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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. |
How can the efficiency of a reversible engine be 100%? |
| Answer» Solution :the efficiency of a REVERSIBLE ENGINE becomes 100% when the temperature of he sink is ZERO KELVIN . | |
| 2. |
The velocities of a body executing SHM at displacement 'a' and 'b' are 'b' and 'a' respectively. The amplitude of S.H.M will be |
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Answer» `a+B` |
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| 3. |
A bullet is fired with a velocity of 900 km/h in the horizontal direction from a gun kept 20 m above the ground . Take g=10"" m//s^2.Find out the time of flight |
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Answer» |
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| 4. |
A small body is projected from the earth's suface vertically up with the escape velocity on the earth. Out of the following curves the one that represents the variation of KE with altitude h is |
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| 5. |
A metal piece of density 8g*m^(-3) is suspended from a wooden hook by an weightless string. The tension in the string is 56g times g. What will be the tension in the string, if the system is immersed in a liquid at 40^(@)C? The surrounding temperature during the experiment is 20^(@)C. At 20^(@)C the specific gravity of the liquid is 1.24. The coefficients of volume expansion of the liquid and the metal are 4 times 10^(-5@)C^(-1) " and " 8 times 10^(-4@)C^(-1) respectively. |
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Answer» Solution :Volume of the metal piece at `20^(@)C`, `V_(20)=56/8=7cm^(3)` `therefore` Volume at `40^(@)C,` `V_(40)=V_(20)[1+8 times 10^(-4) times 20]` `""=7(1+8 times 10^(-4) times 20)cm^(3)` Volume of the displaced liquid `=V_(40)` `therefore` Mass of displaced liquid `=V_(40) times rho_(40)` `""[rho_(40)=` density of the liquid at `40^(@)C]` i.e., weight of the displaced liquid `=V_(40) times rho_(40) times g` `therefore` upthrust `""=V_(40) times rho_(40) times g` `""=7[1+8 times 10^(-4) times 20] times 1.24/(1+20 times 4 times 10^(-5)) times 980` `""=7 times 1.24[1+0.016][1+8 times 10^(-4)]^(-1) times 980` `""=7 times 1.24 times 1.016[1-0.0008] times 980` `""=7 times 1.24 times 1.016 times 0.9992 times 980=8.81 times 980 dyn.` `therefore` Tension in the string `"" =(56-8.81) times 980=4.625 times 10^(4)dyn.` |
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| 6. |
An iron rod of length 50cm is joined at an end to copper rod of length 100 cm at 20^(0)C. Find the length of the system at 100^(0)C and average coefficient of linear expansion of the system. (alpha_(iron) = 12 xx 10^(-6)//^(0) C and alpha_("copper") = 17 xx 10^(-6)//^(0) C . ) |
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Answer» Solution :Increase in length of COMBINED ROD = increase in length of IRON rod + Increase in length of COPPER rod `(Delta l) = (Delta l )_(Fe) + (Delta l)_(Cu)` `(l_(Fe) + l_(Cu)) prop_("Avg") Delta t = l_(Fe ) prop_(Fe) Delta + l_(Cu) prop_(Cu) Delta t` `prop_("Aug") = (l_(Fe) prop_(Fe)+l_(Cu) prop_(Cu))/((l_(Fe) + l_(Cu)))` `= (50 XX 12 xx 10^(-6) + 100 xx 17 xx 10^(-6))/(150) = 15.33 xx 10^(-6) //^(0)`C |
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| 7. |
Derive equation for heat flow rate in rectangular block of solid. |
Answer» Solution :Figure shows a rectangular block of solid of cross section .A..  Let us consider two cross sections ABCD and EFGH, at a distance x and `Deltax` from one end of a block having temperature `T+DeltaT` and T respectively. Here, the distance between two cross section is `Deltax` and temperature difference is `DeltaT`. The ratio `DeltaT//Deltax` is called temperature gradient Temperature gradient : The difference in temperature per unit distance between two parts of a solid is called distance between two parts of a solid is called as temperature gradient parts of a solid is called as temperature gradient. Its unit is Kelvin/meter and dimensional formula is `M^(0)L^(-1)T^(0)K^(1)`. Experiment shown that for a small change in temperature, the amount of heat passing between these two cross sectional planes in direction perpendicular to these planes in time `Deltat` DEPENDS on the following factors : `DeltaQprop(DeltaT)/(Deltax)"(For given "Deltat" and A)"` `DeltaQpropDeltat" (For given "DeltaT//Deltax" and A)"` `DeltaQpropA" For given "Deltat " and "DeltaT//Deltax")"` `:.DeltaQ=-kA(DeltaT)/(Deltax)Deltat` `(DeltaQ)/(Deltat)=-kA(DeltaT)/(Deltax)` . . .(1) Here, k is constant of proportionality called THERMAL conductivity of the material of the block. Its VALUE depends on the type of the material and to some extent on the temperature. The materials with higher value of thermal consuctivity are good conductors of heat. Normalally, if the differnces in the temperatures of different parts of solid are not very large, k can be considered a constant. Negative sign in above equation indicates that with the increase in x the temperature decreases. Taking limits `Deltax to 0` and `Deltat to0` in equation (1) `(dQ)/(dt)=-kA(dT)/(dx)`. . . (2) `:.H=-ka(dQ)/(dt)` `(dQ)/(dt)=H` is called heat current. Heat current is the RATE of flow of heat through any cross section. Its unit is cal/s or J/s, its dimensional formula is `M^(1)L^(2)T^(-3)`. |
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| 8. |
If velocity (v), acceleration (a) and force (F) are taken as fundamental quantities, the dimensions of Young's modulus (Y) would be |
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Answer» `[Fa^(2)v^(-2)]` where K is a DIMENSIONLESS CONSTANT `:. [ML^(-1)T^(-2)]= [LT^(-1)]^(x)[LT^(-2)]^(y)[MLT^(-2)]^(z)= [M^(z)L^(x+y+z)T^(-x-2y-2z)]` EQUATING the powers of M, L and T we get z= 1, x+y+z=-1, -x-2y-2z= -2 Solving we get x= -4, y=2, z=1 `:. Y= v^(-4)a^(2)F^(1)` or `[Y]= [Fa^(2)v^(-4)]` |
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| 9. |
Three bodies , a ring , a solid cylinder and a solid sphere roll down the same inclined plane without slipping . They start from rest . The radii of the bodies are identical . Which of the bodies reaches the ground with maximum velocity ? |
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Answer» RING |
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| 10. |
A horizontal pipe line carries water in a streamline flow. At a point along the pipe where the cross sectiona area is 10cm^(2), the velocity of water is 1m/s and the pressure is 2000Pa. What is the pressure at another section where the cross-sectional area is 5cm^(2). |
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Answer» SOLUTION :ACCORDING to equation of continuity `A_(1)V_(1)=A_(2)V_(2),V_(2)=10//5xx1=2m//s` Now according to Bernoull.s equation `P_(1)+1/2rhoV_(1)^(2)=P_(2)+1/2rhoV_(2)^(2)` ( `:.` HORIZONTAL PIPE) `2000+1/2 10^(3)(1)^(2)=P_(2)+1/2(10^(3))(2^(2))` `:.P_(2)=500Pa` |
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| 11. |
Find the work done by a gas when it expands isothermally at 37^@Cto four times its initial volume |
| Answer» Answer :B | |
| 12. |
A particle moving with shm has a period 0.001 s and amplitude 0.5 cm. Find the acceleration, when it is 0.2 cm apart from its mean position. |
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Answer» |
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| 13. |
A river of salty water is flowling with a velocity 2 m/sec. If the density of th water is 1.2 gm/c.c., then th kinetic energy of each of cubic metre of water is |
| Answer» Answer :D | |
| 14. |
If' a_(m) ' and' Deltaa_(m) are true value and mean absolute error respecively, then the magnitude of the quantity may lie between ....... |
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Answer» `a_(m) + a_(N) "to" a_(m) - a_(n)` |
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| 15. |
A person walks along a straight road from his house to a marked 2.5kms away with a speed of 5 km/hr and instantly turns back and reaches his house with a speed of 7.5 kms/hr. The average speed of the person during the time interval 0 to 50 minutes is (in m/sec) |
| Answer» Answer :B | |
| 16. |
Assertion : Escape velocity from the surface of black hole is greater than speed of light Reason : Light cannot escape from black hole |
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Answer» Both Assertion and Reason are TRUE and Reason is the CORRECT EXPLANATION of Assertion |
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| 17. |
A particle of mass m moves along a circular path of radius r with a centripetal acceleration a_(n) changing with time t as a_(n)=kt^(2), where k is a positive constant. The average power developed by all the forces acting on the particle during the first t_(0) seconds is |
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Answer» at any TIME 't' force ACTING on particle is `msqrt(kr+ K^(2)t^(4))` |
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| 18. |
What is the work efficiecy coefficient in above question? |
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Answer» `beta = (Q_(2))/(Q_(1)-Q_(2)) = (800 xx 10^(3))/((879-800)xx10^(3)) = 10.13` |
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| 19. |
The resultant of two forces 2 N and 3 N is sqrt(19)N. The angle between the forces is |
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Answer» `30^(@)` |
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| 20. |
A body is acted on by a force given by F = (10 + 2t) N. The impulse received by the body during the first four second is |
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Answer» 40 N s |
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| 21. |
Statement I : A vibrating body always moves to and fro about an equilibrium position. Statement II : Due to inertia of motion a vibrating body does not stop at its equilibrium position. |
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Answer» Statement I is TRUE, statement II is true , statement II is a correct EXPLANATION for statement I. |
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| 22. |
What is turbulent flow ? Give its illustration. |
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Answer» Solution :When the flow of fluid is very large , the flow obtained is known as turbulent flow. Illustrations of such type of flow : (i) Sea CURRENTS , (ii) Smoke RISING due to burning of wood , (iii)Stars twinkle because of turbulence in earth atmosphere , (IV) The flow in boat WAVES and around aircraft wing tips. |
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| 23. |
What is solar year? |
| Answer» Solution :The TIME TAKEN by the EARTH to complete one revolution around the SUN. | |
| 24. |
A uniform rod is 4m long and weight 10kg if it is supported on a knife edge at one metre from the end, what weight placed at that end keeps the rod horizontal ? |
| Answer» Answer :B | |
| 25. |
A metal rod AB of length 10x has its one end A in ice at 0^(0)C and the other end B in water at 100^(0)C. If a point P on the rod is maintained at 400^(0)C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of melting of ice is 80 cal/g and latent heat of evaporation of water is 540 cal/g. If the point P is at a distance of lambda_(x) from the ice, find the value of lambda. (neglect any heat loss to the surroundings) |
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| 26. |
A rectangular block has a square base measuring axa, and its height is h, It moves with a speed v on a smooth horizontal surface |
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Answer» It will topple if `V gt sqrt(2gh)` |
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| 27. |
A light ray is incidenton a plane glass slab of thickness t at an angle of incidence t as shown in the figure . If mu is the refractive index of glass. Then find time taken by the light ray to travel through the slab. |
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Answer» Solution :As shown in the above figure distance travelled by the light ray through the SLAB is d From the figure `cos R=t/d, d= t/ (cos r)` Velocity of light in glass = `(Distance travell ED through the glass)/ (time) ` `C/mu= d/(time), time = (d mu)/c= (t mu )/ (cos r times c) = (mu^2 t)/ ( c sqrt(mu^2- sin^2 i))` 
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| 28. |
A force of (4x^(2)+3x)N acts on a particle which displaces it from x=2m to x=3m. The work done by the force is |
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Answer» 32.8 j |
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| 29. |
A body is projected from earth.s surface with a velocity equal to half the escape velocity of earth. The maximum height it will reach is (R —> Radius of earth) |
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Answer» 3R |
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| 30. |
A particle moving along a straight line with initial velocity u and accelerationa continues its motion for n seconds. What is the distance covered by it in the last n^(th) second? |
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Answer» SOLUTION :`S=ut+(1)/(2)at^(2)` Displacement in n seconds `=un+(1)/(2)an^(2)` Displacement in `(n-1)` seconds `=u(n-1)+(1)/(2)a(n-1)^(2)` Displacement in `n^(th)` SECOND = Displacement in n SECONS - displacement in `(n-1)` seconds. `therefore S_(n)=u+a(n-(1)/(2))`. |
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| 31. |
The main scale of vernier callipers has 10 equal divisions in 1 cm and 10 vernier scale divisions coincide with 9 main scale divisions, when both the jaws are in contact. Now, the above instrument is used to measure the external diameter of a cylinder. The 43rd division on the main scale is observed to the left of the zero of the vernier, when the object is held tightly in between the two jaws and the seventh division of the vernier coinsides with a division on the main scale, then |
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Answer» The pitch of the SCALE is 0.1 cm |
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| 32. |
Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side a. At what speed must they move if they all revolve under the influence of one another's gravitation in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ? |
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Answer» Solution :The FORCE of attraction on body C due to bodies at A and B are `F_(1) = (Gm^(2))/(a^(2))` along CA and `F_(2) = (Gm^(2))/(a^(2))` along CB The resultant force on the body at C is `F = sqrt(F_(1)^(2) + F_(2)^(2) + 2F_(1)F_(2) cos 60^(@))` The resultant force acting along CD, Here `OC = (a)/(sqrt(3))` When each body is describing a circular ORBIT with centre of orbit at O, the force F provides the required centripetal force. The radius of the circular orbit is `OC = a//sqrt(3)`, If V is the speed of the body in circular orbit, then Centripetal force = resultant GRAVITATIONAL force `(mV^(2))/(a//sqrt(3)) = (sqrt(3)Gm^(2))/(a^(2))` or `V = sqrt((Gm)/(a))` 
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| 33. |
An impulsive force gives an initial velocity of -1.0 ms^(-1)to the mass in the unstretched spring position. What is the amplitude of motion ? Give x as a function of time for the oscillating mass. Given m=3 kg, k= 1200 Nm^(-1) |
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| 34. |
Potential energy (U) of a body of unit mass moving in a one-dimension conservative force field is given by, U=(X^(2)-4X+3). All units are in S.I. |
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| 35. |
A meteorite of mass m collides with a satellite which was orbiting around a planet in a circular path of radius R. Due to collision the meteorite sticks to the satellite (mass = 10 m) and the satellite is seen to have into an orbit whose minimum distance from the planet is R/2. Determine the velocity v of hte meterite before collision, Mass of the planet is M. |
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Answer» `SQRT((49GM)/(5R))` |
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| 36. |
In a spring gun having spring constant 100 N/m, a small ball of mass 0.1 kg is put in its barrel by compressing the spring through 0.05 mas shown in Figure. The ball leaves the gun horizontally at a height of 2 above the ground. Find (a) The velocity of the ball when the spring is released. (b) Where should a box be placed on the ground so that the ball falls in it. (g = 10 m//s^(2)) |
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Answer» Solution :(a) When the spring isreleased its elastic potential energy `(1//2) kx^(2)` is converted into kinetic energy `(1//2)mv^(2)` of the ball ,so, by conservation of mechanical energy `(1)/(2)mv^(2)=(1)/(2)kx^(2)rArrv=xsqrt((k)/(m))` So `v=0.05sqrt((100)/(0.1))m//s=sqrt((5)/(2))m//s` (b) As initial VERTICAL component of VELOCITY of ball is ZERO, time taken by the ball to reach the ground.  `t=sqrt((2H)/(g))=sqrt((2xx2)/(10))=sqrt((2)/(sqrt(5)))s,` So the horizontal distance travelled by the ball in this time `d=vt=sqrt((5)/(2))xxsqrt((2)/(5))=1m` |
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| 37. |
A man is standing on a rotating table and he drops a heavy mass from his hand outside the table. How will the angular speed of the table change? |
| Answer» Solution :When the MAN drops the heavy mass from his hand OUTSIDE the table moment of inertia of the system about the axis of rotation decreases. We know that angular momentum = moment of inertia `xx` angular VELOCITY. Since no torque is applied from outside according to the principle of conservation of angular momentum angular velocity of the system will increase dueto DECREASE of its moment of inertia. | |
| 38. |
Assertion : When water flow from wider tube to narrow tube ,its velocity increases.Reason : According to equation of continuity multiplication of area and velocity remain constant. |
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Answer» Both ASSERTION and REASON are TRUE and the reason is the CORRECT EXPLANATION of the assertion. |
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| 39. |
An object takes 1 second to slide down a rough 45^(@) inclined plane. The time taken to slide down a smooth 30^(@) inclined plane having the same slope length is (mu = 0.5) |
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Answer» `SQRT(2)s` |
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| 40. |
Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it. |
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Answer» During oscilating at ny positin x below the equlibrium positon let tehh velocilty of m be v and angular velocity of the puley be OMEGA. If r is the radius of the pulley, then `v=romega` At any instant total energy =constant (For SHM) `:. 1/2 mv^2+1/2Iomega^2+1/2k[(x+delta)^2-delta^2]-mgx` =constant `rarr(1/2mv^2+(1/2)Iomega^2+(1/2)kx^2+kxd-mgx` ltbrgeconstant `rarr (1/2)mv^2+(1/2)1(v^2+r^2)+(1/2)kx^2` =constant `(delta=mg/k)` Taking derivative ofboth sides with respect of t `mv.(dv)/(dt)1/r^2v.(dv)/(dt)+kx(dx)/(dt)=0` `rarr a((m+1)/r^2)=-kx` `(:.=(dx)/(dt)and a =(dv)/(dt))` `rarr a/r=k/((m+1)/r^2)=omega^2` `rarr T=2pisqrt((m+1/r^2)/k)` |
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| 41. |
The stress required to double the length of a wire of Young's modulus Y is |
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Answer» `Y/2` |
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| 42. |
Consider a disc rotating in the horizontal plane with a constant angu lar speed omega about its centre O. the disc has shaded region on one side of the diameter and an unshaded region on the other side as shown in Fig. 5.111. when the disc is in the orientation as shown, two pebbles P and Q are simu ltaneously projected at an angle towards R. the velocity of projection in the y-z plane and is same for both pebbles with respect to the disc. Assume that i they land back on the disc before the disc has completed 1/8 rotation, ii their range is less than half the disc radus, and iii omega reamisn constant throughout. then |
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Answer» P lands inthe shaded REGION and Q in the unshaded region. |
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| 43. |
In an astronomical telescope, the focal lengths of the objective and the eye piece are 100cm and 5cm respectively . If the telescope is focussed on a scale 2m from the objective, the final image is formed at 25cm from the eye. Calculate (i) the magnification and (ii) the distance between the objective and the eyepiece. |
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Answer» Solution :`f_0=100cm ,f_e=5cm` To find the image distance due to objective `mu_0=-2m=-200cm, v_0=?` For a lens `1/f=1/v-1/u, 1/+100=1/v_0-1/(-200)` `1/v_0=1/100-1/200=1/200, v_0=200cm` MAGNIFICATION of objective `m_0=v_0/v=200/(-200)=-1` To find the object distance for the eyepiece `v_e=-25cm, u_e=?` For a lens `1/f=1/v-1/u,1/(+5)=1/(-25)-1/u` `1/u_e=-1/25 -1/5=- 6/25, u_e=- 25/6cm` Magnification of the eyepiece `m_e=v_e/u_e= (-25 TIMES 6)/(-25)=6` (i) Magnification of the telescope `=m_0 times m_e=-1 times 6=-6` (II) Distance between the objective and the eyepiece= `v_0+|u_e|=200+25/6 =204.2cm` |
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| 44. |
Angle between velocity and acceleration vectors in the following cases |
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Answer» a-h, b-g, c-f, d-e |
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| 45. |
Air is streaming past a horizontal air plane wing such that its speed in 120m//s over the upward surface and 90m//s at the lower surface . If the density of air is 1.3kg//m^(3) and the wing is 10 m long and has an average width of 2m , then the difference of the pressure on the two sides of the wing of ...... Pascal . |
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Answer» `4095.0` `P_(1)+(1)/(2)rhov_(1)^(2)=P_(2)+(1)/(2)rhov_(2)^(2) "" [becauseh_(1)=h_(2)]` `P_(2)-P_(1)=(1)/(2)RHO(v_(1)^(2)-v_(2)^(2))` `=(1)/(2)xx1.3[(120)^(2)-(90)^(2)]` `=(1)/(2)xx1.3[14400-8100]` `=(1)/(2)xx1.3xx6300` `thereforeP_(2)-P_(1)=4095Pa` |
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| 46. |
If n small droplets of water each of radiur r, coalesce to form a single drop of radius R. The amount of heat generated is |
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Answer» `4PIT(NR^(2)-R^(2))` |
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| 47. |
The coordinates of the centre of mass of a system of three particles of masses 1 mg , 2 mg and 3mg are (2 ,2 ,2) . The coordinates of the fourth particle of mass 4 mg to be positioned sothat the center of mass of the four particles system is at the origin of the three – dimensional rectangular coordinate system are |
| Answer» SOLUTION :CLOSE to the MASSIVE PARTICLE. | |
| 48. |
In action force acting on a body is gravitational in nature, the reaction force |
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Answer» may be a contact FORCE |
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| 49. |
The ratio between kinetic and potential enerdies of a body executing simple harmonic motion,when its is at a distance of 1/N of its amplitude from the mean potion is, |
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Answer» `N^(2)+1` |
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