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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12551. |
(A): If radius of a sphere attached to a wire is doubled, elongation of the wire reduces to one forth of original value.(R): Elongation in a wire is directly proportional to area of cross section. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 12552. |
A particle of mass m moving in a circular path of constant radius r such that its centripetal acceleration a_(c) is varying with time t as a_(c)= k^(2)rt^(2), where k is a constant. |
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Answer» Solution :As `a_(c )=(V^(2)//R)` so `(v^(2)//r)=k^(2)rt^(2)` KINETIC energy K `= (1)/(2)mv^(2)=(1)/(2)mk^(2)r^(2)t^(2)` Now by work - Energy Theorem `W=Delta K = (1)/(2)mk^(2)r^(2)t^(2)-0rArr P =(dW)/(dt)` `RARR P=(d)/(dt)=(1)/(2)mk^(2)r^(2)t^(2)=mk^(2)r^(2)t` |
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| 12553. |
In column-I there are two graphs and in Column-II whose graph is for this are given . Joun them appropriately : |
| Answer» SOLUTION :`(a-ii), (b-iii)` | |
| 12554. |
Give the magnitude and direction of the net force acting on a car moving with a constant velocity of 30km/h on a rough road |
| Answer» Solution : Force is being applied to OVERCOME the force of friction . But as VELOCITY of the car is constant , it ACCELERATION . A = 0 . HENCE net force in the car F = ma = 0 | |
| 12555. |
If force is the cause then the effect is |
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Answer» mass |
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| 12556. |
Figure shows a siphon. The liquid shown in water. The pressure difference P_(B)-P_(A) between the points A and B is |
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Answer» `400N//m^(2)` |
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| 12557. |
A smooth sphere of radius R is fixed on a horizontal surface. A steel, moving horizontally with velocity v_(0)=5sqrt(gR), collides with the sphere, changes its direction and moves with velocity v as shown. Then |
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Answer» `4sqrt(2)v_(0)=5sqrt(3)V` |
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| 12558. |
(A): Displacement can decrease with time, but distance can never decrease with time. (R) : Distance can be many valued funciton, but displacement can be single valued function. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 12559. |
A river flows at the rate of 3 km/hr and a person can row a boat at a speed of 5 km/hr is still water. If the difference between the times taken to cross the river by the shortest path the quickest time be 4 minutes, find the width of the rivers. |
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Answer» `2 1/3 KM` |
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| 12560. |
A cylinder with a movable piston contains3 moles of hydrogen at standard tempera-ture and pressure. The walls of the cylinder arc made of heat insulator, and the piston is insulated by having a pile of sand on it. The factor of increase of the pressure when the gas is compressed to half its original volume is [given 2^(1.4)=2.64] |
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Answer» 2.64 |
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| 12561. |
A soap bubble of radius 'r' and surface tension 'T' is given a potential of 'V' volt. If the new radiys 'R' of the bubble is related to its initial radius by equation. P_(0)[R^(3) - r^(3)] + [lambda T [R^(2) - r^(2)] - epsilon_(0) V^(2)R//2 = 0, where P_(0) is the atmospheric pressure. Then find lambda |
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Answer» [`sigma^(2)/(2epsilon_(0))` is EXCESS pressure due to uniform charge distribution on the surface of a bubble pressure is larger than OUTSIDE]  Clearly `P_(A) XX (4)/(3)pir^(3) = P_(A)^(') xx (4)/(3) piR^(3) rArr P_(A) = P_(A) ((r)/(R))^(3)` Now `P_(A) - P_(0) = (4T)/(r)` & `P_(A) - P_(0) = (4T)/(R) - (sigma^(2))/(2epsilon) .......(3)` so form `P_(A)((r)/(R))^(3) - P_(0) = (4T)/(R) - (sigma^(2))/(2epsilon) .....(4)` `[Eq^(N)(2)xx((r)/(R))^(3)-Eq^(n)(4)]` `P_(0) - P_(0)((r)/(R))^(3)=(4T)/(r){((r)/(R))^(3)-(r)/(R)}+(V^(2)epsilonR)/(2)` `rArr P_(0) (R^(3) - r^(3)) + 4T (R^(2) - r^(2)) - (V^(2)epsilon_(0)R)/(2) = 0`., (Hence provide) Ans. `lambda = 4` |
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| 12562. |
Diatomic gas at pressure 'P' and volume 'V' is compressed adiabatically to 1/32 times the original volume . Then the final pressure is |
| Answer» ANSWER :C | |
| 12563. |
Upon heating, the length of the side of a cube changes by 2%. The volume of the cube changes by |
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Answer» 0.01 |
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| 12564. |
100g of water is supercooled to -10^(@)C. At this point, due to some disturbance mecha-nised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? [S_(w)=1"cal"//g//""^(@)CandL_("Fusion")^(w)=80"cal"//g] |
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Answer» `0^(@)C,12.5g` |
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| 12565. |
A spring is cut into two halves. What happens to its period of oscillation of a mass suspended from it, before cutting and after cutting ? |
| Answer» Solution :Before cutting LET the period be `T_1`and after cutting `T_2`When it is cut into two EQUAL parts the force constant of each PART becomes 2K and hence new period `T_2 = 1/sqrt2T_1` | |
| 12566. |
To a man running upwards on the hill, the rain appears to fall verttically downwards with 4 m/s. The velocity vector of man with respect to the Earth is(2i+3j) m/s.If the man starts running down the hill with the same speed. The relative speed of the rain with respect to the man is |
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Answer» `4I +2J` |
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| 12568. |
Obtain the Coefficient of performance (COP). (beta). |
| Answer» Solution :It is defined as the ratio of heat extracted from the cold BODY (sink) to the external WORK done by the COMPRESSOR W. | |
| 12569. |
Stationary waves produced in a string of length 60 cm is described by y = 4 sin ((pix )/(15))cos (96pi t)(where x and y are in cm and is in s). Find (1) position of nodes (il) positions of antinodes (iii) maximum displacement of a particle at * 5 cm (iv) equations of component waves of given stationary wave. |
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Answer» Solution :Comparing `y = 4 sin ((pix )/(15)) cos (96 pi t )` with `y = 2 A sin (KX) cos (omega t ) ` we get, `2A = 4 implies A = 2 cm` `K = (pi)/(15) (rad)/(cm) = (2pi)/(lamda ) implies lamda = 30 cm` `omega = 96 pi rad//s` (i) Nodes are located at `0, (lamda)/(2) , lamda, (3 lamda)/(2), 2 lamda,...` `=15 cm, 30 cm, 45 cm, 60 cm,...` (ii) Antinodes are located at `(lamda)/(4) , (3lamda)/(4), (5lamda)/(4), (7lamda)/(4),...` `=7.5 cm , 22.5 cm, 37.5 cm 52.5 cm,...` (iii) Amplitude of stationary wave at `x = 5 cm = 2 A sin (kx)` `= 4 sin ((pi)/(15) xx 5)` `= 4 sin ((pi)/(3)) = 4 xx 0.8660 = 3.464 cm` (iv) We have `y = 4 sin ((pix )/( 15)) cos (96 pi t )` `THEREFORE y = 2 xx 2 sin alpha cos beta` (Where `alpha = (pi x )/(15) and beta = 96 pi t )` `=2 {sin (alpha + beta) + sin (alpha -beta)}` (By formula) `=2 { sin ((pi x )/( 15) + 96 pi t ) + sin ((pi x)/( 15) - 96 pi t )}` `=2 sin ((pi x )/( 15) + 96 pi t ) + 2 sin ((pi x )/(15) - 96 pi t )` `= y _(1) + y _(2)` Where `y _(1) = 2 sin ((pi x )/(15) + 96 pi x ) cm` `y _(2) = 2 sin ((pi x )/( 15) - 96 pi t ) cm` are the two component waves of given stationary wave. |
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| 12570. |
A time varying power and velocity of particle as function of mass m. find. (a) kinetic energy and velocity of particle as function of time. (b) average power over a time intrval from t=0 to t=t. |
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Answer»
`:. v=sqrt((2)/(m))t` (B) `P_(AV) =W/t =t^(2)/t =t` |
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| 12571. |
Two particles A and B are thrown simu ltaneously from the same point at the same angle of projection but with the two different initial velocities (v+u) and (v-u) respectively. Which of the following statements will b e true in respect of their motions? |
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Answer» The difference in their maximum heights is (2uv/g) `therefore Deltat=t_(1)-t_(2)=(2u sin theta)/(g)` |
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| 12572. |
(A): A refrigerator transfers heat from lower temperature to higher temperature (R ): Heat cannot be transferred from lower temperature to higher temperature without doing any external work |
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Answer» Both (A) and(R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 12573. |
Ball B, of mass mg, is suspended from a string of length /attatched to cart A, of mass m, which may roll freely on a frictionless horizontal surface. If the ball is given an initial horizontal velocity v, while the cart is at rest, determine (a) the velocity of B as it reaches its maximum height, (b) the maximum heighth through which В will rise. |
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Answer» |
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| 12574. |
If K_i and K_f are the initial and final values of kinetic energy of a body respectively, then the work done by the net force on the body is equal to |
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Answer» `(K_f K_i)/(K_f - K_i)` `W = K_f - K_i`. |
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| 12575. |
A body (A) of mass 100 g and another body (B) of mass 400 g approach each other with velocities of 100 cmcdot s^(-1) and 10cm cdot s^(-1) respectively. After a head on collision, the bodies coalesce. After collision, find the displacement of the combined mass in 10 s . |
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Answer» |
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| 12576. |
Statement - A: If a uniform metal disc is remoulded into a solid sphere, then the moment of inertia about the axis of symmetry increases than that before. Statement - B: For a given body and for a given plane, the moment of inertia is minimum about an axis passing through the centre of mass. |
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Answer» Both A and B are WRONG |
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| 12577. |
If length and breadth of a plane are (40 + 0.2) and (30 +- 0.1) cm, the absolute error in measurement of area is |
| Answer» Answer :A | |
| 12579. |
A cyclist is moving on a smooth horizontal curved path of radius of curvature 10m. With a speed 10 ms^(-2). Then his angle of leaning is |
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Answer» `15^(@)` |
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| 12580. |
The front wind screen of a car is inclined at an 60^@ with the vertical. Hailstones fall vertically downwards with a speed of 5 sqrt(3) ms^-1. Find the speed of the car so that hailstones are bounced back by the screen in vertically upward direction with respect to car. Assume elastic collision of hailstones with wind screen. |
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Answer» `vec v_h = -(v_1)/(2) hat j + (v_0 - v_1 (sqrt(3))/(2)) hat i` Now `v_0 - v_1 (sqrt(3))/(2) = 0 rArr v_1 = (2 v_0)/(sqrt(3))` Also `((v_1)/(2))^2 + (v_0 - v_1 (sqrt(3))/(2))^2 = (5 sqrt(3))^2` `rArr v_1^2 = 75 xx 4 rArr (4 v_0^2)/(2) = 75 xx 4 rArr v_0 = 15 ms^-1`.  .
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| 12581. |
When the strong wind passing over the building the force acting on building is in …… |
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Answer» |
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| 12582. |
A thermodynamics system is taken through the cycle ABCS as shown in figure. Heat rejected by the gas during the cycle is |
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Answer» PV |
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| 12583. |
A particle moving with certain velocity collides elastically with another particle at rest. If it were head on collision, the K.E. transferred by the colliding particle is 100% when its mass is equal to |
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Answer» the MASS of the STATIONARY particle |
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| 12584. |
A whistle of frequency 540 Hz rotates in a circle of radius 2 m at an angular speed of 15rads^(-1). What is the lowest and highest frequency heard by a listener a long distance away at rest with respect to centre of the circle ? Can the apparent frequecny be ever equal to the actual frequency? Take V=330ms^(-1). |
Answer» Solution :The situation is shown in the figure.  Speed of source (WHISTLE), `v_(s)=romega=2xx15=30ms^(-1)` Actual frequency, `v=540Hz,` Speed of sound, `V=330ms^(-1)` When the whistle (at position B) is moving away from the observer, the apparent frequency is lowest. It is GIVEN by `v'=(V)/(V+v_(s))xxv=(330)/(330+30)xx540` =495Hz When the whistle (at position D) is moving towards the observer, the apparent frequency is highest. It is given by `v''=(V)/(V-v_(s))xxv=(330)/(330-30)xx540` =594 Hz. When the whistle is at A (or) C, its speed along. `OL=v_(s)cos09^(@)=0` `:.` Apparent frequency= Actual frequency. |
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| 12585. |
A small bottle of an ideal gas is brought into a thermally insulated room and it is broken into pieces in the room. What happens to the temperature of the room? |
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Answer» Solution :It remains constant, as free expansion of the GAS does not WORK, DW = 0 and no heat is SUPPLIED dQ = 0. ` thereforedU =o or U =` constant |
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| 12586. |
The escape velocity from earth is v_(e). A body is projected with velocity 2v_(e). With what constant velocity will it move in the inter planetary space ? |
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Answer» `v_(e)` `rArr-(GMm)/(R)+(1)/(2)m(2v_(e))^(2)=0+(1)/(2)MV^(2)` or `-(GM)/(R)+2v_(e)^(2)=(1)/(2)V^(2)` or `-(2GM)/(R)+(8GM)/(R)=v^(2)` or `v=sqrt((6GM)/(R))=sqrt(3((2GM)/(R)))=sqrt(3(2gR))=sqrt(3)v_(e)`. |
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| 12587. |
Two particles P and Q start from origin and execute SHM along X-uxis with same amplitude but the periods 3 seconds and 6 seconds respectively. The ratio of the velocities P and Q when they meet is |
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Answer» 1:2 |
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| 12588. |
Ten small planes are flying at a speed of 150 km//h in total darkness in an air space that is 20 xx 20 xx 1.5 km^(3) in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are, On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximately by a sphere of radius 10 m. |
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Answer» 125 h `V=20xx20xx1.2Km^(3).` Dimeter of plane, `d=2R=2xx10` `= 20 m 20xx10^(-3)km` `n=N/V=(10)/(20xx20xx1.5)=0.067Km^(-3)` Mean free path of a plane `lamda =(1)/(sqrt2pid^(2)n)` Time elapse before collision of TWO planes randomaly, `t=(lamda)/(v)=(1)/(sqrt2pid^(2)NV)` `=(1)/(1.414xx3.14xx(20)^(2)xx10^(-6)xx(0.0167)xx(150))` `=(10^(6))/(4449.5)=224.74h~~225h` |
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| 12589. |
When a metallic sphere falls through castor oil, its velocity become uniform, called terminal velocity. Write the expression for terminal velocity. |
| Answer» SOLUTION :`v=2/9 (a^2(rho-sigma))/ETA G` | |
| 12590. |
When a metallic sphere falls through castor oil, its velocity become uniform, called terminal velocity. The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20^@C is 6.5 cm/s. Compute the viscosity of the oil at 20^@. Density of oil is 1.5xx10^3kg//m^3. |
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Answer» SOLUTION :`a=2xx10^-3m, v=6.5xx10^-2ms^-1, rho=8.9xx10^3kgm^-3` `sigma=1.5xx10^3kgm^-3, ETA` = ? `v=2/9 (a^2(rho-sigma))/eta G` `v=2/9 (a^2(rho-sigma))/eta g = 2/9 ((2xx10^-3)^2 (8.9xx10^3-1.5xx10^3)9.8)/(6.5xx10^-2)=0.99 kgm^-1s^-1` |
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| 12592. |
The mass and radius of a planet are double that of the earth. The time perod of simple pendulum on the planet, if T is the time period of simple pendulum on earth. |
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Answer» `2T` |
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| 12593. |
These questions have statement I and statement II. Of the four choices given below, choose the one that best describes the two statements. Statement I : If a body tries to slip over a surface then friction acting on the body is necessarily equal to the limiting friction. Statement II :Static friction can be less than the limiting friction. |
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Answer» STATEMENT I is TRUE, statement II is true, statement II is a correct EXPLANATION for statement I. |
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| 12594. |
sqrt(58.97)= |
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Answer» 7.679 |
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| 12595. |
A car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. Which one will have greater momentum? |
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Answer» Solution :`K=P^2/(2M), P=sqrt(2mK)` `P PROP sqrtm` for the same KE. So TRUCK will have greater momentum. |
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| 12596. |
A car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads.If the mass of turck is 100 times greater than that of the car, find the ratio of velocity of the truck to that of the car . |
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Answer» SOLUTION :`1/2mv_c^2=1/2(100M)v_t^2` `v_t^2/v_c^2=1/100` `v_t/v_c=1/10` |
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| 12597. |
A plane mirror is placed at origin parallel to y-axis. Facing the posistive x-axis. An object starts from (2m,0,0) with a velocity of (2hat i+2hat j)m//s. Find the relative velocity of image with respect to object. |
Answer» SOLUTION : The RELATIVE velocity of image with respect to object along nomral =`-4 hat i` The relative velocity image with respect to object along plane of MIRROR =0. Hence the relative velocity of image with respect to object= `-4 hat i` |
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| 12598. |
The maximum value of magnitude of (vecA-vecB) is |
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Answer» A-B |
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| 12599. |
Obtain the equation of frequency observed by moving observer and stationary source. |
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Answer» Solution :When the observe is MOVING with velocity `v _(0)` towards the source and the source is at rest, we have to proceed in a different manner. We work in the reference frame of the moving observer. In this reference frame the source and medium are approaching at speed `v _(0)` and the speed withwhich the wave approaches is `v _(0)+v .` Following a similar PROCEDURE as in the previous case, we find that the time interval between the arrival of the FIRST and the `(N+1) ^(th)` crests is, `t _(n+1) -t_(1) = n T_(0) -(n v _(0) T_(0))/(v _(0) +v)` To observer thus, measures the period of hte wave to be, ` T = (t _(n +1) -t _(1))/(n)` `T_(0) (1- (v _(0))/(v _(0) + v)) = T _(0) ((v)/(v _(0) + v ))` `= T _(0) ((1)/(v _(0) //v + 1))` `= T _(0) (1 + (v _(0))/( v))` giving, `v = v _(0) [ 1 + (v _(0))/(v)]` When observer moves toward stationary source then observed FREQUENCY is less than original frequency. When observe moves away from stationary source then `v _(0)` should be used in place of `-v _(0)` in equation `therefore` Frequency observed by observer, `v =v _(0) [1- (v _(0))/(v)]` Thus, frequency observed by observer will be greater than original frequency. |
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| 12600. |
If two waves ofsame amplitude produce a resultant wave ofsame amplitude a phase difference between the two is |
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Answer» `60^0` |
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