Saved Bookmarks
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.
| 12001. |
For a system to be in equilibrium, the torques acting on it must balance. This is true only if the torques are taken about |
|
Answer» the CENTRE of the SYSTEM |
|
| 12002. |
A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches upto to maximum height of (3v^(2))/(4g) with respect to the initial position. The object is |
|
Answer» ring Given, `H=(3v^(2))/(4g)` `v^(2)=(2gh)/(1+(K^(2))/(R^(2)))=(2g 3v^(2))/(4g(1+(K^(2))/(R^(2))))=(6 gv^(2))/(4g(1+(K^(2))/(R^(2))))` `1=(3)/(2(1+(K^(2))/(R^(2))))or 1+(K^(2))/(R^(2))=(3)/(2)or (K^(2))/(R^(2))=(3)/(2)-1=(1)/(2)` `K^(2)=(1)/(2)R^(2)` (EQUATION of disc) Hence, the object is disc. |
|
| 12003. |
Flux passing through the shaded surface of a sphere when a point charge q is placed at the center is (radius of the sphere is R) |
|
Answer» Solution :K.E = K.E of sphere DUE to rolling + K.E due to rotation about 0 `=[(1//2)mv^(2) +1//2 IOMEGA^(2)] + 1/2 I xx v^(2)//R^(2) = (7//10)mv^(2) + 1/2 xx 2/5 mr^(2) xx v^(2)//R^(2) =(7//10) mv^(2) [1+(2R^(2))/(7R^(2))]` |
|
| 12004. |
A body is projected with an initial velocity of 58.8m/s at angle 60^(@) with the vertical. Find the vertical component of velocity after 2 sec. |
|
Answer» |
|
| 12005. |
Two metal balls of densities d and 3d and radii r_(1) and r_(2) are travelling in the same liquid. The density of the liquid is d//2. The ratio of their terminal velocities is 4:5 respectively. Then the ratio of their radii is |
|
Answer» `3:4` |
|
| 12006. |
A tuning fork of frequency 256 Hz and an open organ pipe of slightly lower frequency are at 17^(@) C. When sounded together , they produce4 beats. On altering that the number of beats per second first diminshes to in what direction has the temperature of the air in thepipe been altered ? |
|
Answer» Solution :`n = ( c_(17))/( 2I)` where `I = "length of the PIPE"` `:. 256 - ( c_(17))/( 2I) = 4 or ( c_(17))/(2 I) = 252` Since beats DECREASE first and then INCREASE to `4` , then the frequency of the pipe increases . This can happen only if the temperature increases. Let `t` be the final temperature , in Celsius. Now ` (c_(t))/( 2 I) - 252 = 4 or (c_(t))/( 2I) = 260` Dividing `( c_(t))/( c_(17)) = (260)/(252)` or `SQRT ((273 + t)/( 273 + 17)) = (260)/(252)``( C prop sqrt(T))` or ` t = 308.7 - 273 = 35.7^(@) C` `:.` Rise in temperature ` = 35.7 - 17 = 18.7 ^(@) C` |
|
| 12007. |
Figure shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system canslide on a horizontal road. The system is started towards right with an initia velocity v. The friction coefficient between the road and the bigger block is mu and that between the block is mu/2. Find the time elapsed before the smaller blocks separates from the bigger block. |
|
Answer» Here `a_1gta_2` so that moves on M. Suppose after time t, m is seperted fromM. In this time , m covers `vt+1/2 a_1t^2 and S_v=vt+1/2a_2t^2` For m to separate from M `vt+1/2a_1t^2=vt+1/2a_2t^2+1`...ii Again from the free body DIAGRAM `ma_1+mu/2R=0` `rarr ma_1=-(mu/2)MG=(mu/2)mxx10` `rarr a_1=-5mu`  Again `Ma_2+mu(M+m)g-(mu/2)mg=0` `rarr 2Ma_2=mumg-2muMg-2mumg` `rarr a_2=(-mumg-2muMg)/(2M)`  PUTTING values of `a_1` and `a_2in EQUATION i. we get, `t sqrt((4Ml)/((M+m)mug)` |
|
| 12008. |
A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocityof 3hati" m s"^(-1). And the other with a velocity of 4hatj" m s"^(-1). If the explosion occurs in 10^(-4)" s", the average force acting on the third piece in newton is |
|
Answer» `(3HATI + 4haj)XX 10^-4` |
|
| 12009. |
Give below figure,depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)? |
|
Answer» SOLUTION :(i) figure, (a) does not represent periodic MOTION, as the motion neither repeats not comes to mean POSITION. (ii) figure, (b) represents periodic motion with period equal to 2S. (iii) figure(c)does not represent periodic motion, because it is not IDENTICALLY repeated. (iv) figure,(d0 represents periodic motion with period equal to 2s. 
|
|
| 12010. |
A 2 kg mass is suspended by a rubber cord 2 m long and of cross-section 0.5 cm. It is made to describe a horizontal circle of radius 50cm in 4 times a second. Find the extension of the cord.(Young's modulus Y=7xx10^(8) Nm^(-2)) |
Answer» Solution :Given, m=2kg, `omega=2pif=2pixx4=8pi "rads"^(-1),r=0.5` m  From the FIGURE, `T=" cos"THETA=mg` and `T" sin"theta=(mv^(2))/(r)=momega^(2)` Tension in cord `T=sqrt((T" sin"theta)^(2)+(T" cos"theta)^(2))` `=sqrt((momega^(2)r)^(2)+(mg)^(2))` `T=sqrt([2xx(8pi)^(2)xx0.5]^(2)+[2xx10]^(2))=631 N` The EXTENSION of cord, `DELTAL=(TL)/(AY)=(631xx2)/(0.5xx10^(-4)xx10^(8))` `=360xx10^(-4)m=3.60cm` |
|
| 12011. |
A body of mass 1 kg is thrown upwards with a velocity 20 ms^(-1). It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction ?(Take g = 10 ms^(-2)) |
|
Answer» 20J |
|
| 12012. |
At what angle must a railway track with a level of 300 m radius be banked for safe running of trains at a speed of 90 km h^(-1) ? |
|
Answer» SOLUTION :`r=300 m, v= 90 KM h^(-1) = 25 MS^(-1), theta` = ? `tan theta = v^(2)/(rg) = (25 xx 25)/(300 xx 9.8)= 0.2126, theta = tan^(-1) (0.2126) = 12^(@)` |
|
| 12013. |
The figure on right shows a cooling curve for a substance which, .c) starting as a liquid, eventually solidifies. The specific heat capacity of the liquid is 20 xx 10^(3)Jkg K. The rate of temperature drop just before solidification at 60^@Cis 5 k/min. The latent fusion of the substance is |
|
Answer» `1.0 xx10^5 ` J/kg |
|
| 12014. |
Explain what is phase and draw in a single graph of different phases of simple harmonic motions. |
|
Answer» Solution :The position of a particle is determined by the equation `x(t)= A cos (omega t+PHI)` in time t. Where `(omega t+ phi)` is called PHASE of the motion or phase in the time t. It shows the position of motion of OSCILLATOR at that time. Phase-constant (Phase angle) : ..At time to t=0, the phase of simple harmonic oscillator is known initial phase or phase-constant or phse angle... If the amplitude is fixed, then at time t=0, initial phase `phi` can be determine from displacement of particle. `x(t) = A cos phi ""[therefore t=0]` `therefore cos phi = (x(t))/(A)` `therefore phi = cos^(-1) ((x(t))/(A))` The curves 3 and 4 are for `phi = 0" and "phi = -(PI)/(4)` respectively. The amplitude A is samefor both the PLOTS as shwon in figure. 
|
|
| 12015. |
10g of ice of 0^(@)Cis mixed with 100 g of waterat 50^(@)C in a calorimeter.The final temperatureof the mixtureis [Specificheat of water= 1cal g^(-1).^(@)C^(-1),letent offusionof ice= 80 cal g^(-1)] |
|
Answer» `31.2^(@)C` Mass of WATER, `m_(w) = 100 g` Mass of ice, `m_(i) = 100g` Specific heat of water , `s_(w) = 1 CAL g^(-1) .^(@)C^(-1)` Latent heat of fusion of ice, `L_(fl) = 80 cal g^(-1)` Let T be the final temperature of the mixture Amount of heat lost by water `m_(w)s_(w)(DeltaT)_(w) = 100 XX 1 xx (50 - T)` Amount of heat gained by ice `m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)` ACCORDING of heat lost by water `m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)` According to principle of CALORIMETRY Heat lost = Heat gained `100 xx 1 xx (50 - T) = 100 xx 80 + 10 xx 1 xx (T - 0)` `500 - 10T = 80 +T` `11T = 420` or `T = 38.2 .^(@)C` |
|
| 12016. |
(I) Perpendicular Axis theorem holds good only for plane Laminar objects. (II) The perpendicular distance between the Axis of Rotation of a body and the centre of gravity of the body is called Radius of Gyration. Which one is correct ? |
|
Answer» I only |
|
| 12017. |
A man turns on rotating table with an angular speed omega. He is holding two equal masses at arms length. Without moving his arms, he just drops the two masses. How will his angular speed change? |
|
Answer» LESS than `omega` |
|
| 12018. |
In the shown planar frame made of thin uniform rods, the length of sections AB and EF is L_1 and its thermal coefficient of expansion is alpha_(1). The length of section CD is L_(2) and its thermal coefficient of expansion is alpha_(2). CB and DE are of same length having linear thermal coefficient of expansion alpha_(2). It is given that (alpha_1)/( alpha_2) = 3.5 Points A, B, E and F reside on same line, that, is, sections, AB and EF overlap. Then find the ration of (I_2)/( I_1) for which distance between end A and end F remain the same at all temperature. |
|
Answer» |
|
| 12019. |
A particle of mass m describes a circular path of radius 'r' such that speed V = alphasqrt(S) (S is distance traveled). Then power is proportional to: |
|
Answer» S |
|
| 12020. |
An object is droppedfrom a height h from the ground. Every time it it hitsgroundit loses50 %of its kineticenergy . The total distancecoveredast to oo is : |
|
Answer» 3H Kinetic energy of object = potentialenergy of object . K.E = mg(h) Thisobjecthitethe groundbut loose50 % ofinitial K.E K.E =mg(h/2) Every time object halfof the heightfrom which it was DROPPED. Total distancecovered by object hiting the ground toll `t to oo` `H = h+2 (h/2 +h/(2^(2)) +....+h/(2^(n))) = h+h (1+1/2 +1/(2^(2))+....oo)` `= h + h (1/(1-1/2))` H = 3 h |
|
| 12021. |
Two sitar strings A and B playing the note .Ga. are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz. What is the of B ? |
|
Answer» |
|
| 12022. |
A bullet of mass 25 g moving withvelocity of 500 ms^(-1) enters into a wooden block and comes out of it with a velocity of 100 ms^(-1). Find the work done by the bullet while passing through the wooden block. |
|
Answer» Solution :`m=25xx10^(-3)KG, U=500 m//s, v=100 m//s` Work energy THEOREM `W=(1)/(2) mv^(2)-(1)/(2)m u^(2)` `=(1)/(2)xx25xx10^(-3)[(100)^(2)-(500)^(2)]=3000J` |
|
| 12023. |
Discuss how ripples are formed in still water. |
| Answer» Solution :Suppose thedrop a stone in a trough of still water, we can see a disturbance produced at the PLACE where the stone strikes the water surface. We find that this disturbance spreads out( diverges out) in the form of concentric circles of ever INCREASING radii (ripples) and STRIKE the { boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the v,after molecules on the surface. Actually the particles of the water (medium) themselves do not moYe outward ,Yith the disturbance. This can be obsen-ed by keeping a PAPER strip on the watersurface. The strip MOVES up and down , then the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their mean positions. | |
| 12024. |
The volume of a piece of floating wax is 22cm^(3). 20cm^(3) of its volume is in water. Determine the mass and specific gravity of the piece of wax. |
|
Answer» |
|
| 12025. |
The plane which can be formed with the vectors overline a=3overline i-4overline j+2overline k,overline b=2overline i-overline j+6overline k,overline c=5overline i-5overline j+4overline k,is. |
|
Answer» Quadrilateral |
|
| 12026. |
Each of two persons has a whistle of frequency 500 Hz . One person is at rest at a particular place nad the second person recedes from him with a velocity of1.8 m * s^(-1) If both of them blow whistles , how many beats will be heard by each of them ? Velocity of sound= 330 ms^(-1) |
|
Answer» Solution :Both of them listen to the sound of frequency 500 Hz form their own WHISTLES . In the first case , let us suppose that the first person is listenting to the sound coming from the whistle of the second person . Here the velocity of the listener , ` u_(o) = 0 ` Since the source is receding , the velocity of the source , ` u_(s) = - 1.8 * s^(-1)` ` therefore` Apparent frequency , ` n. = (V)/(V - u_(s)) XXN = (330)/(330 - (-1.8)) xx 500 = 497.29` Hz ` therefore ` NUMBER of beats per second ` = n- n. = 500 - 497.29` ` = 2. 71 ~~3 ` In the second case , let us suppose that the second person is listening to the sound coming from the whistle of the first person. Here the velocity of the source , ` u_(s) = 0 ` Since the listener is receding , The velocity of the listener , ` u_(0) = - 1.8 * s^(-1)` ` therefore ` Apparent frequency , ` n. = (V + u_(o))/(V) xxn = (330 - 1.80/(330 ) xx 500 = 497.27 Hz` ` therefore ` Number of beats per second = n - n. = 500 - 497.27 ` = 2.73 ~~ 3` |
|
| 12027. |
A uniform metre rod is bent at the middle in the form of L shape with the bent arms at 90^(0) to each other. The distance of the centre of mass from the bent point is |
|
Answer» `(1)/(4 SQRT(2))m` |
|
| 12028. |
The co-efficient of friction between the floor and the body B is 0.1 The co-efficient of friction between the bodies BThe mass of A is m//2 and of B is m Which of the following statements are ture ? . |
|
Answer» The bodies will move together if `F = 0.25mg` (a) Force of friction between `A` and `Bf_(1) = 0.2 ((m)/(2)) g = 0.1 mg` Force of friction between `B` and FLOOR `f_(2) = 0.1 (m + (m)/(2)) g = (0.3)/(2) mg = 0.15mg` The bodies will move together, if `F = f_(1) + f_(2) = 0.1 mg + 0.15mg = 0.25mg` If `F= 0.5mg` force between `A` and `B` will exceed `0.1mg` and the the body A will slip The bodies cannot move together when `F= 0.5mg` If `F =0.1mg` the bodies will be at rest as `Flt(f_(1) + f_(2))` The two bodies will move together till max accelerating force on `A = 0.1 mg = f_(1)` acceleration of `A = (0.1mg)/(m//2) = 0.2g` acceleration of `B = 0.2 g` max ACC. force on `B = m xx 0.2g = 0.2 mg` As `F_(2) =0.15mg` `:.` Maximum value of `F` for with `A` and `B` will move together = `0.1 mg + 0.1 mg + 0.2 mg + 0.15mg = 0.45mg`  .
|
|
| 12029. |
A hollow cylinder of diameter 20 cm is rotating about its own vertical axis. If the angular velocity of the cylinder is not less than 180 rpm, then a body remains attached on the inside wall of the cylinder and keeps rotating with the cylinder. What is the coefficient of rotating with the cylinder. what is the coefficient of friction between the body and the wall of the cylinder? Given, pi^(2) = 9.8. |
|
Answer» |
|
| 12030. |
One mole of a monatomic ideal gas is taken througlh a cycle ABCD as shown in the P-V diagram Column II gives the charctierstic involved in the cycle. Math them with each of the processes given in Column-I {:("Column"-I,"Column"-II),((A)"Process" A to B,(p)"Internal energy decreases"),((B) "Process" B to C,(q) "Internal energy incerease"),((C) "Process" C to D,(r)"Heat is lost"),((D) "Process" D to A,(s) "Heat is gained"),( ,(t) "Work is done on the gas"):} |
|
Answer» <P> |
|
| 12031. |
Check the wrong statement if object is real (i) A concave mirror can give a virtual image (ii) A concave mirror can give a diminshed virtual image (iii) A convex mirror can give a real image (iv) A convex mirror can give a diminshed virtual image. |
|
Answer» I,ii |
|
| 12032. |
The area under a .force - displacement. curve gives : |
| Answer» ANSWER :C | |
| 12033. |
The following are the p-V diagrams for cyclic processes for a gas. In which of these processes heat is not absorbed by the gas ? |
|
Answer»
|
|
| 12034. |
Two cars, A and B move along the x-axis. Car A starts from rest with constant accelertion while car B moves with consstant velocity. a. At what time s, t s, if any, do A and B have the same position? . b. At what time s if any, do A and B have the same velocity? What is the velocity of car B at this time. c. Graph velocity versus time for both A and B. d. At what time s. If any, does car A pass car A? e. At what time s, if any does car B pass car A? |
|
Answer» SOLUTION :The two cars have the same position at time when when their `x-t` graphs cross. The firure shows this occurs at `t=1 s` and `t=3 s`. b. `A` and `B` have the same velpcoty when slpe of the tangent on parabola is equal to the slope of the straight line.From position-time graph, it is clear it occurs at `t=2 s`. Car `B` travels with constant velocitly. `v_(B)=(20-0)/(4-0) =5 MS^(-1)` C.  d. Car `A` passes car `B` when`x_(A)` MOVES above `x_(B)` in the `x-t` graph This happens at `t=3 s` E. Car `B` passes car `A` when `x_(B)` moves above `x_(A)`, x-t graph. This happens at `t= 1 s`. |
|
| 12036. |
An eye specialist prescribes spectacles having a combination of a convex lens of focal length 40cm in contact with a concave lens of focal length 25cm. The power of this lens combination is |
|
Answer» `+1.5D` |
|
| 12037. |
An uniform rod of mass M. length L stands vercany on ground. When it is set into rotation, about the bottom most point, at the time when it hits the ground. |
|
Answer» Angular VELOCITY of centre of mass is `sqrt((6G)/(l))` |
|
| 12038. |
The spectral energy distribution of the sun has maxima at 4753 Å. Find the temperature of a star for which spectral distribution has maxima at 10350 Å. [Temperature of sun is 6000 K] |
|
Answer» SOLUTION :According to Wien's displacement LAW `lambda_(m) alpha1/T` `THEREFORE lambda_(2)/lambda_(1)=T_(1)/T_(2) rArrT_(2)=T_(1)lambda_(1)/lambda_(2)` `RARR T_(2) = 6000 xx 4753/10350` `rArr T_(2)=2755.4 `K |
|
| 12039. |
A particle of mass 100gm is thrown vertically upwards with a speed of 5m/s. The work done by the force of gravity during the time the particle goes up is (g= 10 ms^(-2)) |
|
Answer» `-0.5J` |
|
| 12040. |
A disc is rotatin g aro u n d its cen tre in a horizontal plane at the rate of 60 rotation/ minute. A coin (1A^(st)) is placed at a distance of 18 cm and 2^(nd) similar coin 20 cm from its centre.The co-efficient of static friction between the disc and the coins is 0.2. Which coin will be thrown away from the disc ? Which coin will keep rotating with the disc ? |
|
Answer» both COINS THROWN away for the circularmotioncentrifugalforce`le ` maximumstaticfriction. `(mv^(2))/(r ) le f _(s)` Timerequiredfor 60rotationis 60second Timerequiredfor 1rotationis `T = (60)/(60)=1` Now linearvelocity`v=(2pi r)/( T )` and`f_(s(max))= mu_(s ) N = mu_(s ) MG` fromequation(1) `(m )/(T) xx (2pi r)/(T ) le mu_(s ) mg` `le((0.2).(10).(1)^(2))/(4xx 10)` `le0.05 m` `r le m` boththe coin from the 18 cm and 20 cmdistancefrom the centreare thrownaway . |
|
| 12041. |
Define Average speed. Write it equations. |
|
Answer» Solution :Average speed is defined as the mean (or) average of all the speeds of molecules. `barv=(v_(1)+v_(2)+v_(3)+.....+v_(N))/(n)= SQRT((8RT)/(piM))= sqrt((8kT)/(pi m)) rArr barv =1.60 sqrt((kT)/m)` |
|
| 12042. |
Five kg of air is heated at constant volume. The temperature of air increases from 300K to 340 K. If the specific heatat constant volumeis 0.169 kcal/kg K, find the amount of heat absorbed. |
|
Answer» Solution :`m = 5kg, DT = 340-300=40K`, `C_(v)=0.169 "kcal"//KG K`, Principal specific heat at constant volume `C_(v)=(dQ)/(MDT)` The amount of heat absorbed, `dQ=mC_(v) dT` `=5xx0.169 XX 40 = 33.8` kcal. |
|
| 12043. |
A thin square plate is of mass m and the length of each of its sides is I. Find the moment of inertia of the plate when it is rotating about an axis (i) passing through its centre and perpendicular to its plane and (ii) passing through one of its corners and perpendicular to its plane. |
|
Answer» Solution :Passing through its centre and perpendicular to its plane : `I=(m(l^(2)+l^(2))/(12)=(2ML)/(12)=(ml^(2))/(6))` (ii)Passing through one of its CORNERS and perpendicular to its plane `I=I_(G)+Mr^(2)` `I=(ml^(2))/(6)+(ml^(2))/(2)=(ml^(2)+3ml^(2))/(6) rArr I=(4ml^(2))/(6)=(2ml^(2))/(3)` |
|
| 12044. |
In the figure a truck is moving on a horizontal surface with acceleration a. Two blocks of equal masses m are supported on the truck as shown in figure. Given that when the block at the top surface is just about to slide, other block remains hanging at 30^(@) from the vertical. In this system. |
|
Answer» `a=(g)/(sqrt(3))` `T cos 30^(@)=mg …….(2)`  dividing equation `(1)` by equation `(2)` `tan 30^(@)=(a)/(g)` `rArr a=gtan 30^(@)` `rArr a=(g)/(sqrt(3))` Ans. From `(2)T=(mg)/(cos 30^(@))=(2MG)/(sqrt(3))`Ans. and `mu mg-T=ma` `rArr mu mg=T+ma=(2mg)/(sqrt(3))+ma` `=(2mg)/(sqrt(3))+(mg)/(sqrt(3))` `rArr mumg=(3MG)/(sqrt(3))=sqrt(3)mg` `=mu=sqrt(3) Ans.` |
|
| 12045. |
Is a blackbodyblack ? |
| Answer» SOLUTION :No. ,.Blackness , referto idealpropertyof absorbingall theradiationincidenton the BODY . If thebodyabsorbs all the radiationit is described as the BLACK body . A blackcouldappear POSSIBLY as red, whiteor any colour dependingon itstemperatur e. | |
| 12046. |
To simulate car accidents , auto manufactures study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth roadand colliding with a horizontally mounted spring of spring constant 6.25 xx10^(3) Nm^(-1) .What is the maximum compression of the spring ? |
|
Answer» Solution :At maximum compressionthe kinetic energy of the car is converted entirelyinto the potentialenergy of the spring. The kinetic energy of the moving car is , `K = 1/2 mv^(2)` `=1/2 xx10^(3)xx(5)^(2)` ` = 1.25 xx10^(4)J` `V = 18 "kmh"^(-1)` ` = (18 xx1000)/(3600)` ` = 5 ms^(-1)` From the law of the conservation of MECHANICAL energy potential energy of spring = kineticenergy of moving car `V = 1/2 kx_(m)^(2)` Where k = spring CONSTANT , `x_(m) ` = maximum COMPRESSION ` 1.25 xx10^(4) =1/2 xx6.25 xx10^(3) xxx_(m)^(2)` ` =(2xx1.25 xx10^(4))/(6.25 xx10^(3)) =x_(m)^(2)` ` :. x_(m) = sqrt((2.5xx10^(4))/(6.25xx10^(3))` ` sqrt(4)` ` :. x_(m) = 2m ` ` :. ` Since spring will be compressed by 2 m . |
|
| 12047. |
A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of 25 N . The bird (mass m = 0.5 kg ) flies upward in the cage with an acceleration of 2m//s^(2) . The spring balance will now record a weight of |
|
Answer» 24 N |
|
| 12048. |
If humans were to settle on other planets, which of the fundamentalquantities will be in trouble? Why? |
| Answer» SOLUTION :If humans were to settle on other planets the fundamental quantity .mass. will be in trouble as the law of gravitation is DISTURBED which UNIFIES the fundamental FORCES of nature. | |
| 12049. |
In a stationary wave patternn that forms as a result of reflection of waves from an obstacle , the ratio of the amplitude at an antinode and a node isb = 1.5 . What percentage of the energy passes across the obstacle ? |
|
Answer» Solution :As we have STUDIED that when incident WAVE and reflected wave superimpose to produce stationary wave , the ratio of amplitudes at ANTINODE and at node is given by `(A_(MAX))/(A_(min)) = (A_(i) + A_(r))/( A_(i) - A_(r))` This ratio is given as ` 1.5 or 3//2`. `(A_(i) + A_(r))/( A_(i) - A_(r)) = (3)/(2) or (1 + (A_(r))/(A_(i)))/(1 - (A_(r))/(A_(i))) = (3)/(2)` Solving this equation , we get `(A_(r))/(A_(i)) = (1)/(5) rArr (( A_(r))/( A_(i)))^(2) = (1)/(25) or , I_(r) = 0.04 I_(i)` this MEANS `4%` of the incident energy is reflected or `96%` energy passes across the obstacle. |
|
| 12050. |
Velocity - time graph of a particle executing SHM is as shown in fig. Select the correct alternatives. A: at position 1, displacement of particle may be +ve or -ve B: at position 2, displacement of particle is -ve C: at position 3, acceleration of particle is +ve D: at position 4, acceleration of particle is +ve |
|
Answer» A,B |
|
