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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 certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then |
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Answer» ENERGY `=4VT((1)/(R )-(1)/(R ))` is RELEASED |
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| 2. |
The potential energyof a system under theinfluenceof conservativeforce at positionx isV(x) =[3x^(2)+4x+5].At x = 2 m the conservativeforce actingon the system is ………. |
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Answer» `-20` N ` :. F = (dU)/(DX) = d/(dx) [3x^(2) +4x+5] ` ` :. F = 6x + 4` If x = 2, them `F = 6 xx2+4 = 16 N ` F = 16 N ` :. F = 16 N ` |
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| 3. |
What is power ? Give its dimensional formula ? |
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Answer» Solution :The RATE of work done is called power. Dimensional formula of power is `ML^(2)T^(-3)` |
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| 4. |
A vertical glass tube, closed at the bottom contains mercury to a height of 100cm at 30^(@)C. The height of mercury column at 50^(@)C is. (gamma" for "Hg=1.8xx10^(-4)//""^(@)C,alpha_(g)=8xx10^(-6)//""^(@)C) |
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Answer» 100.329 cm |
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| 5. |
Two kg of water is converted into steam byboiling at atmospheric pressure. The volume changes from 25xx10^(-3) m^(3)to 3.34 m^(3) . The work done by the system is about |
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Answer» `-340kJ ` |
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| 6. |
The moment of inertia of a circular ring about one of its diameters is I . What will be its moment of inertia about a tangent parallel to the diameter? |
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Answer» 4 I ACCORDING to theorem of parallel axes , the moment of inertia about the given axis is `I. = (1)/(2) MR^(2) + MR^(2) = (3)/(2) MR^(2) = 3I ""` (Using (i)) |
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| 7. |
It is given that the mass m of the largest stone that can be moved by the folowing river depends upon the velocity v, density rho and acceleration due gravity g. Using dimensions show that m=(kv^(6)rho)/(g^(3)). |
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Answer» Solution :`m=kv^(X)rho^(y)g^(Z)` Taking dimensions on both sides …I `L^(0)T^(0)M=[LT^(-1)]^(x)[ML^(-3)]^(y)[LT^(-2)]^(z)` `L^(0)T^(0)M=L^(x-3y+z)M^(y)T^(-x-2z)` Equating the dimensions of M,T and L on both sides `y=1` ……ii `-x-2z=0` or `x+2z=0` `x=3y=z=0` `x+z=3y=3`.......iii EQUATION ii minus equqtion (iii) `x+2z-x-z=0-3,z=3,x=-2z=6` Substituting the value of x, y and z in equation (i) `m=kv^(6)rho^(1)g^(-3)=K(v^(6)beta)/(g^(3))` |
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| 8. |
(a) Algebaric sum of moments of masses about centre of mass is zero (b) For small bodies centre of mass coincides with centre of gravity (c ) Position of centre of mass depends on co-ordinate system (d) Position of centre of mass is independent of mass distribution |
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Answer» a & B are correct |
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| 9. |
Pick up the most acurate and most precise measurements out of the following |
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Answer» 50.00m is the most ACCURATE MEASUREMENT |
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| 10. |
The force required to stretch a spring varies with the distance as shown in figure. If the experiment is performed with the above spring of half the length, the line OA will |
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Answer» shift TOWARDS F-axis |
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| 11. |
The length of a string tied to supports is 40 cm.Maximum wave length of the stationary wave produced is |
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Answer» 20 cm |
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| 12. |
Which one of the following displacement-time graphs represents one dimensional motion of a particle? |
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Answer»
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| 13. |
A block of 10kg is pulled at a constant speed on a rough horizontal plane by a force of 20N. Calculate the coefficient of friction ? |
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Answer» |
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| 14. |
When heat energy of 1500 J is suppliedto a gas the external work done on the gas is 525 J what is the increase in its internal energy |
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Answer» Solution :Heat energy SUPPLIED `DELTAQ= +1500J` EXTERNAL work done `Deltaw=-525J` By `1^(st)` law of THERMODYNAMICS `DeltaQ = DeltaU +Deltaw` `:. DeltaU =DeltaQ-Deltaw=1500+525=1025J` |
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| 15. |
The relation between acceleration of gravity at pole and at an equator of earth is ....... |
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Answer» `g_p LT g_e` |
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| 16. |
A conical glass capillary tube of length 0.1 m has diameter 10^(-3) and 5xx10^(-4)m respectively at its ends. When it is just immersed in a liquid at 0^(@)C with larger diameter in contact with liquid the liquid rises to 8xx10^(-2) m in the tube. if another cylindrical glass capillary tube B is immersed in the same liquid at 0^(@)C the liquid rises to 6xx10^(-2) m height. The rise of liquid the tube B is only 5.5xx10^(-2)m when the liquid is at 50^(@)C. density of the liquid is ((1)/(14))xx10^(4)(kg)/(m^(3)) and angle of contact is zero. Effect of temperature on the density of the liquid and glass is negligible. The rate at which the surface tension changes with temperatrure considering the change to be linear is given by -1.4xx10^(-n)(N)/(m^(@)C. what is the value of n? |
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Answer» `r=r_(2)-((r_(2)-r_(1))/(l))XXH,T_(0)=(hrrhog)/(2)` `(DeltaT)/(DELTATHETA)=(T_(50)-T_(0))/(Deltatheta)=(7.7xx10^(-2)-8.4xx10^(-2))/(50)` `=-1.4xx10^(-4)(N)/(m^(@)C)` `=-1.4xx10^(-n)(N)/(m^(@)C)` (given) HENCE `n=4` |
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| 17. |
The figure shows an instantaneous profile of a rope carrying a progressive wave moving from left to right, then (a) the phase at A is greater than the phase at B (b) the phase at B is greater than the phase at A (c ) A is moving upwards (d) B is moving upwards |
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Answer» `a` and `C` |
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| 18. |
A ray of light undergoes deviation of 30^@ when incident on the equilateral prism of refractive index sqrt2. What is the angle subtended by the ray inside the prism with the base of the prism? |
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Answer» Solution :Here `delta=30^@ and A=60^@` So if the prism had been in MINIMUM deviation, `mu= (sin[30^@+60^@]//2)/(sin (60^@//2))=sin 45^@/sin 30^@=1/sqrt2 times 2=sqrt2` And as `mu` of the prism is given to be `sqrt2`  The prism is in the position implies that `r_1=r_2=r=(A/2)=(60^@/2)=30^@` Therefore, angle subtended by the ray INSIDE the prism with the surface `AB(90^@-r)=(90^@-30^@)=60^@` and as base also SUBTENDS an angle of `60^@` with the face AB, the ray inside the prism in parallel to the base i.e., the angle subtended by the ray inside the prism with the base is zero. |
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| 19. |
A diver in a swimming pool bends his head before diving. It |
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Answer» INCREASES his linear velocity Moment of inertia `I=MR^(2)` Since R decreases, moment of inertia ALSO decreases. |
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| 20. |
Set the following vectors in the increasing order of their magnitude. (a) 3hat(i) + 4hat(j)(b) 2hat(i) + 4hat(j) + 6 hat(k) (c) 2hat(i) + 2hat(j) + 2hat(k) |
| Answer» Answer :B | |
| 21. |
A 2.0 kg particle undergoes SHM according to x = 15 sin ((pi t)/(4) + (pi)/(6)) (in SI units) (a) What is the total mechanical energy of the particle ? (b) What is the shortest time required for the particle to move from x = 0.5m to x = - 0.75m ? |
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Answer» `= (1)/(2) xx 2.0 xx ((pi)/(4))^(2) xx (1.5)^(2)` `= 1.39 J` (b) `0.5 = 1.5sin ((pi t_(1))/(4) + (pi)/(6))` From here find `t`. Then,`- 0.75 = 1.5sin((pi t_(2))/(4) + (pi)/(6))` From here find `t_(2)`. Now, `t_(1) ~ t_(2)` is the required TIME. |
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| 22. |
While performing Searle's experiment, a weight of 50N is suspended from a wire. The extension produced is 0.121cm and is measured by a micrometer of least count 0.001cm. The diameter of the wire is 0.700cm, measured by a screw gauge of least count 0.001cm. The length is 100 cm measured with the help of scale of least count 0.1 cm. Calculate the young's modulus of material of wire, given by Y=(FL)/(Al) where F is weight, L is length of wire, A is area of wire and l is extension produced. |
| Answer» SOLUTION :`Y= (1.07xx10^(9) PM 0.013xx10^(9))N"/"m^(2)` | |
| 23. |
Choose the correct statement A: It is impossible to derive continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings. B: Heat engine can convert whole of the heat energy supplied to it into useful work. |
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Answer» only A |
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| 24. |
Which of the following is in increase order? |
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Answer» EXA, TERA, HECTO |
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| 25. |
A particle of mass m executes SHM with amplitude a and frequency v. The average kinetic energy during motion from the position of equilibrium to the end is |
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Answer» `2pi^(2)ma^(2)V^(2)` |
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| 26. |
Which of the following functions represent SHM : sin^(2)omegat |
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Answer» Solution :A MOTION will be SHM if acceleration is directely proportional to its displacement `a=- OMEGA^(2)y` `y= sin^(2) omega t` `rArr (dy)/(dt)=2(sin omega t)(omega COS omega t)` `= omega sin 2 omega1`, `(d^(2)y)/(dt^(2))=-2 omega^(2)cos 2omega t, rArr (d^(2)y)/(dt^(2)) prop y` `:.` It is oscillatory but not SHM. |
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| 27. |
A block of mass 2.20 kg is acceleratedacross a rough surface by a rope passing over a pulley as shown in Fig. 7.91 . The tension in the rope is 10.0 N , and the pulley is 10.0 cmabove the block . The coefficientof friction is 0.4 . (a) Determine the acceleration of blockwhen x = 0.4 m . (b) Find the value of x at which the acceleration becomes zero . |
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Answer» cos `theta = ((4)/(sqrt(17)))`  As T cos`theta - MU mg = m xx a` (B) T cos`theta' = mu N` |
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| 28. |
A cylinderical tank 1 m in radius rests on a platform 5m high. Initially the tank is filled with water upto a height of 5m. A plug whose area is 10^(-4)m^(2) is removed from an orifice on the side of the tank at the bottom. Calculate (a) Initial speed with which the water flows from the orifice (b) initial speed with which the water strikes the ground (c ) the time taken to empty the tank to half its original value. g = 10ms^(-2). |
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Answer» Solution :(a) Speed of efflux, `u = sqrt(2gh)` `= sqrt(2 xx 10 xx 5) = 10 ms^(-1)` (b) the initial vertical velocity of WATER flowing out of orifice is zero. Its vertical velocity when it hits the GROUND is: `upsilon_(0) = sqrt(u^(2)+UPSILON^(2)) = sqrt(10^(2)+10^(2)) = 10sqrt(2)` `= 14.14 ms^(-1)`. (c ) Let A, `A_(0)` be the area of cross-section of the cylindrical tank an plug fitted to orifice respectivley. Then `A = pixx (1)^(2)=pi m^(2)` and `A_(0)= 10^(-4)m^(2)`. When the height of water level above the hole is y, the velocity of flow through the orifice is `upsilon=sqrt(2gy)`. If dV volume of water FLOWS out in time dt and dy. be the decreases. in height of water in tank the rate of flow of water is : `(dV)/(dt)=A_(0)upsilon = A_(0)sqrt(2gy) or (-Ady)/(dt) =A_(0) sqrt(2gy)` or `dt = (-A)/(A_0) (dy)/(sqrt(2gy)) [ :' dV=-A dy]` on intergration it within the conditions of flow of water `int_(0)^(t) dt = int_(h)^(h//2) (-A)/(A_(0)) (dy)/(sqrt(2gy)` `t - (A)/((A_0) sqrt((2)/(G)) (sqrt(h)-sqrt(h//2))` `=(pi xx (1)^(2))/(10^(-4)) sqrt(2/10) (sqrt(5)-sqrt(5//2))` `= 9.2 xx 10^(3) s ~~2.5 h`. |
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| 29. |
A girl throws a ball with initial velocity v at an inclination of 45^(@). The ball strikes a smooth vertical wall at a horizontal distance d from the girl and after rebouncing return to her hand. The coefficient of retitution between the wall and the ball is (k g d)/(v^(2) - g d) where 'k' is |
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Answer» |
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| 30. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for v_(o) |
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Answer» Solution :There are two external forces on the bob : gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the SYSTEM is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A , `E= (1)/(2)mv_(0)^(2)""(6.12)` `T_(A)-mg= (mv_(0)^(2))/(L)` [Newton.s SECOND LAW] where `T_(A)` is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string `(T_C)` becomes zero. Thus, at C `E= (1)/(2)mv_(c )^(2)+ 2mgL""(6.13)` `mg= (mv_(c )^2)/(L)` [Newton.s Second Law] (6.14) where `v_(c )` is the speed at C. From Eqs. (6.13) and (6.14) `E= (5)/(2)mgL` Equatting this to the energy at A `(5)/(2)mgL= (m)/(2)v_(0)^(2)` or, `v_(0)= sqrt(5gL)`. |
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| 31. |
The rms velocity of the molecules in a sample of helium is 5//7 times that of molecules in a sampleof hydrogen . If the tempertaure of hydrogen is 0^(@)C . Then, what is the temperature of helium? |
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Answer» Solution :From RMS SPEED , `SQRT((3RT_(H_(e)))/M_(H_(e)))=5/7(sqrt((3RT_(H_(2)))/M_(H_(2))))` `therefore T_(H_(e))=25/49(M_(H_(e))/M_(H))T_(H_(2))=25/49xx4/2xx273=273K` `T_(H_(e))=0^(@)C` |
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| 32. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The ratio of the kinetic energies (K_(B)"/"K_(C )) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C. |
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Answer» Solution :The ratio of the kinetic energies at B and C is : `(K_B)/(K_C)=((1)/(2)mv_(B)^(2))/((1)/(2)mv_(c )^(2))=(3)/(1)` At point C, the string becomes slack and the velocity of the bob is HORIZONTAL and to the left. It the connecting string is cut at this INSTANT, the bob will execute a projectile motion with horizontal projection akin to a rock KICKED horizontally from the EDGE of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. |
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| 33. |
State whether the fundamental law of calorimetry is applicable in the following cases : (i) sugar is added to water taken in a calorimeter |
| Answer» Solution :I While DISSOLVING in water, sugar will ABSORB necessary heat of solution. If it is not TAKEN into consideration, the fundamental LAW of calorimetery cannot be APPLIED. | |
| 34. |
A cylinder of mass 5kg and radius 30cm, free to rotate about its axis, receives an angular impulse of 3kg ms^(-1) initially followed by a similar impulse after every 4s. What is the angular speed of the cylinder 30s after the initial impulse if the cylinder is at rest initially ? |
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Answer» Solution :In `30S` the cylinder RECEIVES 8 angular impulses including the intial impulse `sumtauxxt=I(omega_(2)-omega_(1))` `8xx3=(MR^(2))/(2)omega_(2)`(since `omega_(1)=0`) `8xx3=(5xx(0.3)^(2))/(2)OMEGA^(2)` On solving `omega_(2)=106.7rads^(-1)` |
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| 35. |
A machine gun fires 240 bullets per minute with certain velocity. If the mass of each bullet is 10^(-2) kg and the power of the gun is 1.8 kW, find the velocity with which each bullet is fired. |
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Answer» |
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| 36. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The speeds at points B and C. |
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Answer» Solution :It is clear from Eq. (6.14) `v_(C )= sqrt(GL)` At B, the energy is `E=(1)/(2)mv_(B)^(2)+mgL` Equatting this to the energy at A and employing the result from (i), namely `v_(0)^(2)= 5gL`, `(1)/(2)mv_(B)^(2)+mgL= (1)/(2)mv_(0)^(2)` `(5)/(2)mgL` `v_(B)= sqrt(3gL)`. |
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| 37. |
Negative sign in W = -mgh denotes what ? |
| Answer» SOLUTION :It DENOTES the WORK DONE against the FORCE . | |
| 38. |
The diameter of a rotating fly wheel is R. Its coefficient of linear expansion is alpha.. If the temperature is increased by DeltaT, the percentage change in its rotational KE would be |
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Answer» `ALPHA,Deltat.100` |
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| 39. |
A particle of mass M is placed at the centre of a uniform spherical shell of equal mas and radius a. The gravitational potential at a point P at a distance a/2 from the centre. |
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Answer» `-(GM)/(3A)` |
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| 40. |
The amplitude of a vibrating body situated in a resisting medium…………… |
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Answer» DECREASES LINEARLY with time |
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| 41. |
The binding energy of particle (mass = 50 kg) and earth system is [G = 6.6 xx 10^(-11) N-m^(2)//kg^(2),M_(E ) = 6 xx 10^(24)kg, R_(E ) = 6400 km] |
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Answer» `3.86 XX 10^(12)` J |
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| 42. |
A spring is elongated by 2 cm due to a 80 g mass attached to it. Another body of mass 600g is attached to the end of the spring and it is displaced by 8 cm from its equilibrium position. Calculate the energy of the system in this position. Considering the principle of conservation of energy, determine the velocity of the body when it is at a distance of 4 cm. |
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Answer» SOLUTION :FORCE constant of the spring, `k=(80xx980)/2=40xx980"dyn"*cm^(-1)` Mass, m = 600 g , amplitude, A = 8 cm. Total energy, E = MAXIMUM potential energy = potential energy at the ends of the path of motion = `1/2kA^(2)=1/2xx40xx980xx(8)^(2)=1254400"erg"=0.12544J`. Even for x = 4 cm, the total energy REMAINS unchanged. If v is the velocity at this POSITION, then `1/2mv^(2)+1/2kx^(2)=1/2xx40xx980xx64` or, `1/2mv^(2)=1/2xx40xx980xx64-1/2xx40xx980xx4^(2)` = `1/2xx40xx980xx(64-16)` = `1/2xx40xx980xx48` or, `v^(2)=(40xx980xx48)/m=(40xx980xx48)/600=4xx49xx16` or, `v=sqrt(4xx49xx16)=2xx7xx4=56cm*s^(-1)`. |
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| 43. |
Two physical quantities are having the same dimensions. Does it mean that they are the same? Give an example supporting your answer. |
| Answer» SOLUTION :No . TORQUE and WORK. | |
| 44. |
A sample of gas expands from volume V1 to V2 the amount of work done by the gas is greatest . When the expansion is |
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Answer» ISOTHERMAL PROCESS |
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| 45. |
A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When the particle move from one point to a diametrically opposite point, its |
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Answer» MOMENTUM does not CHANGE |
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| 46. |
When an ideal gas is compressed adiabatically, its temperature rises, the moolecules on the average have more kinetic energy than before. The kinetic energy increases |
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Answer» because of COLLISIONS with MOVING parts of the WALL only |
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| 47. |
Two bars of masses m, and m, connected by an light spring (intially at its natural length) rests on a horizontal plane. The coefficient of friction between the bars and the surface is mu. The mnimum costant force that bar to be applied in the horizontal direction (along the length of spring) to the bar of mass m_(1) in order to shift the other bar of mass m_(2) Acceleration due to gravity is g) will be |
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Answer» `mu(m _(1) + m_(2))G ` |
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| 48. |
Rain, pouring down at an angle alpha with the vertical has a speed of 10 m/s. A girl runs against the rain with a speed of 8 m/s and sees that the rain makes an angle beta with the vartical, then relation between alpha and beta is |
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Answer» `TAN BETA =(8+10sin ALPHA)/(10COS alpha)` |
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| 49. |
An open tank 10m long and 2m deep is filled up to 1.5m height of oil of specific gravity 0.82. The tank is uniformly accelerated along its length from rest to a speed of 20 m/s horizontally. The shortest time in which the speed may be attained without spilling any oil is g=2 is |
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Answer» 20s |
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| 50. |
(A) : A work done in moving a body over a closed loop is zero for every force in nature.(R ) : Work done independ on nature of force. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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