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. |
A particle projected withvelocity v_(0), strikes at right angles alpha plane through the point of projection and of inclination beta with the horizontal. Find the height of the point struck, from horizontal plane through the point of projection. |
|
Answer» Solution :Let `alpha` be the ANGLE between the velocity of projection and the inclined plane. `v_(ox.)=v_(0)cos alpha, v_(oy.)=v_(0)sin alpha` `a_(x.)=-g beta""a_(y.)=-g cos beta` `implies v_(x.)(t)=v_(0)cos alpha-g sin beta t` At the point of impact `v_(x.)=0impliest=(v_(0)cos alpha)/(g sin beta)`........i  At y. at the point is zero `impliesv_(0)sin alpha t -1/2 g cos beta t^(2)=0impliest=(2v_(0)sin alpha)/(g cos beta)`......ii From IA nd ii `(v_(0)cos alpha)/(g sin beta)=(2v_(0)sin alpha)/(g cos beta)` `tan alpha=1/2cot beta` .................iii `x=v_(0)cos(alpha+beta)t` `=V_(0)[cos alpha cos beta-sin alpha beta](V_(0)cos alpha)/(g sin beta)=(V_(0)^(2))/g[cos^(2)alpha cot beta-sin alpha cos alpha]` `=(v_(0)^(2))/g[(2/(sqrt(4+cot^(2)beta)))^(2)cotbeta-(cot beta)/(sqrt(4+cot^(2)beta))2/(sqrt(4+cot^(2)beta))]` (using `tan alpha=1/2cot beta)=(v_(0)^(2))/g .(2cot beta)/(4+cot^(2)beta)` From figure `:.y=x tan beta=(V_(0)^(2))/g . (2 cot beta)/(4+cot^(2)beta) tan beta impliesy=(2V_(0)^(2))/(g (4+cot^(2)beta))` |
|
| 2. |
Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end? |
|
Answer» Solution :Let the length of the rod = `l`cm, its weight = W = Mg , where M is the mass of the rod. When the support at one end is removed suddenly, centre of gravity of the rod FALLS downwards with an ACCELERATION a. Let R = reaction force at the end with the suppot. Hence if the C.G. now falls with an acceleration a, the rod will turn about the point P. The torque on the rod = `Mg.(l)/(2)` ALSO, Mg-R = Ma or, `a = ("Mg-R")/(M)` Here MOMENT of inertia `I= (I)/(3) Ml^(2)` = moment of inertia of the rod about the perpendicular axis passing through the end of the rod and the angular acceleration , `ALPHA=(alpha)/(l//2)=(2a)/(l)` `:. (1)/(3) Ml^(2)alpha=Mg""(l)/(2)[because tau=Ialpha]` or,`(1)/(3)Ml^(2).(2a)/(l)="Mg"(l)/(2)or,(2)/(3)a=(g)/(2)or, (2)/(3)(("Mg-R)/M)=(g)/(2)` `:. R= ("Mg")/(4)=(W)/(4)` Therefore, when one support is removed the support at the other end will exert a reaction force of `(W)/(4)`. 
|
|
| 3. |
Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50cm. Assuming that only gravitational forces acting on the particles mutually , the initial acceleration of the first particle. |
|
Answer» SOLUTION :Gravitational force between the two PARTICLES `F = (Gm_(1)m_2)/r^2` `F = (6.67xx10^(-11)xx1xx2)/((0.5)^2)=5.3xx10^(-10)N` The acceleration of 1.0 Kg PARTICLE is `a_(1)=F/m_1=(5.3 xx10^(-10))/1=5.3xx10^(-10)m//s^2` It is towards the 2.0 Kg particle. |
|
| 4. |
Which of the following are correct for rolling friction |
|
Answer» The extent of deformation of the surfaces in CONTENT |
|
| 5. |
A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) =A_(2) sin [omega t+(pi)/(3)], then find the maximum speed of the particle. |
| Answer» SOLUTION :Maximum SPEED of the PARTICLE `v_("max") = A omega`. | |
| 6. |
The gravitation force exerted on rocket a time toheight h from the surface of earth is 1/3 time to the gravitation force exerted at the surface of sea. Then obtain the relation between h and R_e(radius of earth). |
|
Answer» SOLUTION :`IMPLIES (GM_em)/((R_E+H)^2)=1/3((GM_em)/r^2)` `:. (R_e+h)^2 =3R_e^2` `:. R_e + h = SQRT3 = 1.732 R_e` `:. h = 0.732 R_e` |
|
| 7. |
This question has statement 1 and statement 2. Of the four choice given after the statements, choose the one that best describes the two statements. If two springs S_(1) and S_2 of the force constant K_(1) and K_2 respectively, are stretched by the same force, it is found that more work is done on spring S_(1) than on spring S_2 Statement-1: If stretched by the same amount, worked done on S_(1) will be more than that of S_2 Statement-2 : k_(1) lt k_(2) |
|
Answer» STATEMENT-1 is false, statement 2 is true |
|
| 8. |
A particle executes two types of SHM. x_(1) = A_(1) sin omega t " and "x_(2) =A_(2) sin [omega t+(pi)/(3)], then find the maximum acceleration of the particle. |
|
Answer» Solution :MAXIMUM ACCELERATION `= -OMEGA^(2) x` `= -omega^(2) XX (sqrt(3)A_(2))/(2)`. |
|
| 9. |
A long cylindrical metal vessel, having a linear coefficient (alpha), is filled with a liquid upto a certain level. On heating it, it is found that the height of liquid column in the cylinder remains the same. What is the volume coefficient of expansion of the liquid? |
|
Answer» Solution :Volume of liquid V = AL , Increase in volume `Delta V = (Delta A) l (THEREFORE l ` is constant ) `IMPLIES V gamma _(l) Delta T = (2 alpha) A.l Delta T implies V gamma _(l) DeltaT = 2 alpha V Delta T` `therefore V = Delta l implies gamma _(l) =2 alpha ` `implies `VOlume COEFFICIET of expansion of liquid `= 2 alpha` |
|
| 10. |
Which of the following is scalar quantity? |
|
Answer» momentum |
|
| 11. |
If .g. on the surface of the earth is 9.8 ms , find its value at a depth of 3200km (radius of the earth = 6400km) |
|
Answer» SOLUTION :With d = 3200 KM `g_(d)=G[1-d/R]impliesg_(d)=4.9ms^(-2)` |
|
| 12. |
The work done by a gravitational force acting on a free fall body will be positive , negative or zero ? |
| Answer» Solution :POSITIVE , Because the displacement of free FALL body is I the DIRECTION og GRAVITATIONAL force . | |
| 13. |
Once Amit was going to his house. He was listening music on mobile with earphone while crossing the railway line and he did not hear the sound of approaching train though the train was blowing horn. A person nearby ran towards him and push away just as the train reached there. amit realised his mistake and thanked the person. (a) Describe the value possessed by the person. (b) Name the phenomenon of change in frequency of sound when there is relative motion between the observer and source of sound. |
|
Answer» Solution :(a) The VALUES possessed by the person are - (i) pres-ence of mind, (II) general AWARENESS, (iii) good understanding, (IV) prompt decision-making ABILITY, (v) concern for other people.s safety and well-being, (vi) helping and caring nature. (b) The phenomenon is Doppler.s effect in sound. |
|
| 14. |
A mercury drop shape as a round tablet of radius R and thickness h is located between two horizontal glass plates. Assuming that hltltR, find the mass m of a weight which has to be placed on the copper plate to diminish the distance between the plates by n-times the contact angle is equal to theta. calculate m if T is surface tension of the liquid. |
|
Answer» `m=(2piRT^(2)|costheta|)/(gh)(n^(2)-1)` Difference `P=T[(1)/(r_(1))+(1)/(r_(2))]=T[(1)/((h)/(2|costheta|))+(1)/(R)]` As h is small in comparison to R, so `(1)/(R)ltlt(1)/(h),:P=(2T|costheta|)/(h)` Let `R^(')` becomes the new radius of curvature when the distance between the plates is decreased by `n-` times ASSUMING mercury to be in compressible, `piR^(2)h=pi(R^('))^(2)(h)/(n),(R^('))^(2)=nR^(2)` The force EXERTED by the mercury drop now becomes `F^(')=(2pi(R^('))T|costheta|)/(((h)/(n)))=n^(2)F` If mg be the weight placed on the upper plate, then `F^(')=F+mg,m=(2piR^(2)T|costheta|)/(gh)(n^(2)-1)` |
|
| 15. |
A body falls from height 20m. If co-efficient of restitution is(1)/(2),what is the time interval between starting pointand second bounce ? [g = 10 ms ^(-2)] |
|
Answer» |
|
| 16. |
What provides the restoring force in the following cases ? Compressed spring becomes force for oscillation. |
| Answer» SOLUTION :ELASTICITY of MATERIAL of SPRING. | |
| 17. |
The specific heat capacity of a gas at constantpressure is greater than that at constant volume because |
|
Answer» at constant volume, all the HEAT suppliedgoes to increase the internal ENERGY of the GAS |
|
| 18. |
What is meant by angular frequency ? |
| Answer» SOLUTION :Angular frequency is the number of oscillations PER SECOND of a PARTICLE executing simple HARMONIC motion. | |
| 19. |
A ball is thrownin a parabolic path. Is there any point at which the acceleration is perpendicular to the velocity . |
| Answer» Solution :Yes, the ACCELERATION of the ball is PARPENDICULAR to the velocity at the HIGHEST point of the PARABOLIC path . | |
| 20. |
The gravitational field in a region is given by vec(E) = (4hat(i) + 3hat(j)) N//Kg. The gravitational potential at the points (3m, 0) and (0, 4m). If the potential at the origin is taken to be zero. |
|
Answer» Solution :`VEC(V) = -(vec(E).vec(r))` At (3m, 0) `vec(V) = -(4hat(i) + 3HAT(J)) .(3hat(i)) = -12J//kg` At (0, 4m) `vec(V) = -(4hat(i) + 3hat(j)) .(4hat(j)) = -12J//Kg` |
|
| 21. |
A particle of mass M moves in a circular path if radius r with a constant speed equal to V. Then its centripetal acceleration is |
| Answer» Answer :A | |
| 22. |
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R= R_(0)[ 1 + alpha (T - T_(0))] . The resistances is 101.6 Omega at the triple point of water 273.16 K, and 165.5 Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Omega? |
|
Answer» SOLUTION :Here, `R_(0) = 101.6 OMEGA, T_(0) = 273.16` K Case (i) `R_(1) = 165.5 Omega, T_(1) = 600.5K ` Case (ii) `R_(2) = 123.4 Omega , T_(2) = ?` USING the relation R = `R_(0) [1 + ALPHA (T - T_(0) )]` Case (i) `165.5 = 101.6 [1 + alpha (600.5 - 273.16) ]` `alpha = (165.5 - 101.6 )/(101.6 xx (600.5 - 273.16)) = (63.9)/(101.6 xx 372.34)` Case (ii) 123.4 = 101.6`[ 1 + alpha (T_(2) - 273.16)]` or `123.4 = 101.6 [ 1 + (63.9)/(101.6 xx 327.34) (T_(2) - 273.16)] ` = 101.6 + `(63.9)/(327.34) (T_(2) - 273.16)` or `T_(2) ((124.4 - 101.6)xx 327.34)/(63.9) + 273.16` = 111.67 + 273.16 = 384.83K |
|
| 23. |
A particle of mass m is executing oscillations about the origin of the x-axis with amplitude A. Its potential energy U(x)=ax^(4), where a is a positive constant. What is the x - coordinate of the particle, where potential energy is one-third of the kinetic energy? |
|
Answer» |
|
| 24. |
A train of mass 400 tons climbs up an incline of 1//49 at the rate of 36kmh^(-1) . The force of friction is4kg' at per ton. Find the power of the engine. |
|
Answer» |
|
| 25. |
An athIete covers a circuIar path of radius 70 m in 55 s and returns to the starting point. Find the magnitude of his aevrage speed and averge velocity. If he compIetes the Iast 40 m of the path at the same rate in 4 s, find the vaIue of instantaneous speed and velocity in that intervaI of time. |
|
Answer» |
|
| 26. |
A block of massm = 1 kg moving on a horizontal surface with speed v_(1) = 2ms^(-1) enters a rough patch ranging from x = 010 m to x = 2.01. m. The retarding force F_(r) on the block in this range is inversely proportional to x over this range. F_(r) = (-K)/(x) for 0. 1 lt x lt 2.01m = 0 for x lt 0.1 and x gt 2.01m Where k = 0.5 j. What is the final kinetic energy and speed v_(f) of the block as it crosses this patch ? |
|
Answer» Solution :From work energy theorem `K_(f) - K_(i) = intFdx` `rArrK_(f)=K_(1)+underset(0.1)overset(2.01)INT((-k)/(X))DX=(1)/(2)mv_(i)^(2)-kln(x)|(2.01),(0.1)|` `=(1)/(2)mv_(1)^(2)-kln(2.01//0.1)=2-0.5 log_(e) (20.1)` `2 - 1.5 = 0.5j` `v_(f)=sqrt(2K_(f)lm)=1ms^(-1)` |
|
| 27. |
Two metal rods A and B are having their initial lengths in the ratio 2:3, and coefficients of linear expansion in the ratio 3:4. When they are heated through same temperature difference the ratio of their linear expansions is …………….. . |
|
Answer» SOLUTION :`Deltal_(A)=l_(A)alpha_(A)Deltat, Deltal_(B)=l_(B)alpha_(B)Deltat` `(Deltal_(A))/(Deltal_(B))=(l_(A)alpha_(A))/(l_(B)alpha_(B))=(2)/(3)xx()/(4)=(1)/(2)rArr Deltal_(A):Deltal_(B)=1:2` |
|
| 28. |
Tone A has frequency of 240 Hz. Of the following tones, the one which will sound least harmonious with A is |
|
Answer» 240 |
|
| 29. |
Statement I: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. Statement II: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. |
|
Answer» Statement I is true, statement II is true , statement II is a correct EXPLANATION for statement I. |
|
| 31. |
A point charge q is placed at one corner of a cubicalbox. Find the total flux associated with that box . |
|
Answer» Solution :If q is placed as hsown , seven moreidentical cubes are required to enclose , that charge completely. Then the total fluxthrough all faces of 8 cubes is`(q)/(in_(0))`. Nowcontribution of thisflux through each cube( INCLUDING thegiven one ) is `(q)/(8 in_(0))`. Here it is interesting to note thatthe faces whichhave common edgeon whichthegiven chargeis kepthave no flux and the flux `(q)/(8in_(0))` EMERGES through theremaining three faces fo the cube . |
|
| 32. |
A sphere contracts in volume by 0.01% when taken to the bottom of sea 1km deep. If the density of water is lgm /cc Then the bulk modulus of water is |
|
Answer» `9.8xx10^(5) NM^(-2)` |
|
| 33. |
A capacitor of capacitance C = 3.0 pm 0.1 muF is charged to a voltage of V = 18 pm 0.4 Volt. Calculate the charge Q[Use Q = CV]. |
|
Answer» SOLUTION :`(C +DeltaC) = (3.0 PM 0.1) muF` `(V+DeltaV) = (18pm0.4)V` Q = CV `Q = 3.0 xx 10^(-6) xx 18 = 54 xx 10^(-6)` COULOMB Error in `V = (DeltaV)/(V) xx 100 = (0.4)/(18) xx 100 = 2.2%` Error in Q = Error in C + Error in V = 3.3% + 2.2% = 5.5% `:.` Charge Q = `(54 xx 10^(6) pm 5.5%)` coulomb |
|
| 34. |
A particle is projected from the origin in X-Y plane. Acceleration of particle in Y direction is alpha. If equation of path of the particle is y=ax-bx^(2), then find initial velocity of the particle. |
|
Answer» SOLUTION :`y=ax-bx^(2)""y=xtantheta-(alphax^(2))/(2U^(2)cos^(2)theta)` `tantheta=a and (alpha)/(2u^(2)cos^(2)theta)=b""usqrt((alpha(1+a^(2)))/(2B))`. |
|
| 35. |
When gases are mixed, the internal energy will remain conserved. When external energy provides heat energy there may be temperature variation leading to internal energy change and/or volume variation spending/gaving work/energy. The change is internal energy is independent of the path followed, while work and heat energy depends on the path followed. Internal energy depends onthe number of degrees of freedom which in turn affects the ratio of specific heat capacities.Two perfect gases of n_1 and n_2moles at absolute temperatures T_1 and T_2are mixed. The temperature of the mixture will be, |
|
Answer» `T = (n_1T_2 + n_2T_1)/((n_1+n_2))` |
|
| 36. |
When gases are mixed, the internal energy will remain conserved. When external energy provides heat energy there may be temperature variation leading to internal energy change and/or volume variation spending/gaving work/energy. The change is internal energy is independent of the path followed, while work and heat energy depends on the path followed. Internal energy depends onthe number of degrees of freedom which in turn affects the ratio of specific heat capacities.If a gas has a ratio of y the number of degrees of freedom will be |
|
Answer» `(1)/(GAMMA - 1)` |
|
| 37. |
When gases are mixed, the internal energy will remain conserved. When external energy provides heat energy there may be temperature variation leading to internal energy change and/or volume variation spending/gaving work/energy. The change is internal energy is independent of the path followed, while work and heat energy depends on the path followed. Internal energy depends onthe number of degrees of freedom which in turn affects the ratio of specific heat capacities. In a process with a temperature change of dT, internal energy changes by dU = nC_vdT . The processis applicable for |
|
Answer» CONSTANT volume |
|
| 38. |
An aeroplane is flying with a velocity 800kmph relativeto air wind is belong from west to east with a velocity 15ms^(-1). Find velocity of aeroplane relative to the earth to fly south ward |
|
Answer» |
|
| 39. |
A merchant claims to sell pure gold coins of relative density 19.3. He sells gold coins of weight 26.075 gf to a customer. In order to check the purity of gold coins, the customer weighs the coins when immersed in pure water and funds that they weigh 24.225 gf in, water. Was the merchant lying? |
|
Answer» Solution :Weight of GOLD coins in AIR =26.075 gf Weight of gold coins in water =24.25gf LOSS of weight of coins in water =26.075-24.225 =1.85gf Relative DENSITY of the coins `=("weight of coins in air")/("loss of weight in water")=(26.075)/(1.85)` =14.095 Since the relatively density of the gold coins is much less than the CLAIMED by merchant (which is 19.3), so the merchant is lying. |
|
| 40. |
Obtain an expression for final velocities of two colliding bodes initially in motion. |
|
Answer» Solution :From the law of conservation of LINEAR MOMENTUM, `m_(1)v_(1I)+m_(2)u_(2i)=m_(1)v_(1f)+m_(2)v_(2f)` From the law of conservation of total kinetic energy, `(1)/(2)m_(1)v_(1i)+(1)/(2)m_(2)v_(2i)=(1)/(2)m_(1)v_(1f)^(2)+(1)/(2)m_(2)v_(2f)^(2)` i.e., `m_(1)v_(1i)^(2)+m_(2)v_(2i)^(2)=m_(1)v_(1f)^(2)+m_(2)v_(2f)^(2)` `m_(2)(v_(2i)+v_(2f))(v_(2i)-v_(2f))=m_(1)(v_(1f)-v_(1i))(v_(1f)+v_(1i))` `m_(2)(v_(2i)+v_(2f))(v_(2i)-v_(2f))=m_(1)(v_(1f)-v_(1i))(v_(1f)+v_(1i))` From (1) `m_(1)(v_(1f)-v_(1i))=m_(2)(v_(2i)-v_(2f))`, using (5) in (1) we get, `m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)(v_(1f)+v_(1i)-v_(2i))` `m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)v_(1f)+m_(2)v_(1i)-m_(2)v_(2i)` i.e., `v_(1i)(m_(1)-m_(2))+2v_(2i)(m_(2))=v_(i f)(m_(1)+m_(2))` `therefore v_(1f)=((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)+((2m_(2)v_(2i))/(m_(1)+m_(2)))` Similarly `v_(2f)=((m_(2)-m_(1))/(m_(1)+m_(2)))v_(2i)+((2m_(1)v_(1i))/(m_(1)+m_(2)))` |
|
| 41. |
Ship A is 10 km due west of ship B. Ship A heading directly north at a speed of 30 kmph while ship B is heading in a direction 60^(@) west of north at a speed 20 kmph. Their closest distance of approach will be……. |
|
Answer» Solution :`bar(V_(A)) = 30 j bar(V_(B)) = 20 Sin 60^(@)(hat(i)) + 20 Cos 60 (hat(j))` `= -10 SQRT(3)hat(i) + 10 hat(j)` `bar(V_(BA)) = bar(V_(B)) - bar(V_(A)) = -20 hat(j) - 10 sqrt(3)hat(i)` `Tan phi = (20)/(10sqrt(3)) = (2)/(sqrt(3))` and `Sin phi = (2)/(sqrt(7))` From `Delta ABC , (x)/(10) = (2)/(sqrt(7))` `x = (20)/(sqrt(7)) = 7.56 Km`  |
|
| 42. |
A uniform ladder is standing against a vertical smooth wall, resting on a horizontal floor. The ladder makes an angle of 45^(@) with the floor when its lower end is about to slip. What is the value of the coefficient of friction between the ladder and the floor? |
|
Answer» |
|
| 43. |
A cyclinder is completely filled with water. If 1/4 of the volume of water leaks out, its centre of mass, |
|
Answer» MOVES up |
|
| 44. |
Boltzmann Constant, K = R/N where N is Avogadro. number. Which quantity does PV/kT represent? |
| Answer» SOLUTION :NUMBER of MOLECULES in the GAS. | |
| 45. |
A block of mass m is thrown upwards with some initial velocity as shown in figure. On the block {:("Column I","Column II"),("(A) Net force in horizontal direction","(p) Zero"),("(B) Net force in vertical direction",(q) m(g sin theta+mug cos theta)),("(C) Net force along the plane",(r)m(g sin theta cos theta+mu g cos^(2) theta)),("(D) Net force perpendicular to plane",(s)m(g sin^(2) theta+mug sin theta cos theta)):} |
|
Answer» `THEREFORE` Acceleration of the block is `a=(g sin theta+mu g cos theta)` down the plane Now, net force in any direction is equal to, F=m(component of acceleration in that direction) |
|
| 46. |
Why do spring balances show wrong readings when it has been used for a long time? |
| Answer» Solution :Due to repeated stress and strain the ELASTIC strength of the BRIDGE gets REDUCED after a long TIME the bridge develops elastic fatigue and ultimately MAY collapse. | |
| 47. |
A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5) The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO=OQ. The distance PO is equal to |
|
Answer» 5R |
|
| 48. |
A mason worker throws the bricks to a height of 10m, at which the bricks reach with a velocity of 6ms^(-1). The percentage of energy he is wasting is nearly (g = 10ms^(-2)) |
|
Answer» 0.15 |
|
| 49. |
A standing wave is set up in a string of a variable length and tension by a vibrator of variable frequency . Both ends of the string are fixed . When the vibrator has a frequency f , in a string of length L and under tension T , n antinodes are set up in the string is doubled , by what factor should the frequency be changed so that the same number of antinodesis produced?(b) If the frequency and lenght are held constant , what tension will producen + 1 antinodes ? ( c) If the frequency is tripled and the length of the string is halved , by what factor should the tension be changed so that twice as many antinodes are produced ? |
|
Answer» Solution :In (a) , we except a lower frequency to go with a longer WAVELENGTH . In (b) , lower tension should go with lower wave speed for shorter wavelength at constant frequency . In ( C) , we will just have to divide it out. a. We have ` f_(n) = (n)/( 2L) = sqrt ((T)/(alpha))`(i) Keeping ` n , T and mu ` constant , we can create twoequations . ` f_(n) L = (n)/(2) sqrt((T)/(mu)) and F'_(n) L' = (n)/(2) sqrt((T)/(mu))` Dividing the equations gives `(f_(n))/( f'_(n)) = (L')/(L)` If `L' = 2 L , then f' _(n) = 1//2 f_(n)` THEREFORE , in order to double the length but keep teh same number of ANTINODES , the frequency should be halved . b. From Eq. (i), we can hold `L and f_(n)` constant to get `(n')/(n) = sqrt((T)/(T'))` From this relation , we see that the tension must be decreased to `T' = T ((n)/(n + 1))^(2)` to produce ` n +1` antinodes c. The time , we rearrange Eq. (i) to produce `( 2 f_(n) L)/(n) = sqrt ((T)/(mu)) and ( 2 f'_(n) L')/(n') = sqrt ((T')/(mu))` Then dividings gives `(T')/(T) = (( f_(n))/( f_(n)) xx (n)/(n') xx (L')/(L))^(2) = (( 3 f_(n))/( f_(n)))^(2) xx ((n)/( 2n))^(2) xx ((L//2)/(L))^(2) xx ((L//2)/(L))^(2) = (9)/(16)` |
|