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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 siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m cdot s^(-1). Then the frequency of sound that the observer hears in the echo reflected from the cliff is (take velocity of sound in air = 330 m cdot s^(-1)) |
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Answer» 800 Hz `n.=n(V/(v-v_(s)))=800(330/(330-15))` `=(800xx330)/315=838` Hz The option (B) is correct. |
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| 2. |
Moment of inertia of a straight wire about an axis perpendicular to the wire and passing through one of its end is I. This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would be : |
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Answer» `((3)/(PI^(2)))I` |
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| 3. |
For the case of an ideal gas, find the equation of the process in the variable T, Vin which the molar specific heat capacity varies as (a)C=C_v + alpha T ,(b)C=C_v +beta V , (c)C=C_v + aP where alpha , beta a are constants |
| Answer» SOLUTION :`(V )/(e^x) = ` CONSTANT, where` x=( alpha T)/(R ) ` (b) `Te ^(R//V BETA )` = constant ,(c) V-aT= constant | |
| 4. |
Is it necessary that any mass should be present at the centre of mass of a systl |
| Answer» Solution :No. It is not NECESSARY to have some mass at centre of mass of the SYSTEM Ex , In case of a HOLLOW SPHERE (or) a ring centre of mass is at its centre. But there is no mass at its centre. | |
| 5. |
A space station is set up in space at a distance equal to earth's radius from earth's surface. Suppose. Let V_(1) and V_(2) be the escape velocities of the satellite on earth's surface and space station respectively. Then |
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Answer» `V_(2) = V_(1)` |
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| 6. |
The breaking stress of wire dpends on |
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Answer» material of the WIRE |
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| 7. |
A block placed on a horizontal surface is being pushed by a forceF making an angle 0 with the verticle If the friction coefficient is mu how much force is needed to get the block just started Discuss the situation when tantheta lt mu. |
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Answer» Solution :In limitting equilibrium force of friction `F = mu R` In equilibrium ALONG the HORIZONTAL `F SIN theta = mu R` and along the verticle `F cos theta + mg = R = (F sin theta )/(mu` `:. Mg = F(( sin theta )/(mu) - cos theta)` or `F = (mu mg)/(sin theta - mu cos theta)` If tan `theta lt mu` `(sin theta - mu cos theta lt theta)` `:. F` is negative So for angles less angles less than `tan^(-1) mu` on cannot push the BLOCK ahed howsoever large the force may be . |
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| 8. |
A capillary tube of radius r is immersed in waer and water rises in it to a height h. The mass of water in the capillary tube is 5g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is |
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Answer» 2.5 g |
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| 9. |
A simple pendulum of length 'l' carries a bob of mass 'm'. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the of suspension. When the string is horizontal, the net force on the bob is |
| Answer» ANSWER :C | |
| 10. |
At a certain temperature, the r.m.s. velocity forO_2 is 400m/sec. At the same temperature. the r.m.s velocity for H_2 molecule will be |
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Answer» 100 m/sec |
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| 11. |
The fundamental unit which has same power in the dimensional formula of surface tension and coefficient of viscosity is |
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Answer» mass The dimensional formula for coefficient of viscocity is `[ML^(-1)T^(-1)]` |
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| 12. |
A cubical wood piece of mass 13.5 gm and volume 0.0027 m^3 float in water. It is depressed and released, then it executes simple harmonic motion. Calculate the period of oscillation ? |
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Answer» |
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| 13. |
What is meant by zero isothermal ? |
| Answer» SOLUTION :Suppose we draw an isothermal at zero kelvin then that isothermal is called zero isothermal. This is the lowest POSSIBLE isothermal that could be DRAWN on a P-V diagram, when we formulate Kelvin.s thermodynamic SCALE of TEMPERATURE. | |
| 14. |
A load of fles, groduces an extension of 2 mm in a wire 3m long and 1 mm in diameter. Calculate Young.s modulus of the material of the wire. |
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Answer» SOLUTION :SINCE `M=4kg, l =3 m, r=1//2 mm and Delta l=2mm` `Y=(Mgl)/(pir^(2)(Deltal))=(4xx9.8xx3)/((22)/(7)xx((1)/(200))^(2)((2)/(10^(3))))=7.48xx10^(10)N//m^(2)` Young.s modulus of the material of the wire `=7.48xx10^(10)N//m^(2)` |
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| 15. |
The position vector of a particle varies with time as vec(r)=vec(r_(0)) ( 1 - alpha t) t where vec(r_(0)) is a constant vector & alpha is a positive constant then the distance covered doring the time interval in which particle returns to its initial position is : |
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Answer» `r_(0)//ALPHA` |
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| 16. |
Two beads of massesm_(1) and m_(2) are closedthreaded on a smoothcircularloop of wireof radius R. Intially both the beads arein positionA and B in verticalplane . Now , the beadA is pushed slightly so that itslide on the wire and collideswith B . if B risesto the height of the centre ,the centre of the loop Oon the wire and becomes stationaryafterthe collision , provem_(1) : m_(2) = 1 : sqrt(2) |
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Answer» Solution :Let `v_(1)`be the velocityof head A of mass `m_(1)` when it slideon the wireand collideswith head B , `1/2 m_(1)v_(1)^(2) =m_(1)g (2R) "" ( :. h = 2R)` ` :. V_(1) = sqrt(4g)R ""……..(1)` COLLISION being elastic , so the lawof momentum is , ` m_(1)v_(1) =m_(2)v_(2)` ` :. v_(2) =(m_(1))/(m_(2)) v_(1) =(m_(1))/(m_(2)) (sqrt(4g)R) "".....(2)` After, collision bead of mass `m_(2)` reaches to the HEIGHT R , ` :. 1/2 m_(2)v_(2)^(2) =m_(2)gR "" ( :. h = R)` ` :. v_(2) = sqrt(2gR) ""......(3)` Substituting the valueof `v_(2)` from EQUATION (3) in euation (2) , `sqrt(2gR) =(m_(1))/(m_(2)) sqrt(4gR)` ` :. (m_(1))/(m_(2)) =1/(sqrt(2))` ` :. m_(1) : m_(2) =1: sqrt(2)` |
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| 17. |
What do you mean by percentage error ? |
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Answer» Solution :The RELATIVE error EXPRESSED as percentage is called percentage error. Percentage error `=(Deltaa_(m))/(a_(m))XX100%`. |
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| 18. |
A baloon of mass M is stationary in air. It has a ladder on which a man of mass mis standing. If the man starts climbing up the ladder with a velocity v relative to ladder, the velocity of balloon is |
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Answer» `(m)/(M)` V UPWARDS |
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| 19. |
An aluminium sphere of diameter 10 cm is submerged in glycerine of density 1.26g*cm^(-3) " at " 20^(@)C. Calculate the change in apparent weight of the sphere when the glycerine is heated to 100^(@)C. Coefficient of volume expansion of glycerine =5 times 10^(-4@)C^(-1) and that of linear expansion of aluminium =2.5 times 10^(-6@)C^(-1). |
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Answer» |
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| 20. |
At what depth below the surface of the earth, acceleration due to gravity g will be half its value 1600km above the surface of the earth? |
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Answer» `4.2 xx 10^(6)m` |
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| 21. |
Which quantity have unit rpm. Show it in rad/s. |
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Answer» Solution :rpm means revolution. PER minute. It is a UNIT of angular speed 1 rpm = `("1 revolution")/("minute")=(2pi" rad")/("60 SECOND")=(pi)/(30)` rad/s |
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| 22. |
A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically with iron strip towards left and copper strip towards right. If this bimetal is heated, it will |
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Answer» REMAIN straight |
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| 23. |
(A) : Total K.E. of 8 litres of helium molecules at 5 atmosphere pressure will be 6078J.(R): In a given gas, the K.E. of gas molecules E=(1)/(3)PV , where P is the pressure and V is the volume of gas |
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Answer» Both (A) and ( R) are TRUE and (R ) is the correct explanation of (A) |
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| 24. |
Universal time is based on : |
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Answer» Joule-Thomson effect |
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| 25. |
Define angle of friction. The inclination theta of a rough plane is increased gradually. The ply on the plane just comes into motion when inclination theta becomes 30^(@). Find coefficient of friction. If the inclination is further increased to 45^(@) then find acceleration of the body along the plane. (g = 10 m.s^(-2)) |
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Answer» Solution :Here, ANGLE of repose `30^(@)` `:.` Coefficient of friction, `mu= tan30^(@) = (1)/(sqrt(3))` If the angle of inclinataion is `45^(@)`, the acceleration of the BODY, `a = g(SIN45^(@)-mu.cos45^(@))= 10xx(1)/(sqrt(2))(1-mu.)` `(10)/(sqrt(2))(1-(1)/(sqrt(3)))~~2.99"m.s"^(-2)[becausemu.~~mu]` |
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| 26. |
As F shown in diagram increases from zero the graph between friction and F will be |
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Answer»
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| 27. |
In the fig, find the acceleration of mass m_(2) |
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Answer» SOLUTION :`l_(1)+2l_(2)=` constant on differentiating `v_(1)+2v_(2)=0` Again differentiating `a_(1)+2a_(2)=0 RARR a_(1)=-2a_(2)`  .-ve. sign indicates that the accelerations are in opposite direction. Suppose acceleration of `m_(2)` is `a_(2)` downward and then acceleration of `m_(1)` will be `a_(1)` UPWARDS. `T-m_(1)g=m_(1)a_(1)` `T=m_(1)g+m_(1)a_(1)` `m_(2)g-2T=m_(2)a_(2)`  `m_(2)g-2(m_(1)g+m_(1)a_(1))=m_(2)a_(2)` `m_(2)g-2m_(1)g=m_(2)a_(2)+4m_(1)a_(2)(THEREFORE a_(1)=2a_(2))` .-. sign should not be substituted `a_(2)=((m_(2)-2m_(1))g)/(4m_(1)+m_(2))MS^(-2)` |
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| 28. |
A particletiedto a stringof lengthl isdescribinga verticalcircleitsspeedat thebottommostpointis sqrt( 7gl ) .Thenitsspeedat thetopmostpointwillbe |
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Answer» ` SQRT(gl)` The heightof theparticleis measuredfromthe bottom mostpointand itspotentialenergyat thispointcan bemostpointand itspotentialenergyat thispointcan betakento be`= mg* 0=0 ` ` therefore `totalenergyof theparticleat thebottommostpoint `=(7 )/(2)mg l + 0 =(7)/(2)m gl ` if thevelocityof theparticleat thetopmostpointis v . itskineticenrgy`=(1) /(2)MGL ` ifthevelocityof theparticleat thetopmostpointis ` V ,`itskineticenergy`=(1)/(2)m v^(2)` andpotentialenergy`= mgh = mg * 2l2mgl` `therefore ` TOTALENERGY`=((1)/(2) m v^(2) + 2mgl) ` so `(1)/(2) mv^(2)+ 2mgl= (7) /(2)mgl` or ` (1)/(2)mv^(2)= (3)/(2)mgl ` or`v= sqrt(3gl ) ` |
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| 29. |
Consider the quantities, pressure, power, energy, impluse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are |
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Answer» IMPULSE, PRESSURE and AREA |
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| 30. |
A light stright fixed at one end to a wooden clamp on a ground passes over a fixed pulley and hangs at the other side. It makes an angle 30^(@) with the ground. A boy weighing 60 kg climbs up the rope. The wooden clamp can tolerate upto vertical force 360N. Find the maximum acceleration in the upward direction with which the boy can climb safely. The friction of pulley and wooden clamp may be ignored (g = 10 m//s^(2)) |
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Answer» |
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| 31. |
The volume of a liquid at 10^(@)C is 50 c.c. Find its volume at 100^(@)C? (gamma_(R) of liquid is 7xx10^(-4)//""^(@)C). |
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Answer» SOLUTION :Initial volume of the liquid `(V _(1)) = 50 ` c.c Initial temperature `(t _(1) ^(@)C) = 10 ^(@)C:`Final temperature `(t _(2) ^(@) C) = 100^(@) C` Coefficient of real expansion of the liquid `(gamma _(R )) = 7 xx 10 ^(-4) //^(@)C ` from `v _(2) = v _(1) [1 + gamma _(R) (t _(2) - t _(1)) ]` `= 50 [ 1+ 7 xx 100 ^(-4) xx 90] = 53.1 5 cm ^(3)` `therefore v _(2) = 53.15 cm ^(3)` Volume of liquid at `100^(@)C=53.15 cm ^(3)` |
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| 32. |
A boy has 60 kg-wt. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6. What is the minimum volume of wooden log ("Density of river water is "1000 kg//m^(3)) |
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Answer» `6.66m^(3)` `(("weight of LIQUID"),("DISPLACED by body"))="up thrust"` |
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| 33. |
hatiandhatj are unit vectors along x - and y- axis respectively . Whatis the magnitude and direction of the vectors hati+hatjandhati-hatj? What are the components of a vector A=2hati+3hatj along the directions of hati+hatjand hati-hatj? [ You may use graphical method] |
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Answer» |
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| 34. |
Water from a tap falls vertically with a velocity of 3 m.s^-1. The area of cross section of the mouth of the tap is 2.5 cm^2. If the flow of water is uniform and steady throughout, then determine the cross section area of the tube of flow of water at a depth of 0.8 m from the mouth of the tap. [g=10m.s^-2] |
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Answer» Solution :The area of cross section of the mouth of the TAP `A_1=2.5 cm^2` and velocity of water FLOW there, `v_1=3m. S^-1`. LET the area of cross section of the tube of flow of water at a depth of 0.8 m below the tap be `A_2 ` and velocity of water of flow there be `v_2`. `therefore "" v_2^2=v_1^2+2gh` `=(3)^2+2xx10xx0.8` `=9+16=25` or, `""v_2=5m.s^-1` We know that , `A_1v_1=A_2v_2` `or,"" A_2=(A_1v_1)/(v_2)=(2.5xx3)/(5)=1.5 cm^2`. |
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| 35. |
An ice cube of mass 0.1 kg of 0^@Cis placed in an isolated container which is 227^@CThe specific heat capacity S of the container varies with temperature T according to the empirical relation S = A+BT, where A = 100 cal/kg -K and B = 2xx10^(2) cal//kg - K^(-2).If the final temperature of the container is 27^@C Latent heat of fusion of water is 8x10^(4) cal/kg, specific heat of water is 10^(-3)cal/kg-k. If the coefficient of lincar expansion for ice varies with temperature as alpha (T) . If L_0is initial length of the edge of ice cube at T, and the temperature of ice changed from T_0toT(T_(0) gt T)then: |
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Answer» `L=L_0int_(T_0)^(T)ALPHA(T)DT` |
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| 36. |
An ice cube of mass 0.1 kg of 0^@Cis placed in an isolated container which is 227^@CThe specific heat capacity S of the container varies with temperature T according to the empirical relation S = A+BT, where A = 100 cal/kg -K and B = 2xx10^(2) cal//kg - K^(-2).If the final temperature of the container is 27^@C Latent heat of fusion of water is 8x10^(4) cal/kg, specific heat of water is 10^(-3)cal/kg-k. Heat gained by the ice will be : |
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Answer» `10xx10^(13)` cal |
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| 37. |
An ice cube of mass 0.1 kg of 0^@Cis placed in an isolated container which is 227^@CThe specific heat capacity S of the container varies with temperature T according to the empirical relation S = A+BT, where A = 100 cal/kg -K and B = 2xx10^(2) cal//kg - K^(-2).If the final temperature of the container is 27^@C Latent heat of fusion of water is 8xx10^(4) cal/kg, specific heat of water is 10^(-3)cal/kg-k. The mass of th container will be : |
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Answer» 0.495 KG |
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| 38. |
A source of sound of frequency 500 Hz is moving towards an observer with velocity 30 m/s.The speed of the sound id 330 m/s.The frequency heard by the observer is |
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Answer» 545 Hz |
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| 39. |
In a given process of an ideal gas, dW=0 and dQlt0. Then for the gas |
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Answer» The TEMPERATURE will decrease |
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| 40. |
For a simple pendulum the graph between length and time period will be a…………… |
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Answer» hyperbola This shows that the graph between T and L will be a Parabola. |
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| 41. |
A particle moves in a circular path such that its speed v varies with distance as V= alphasqrt(s)where alpha is a positive constant. Find the acceleration of the particle after traversing a distance s. |
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Answer» `alpha^(2)SQRT((1)/(4)+ (s^(2))/(R^(2)))` |
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| 42. |
If same force is acting on two masses m_1 and m_2and the accelerations of two bodies are a_1 and a_2 respectively, then |
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Answer» `a_2/a_1 = m_2/m_1` |
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| 43. |
Kepler's second law regarding constancy of aerial velocity of a planet is consequence of the law of conservation of ............... |
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Answer» energy |
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| 44. |
The periodof revolution of a satellite depends upon the radius of the orbit, the mass of the planet and the gravitational constant. Prove that the square of the period varies as the cube of the radius. |
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Answer» Solution :The period of revolution of t of the satellite depends upon the radius of the orbit rthe mass m of the PLANET and the grativational constant G `t prop r^(x)m^(y)G^(z)` Substituting the dimensional formula of t,r,m and G `T=L^(x)M^(y)(M^(-1)L^(3)T^(-2)]^(z)` `M^(0)L^(0)T=L^(x+3z)M^(y-z)T^(-2z)` Using the principle of homogeneity of DIMENSIONS `x+3z=0, y-z=0,y-z=0, -2z=1` Solving `z=-1/2,y=-1/2,x=3/2` `t prop r^(3//2)m^(-1//2)G^(-1//2)` m and G are constants `t^(2)PROPR^(3)` i.e. the square of the period VARIES as the CUBE of the radius |
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| 45. |
An ideal fluid flows through a pipe of circular cross -section made of two sections with diameters 2.5 cm and 3.75 cm. The ratioof the velocities in the two pipes is |
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Answer» `9:4` |
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| 46. |
Explain the behaviour of the oscillator when the driving frequency is far from natural frequency in small damped oscillations. |
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Answer» Solution :For small damping and `omega_(d)B lt lt m(omega^(2)-omega_(d)^(2))` then `omega_(d)b` is NEGLECTED compared to `m(omega^(2)- omega_(d)^(2))` Amplitude of damped oscillation `A= (F_0)/(m(omega^(2)-omega_(d)^(2)))` and for `omega= omega_(d)` amplitude is `A= (F_0)/(omega_(d)b)`. Below figure shows the dependence of the displacement amplitude of an oscillator on the ANGULAR frequency of the driving force for different amount of damping present in the SYSTEM.  The curves in this figure show that, if we go on changing the driving frequency `(omega_d)`, the amplitude tends to infinity when it EQUAL to the natural frequency `(omega)`. But this the ideal case of zero damping which never arises in a real system because the damping is never perfectly zero. The curves for amplutude `(omega)/(omega_0)` corresponding to different value of b are shown in figure. The amplitude becomes infinite for `omega_(d)= omega` when b=0. As damping increases, the value of maximum amplitude in graph displaced towards left side. |
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| 47. |
A pin is placed 10cm in front of a convex lens of focal length 20cm and refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature of 22cm. Determine the position of the final image. Is the image real or virtual. |
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Answer» Solution :As radius of curvature of silvered surface is 22cm, so `f_M=R/2=(-22)/2=-11cm` and Hence `P_M=-1/ f_M=- 1/(-0.11)=1/0.11 D` Further as the focal length of lens is 20cm, i..E, 0.20m its power will be given by `P_L=-1/f_L=1/0.20 D` Now as in image formation, LIGHT after passing through the lens will be reflected back by the curved MIRROR through the lens again. `P=P_L+P_M+P_L=2P_L+P_M` i..e., `P=2/0.20+1/0.11=210/11 D` So the focal length of equivalent mirror `F=-1/P=-11/210 m=-110/21 cm` i..e, the silvered lens behaves as a concave mirror of focal length `(110//21) cm`. SO for object at a distance 10cm in frontof it `1/v+1/(-10)=- 21/110` i..e, `v=-11cm` i..e, image will be 11cm in FRONT of the silvered lens and will be real as SHOWN in figure. 
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| 48. |
While launching a rocket of mass 2xx10^(4) kg, a force of 5xx10^(5)N is applied for 20s. What is the velocity attained by the rocket at the end of 20s? |
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Answer» 300m/s `therefore F XX t= m(V-u), " where " u=0` `5XX10^(5) xx 20 = 2XX10^(4) xx v` `thereforev= (100xx10^(5))/(2xx10^(4))= 500m//s` |
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| 49. |
When the hands are stretched out from the body, the moment of inertia of the ice dancer |
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Answer» DECREASES |
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| 50. |
Magnetic induction and magnetic flux differ in the dimensions of |
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Answer» Mass |
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