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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12201. |
A sphere is rotatiing between a a rough wall and a smooth wedge as shown in figure If mu is the cofficent of frication between wall and sphere then normal reation between wall and sphere will |
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Answer» INCREASE if `MU` INCREASES |
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| 12202. |
If A = 2i - 3 j + 4k its components in yz plane and zx plane are respectively |
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Answer» `SQRT13 and 5` |
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| 12203. |
The volume of air at 30^(@)C" is "75 times 10^(3)mm^(3). When the temperature is raised to 98^(@)C at constant pressure, the air occupies a volume of 92 times 10^(3)mm^(3). Calculate the volume coefficient of air. |
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| 12204. |
The angle of contact at the interface of water glass is 0^(@) ethylalcohol-glass is 0^(@) mercury glass is 140^(@) and methyliodide-glass is 30^(@) A glass capillary is put in a through containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the through is |
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Answer» Water |
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| 12205. |
How much steam at 100^(0)C is to be passed into water of mass 100 g at 20^(0)C to raise its temperature by 5^(0)C ? (latent heat of steam is 540 cal/g and specific heat of water is I cal//g^(0)C) |
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| 12206. |
An iron shot of mass 50g moving with a velocity of 150m/s, hits a fixed lead target of mass 2.5kg, and comes to rest by getting embedded in it. Assuming that 25% of the heat generated is lost to the surroundings, and that the remaining heat goes to increase the temperature of the iron shot as well as the lead target equally, find the rise in the temperature of the system. Specific heat capacities of iron and lead are 470 J//kg-^(@)C and 130 J//kg-^(@)C respectively. |
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| 12207. |
A mass m moves with a velocity v and collides inelastically with another identical mass at rest. After collision, the first mass moves with velocity v/(sqrt3) in a direction perpendicular to Before collision the initial direction of motion. The speed of the 2^(nd) mass after collision |
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Answer» `(2v)/(sqrt3)`  In x-direction `mv + 0 = 0 + mv_1 cos theta "" ….(i)` where `v_1` is the velocity of `2^(ND)` MASS. In y-direction `0 = (mv)/(sqrt3) - mv_1 SIN theta` or `mv_1 sin theta = (mv)/(sqrt3) "" …….(ii)` Squaring and adding equs. (i) and (ii), we GET `v_1^2 = v^2 + (v^2)/(3) = (4v^2)/(3) :. v_1 = (2v)/(sqrt3)` . |
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| 12208. |
Two cars A and B start racing at the same on a flat race track which consists of two strainght sections each of length 100 pi and one circular section as shown in figure The rule of the race is that each car must travel at constant speed at all time without even skidding |
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Answer» Car A completes its journey before car B `v_(A, max)= sqrt(MU gr) = sqrt(0.1 xx 10 xx 100) = 10 ms^(-1)` `v_(B, max) = sqrt(mu_(B)gR_(B)) = sqrt(0.2 xx 10 xx 200) = 20 ms^(-1)` Time taken by car A `t_(A) = (200 PI + pi. 100)/(100) = 3 pi s` `t_(B) = (200 + pi xx 200)/(20) = 2 pi s` |
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| 12209. |
The coherent system of units ....... |
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Answer» CGS |
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| 12210. |
Explain with illustration the pure translation and combination of translation and rotation motion of rigid body. |
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Answer» Solution :Above two figures shows the different motion of the same body. Suppose P is any point and its centre of mass is at O. The trajectories of O are the translational trajectories `Tr_(1) and Tr_(2)` of the body. The positions O and P at three different instants of time are shown by `O_(1),O_(2),O_(3) and P_(1),P_(2),P_(3)` respectively in both the figure. In figure (a) it is seen that position does not changes at different position. Line like OP have no orientation, the angle makes by OP in horizontal remains same. `THEREFORE alpha_(1)=alpha_(2)=alpha_(3)` Such motion is a pure translation. In pure translation motion particles like O and P have velocity at different position. In figure (b) in case of COMBINATION of translation and rotation, the velocity of O and P differ. `therefore alpha_(1)nealpha_(2)nealpha_(3)`. Such type of motion is the combination of pure translation and combination of translation motion. Another ILLUSTRATION of such type of motion : ROLLING motion of a cylinder is another illustration of this motion. Here the motion of rolling of cylinder on a SLOPE about fixed axis the combination of rotation and translational motion. If a motion of rigid body is not about any axis or not stationary it is pure translation or combination of translation and rotation motion. If the motion of body is pivoted or stationary by any way then this motion is rotational motion. Rotational motion may be about stationary and variable axis. |
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| 12211. |
A unit vector is used to specify |
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Answer» only MAGNITUDE |
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| 12212. |
A circular plate of diameter 10 cm is kept in contact with a square plate of side 10 cm. The density of the material and the thickness are same everywhere. The center of mass of the system will be |
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Answer» INSIDE the circular plate |
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| 12213. |
A ball of mass 'm' is rotated in a vertical by a string. The difference in tension at the top and bottom would be |
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Answer» 6mg |
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| 12214. |
A cylinder is completely filled with water. It 1//4 of the volume of water leaks, out its centre of mass |
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Answer» MOVES up |
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| 12215. |
State the limitation of Newton.s third law briefly. |
| Answer» Solution : Newton.s THIRD LAW is not applicablei) when velocities of moving bodies is nearly equal to LIGHT velocity .c.. ii) NEAR very strong gravitational fields (Ex: Gravitational field between objects whose masses are GREATER than mass of sun). | |
| 12216. |
How much force of gravitation acted on a body at infinite distance from the earth ? |
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| 12217. |
Definemolarspecificheatcapacities. |
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Answer» Solution : (a)Linearmolecule : Energyofonemole ` = ( 7)/(2 )kTxx N_(A)= (7)/(2) RT ` ` C_V= ( DU )/(dT )= ( d )/(dT )[ 7/2 RT ]` `C_v =7/2 R ` `C_p= C_v+ R= 7/2R+R =(9R )/(2)` ` thereforegamma =( C_p )/( C_(V )) = ((9)/(2)R )/((7)/(2)R) = 9/7 = 1.28 ` (b )Non linear molecule : Energyof amole`=6/2kTxx N_(A)= 6/2RT= 3RT ` `C_V= (dU )/(dT ) = 3R ` `C_p = C_v+R = 3R+R = 4R ` ` thereforegamma= (c_p )/(C_(v)) = ( 4R) /( 3R ) = (4)/(3 ) = 1.33` NOTE that according to kin e tic theory model of gases the SPECIFIC heat capacity at constant volume and constant pressure are independent of TEMPERATURE. But in reality it is not sure. The specific heat capacity varies with the temperature. |
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| 12218. |
A vessel, whose bottom has round holes with diameter of 1mm, is filled with water. Assuming thatsurface tension acts only at holes, find the maximum height (in cm) to which the water can be filled in the vessel without leakage, given that surface tension of water is 75 xx 10^(-3)N/m and g = 10m/s^2 |
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| 12219. |
A person in lift is holing a water jar, which has a small hole at the lower end of its side . When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance (d) of1.2 m from th person. In the following state of the lifts motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in ListII and select the correct answer suwer using th code giv en below the lists. List I, List II.Lift si accelerating vertically1. d= 1.2 mup Q. Lift is accelertiong verticallydown with an accelertion less than the gravitational acceelration R. Lift is moving vertically3. dlt 1.2 mup with constant speed S. Lite is falling freely4. No water leaks out of hte jar . |
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Answer» ` P-2, Q-3 , R-2 S-4` Refer to Fig. 2 (CF) . 58`. Wvlocity of effux, ` v-x = sqrt(2 (g+a) H)` time taken by water in reaching the floor, ` t= sqrt ( 2 h)/( (g+a))` :. Horizontal range, ltbRgt ` d= v_x t= sqrt (2 (g +a) H) XX sqrt (2 h)/(g+a) = 2 sqrt (H//h)`  . , If lift is acceleration vetically downwards, with accleration `a` then horizontal range, ` d= sqrt(2 (g-a)H) xx sqrt (2 h)/(g-a) = 2 sqrt Hh` When lift is moving vertically up with constant speed then ` d= sqrt (2 gH) xx sqrt (2h)/g` = 2 sqrt H h` when lift is falling FREELY, then the warter jar is in weightlessness state. Hence, no water leaks out of jet.theus, option (c) is TRUE. |
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| 12220. |
In incompressible flow the density of a fluid changes with time and position . |
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| 12221. |
A rod of mass m and length l is hinged at one of its end A as shown in figure. A force F is applied at a distance x from A. The acceleration of centre of mass a varies with x as |
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| 12222. |
A body of mass .m. is dropped and another body of mass M is projected vertically up with speed .u. simultaneously from the top of a tower of height H. If the body reaches the heightst point before the dropped body reaches the ground, then maximum height raised by the centre of mass of the system from ground is |
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Answer» `H+(U^(2))/(2g)` |
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| 12223. |
A person can jump safely from a height of 2m on the earth. On a planet where acceleration due to gravity is 2.45 ms^2, the maximum height from which he can jump safely |
| Answer» ANSWER :A | |
| 12224. |
A ball A moving with certain velocity collides with another body B of the same mass at rest. If the coefficient of restitution is 2/3. The ratio of velocity of A and B after collision is |
| Answer» Answer :B | |
| 12225. |
The equation of a wave is y = 3 cospi(100t-x) cm.its wavelength is |
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Answer» 3 cm |
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| 12226. |
The ratio of the lengths of two rods is 4:3. The ratio of their coefficients of cubical expansion is 2:3. Then the ratio of their linear expansions when they are heated tnrough same temperature difference is |
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Answer» `2:1` |
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| 12227. |
The refractive indices of crown and flint glasses for violet and red light are 1.523, 1.513, 1.773 and 1.743 respectively. Find the dispersive powers of the glasses. |
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| 12228. |
(A) : At constant temperature, the elongation in a liquid surface does not obey Hooke's law. (R ) : During increase in surface area, the liquid molecules initially present just below the surface will come on to the surface and this is not real elongation. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 12229. |
A simple pendulum is suspended from the roof of a school bus which movies in a horizontal direction with an acceleration a, then the time period is : |
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Answer» `T=prop(1)/(g^(2)+a^(2))` `Tprop(1)/(sqrt(g))` From phythagoras THEOREM, `g.=g^(2)+a^(2)` `g.^(2)=sqrt(g^(2)+a^(2))` `Tprop(1)/(sqrt((sqrt(g^(2)+a^(2))))),Tprop(1)/(g^(2)+a^(2))` 
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| 12230. |
An organ pipe closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is |
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Answer» Solution :In case of closed pipe, `f = (2n + 1) f _(1)` (Where `f _(1)=` fundamental frequency, `2n +1= ` order of harmonic, `n = 0,1,2,3,…=` order of overtone) Note : If we take `f =(2n -1) f _(1),`then `2n -1=` order of harmonic, `n = 1,2,3…., n-1 =` order of overtone) Here `f = 20000 HZ and f _(1) = 1500 Hz and so 20000 = (2n +1) xx 1500` `therefore 2n +1 = 13.33` `therefore 2n +1 =13 (because ` here (2n +1) is less than 15 and an odd integer) `therefore n =6` `6^(th)` overtone is the MAXIMUM order os overtone (or highest overtone) that can be heard in this case. (Because if we take n = 7 then (2n +1) `f _(1) = 15 xx 1500 =22500` which would be GREATER than 20000 Hz and so it would be wrong). |
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| 12231. |
A solid metal sphere and a solid wooden sphere are having the same mass. If they are spinning with same angular velocity , then |
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Answer» METAL SPHERE POSSESSES more angular momentum |
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| 12232. |
(A) : In one dimensional motion constant acceleration, the power delivered is proportional to time.(R ) : In general, power is equal to product of force applied and its velocity. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 12233. |
A motor pump is delivering water at certain rate. In order to increases the rate of delivery by 100%, the power of the motor is to be increased by |
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Answer» 0.3 |
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| 12234. |
A 70 kg man leaps vertically into the air from a crouching position. To take the leap the man pushes the ground with a constant force F to raise himself. The center of gravity rises by 0.5 m before he leaps. After the leap the e.g. rises by another lm. the maximum power delivered by the muscles is (Take g=10 ms^(-2)) |
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Answer» `6.26 xx 10^(3)` WATTS at the START |
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| 12235. |
Select the incorrect statement from the following statements. |
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Answer» Pressure of a gas is directly proportional to MEAN square speed `lamda= (KT)/(sqrt2pid^2P)` `:. lamda prop T ` |
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| 12236. |
A ball is thrown vertically upwards and comes back.Which of the following graph represents the velocity-time graph of the ball during its flight? |
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| 12237. |
The power of the motor placed near a water pond is 3 kW. Find how much of water can be lifted to a height of 20 m in one minute (g = 10 ms^(-2)) |
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| 12238. |
The power of the motor placed near a water pond is 3 kW. Find how much of water can be lifted to a height of 10 m in one minute (g = 10 ms^(-2)) |
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| 12239. |
What is the value of Reynolds number for streamline flow ? |
| Answer» SOLUTION :The value of REYNOLDS number for streamline is between 0 to 1000. | |
| 12240. |
100 g of water is supercooled to -10^(@)C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice What will be the temperature of the resultant mixture and how much mass would freeze ? [s_(w)=1" cal g"^(-1)" "^(@)C^(-1)" and "L_(" fussion")^(W)=80" cal g"^(-1)] |
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Answer» SOLUTION :Mass of water `m=100g` Change in temperature `DELTAT=0-(-10)=10^(@)C` Specific heat of water `s_(w)=1" calg"^(-1)" "^(@)C^(-1)` Latent heat of melting of water `L_(F)=80" calg"^(-1)` Heat required to convert ice at `-10^(@)C` to `0^(@)C` water. `Q="ms"_(w)DeltaT` `=100xx1xx10` = 1000 cal Suppose, .m. GRAM ice melt, `:.Q=" mL"` `m=Q/L` `(1000)/(80)` `=12.5g` As only `12.5g` ice is melted from 100g, the temperature of mixture will be `0^(@)C`. |
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| 12241. |
A particle, starting from rest, moves with constant acceleration 4 ms^(-2). The distance travelled by the particle in 5^(th) second is. |
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Answer» 20 m `d _(n) = v _(0)+a/2 (2n -1) , ` here `v _(0) =0, a = 4 ms ^(-2) , n =5` `THEREFORE d _(5) =0+ 4/2 (2 xx 5-1)` `therefore d _(5) = 2 (9) =18m` |
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| 12242. |
A block of mass 2 kg is on a horizontal surface. The co-efficient of static & kinetic frictions are 0.6 & 0.2. The minmimum horizontal force required to start the motion is applied by the body at the end of the 2nd second is (g=10ms^(-2)) |
| Answer» Answer :A | |
| 12243. |
Which of the following expressions have no meaning ? |
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Answer» `(HATI .hatj)XX hatj` |
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| 12244. |
A particle of mass 'm' is attached to the rim of a uniform disc of mass 'm' and radius R. The disc is rolling wihtout slipping on a stationery horizontal surface as shown in the figure. At a particular instant, the particle is at the topmost position and centre of the disc has speed v_(0) amd its angular speed is omega. Choose the correct option (s). |
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Answer» `v_(0)=OMEGAR` |
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| 12245. |
A string of length 2 m is fixed at both ends. If this string vibrates in its fourthnormal mode with a frequency of 500 Hz. Then the waves would travel on it with a velocity of |
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Answer» `125 m//s` |
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| 12246. |
The potential energy of a conservative system is given by V(x) = (x^2 – 3x) joule. Then its equilibrium position is at |
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Answer» `X= 1.5 m` For a conservative field, FORCE, `F = - (dV)/(dx)` `:. F = - (d)/(dx) (x^2- 3x) = - (2x - 3) = -2x + 3` At equilibrium position, `F = 0` `:. -2x + 3 = 0 or x = 3/2 m = 1.5 m`. |
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| 12247. |
What is the shape of graph of stress tostrain upto elastic limit? |
| Answer» SOLUTION :STRAIGHT LINE (LINEAR) | |
| 12248. |
Which types of energies are present in streamline flow of fluid ? |
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Answer» SOLUTION :(1) KINETIC ENERGY (2) Potential energy (3)PRESSURE energy |
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| 12249. |
A rope thrown over a pulley has a on one of its ends and a counterbalancingmass M on its other end. The man whose mass is m, climbs upwards by vec/_\r relative to the ladder and then stops. Ignoring masses of the pulley and the rope, as well as the friction the pulley axis, find the displacement of the centre of mass of this system. |
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Answer» Solution :Let displacement of counterweight be `/_\y` therefore the dispacement of ladder will be `(-/_\y)`. Displacement of man `/_\_(y_m)=/_\_(y_(m,l))+/_\_(y_l)` `Y_(mj)=` displancement of man w.r.t of ladder `=/_\vecr` `y_m=(/_\r-/_\y)` HENCE displacement of centre of mass of the system `/_\vecr_c=(M_("mass")/_\vecr_("mass")M_("ladder")+/_\vecr_("ladder")+M_("mean")/_\vecr_("man"))/(M_("mass")+M_("ladder")+M_("man"))` `/_\r_(c)=(M/_\y+(M-m)-(/_\y)+m(/_\r-/_\y))/(M+(M-m)+m)` `implies /_\r_(c)=(m/_\r)/(2m)("upward")` Method 2: All the bodies of the system are initially at rest. The rope tension is the same both as on the left and the right hand side, at every INSTANT and consequently the momenta of the counterbalancing mass `(vecp_(1))` and the ladder with the Man `(vecp_(2))` are equal at any moment of time. i.e.,  `vecp_(1)=vecp_(2)` or `Mvecv_(1)=mvecv+(M-M)vecv_(1)`............i Here `vecv_(1)` and `vecv` are the velocities of the mass and the man respectively. Velocity of man`vecv_("man") vecv_("man, ladder")+vecv_("ladder")=v'-v_(1)`...........ii Where `vecv'` is the man's velocity relative of the ladder. From eqn i and ii we OBTAIN `vecv_(1)=(m/(2M))vecv'` on the other hand the momentum of the centre of mass is `vecp=vecp_(1)+vecp_(2)=2vecp_(1)` or `2Mvecv_(C)=2M.vecv_(1)` or `vecv_(C)=vecv_(1)=(m/(2m))vecv'` are finally the desired displacement is `/_\vecr_(C)=intvecv_(C).dt=m/(2m)intvecv'dt=m/(2M)/_\vecr` or displacement of centre of mass of system is `/_\vecr_(C)=m/(2M)/_\vecr` As system I and system II both are having same mass and experiene equal force. Both starts from rest. hence displacement of centre of masses FO systems should be equal. As the system I has only one object, the diaplacement of the object will be same as the displascement of its centre of mass which is `/_\y` upward. hence the displacement of centre of mass of system II will so be `/_\y` upward `(/_\vecy_(cm))_("systemI")=(/_\vecy_(cm))_("systemII")=/_\y("upward") ` `implies /_\y=(m(/_\r-/_\y)+(M-m)(-/_\y))/(m(M-m))` `implies /_\y=(m/_\r)/(2M)("upward")` 
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| 12250. |
Form the centre of a large concave mirror of radius of curvature 3 sm , a small sperical steel ball is placed at a little distance on the mirror itself . The ball is then released to execute oscillatory motion on the mirror . The time period of motion of the ball is (Neglect friction and take g=10m/s^(2) |
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Answer» 2.38s  The ball is PLACED at A, a small distance O from the centre of the mirror of radius of curvature R ( = OC) `angleACO = theta`, MASS of the ball = m Resolving weight mg into TWO components, we get `mgsin theta` providing the restoring force for S.H.M. `therefore F = -mg sin theta ` ` = -mg theta` (`theta` is small so x/R is very small) where `x = OA = Rtheta` `rArr F = -mg(x)/(R)` Also, `F = -kx`, on COMPARING `therefore k = (mg)/(R )` SO, the time period is `T = 2pi sqrt((m)/(k)) = 2pisqrt((R)/(g))` `=2pi XX sqrt((3)/(10)) = 2pisqrt(0.3)` = 3.44 s |
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