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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.
| 2. |
What is circular motion? Give example. |
| Answer» Solution :Circular MOTION is DEFINED as a motion described by an OBJECT traversing a circular path Eg The WHIRLING motion of a stone attached to a STRING. | |
| 3. |
Two block of masses m and 2m are kept on a smooth inclined plane and the system is pushed using force 3mg as shown. Find the contact force between those two blocks |
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| 4. |
On a two-lane car A is travelling with a speed of 36 "km.h"^(-1) . Two cars B and C approach car A in opposite directions with a speed of 54 "km.h"^(-1) each. At a certain instant when the distance AB is equal to AC both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident? |
Answer» Solution :Velocityof car A, `v_A =36 km *h^(-1) = 10 m*s^(-1)` [Fig.2.92] . Let `v_B and v_C` be the velocities of CARS B and C. `therefore v_B =v_C =54km *h^(-1) =5 m *s^(-1)` Velocity of car Brelative to A, `v_(BA)=v_B-v_A =15-10=5 m*s^(-1)` Velocity of car C relative to A , `v_(CA)=v_C-(-v_A)=v_C+v_A=15+10=25 m*s^(-1)` Also the distance of B and C from A is 1 km =100m. Let t=time taken by car C to travel a distance 1 km towards A . `therefore S=v_(CA)t or, 1000 =25t or, t=40s ` SUPPOSE a =minimumu ACCELERATION required for car B to avoid the accident . With this acceleration , car B overtakes car A in 40 s . So, `S=v_(BA)t+1/2at^2 or, 1000 =5xx40 +1/2axx(40)^2` `therefore a=((1000-200)xx2)/((40)^2)= 1m *s^(-2)` |
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| 5. |
In a hardbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the hardbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ? |
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Answer» EAST (APPROXIMATELY) |
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| 6. |
A body starting with a velocity 'v' returns to its initial position after 't' second with the same speed, along the same line. Acceleration of the particle is |
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Answer» `(-2V)/t` |
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| 7. |
What are fluids? |
| Answer» SOLUTION :A fluid is a substance that can FLOW. It ultimately ASSUMES the shape of the CONTAINER. | |
| 8. |
A Particle of mass M is moving in a straight line with uniform speed v parallel to x - axis in X - Y plane at a height .h. from the X - axis. Its angular momentum about the origin is |
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Answer» Zero |
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| 9. |
Assertion: The bodies in the satellites revolving round the earth are in the state of weightlessness Reason : The net acceleration due to gravity due to the earth is zero in the satellite |
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Answer» Both Assertion and REASON are TRUE and Reason is the correct explanation of Assertion |
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| 10. |
A constant torque acting on a uniform circular wheel changes is angular momentum from A to 4A in 4S. The torque acted on it is |
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Answer» `(3A)/(4)` |
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| 12. |
A speck of negligible mass begins to fall and gathers water vapour at a constant rate beta. Calculate its acceleraton and its mass when it falls 2.45 m. |
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| 13. |
If the displacement of a particle is represented by y(t)=A sin (omega t+phi), then give the expression for the maximum accleration of the particle. |
| Answer» SOLUTION :The instantaneous acceleration of a particle, `a(t)=-omega^(2)A sin(omega t+phi).` The term, `omega^(2)A`, REPRESENTS the maximum acceleration. The acceleration is maximum at the EXTREME position directed towards the mean position. | |
| 14. |
A particle is executing SHM with an amplitude oif 0.2 m. At what distance from themean position the potential energy of the particle will be equal to its kinetic energy? |
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Answer» Solution :P.E at the DISPLACEMENT x `P.E.=1/2m^(2)OMEGA^(2)x^(2)`………..1 K.E at the displacement x is given by `K.E. =1/2m^(2)omega^(2)(A^(2)-x^(2))`………2 But `PE=KE` i.e. `1/2m^(2)omega^(2)x^(2)=1/2 m omega^(2)(A^(2)-x^(2))` `impliesx^(2)=A^(2)-x^(2)` or `implies2x^(2)=A^(2) x^(2)=(A^(2))/2` `:. x=A/(SQRT(2))` But`A=0.2m` `:.x=(0.2)/(sqrt(2))=0.1xxsqrt(2)=0.1414m` |
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| 15. |
List the factors affecting the mean free path. |
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Answer» Solution :(1) Mean free path increases with increasing TEMPERATURE. As the temperature increases, the average speed of each molecule will increase. It is the REASON why the SMELL of hot sizzling food reaches several metre AWAY than smell of cold food. (ii) Mean free path increases with decreasing PRESSURE of the gas and diameter of the gas molecules. |
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| 16. |
A car moving with constant acceleration crosses twopoints A and B with velocities 10 m/s and 20 m/s in its path. Find the velocity of the car midway between A and B. (in m/s). |
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Answer» `sqrt(500)` |
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| 18. |
The displacement of a particle is related to time by the expression s=kt^(3), what is the dimension of the constant k? |
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Answer» |
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| 19. |
If R is the horizontal range for theta inclination and h is the maximum height reached by the projectile, show that the maximum range is given by (R^(2))/(8h)+2h |
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Answer» SOLUTION :We know that horizontal range `R=(U^(2)sin 2 theta)/g` and MAXIMUM height `h=(u^(2)sin^(2)theta)/(2g)` `:.(R^(2))/(8h)+2H=([(u^(2)sin 2 theta)/g]^(2))/(8[(u^(2)sin^(2)theta)/(2g)])+2[(u^(2)sin^(2)theta)/(2g)]` `=(u^(4)(2 sin theta cos theta)^(2))/(g^(2)xx8(u^(2)sin^(2)theta)/(2g))+(u^(2)sin^(2)theta)/g` `=(u^(2))/g(cos^(2)theta+sin^(2)theta)=(u^(2))/g=R_("MAX")` |
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| 20. |
The co-ordinates of centre of mass of particles of mass 10, 20 and 30 gm are (1, 1, 1) cm. The position co-ordinates of mass 40 gm which when added to the system, the position of combined centre of mass be at (0, 0, 0) are, |
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Answer» (3/2, 3/2, 3/2) |
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| 21. |
The following graph shows a V-T graph for isobaric processes at two different pressures. Identify which one occurs at higher pressure. |
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Answer» Solution :From the ideal GAS equation, `V=(MUR)/(P)T` `V-T` graph is inversely PROPORTIONAL to the pressure. If the slope is greater, LOWER is the pressure. The slope `=(muR)/(P)` Here `P_(1)` has larger slope than `P_(2)`. So `P_(2) gt P_(1)`. 
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| 22. |
A boy is running on a circular track of radius 50m. Calculate the displacement of the boy after completing 5 rounds of the track . |
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Answer» Solution :Since FINAL and initialpositions are same HENCE displacement of the BOY will be , `DELTA r = r - r = 0` |
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| 23. |
A and B are the numerical values of a physical quantity and C and D are its units in two system of measurement. If C gt D, then |
| Answer» Answer :B | |
| 24. |
From the following statements select the incorrect statements. |
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Answer» 1 giga watt = `10^(9)` watt |
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| 25. |
When white light passes through the achromatic combination of prism, then what is observed |
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Answer» Only deviation |
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| 26. |
The work done by all the forces (external and internal) on a system equals the change in |
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Answer» TOTAL energy |
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| 27. |
A block of mass 1kg is pushed on a movable wedge ofmass 2kg and height h=30 cm with a velocity u=6m//sec. Before striking the wedge it travels 2m on a rough horizontal portion. Velocity is just sufficient for the block to reach the top of the wedge. Assuming all surfaceare smooth except the given horizontal part and collision of block and wedge is jerkless, the friction coefficient of the rough horizontal part is : An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of 800m and moving vertically upward at a constant velocity of 5m//s. The horizontal displacement of balloon from the observer is 1600m. Shells fired from the gun have an initial velocity of 400 m//s at a fixed angle theta ( sin theta =3//5 and cos theta =4//5 ). The observer having gun waits ( for some time after observing the balloon) and fires so as to destroy the balloon. Assume g=10m//s^(2). Neglect air resistance. |
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Answer» `0.125` Apply WORK energy THEOREM `mu mg (2)+mgh=KE_(i)-KE_(f)...(1)` at the highest point `V_(block)=V_(wedg e)` `2-mu+3=(1)/(2)1(6)^(2)-(1)/(2)3v^(2)` `mu=(54)/(80)=0.675` |
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| 28. |
If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental units, the possible relation to gravitational constant (G) is.............. |
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Answer» `c^(0)GP^(-3)` |
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| 29. |
When is the exchange of energy maximum during an elastic collision. |
| Answer» SOLUTION :ENERGY exchange will be maximum of the two colliding bodies are of equal MASSES. | |
| 30. |
Flow cytometry, illustrated in the figure, is a technique used to sort cells by type. The cells are placed in a conducting saline solution which is then forced from a nozzle. The stream breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets seprate from the stream. Charging the collar polarizes the conducting liquid, causing the droplets to become charged as they break off from the steam. A leser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplats with no desired cell receive no charge. The charged. The charges droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs force that directs them into different collection tubes, depending on air charge. |
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Answer» `+2q, +Q, O, -q, -2q` |
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| 31. |
Flow cytometry, illustrated in the figure, is a technique used to sort cells by type. The cells are placed in a conducting saline solution which is then forced from a nozzle. The stream breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets seprate from the stream. Charging the collar polarizes the conducting liquid, causing the droplets to become charged as they break off from the steam. A leser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplats with no desired cell receive no charge. The charged. The charges droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs force that directs them into different collection tubes, depending on air charge. Another way to sort the droplets would be to given each droplet the same charge, then vary the electric field between the deflection plates. For the apparatus as sketched, this technique will not work because |
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Answer» Several droplets are between the plates at one time, and they would all feel the same force. |
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| 32. |
Passage - II : Uniform rod AB hinged about a fixed point 'P' is initially vertical. Rod is released from vertical position. When rod is in horizontal position. The acceleration of centre of mass of rod is |
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Answer» `-(6G)/(7)hat(i)-(3g)/(7)hat(J)` |
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| 33. |
A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.2. What is the maximum speed in kmph of the car for safe driving ? (g=10 ms^(-2)). |
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Answer» Solution :Maximum speed. `v=sqrt(mu_(s)GR)` `v=sqrt(0.2xx10xx18)=sqrt(36)=6 MS^(-1)` |
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| 34. |
Passage - II : Uniform rod AB hinged about a fixed point 'P' is initially vertical. Rod is released from vertical position. When rod is in horizontal position. The acceleration of end B of rod is |
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Answer» `-(6G)/(7)HAT(i)-(12)/(7)hat(j)` |
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| 35. |
Passage - II : Uniform rod AB hinged about a fixed point 'P' is initially vertical. Rod is released from vertical position. When rod is in horizontal position. Reaction on hinge is |
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Answer» `-(8mg)/(7)hat(i)-(12mg)/(7)hat(J)` |
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| 36. |
What is the effect on (i) the force constant of the spring (ii) the frequency of oscillations? If the amplitude of oscillations of a loaded spring is doubled. |
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Answer» SOLUTION :When amplitude of a loaded spring is doubled then there is no change in the VALUE of force CONSTANT of the spring. (i) Force constant K `=("Force")/("displacement")`. (ii) Time period of a loaded spring `T= 2pi (m/k)` . It is independent of amplitude of oscillation. |
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| 37. |
A block of mass 'm' is tied to one end of string which passes over a smooth pulley 'A' and under a light smooth movable pulley B. The other end of the string is attached to lower end of a string constant K_2. Find the period of small oscillations of mass 'm' about its equilibrium position. |
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Answer» Let T be extra tension T= ma If `x_1 and x_2` are the EXTENSIONS of the springs `T= k_2 x_2 and 2T = k_1 x_1 ` `x= 2(x_2 + 2x) = (T)/(k_2) + (2T)/(k_1) = 2T[(k_1 + 2k_2)/(k_1 k_2)]` from T = ma `((k_1 k_2)/(k_1 + 2k_2)) x/2 = ma` `omega = sqrt((k_1 k_2)/(2(k_1 + 2k_2))) m , T= 2pi sqrt((2M(k_1 + 2k_2))/(k_1 k_2))` |
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| 38. |
A body moving with a uniform acceleration had velocities of 20 m/s and 30 m/s when passing the points P and Q of its path. Find the velocity midway P and Q ( inm/s) |
| Answer» Answer :C | |
| 39. |
The displacement of a particle starts from rest is proportional to the square of time, then the particle travels with |
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Answer» UNIFORM acceleration |
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| 41. |
State Kepler's laws on planetary motion. Explain the way the three laws can be proved. |
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Answer» Solution :1st part: 2nd part: If `r_1 and r_2` are the shortest and the longest distances of the planet from the sun , then the semimajor axis is given by `((r_1+r_2)/2)`. 1st law is proved from angular momentum conservation INA CIRCULAR path. Since, L= `2mxx` are VELOCITY , for equal time, area sweptwill be same . so, 2nd law is proved. From `F=(GM_0m)/(r^2),(mv^2)/r=F and T=(2pir)/(v)` one can prove `T^2prop r^3`. so, 3RD law is also proved. |
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| 42. |
The book titled 'Principia' was written by |
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Answer» Maxwell |
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| 43. |
A body is projected with the velocity u_(1) from the point A as shown in Figure. At the same time another body is projectd vertically upwards with the velocity u_(2) from the point B. What shold be the value of (u_(1))/(u_(2)) for both the bodies to collide. |
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Answer» 2 `(v_(1)^(2)SIN^(2)30^(@))/(2g)=(v_(2)^(2))/(2g)or(v_(2))/(v_(1))=sin30^(@)=(1)/(2)` |
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| 44. |
A rod of length 2 m is kept vertical inside a smooth spherical shell of radius 2m. The rod starts slipping insidethe shell . Mass of the rod is 4 kg. Angular speed of the rod(in rad/s) in theposition when it becomes horizontal is |
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Answer» 4.6 |
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| 45. |
A rod of length 2 m is kept vertical inside a smooth spherical shell of radius 2m. The rod starts slipping insidethe shell . Mass of the rod is 4 kg. Velocity of centre of the rod (in m/s) at the instant is approximately |
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Answer» 5.5 |
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| 46. |
The angular displacement of a particle is given |
| Answer» SOLUTION :`OMEGA=(d THETA)/(DT)=omega_(o)+alphat=4"RAD"//s` | |
| 47. |
An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere: The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant. |
| Answer» ANSWER :A | |
| 48. |
To a person travelling due east, the wind appears to come from north east, but when he doubles his speed it appears to came from a direction Tan^(-1) (2//3) north of east. The direction of wind is tan^(-1)1//k. West of south. Where 'k' is |
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Answer» |
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| 49. |
A uniform disc of mass of 0.50 kg and radius 0.20 m lies on one side initially at rest on a frictionless horizontal surface. A constant force of 3 N is then applied tangentially at its perimeter using a light weight cord wrapped aroutid its edge. What is the (i) angular acceleration (ii) angular velocity and angular displacement after 2s ? |
| Answer» Solution :`tau=Tr = Ialpha = (1//2)mr^(2) xx (a//r) =(1//2) mar, T=(1//2) ma, ma =F-T, F=(3//2)ma, a=2F//3m = 2 xx 3//3 xx 0.5 = 4 m//s^(2), a=a//r = 4//0.2 = 20 "rad/s"^(2) omega = omega_(0) + alphat = 0+20 xx 2 = 40 "rad/s", THETA = (1//2) at^(2) =(1//2) xx 20 xx 2^(2) = 40 "rad"` | |
| 50. |
(A) : Stress is a not vector quantity. (R): Stress can not be assigned a specific direction |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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