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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.
| 1. |
A body at rest explodes and breaks up into 3 pieces. Two pieces having equal mass, fly off perpendicular to each other with the same speed of 30 m/sec. The 3rd pieces has 3 times the mass of each of the other pieces. Find the magnitude and direction of its velocity immediately after the explosion. |
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Answer» Solution :For momentum conservation in each direction of MOTION of the smaller pieces. `{:(3M v cos theta = m xx 30),(3m v sin theta = m xx 30):}}` These two relations GIVE, `TAN theta = 1`, and `v=10 sqrt(2)m//s` The 3RD piece of mass 3m will go making an angle `(180-theta)=135^(@)` relative to either piece of equal masses. 
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| 2. |
For a system to follow the law of conservation of linear momentum during a collision, the condition is (a) total external force acting on the system is zero(b) total external force acting on the system is finite and time of collision is negligible(c ) total internal force acting on the system is zero. |
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Answer» (a) only |
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| 3. |
Maximum range of projectile depend on angle of projectile. |
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| 4. |
A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway, the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s. |
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Answer» Solution :Given, speed of car = speed of truck = 72 km/h `= 72 XX (5)/(18) m//s = 20 m//s` Now, `v = u + a _(t)t ` `0= 20 + a _(t) xx 5` `a _(t) = - 4 m //s ^(2)` For car, `v = u + a _(c)t ` `0= 20 + a _(c) xx 3` `a _(c) = - (20)/(3) m//s ^(2)` Let car be at a DISTANCE x from truck, when truck gives the signal and t be the time taken to cover this distance. As human response time is `0.5s,` therefore, time of retarded motion of car is `(t - 0.5)s.` Velocity of car after time t, `v _(c) =u - at` `=20 - ((20)/(3)) (t - 0.5)` Velocity of truck after time l, `v _(t) = 20 - 4t` To avoid the collision, `v _(c) = v _(t)` `20- (20)/(3) (t -0.5) = 20 - 4t` `4t = (20)/(3) (t - 0.5)` `t = 5/3 (t - 0.5)` `3t = 5t - 2.5` `therefore t = (2.5)/(2) = 5/4 s` Distance travelled by the truck, `x _(t) = u_(t) t + 1/2 a _(t) t ^(2) = 20 xx 5/4 +1/2 xx (-4) xx ((5)/(4)) ^(2) =21.875 m` Distance travelled by car = {Distance travelled by car in `0.5 s` (with constnat speed )} + {Distance travelled by car in `(t - 0.5) s` (with retardation)} `x _(c) = (20 xx 0.5) + 20 ((5)/(4) -0.5) - (1)/(2) ((20)/(3)) ((5)/(4) - 0.4) ^(2) = 23.125 m` `thereforex =x _(c) - x _(t)=23.125 -21.875 =1.250m` Therefore, to avoid the collision, the car MUST maintain a distance more than `1.250m` from the truck. |
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| 5. |
Match Column-I with Column-II : {:("Column-I","Column-II"),("(1) Change in momentum","(a) Force"),("(2) Rate of change of momentum","(b) Impulse of force"),(,"(c) Momentum"):} |
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| 6. |
A solid sphere rolls without slipping on an inclined plane, then which of the following is/are correct ? |
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Answer» Torque of friction about point of contact is zero |
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| 7. |
Whatis breaking stress for a wire of unitcross section called? |
| Answer» SOLUTION :TENSILE STRENGTH. | |
| 8. |
For a uniform flow of an incompressible, non-viscous fluid the Bernoulli's theorem expresses, |
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Answer» CONSERVATION of ANGULAR momentum |
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| 9. |
How can we measure distance of star which is nearly 50 light years away from us? Can we measure the distance from a star 150 light away using the same method? |
| Answer» SOLUTION :We can measure the distance of a star that is 50 light years AWAY using parallex method. The star will be observed from two different locations on earth and by MEASURING the distance between both the locations and angle subteded by the star on both positions, the star distance can be calculated. This method cannot be used to measure a star which is more than 100 light years as the ACCURACY of the result is not achieved. Thus, we cannot measure the star 150 light years away from us using the same method. | |
| 10. |
The mean radius of earth is R, its angular speed on its own axis is omega and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite? |
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Answer» `(R^(2)g"/"omega^(2))^(1/3)` |
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| 11. |
(n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass. |
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Answer» Solution :Let r be the POSITION VECTOR of the centre of MASS of a regular n-polygon. `(n-1)` equal point MASSES are placed at `(n-1)` vertices of the regular n-polygon, therefore, for its centre of mass. `r_(CM)=((n-1)mr+ma)/((n-1)m+m)` `therefore 0=((n-1)mr+ma)/((n-1)m+m)` (`because r_(CM)` is at origin) `therefore (n-1)mr+ma=0` `r=-(a)/((n-1))` |
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| 12. |
An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 "cm"// "sec" and the period is 628 m sec. The amplitude of the motion in centimeter is |
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Answer» 3 |
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| 13. |
A small object of mass m is attached to a light string which passes through a hollow tube. The tube is hold by one hand and the string by theother. The object is set into rotation in a circle of radius R and velocity v. The string is then pulled down shortening the radius of path of r. What is conserved? |
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Answer» angular MOMENTUM |
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| 14. |
If the ice at the poles of the earth melts, how would this affect the length of the day? |
| Answer» Solution :If the polar ICE melts a part of the water thus produced will SHIFT from the POLES towards the equatorial region, and hence this water will shift away from the axis of rotation of the earth. Consequently themoment of INERTIA of the earth will increase. Now according to the principle of conservation of angular momentum with the increase in moment of incertia of the earth its angular VELOCITY will decrease and hence the length of the day will increase. | |
| 15. |
A certain mass of an ideal diatomic gas contained in a closed vessel to heated it is observed that the temperature remains constant. However, half the amount of gas gets dissociated. The ratio of the heat supplied to the gas initial internal energy of the gas will be |
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Answer» `1:2` and `U_(2) = n/2((5R)/(T))T+(2N)/(2)((3R)/(2))T = n((11R)/(4))T` `U_(2) =U_(1)+Q` `:. Q/(U_1) = (U_2)/(U_1)-1 = 11/10 - 1 = 1/10`. |
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| 16. |
The workdone in bowing a soap bubble of radius aunder isothermal conditions is (the surface tension of soap film is T) |
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Answer» `4pia^(2)T` |
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| 17. |
Which component of torque is responsible for the rotation motion about Z-axis? |
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Answer» SOLUTION :Z-component of torque is responsible for the ROTATION motion about Z-axis. Z-component of torque : `tau_(z)=(x_(n)F_(ny)-y_(n)F_(NX))hatk` |
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| 18. |
Two bodies A and B of mass 5 kg and 10 kg respectively in contact with eachother rest on a table against a rigid partition The coefficient of friction between the bodies and the table is 0.15 A force of 200 N is applied horizontally at A Calculate (i) reaction of the partition (ii) action reaction forces between A and B What happens when the partition is removed ? Does the anwer to (ii) change when the bodies are in motion ? Ignore difference between mu_(s) and mu_(k) . |
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Answer» Solution :Here , mass of body A, `m_(1) = 5 kg` mass of body B, `m_(2) = 10 kg` COEFFICIENT of friction between the bodies and the table ` mu = 0.15 ` Horizontal FORCE applied on A , `F = 200 N ` (i) Force of limiting friction acting to the left `f = mu (m_(1) + m_(2) g` `= 0. 15 (5 + 10 ) xx 9.8 = 22 .05 N` Net force exerted on the partition (to the right) `F = F - f = 200 - 22.05 = 177.95 N ` Reaction of partition `= 177 . 95 N` to the left Force of limiting friction acting on the body A `f_(1) = mu m_(1) g = 0 .15 xx 5 xx 9.8 = 7 . 35 N` Net force exerted by body A on body B `F'' = F - f_(1) = 200 - 7 .35 = 192 . 65 N` This force is to the right Reaction of body B on body A = 192 . 65 N to the left Acceleration produced in the system `a = (F)/(m_(1) + m_(2)) = (177 . 95)/(5 + 10) = 11 .86 m//s^(2)` Force producing MOTION in body A `F_(1) = m_(1) a = 5 xx 11.86 = 59.3 N` Net force exerted by body A on B when partition is removed ` = F '' - F_(1) = 192.65 - 59.3 = 133.35 N` Hence reaction of body B on body A when partition is removed`= 133 . 35 N` Thus answers to (ii) do change . |
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| 19. |
A uniform chain in held on a frictionless table with one-third of its length hanging over the edge. If the chain hasa length l and mass m , how much work is required to pull the hanging part back on the table ? |
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| 21. |
Amorphous solids are not true solids . Explain . |
| Answer» Solution :Atoms in amorphous solids are not ARRANGED in a regukar manner . But they cananot move RELATIVE to each otheras in case of LIQUIDS . Thus they ebhave as SUPER cooled liquid of high VISCOCITY. | |
| 22. |
Three identical cylinders have mass M each and are placed as shown in the figure. The system is in equilibrium and there is no contact between B and C. Find the normal contact force between A and B. |
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| 23. |
In a periodic motion when a body moves to and fro about a fixed mean position its acceleration. |
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Answer» Proportional to DISPLACEMENT of BODY from MEAN position and is always directed toward the mean position |
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| 24. |
Two balloons are filled, one with pure He gas and other with air, respectively. If the pressure and temperature of these balloons are same then the number of molecules per unit volume is |
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Answer» More in the He FILLED balloon |
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| 25. |
Mercury does not wet glass. This is the property of liquid known as |
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Answer» adhesion |
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| 26. |
The sound level from a musical instrument playing 50 dB . If three identical musical instruments are played together then compute the total intensity . The intensity of sound from each instrument is 10^(-12) W m^(-2) |
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Answer» Solution :`Delta L = 10 log _(10) |(I_(1))/(I_(0))| = 50` dB `log_(10) |(I_(1))/(I_(0))| = 5` dB `(I_(1))/(I_(0)) = 10^(5) IMPLIES I_(1) = 10^(5) I_(0) = 10^(5) xx 10^(-12) W m^(-2)` `I_(1) = 10^(-7) W m^(-2)` Since THREE musical instruments are played , therefore `I_("total") = 3 I_(1) = 3 xx 10^(-7) W m^(-2)` |
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| 27. |
When a guitar string is sounded with a 440 Hz tuning fork, a beat frequency of 5 Hz is heard. If the experiment is repeated with a tuning fork of 437 Hz, the beat frequency is 8 Hz. The string frequency (Hz) is .... |
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Answer» 445 `f ~440 =5 implies f = 445 Hz OR f = 435 Hz` Also `f ~437 = 8 implies f = 445 Hz OR f = 429 Hz` `implies f = 445 Hz` (`because` If we take `f = 435 Hz` then `435 ~437 =2 ne 8` if we take `f = 429 Hz` then `429 ~440 =11 ne 5).` |
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| 28. |
A sphere of radius 0.1 m and mass 8pi kg is attachedto the lower end of a steel wire of length5.0 m and diameter 10^(-3)m. The wire is suspended from 5.22 m highceiling of a room. When the sphere is made to swing as a simplependulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel =1.994xx10^(11)N//m^(2). |
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Answer» Solution :As the length of the wire is 5m and diameter `2xx0.1=0.2m` andat lowestpoint it grazes the floor which is at a distance 5.22 m from the roof, the increase in length of the wire at lowest point `DeltaL=5.22-(5+0.2)=0.02m` So tension in the wire (due to elasticity) `T=(YA)/(L)DeltaL` `=(1.994xx10^(11)xxpi (5XX10^(-4))^(2)xx0.02)/(5)=199.4piN` and as equation of circular motion of a mass .m. tied to a STRING in a vertical plane is `(mv^(2)//R)=T-mgcos theta` So atlowest point`(mv^(2)//r)=T-mg ""["as" theta =0]` But her `r=5+0.02+0.1=5.12m` So, `(8pi v^(2)//5.12)=(1.99.4pi-8pi xx9.8)` i.e., `v^(2)=(121xx5.12//8)=77.44, " so v" =8.8m//s`. |
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| 29. |
Explain propagation of sound waves in solids. |
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Answer» Solution :Crystalline solids possess lattice STRUCTURE in which atoms or molecules are arranged with some definite geometric periodic pattern. In the normal condition, without any disturbance, all these atoms or molecules are under equilibrium because forces exerted from surrounding balance each other. Now, under this equilibrium condition, if disturbance is produced in any atom or molecule, it gets displaced from its equilibrium position. Here this atom or molecule behaves as it is elastically connected to NEIGHBOURING atoms or molecules. Hence, RESTORING force is developed in such imaginary elastic springs. Such a force BECOMES responsible for producing very SMALL oscillations in the atoms or molecules, turn by turn along the direction of motion of disturbance which ultimately results into propagation of wave in the solid medium from one end to another end. |
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| 30. |
A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the liquid is sigma , the change in energy in this process is : |
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Answer» `PI SIGMA D^2( N^(1//3)-1)` |
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| 31. |
If the depression d at the end of a loaded bar is given by d= (Mgl^3)/(3y i)where M is the mass, l is the length and y is the young's modulus, then i has the dimensional formula |
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Answer» `L^2` |
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| 32. |
In the above problem if the lift moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will be |
| Answer» Answer :D | |
| 33. |
Four mass capilary tubes, a, b, c and d having diameters, 2mm, 1mm, 0.8mm and 0.6mm are dipped in four beakers containing water with a tube vertically, 'b' tube 30^(@) 'c' tube 45^(@) and d tube 60^(@) inclination with the vertical arrange the lengths of water column in the tubes in descending order |
| Answer» Answer :A | |
| 34. |
What are the units and dimensions of the coefficient of thermal expansion of a solid or liquid or gas? |
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Answer» Solution :For Coefficient of thermal expansion of solids, liquids and GASES unit is PER `0^(@)C`. Dimensional formula : `K^(-1)` |
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| 35. |
A nucleus moving with a velocity vec(v) emits an alpha - particle. Let the velocities of the alpha - particle and the remaining nucleus be vec(v)_(1) and vec(v)_(2) and their masses be m_(1) and m_(2). |
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Answer» `vec(v), v_(1)` and `vec(v)_(2)` must be parallel to each other. |
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| 36. |
Distance between two places is 200 km. alpha a of steel is 12xx10^(-6)//""^(@)C. Total space that must be left between steel rails to allow for a change of temperature from 36^(@)F to 117^(@)F is |
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Answer» 1.08 km |
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| 37. |
A solid sphere is rolling without slipping on rough ground as shown in figure. If collides elastically with an identical another sphere at rest. There is nofriction between the two spheres . Radius of each sphere is R and mass is m. What is the net angular impulse imparted to second sphere by the external forces? |
| Answer» Answer :A | |
| 38. |
A solid sphere is rolling without slipping on rough ground as shown in figure. If collides elastically with an identical another sphere at rest. There is nofriction between the two spheres . Radius of each sphere is R and mass is m. Linear velocity of first sphere after it again starts rolling without slipping is |
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Answer» `(2)/(5)OMEGAR` |
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| 39. |
Example for scalar is |
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Answer» distance |
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| 40. |
A body of mass 2 kg starts from rest and moves in a uniform acceleration. It acquires a velocity of 20 ms^(-1) in 4 s. The power exerted on the body at 2 s in watt is |
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Answer» 50 Let a be the uniform accelration. Using , `v = u + at` `:. 20= 0+ a(4) or a = (20)/4 = 5 ms^(-2)` Force , `F = ma = (2 kg)(5 ms^(-2)) = 10 N` The velocity acquired by the body in 2s is `v = at = (5 ms^(-2))(2s) = 10 ms^(-1)` Power exerted on the body is `P = Fv = (10 N)(10 ms^(-1)) = 100 W`. |
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| 41. |
One gram of water becomes 1671 c.c. of steam when boiled at a pressure of one atmosphere. The heat of vaporisation at this temperature is 540 cal per g. Compute the external work and increase in internal energy. |
| Answer» SOLUTION :168.7 J 2099 J | |
| 42. |
A person normally weighing 50kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^(-1) and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time. Will there be any change in weight of the body, during the oscillation? |
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Answer» SOLUTION :Here, oscillations are PERIODIC HENCE acceleration changes. The weight of MAN depend on direction of acceleration and its magnitude in up and down motion. (a) Hence, weight of man changes during oscillations. |
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| 43. |
A person normally weighing 50kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^(-1) and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time. If answer of part (1) is yes, what will be the maximum and minimum reading in the machine and at which position? |
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Answer» Solution :Here, oscillations are periodic hence acceleration changes. The weight of man depend on direction of acceleration and its magnitude in up and down motion. (b) The acceleration is maximum at two EXTREME points of the platform. At UPPER extreme POINT acceleration on platform is in downward, From equation of motion `N= mg - ma` `N= mg - m omega^(2) A` `therefore N= m[g- omega^(2)A]` `=50[9.8-4pi^(2)v^(2)A]` `=50[9.8-4xx(3.14)^(2) xx 4 xx 5xx 10^(-2)]` `=50[9.8-7.88]= 50[1.92]` Weight = 96N and at mean POSITION `N= mg` `=50xx 9.8` Weight = 490 N At lower extreme point, `N= mg+ ma` `= m(g+a)` `=m(g + omega^(2)A)` `=m(g+ 4pi^(2)v^(2)A)` `=50(9.8+4xx(3.14)^(2)xx (2)^(2)xx5 xx 10^(-2))` `=50(9.8+7.88)` `=50(17.68)` Weight = 884 N. 
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| 44. |
The wavelength of two sound notes in air are (40)/(195)m and (40)/(193)m. Each note produces 9 beats per second, separately with a third note of fixed frequency. The velocity of sound in air in m//s is |
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Answer» 360 `impliesn_(2) gt n_(3) gt n_(1)impliesn_(2)-n_(3)=9`...........`(1)` `n_(3)-n_(1)=9`............`(2)` `(1)+(2)impliesn_(2)-n_(1)=18` |
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| 45. |
A molecule in a gas container hits a horizontal wall with a speed of 200 m*s^(-1) at an angle of 30^@ with the normal and rebounds with the same speed . Is momentum conservedin the collision ? Is the collision elastic or inelastic ? |
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Answer» SOLUTION :As there is no external force, linear MOMENTUM is CONSERVED. The collision is elastic , because the magnitude of the MOLECULAR velocity REMAINS the same. |
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| 46. |
A neutron collides, head-on with a deuterium at rest. What fraction of the neutron's energy would be transferred to the deuterium? |
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Answer» `89%` `m_n v_(ni)+m_d xx 0 =m_n v_(nf)+m_d v_(DF)` where `v_(ni)` is the INITIAL velocity of the neutron before collision and `v_(nf)` and `v_(df)` are the velocities of the neutron and deuterium after collision `m_n v_(ni)=m_(n) v_(nf)+m_d v_(df)`....(i) According to conservation of kinetic energy , we get `1/2 m_n v_(ni)^2 =1/2m_n v_(nf)^2 + 1/2 m_d v_(df)^2` `m_n v_(ni)^2 =m_n v_(nf)^2 + m_d v_(df)^2`....(ii) Both Eqs. (i) and (ii), it follows that `m_n v_(ni)(v_(df)-v_(ni))=m_n v_(nf)(v_(df)-v_(nf))` `v_(df)(v_(ni)-v_(nf))=v_(ni)^2 - v_(nf)^2 , v_(df)=v_(ni)+v_(nf)` Substituting this in Eq.(i) , we get `v_(nf)=((m_n - m_d))/(m_n+m_d) v_(ni)`...(iii) and `v_(df)=(2m_n v_(ni))/(m_n+m_d)`...(iv) The initial kinetic energy of the neutron is `K_(ni) = 1/2 m_n v_(ni)^2` Final kinetic energy of the deuterium is `K_(df)=1/2m_d v_(df)^2 =1/2 m_d ((2m_n v_(ni))/(m_n + m_d))`...(iv) The initial kinetic energy of the neutron is `K_(ni)=1/2 m_n v_(ni)^2` Final kinetic energy of the deuterium is `K_(df)=1/2 m_d v_(df)^2 =1/2m_d ((2m_n v_(ni))/(m_n + m_d))^2` (Using (iv)) Fraction of neutron's energy TRANSFERRED to deuterium is `f=K_(df)/K_(ni)=(4m_n m_d)/((m_n + m_d)^2)` For deuterium , `m_d=2m_n` `therefore f=(4(m_n)(2m_n))/((m_n+2m_n)^2)=8/9=89%` |
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| 47. |
An ideal monoatomic gas undergoes a process where its pressure is inversely proportional to itstemperature. a) Calculate the molar specific heat for the process b) Find the work done by two moles of gas if the temperature changes from T_1 to T_2 |
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| 48. |
The diameter of cylinders are d_(1)=(2.02 pm 0.01)cm " and "d_(2)=(3.21 pm 0.03)cm respectively. Find the difference between their dimensions along with the error limits. |
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Answer» Solution :`d_(1)=(2.02 pm 0.01)cm, d_(2)=(3.21 pm 0.03)cm` Difference `d= d_(2)-d_(1)= 3.21-2.02= 1.19cm` Maximum error in difference, `TRIANGLED= pm(triangled_(2)+triangled_(1))= pm (0.03+0.01)` Difference in DIAMETER `=(1.19 pm 0.04) cm`. |
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| 49. |
1 rad = …... Degree. |
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