momentum is valid for any central force. |
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Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) |
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| 49. |
The total mechanical energy of a spring-mass system in simple harmonic motion is E= (1)/(2)m omegaA^(2). Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same, The new mechanical energy will |
| Answer» Answer :D | |
