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.
| 13201. |
A block of mass 2kg is attached with a spring of spring constant 4000Nm^-1 and the system is kept on smooth horizontal table. The other end of the spring is attached with a wall. Initially spring is stretched by 5cm from its natural position and the block is at rest. Now suddenly an impulse of 4kg-ms^-1 is given to the block towards the wall. Approximate distance travelled by the block when it comes to rest for a second time (not including the initial one) will be (Take sqrt(45)=6.70) Displacement of the rod by the time the insect meet the moth is |
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Answer» (a) `L/2` `:. 0=(8m)(2v)-(16m)(v)+(48m)v^'` Here `v^'=` ABSOLUTE speed of rod `=0` `:.` Displacement of rod=0` |
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| 13202. |
Proveimpulse momentum equation |
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Answer» Solution :If aforce `(F )` actson a bodyin a veryshortintervalof time `(Detla t)`then formNewton'ssecondlaw inmagnitudeforms Intergratingover timeforman intialtime`t_(i)` toa finaltime `t_(f)` we get overset(i)UNDERSET(f ) (int) dp=overset(i)underset(f ) (int) Fdtg` `P_(i)- P_(i)= underset(t) overset(t)(int)F dt` `p_(1)=`initialmomentum of the bodyat time `t_(i)` `p_(f)=` finalmomentum of the bodyat time `f_(f)` `p_(f)- p_(i) = Delta p`changein MOMENTUM of thebodyduringthe TIMEINTERVAL Theintegral`underset(t ) overset( t ) (int)F dt = J` iscalledthe impulseand itis equalto change a in momentum of the body . If theforce is constantover the timeintervalthen `underset(t ) overset(t ) (int ) Fdt= Funderset( t) overset( t) dt = F (t_(1) - t_(1)) = F Delta t` `F Delta t= Delta p` Impulse = Changein momentum |
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| 13203. |
If a body of mass m is taken out from a point below the surface of earth equal to half the radius of earth, R, to a height R above the earths surface, then work done on it will be_____mgr |
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Answer» 5/6 |
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| 13204. |
Two particles are projected with same velocity but at angles of projection 35^(@) and 55^(@). Then their horizontal ranges are in the ratio of |
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Answer» `1:2` |
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| 13205. |
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be (a) gaoing up with increasing speed(b) going down with increasing speed(c ) going up with uniform speed(d) going down with uniform speed |
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Answer» a, B are TRUE |
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| 13206. |
A block of wood weighs 12 kg and has a relative density 0.6. It is to be in water with 0.9 of its volume immersed. What weight of a metal is needed (a) if the metal is on the top of wood, (b) if the metal is attached below the wood? [RD of metal =14] |
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Answer» |
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| 13207. |
A small mass m is transferred from the centre of a hollow sphere of mass M to infinity. Find work done in the process. Compare this with the situation if instead of a hollow sphere, a solid sphere of same mass were there. |
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Answer» Solution :We know at infinity gravitional potential is taken ZERO. Thus if `V_(C )` be the GRAVITATIONAL potential at centre of holow sphere then external work required the procedd is `W = m(0-V_(C ))` or `= m(0-((GM)/(R )))=(GMm)/(R )` Here `V_(C )= -(GM)/(R )`, the potential at the centre of a hollow sphere of mass M and radius R. If a so solid sphere is here, we have at its centre `V_(C )= - (3)/(2)(GM)/(R )` Thus work required will be `W = m[0-(-(3)/(2)(GM)/(R ))]=(3)/(2)(GMm)/(R )` We can SEE in second case more work is required for the process. |
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| 13208. |
Solar constant and Stefan's constant have same dimensions in |
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Answer» Mass |
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| 13209. |
A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the inclination the angle between the inclined plane and the horizontal is |
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Answer» SOLUTION :`45^(@)` |
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| 13210. |
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 256 Hz, if the length of the column in centimeter is (velocity of sound inair 340 m/s) |
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Answer» `21.25` |
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| 13211. |
A moving blockhaving mass m , collides with another stationary block having mass 4m , The lighter block comes to rest after collision . When the intial velocity of the lighter block is v, then the value of coefficient of restitution (e ) will be |
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Answer» `0.4` ` " P=p" mv =4mv_(2)"" :. V_(2) = (v_(1))/4` ` :. ` Restitutional VALUE , `(e)= (v_(2)-v_(1))/(v_(1)-v_(2))=((v_(1))/4-0)/(v_(1)-0)=1/4 =0.25`. |
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| 13212. |
Earth (uniform solid sphere of Mass M Radius rotaes about its North pole - south pole diameter at a constant angular velocity omega. A small block of mass m kept at a colatitude of theta doesn'tslip on earth's surface Then which of the following statements is INCORRECT ? |
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Answer» Angular MOMENTUM of the BLOCK about centre of the EARTH is conserved |
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| 13213. |
A uniform disc of mass 2kg and radius 1m is mounted on an axle supported on fixed frictionless bearings. A light cord is wrapped around the rim of the disc and mass of 1kg is tied to the free end. If it is released from rest, |
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Answer» the tention in the cord is 5N |
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| 13214. |
Two bodies having the same mass 2kg each have different surface areas 50 m^(2) and 100 m^(2) in contact with a horizontal plane. If the coefficient of friction is 0.2, the forces of friction that come into play when they are in motion will be in the ratio |
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Answer» `1:1` |
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| 13215. |
The linear magnification is produced by a telescope having an objective of 50cm and eyepiece of 5cm focal length when used to read a scale 3m away is |
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Answer» 1.2 |
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| 13216. |
A car weighing 100 kg is on a slope that makes an angle 30^(@) with the horizontal. The component of car.s weight parallel to the slop is(g = 10ms^(-2)) |
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Answer» 500 N |
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| 13217. |
A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's centre. Their time periods are in the ratio of |
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Answer» `1:2` |
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| 13218. |
Wave equation for one progressive harmonic wave isy = 10 ^(-4) sin (600 t - 2x + (pi)/(3))m and t is in s). Velocity of the wave would be ...... m/s |
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Answer» 300 `y = 10 ^(-4) SIN (600 t - 2x + (pi)/(t))` with `y = a sin (omega t - kx +phi)` we get, `omega = 600 rad//s` `K = 2 rad //m` `therefore v = (omega )/(k) = (600)/(2) = 300 (m)/(s)` |
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| 13219. |
Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m, radius = r) |
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Answer» `(1)/(2) MR^(2)` `I_(CM) + I_("diameter") = mr^(2) + (mr^(2))/(2) = (3)/(2) mr^(2)` |
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| 13220. |
A gas obeys PV^2= constant in addition to PV =RT. If on heating, the temperature is doubled, the volume of the gas is |
| Answer» ANSWER :A | |
| 13221. |
China wares are wrapped in straw paper before packing. Why? |
| Answer» Solution :The STRAW paper between the China ware increases the time of EXPERIENCING the JERK during transportation. Hence impact of FORCE reduces on China wares. | |
| 13222. |
Find the dimensions of (i) force (ii) surface tension and (iii) momentum in terms of frequency, velocity and density as fundamental units. |
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Answer» SOLUTION :Frequency `n=T^(-1),T=1/n` Velocity `v=LT^(-1),L=vT=v/n` Density `d=ML^(-3),M=dL^(3)=(dv^(3))/(n^(3))` (i) Force `=MLT^(-2)=(dv^(3))/(n^(3))xxv/nxxn^(2)=(dv^(4))/(n^(2))` (ii) SURFACE tension `=MT^(-2)=(dv^(3))/(n^(3))xxn^(2)=(dv^(3))/n` (iii) Momentum `=MLT^(-1)=(dv^(3))/(n^(3))xxv/nxxn=(dv^(4))/(n^(3))` |
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| 13223. |
Define unit of a physial quantity. |
| Answer» Solution :A physical quantity is made of 2 parts that is NUMERICAL VALUE and unit. Unit gives the absolute value of 1 numerical value of that physical quantity. | |
| 13224. |
Consider three blocks of masses m_(1), m_(2), m_(3) interconnected by strings which are pulled by a common force F on a frictionless horizontal table as in the figure. The tension T_(1) and T_(2) are also indicated a) T_(2) gt T_(1)" if "m_(2) gt m_(1) b) T_(2) =T_(1)" if "m_(2)=m_(1) c) T_(2) gt T_(1) always d) "acceleration of the system"=(F)/(m_(1)+m_(2)+m_(3)) |
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Answer» a, b |
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| 13225. |
Which is most elastic? |
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Answer» iron |
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| 13226. |
If the acceleration due to gravity on the moon is one-sixth of that on the earth. What will be the change in length of a second pendulum there so that it may beat a second there ? Take acceleration due to gravity on earth surface =9.8 ms^(-2). |
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Answer» Solution :On earth `g=9.87 m//s^(2)`. Let L be the LENGTH of second pendulum . Then `T=2s` `T=2pi sqrt((l)/(g))` (or) `l=(T^(2)g)/(4pi^(2))=(2^(2)xx9.8)/(4XX(22//7)^(2))` `=0.9921 m =99.21 CM` On moon, `g_(m)=(1)/(6)g=(9.8)/(6) ms^(-2)` `l'=(T^(2)g_(m))/(4pi^(2))=(2^(2)xx(9.8//6))/(4xx(22//7)^(2))` `=.1654 m=16.54 m` change4 in length `=l-l'=99.21-16.54` `=82.67 cm` |
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| 13227. |
Express Ok on Fahrenheit scale. |
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Answer» `-459.67^(@)` F |
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| 13228. |
At highest point acceleration of projectile is zero. |
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Answer» |
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| 13229. |
A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be |
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Answer» <P>0.25 Heat supplied =`Q_(AB)+Q_(BC)=C_(V)T_(0)+C_(p)2T_(0)` `=(13)/(2)RT_(0)=(13)/(2)P_(0)V_(0)` `therefore` Efficiency of the cyclic process `((1)/(2) P_(0)V_(0))/((13)/(2)P_(0)V_(0))xx100=(1)/(13)xx100=7.7%` |
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| 13230. |
When an air bubble moves up from the bottom of a lake a) velocity decreases and becomes zero b) acceleration increases and becomes zero c) velocity increases and becomes constant d) acceleration decreases and becomes zero |
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Answer» a, d are correct |
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| 13231. |
A particle of mass m is 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. What is the power delivered to the particle by the forces acting on it? |
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Answer» <P> Solution :As `a_(c )= (V^(2)//r) " so" (v^(2)//r)= k^(2) r t^(2)`Kinetic energy `K= (1)/(2) mv^(2)= (1)/(2) mk^(2) r^(2) t^(2)` Now by work -Enetgy THEOREM `W= Delta K= (1)/(2) mk^(2) r^(2) t^(2)- 0 rArr P= (dw)/(dt)` `rArr P= (d)/(dt) = (1)/(2) mk^(2) r^(2) t^(2)= mk^(2) r^(2) t` |
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| 13232. |
If the coefficient of friction of a surface is sqrt 3, then the angle of inctination of the plane to make a body on it just to slide is |
| Answer» Answer :C | |
| 13233. |
A uniform rope of length .L. and linear density .mu. is on a smooth horizontal table with a lentht .l. lying on the table. The work done in pulling the hanging part on to the table is |
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Answer» `(MUG(L-l)^(2))/(2)` |
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| 13234. |
In the figure shown, the mass of the trolley is 100 kg and it can move without friction on the horizontal floor. It length is 12 cm. The mass of the girl is 50 kg. Friction exists between is 12 m. The mass of the girl is 50 kg. Friction exists between the shoes of the girl and the trolley's upper surface, with mu = 1//3, The girl can run with a maximum speed = 9 m//s on the surface of the trolley, with respect to the surface = 9 m//s on the surface of thr trolley, with respect to the surface. At t=0 the girlm starts running from left to the right. The trolley was initially stationary. (g = 10 m//s^(2)) Suppose the girl accelerates slowly, at a constant rate, and acquires the relative speed of 9 m//s only when it reaches the right end of the trolley, then what must be the earth frame acceleration of the girl ? |
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Answer» `2.5 m//s^(2)` |
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| 13235. |
A body of mass 3 kg is moving along a straight line with a velocity of 24 ms^(-1). When it is at a point ‘P’ a force of 9 N acts on the body in a direction opposite to its motion. The time after which it will be at 'P' again is, |
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Answer» 8S USING `a=(F)/(m), S=ut+(1)/(2)at^(2)` |
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| 13236. |
In the figure shown, the mass of the trolley is 100 kg and it can move without friction on the horizontal floor. It length is 12 cm. The mass of the girl is 50 kg. Friction exists between is 12 m. The mass of the girl is 50 kg. Friction exists between the shoes of the girl and the trolley's upper surface, with mu = 1//3, The girl can run with a maximum speed = 9 m//s on the surface of the trolley, with respect to the surface = 9 m//s on the surface of thr trolley, with respect to the surface. At t=0 the girlm starts running from left to the right. The trolley was initially stationary. (g = 10 m//s^(2)) At a certain moment when the girl was accelerating, the earth frame accelerating of the trolley is found to be 1m//s^(2). At this moment, the friction force between the girl's shoes and the trolley's surface is : |
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Answer» 200 N |
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| 13237. |
Four holes of radius R are cut fro a thin square plate of side 4R and mass M. The moment of inertia of the remaining portion about z-axis is ((x)/(3)-(10pi)/(16))"MR"^(2) where 'x' is |
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Answer» |
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| 13238. |
About what axis would a uniform cube have its minimum moment of inertia? |
| Answer» SOLUTION :About an AXIS joining the opposite comers and PASSING through the centre of the CUBE. | |
| 13239. |
A constant force is applied on a 10kg mass at rest. The ratio of works done is 1st, 2nd and 3rd seconds is |
| Answer» Answer :B | |
| 13240. |
A block of ice starts slilding down from the top of an inclined roof of a house along a line of the greatest slope. The inclination of the roof with the horizontal is 30^(@). The height of the highest and lowest points of the roof are 8.1 m and 5.6 m respectivley. At what horizontal distance from the lowest point will the block hit the ground? Neglect air friction [g=9.8m//s^(2)] |
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Answer» Solution :Acceleration of the block along the greatest slope in equal to `a=g sin 30^(@)` Distance TRAVELLED by the block along the greatest slope is equal to `S=((8.1-5.6))/(sin 30^(@))=5m` If u be the speed of the block when it is just abou to LEAVE the roof then `u^(2)=0+2gsin 30^(@)xx5impliesu=7m//s` If t be the TIME taken on hit the ground then `5.6=u sin 30^(@)t+1/2"gt"^(2)=7/2t+1/2xx9.8t^(2)` `implies7t^(2)+5t-8=0` `t=(-5+-sqrt(25-(4)(7)xx(-8)))/(2xx7)` `impliest=(-5+-15.78)/14s`, -ve value is to be rejected , i.e. `t=(-5+15.78)/14=0.77` SEC Horizontal distance travelled is equal to `x=u cos 30^(@)t=(7sqrt(3))/2xx10.78/14mimpliesx=4.67m` |
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| 13241. |
A particle moves in x-y plane according to the law x = 4 sin 6t and y=4(1-cos6t). The distance traversed by the particle in 4 seconds is (x and y are in meters) |
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Answer» 96 m |
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| 13242. |
A uniform solid cylindrical roller of mass 'm' is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the sylinder is 'a and it is rolling without slipping then the value of 'F' is : |
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Answer» 2MA |
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| 13243. |
Equal torques are applied on acylinder and a sphere. Both have same mass and radius rotates about its axis and sphere rotates about one of its diameter. Which will rotates acquire greater speed and why? |
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Answer» SOLUTION :`TAU=I , alpha, alpha=(tau)/(I)` `alpha` in cylinder, `alpha_(C)=(tau)/(I_(C))` `alpha` in sphere, `alpha_(s)=(tau)/(I_(S))` `(alpha_(C))/(alpha_(S))=(I_(S))/(I_(C))=((2)/(5)MR^(2))/(MR^(2))=(2)/(3)` |
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| 13244. |
Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring? |
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Answer» `(MG)/k` |
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| 13245. |
If the momentum of a body decreas by 30% its kinetic energy decreases by |
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Answer» 0.6 |
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| 13246. |
For the wave discribed in Exercise 15.8, plot the displacement (y) versus (t) graphs for x=0, 2 and 4 cm. What are the shapes of these graphs ? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ? |
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Answer» Solution :Given wave EQUATION is `y = 3 sin (36 t +0.018 x + (pi)/(4)) cm ""…(1)` `implies` Phase of a particle at distance x from origin of wave at time t is `theta = 36 t + 0.018x + (pi)/(4) rad ""…(2)` Now we are given three particles whose displacement are to be found out at different instnats of time. They are LOCATED at `x=0,` x=2 cm and x=4 cm respectively. Let us consider their simple harmonic oscillations one by one. For particle at `x=0,` from equation (1), `y =3 sin (36 t + (pi)/(4)) ""...(3)` `therefore y= a sin ((2pi )/(T) t + (pi)/(4))` (Where `(2pi)/(T) = omega = 36 (rad)/(s)) ...(4)` From above equation we can find out vaues of DISPLACEMENTS y at `t =0, (T)/(8), (2T)/(8), (3T)/(8), (4T)/(8),...,` `(8T)/(8) = T` as flollows. `(1) At t =0, y = a sin "" (pi)/(4) = (a)/(sqrt2)` (2)` At t = (T)/(8), y = a sin ((2pi)/(T) xx (T)/(8) + (pi )/(4))` `=a sin ((pi)/(2)) =a` `(3) At t = (2T)/(8), y =a sin ((2pi)/(T) xx (2T)/(8) + (pi)/(4))` `=a sin ((pi)/(2) +(pi)/(4)) =a cos ""(pi)/(4) = (a)/(sqrt2)` `(4) At t= (3T)/(8), y= a sin ((2pi)/(T) xx (3T)/(8) + (pi)/(4))` `=a sin (pi) =0` `(5) At t = (4T)/(8), y = a sin ((2pi)/(T) xx (4T)/(8) + (pi)/(4))` `= a sin (pi + (pi)/(4)) =a (-sin "" (pi)/(4)) =- (a)/(sqrt2)` `(6)At t = (5t)/(8), y = a sin ((2pi)/(T) xx (5T)/(8) + (pi)/(4))` `= a sin ((3PI)/(2)) =-a ` `(7) At t = (6T)/(t), y = a sin ((2pi )/(T) xx (6T)/(8) + (pi)/(4))` `= a sin ((7pi)/(4)) =a sin (2pi - (pi)/(4)) =- (a)/(sqrt2)` `(8) At t = (7T)/(8), y = a sin ((2pi)/(T) xx (7T)/(8) + (pi)/(4))` `= a sin (2pi) =0` `(9) A t t = (8T)/(8) =T,y = a sin ((2pi )/(T) xx T + (pi)/(4))` `=a sin ((pi)/(4)) = (a)/(sqrt2) ` From above calculations, graph of `y to t` can be plotted as follows : ![]() At `t =0,` initial phase of a particle at distance x is [from equation (2)], `theta = 0.018 x + (pi)/(4)` `implies ` For first particle, `x = 0 implies theta _(1) = (pi)/(4) rad` For SECOND particle, `x =4 cm implies theta_(3) =0.072 + (pi)/(4) rad` For all of above particles wa can draw graph of `y to t ` as shown earlier. For all of them, amplitude, frequency and wavelength are same, but their initial phase are different. |
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| 13247. |
An air bubble of radius 2 mm is at a depth of 5 cm below the free surface of a liquid of surface tension 0.06 N/m and density 1000 kg/m""^(3). The pressure inside the bubble is greater than the atmospheric pressure by n xx 140 "N/m"^(2). Find the value of n. |
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Answer» Solution :`R=2 MM =2 xx 10^(-3)m` `S=0.007 N//m` `h=5 CM =5 xx 10^(-2)m` `p=1000 kg//m^(3)` `[Pe]=(2S)/(r)+hpg` `=(2 xx 0.07)/(2 xx 10^(-3)) +5 xx 10^(-2) xx 1000 xx 9.8` `=70+490=560N//m^(2)` |
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| 13248. |
The shape of an ancient water clock jug is such that water level descends at a constant rate at all time. If the water level falls by 4cm every hour, determine the shape of the jar, i.e. specify x as a function of y. The radius of drain hole in 2mm and can be assumed to be very small compared to x. |
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Answer» Here, `-(dy)/(dt)=(4XX10^(-2))/(3600)=1.11xx10^(-5) m//s` `a=pir^(2)=pi (2XX10^(-3))^(2)` `=1.26xx10^(-5)m^(2)` substituting these values in Eq, (i) we have `(1.26xx10^(-5))sqrt(2xx9.8xxy)=pi (1.11xx10^(-5))x^(2)` or, `y=0.4x^(4)` This is the desired x-y relation. |
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| 13249. |
A body of mass m is pushed up with a velocity u along a plane of inclination theta. If coefficient of friction between the body and the inclined plane is mu displacement of the body before coming to rest is |
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Answer» `(U^(2)MU)/(2gsintheta)` |
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| 13250. |
Calculate the linear acceleration of a particle of a particle moving in a circle of radius 0.4 m at the instant when its angular velocity is 2 rad s^(-1) and its angular acceleration is 5 rad s^(-2). |
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Answer» |
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