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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.
| 14201. |
The mass of a uniform ladder 5m is 20kg. A person ofmass 60kg stands on the ladder at a height of 2m from the bottom. The position of centre of mass of the ladder and man from the bottom nearly is |
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Answer» `1M` |
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| 14202. |
A small sphere is placed inside a concave glass vessel kept on a horizontal table. The radius of curvature of the concave vessel is 1.09 m. If the small sphere is displaced slightly from its position of equilibrium and then released, what will be the time period of the simple harmonic motion that the sphere would execute? g=9.81m*s^(-2) |
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Answer» |
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| 14203. |
A string is rigidly tied at two ends and its equation of vibrationis given by y=cos 2pix. Then minimum length of string is |
| Answer» ANSWER :B | |
| 14204. |
When a rigid body moves all particles that constitute the body follows |
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Answer» same PATH |
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| 14205. |
A body is floating in a liquid. Weights of the body is |
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Answer» More than the weights of the BODY |
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| 14206. |
A body is thrown up with a velocity 40 "ms"^(-1), At same time another body is droped from a height 40 m. Their relative accelaration after 1.3 seconds is |
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Answer» 4g |
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| 14207. |
A small plane mirror is rotating at a constant frequency of n rotation per second. With what linear velocity will a light spot move along a spherical screen of radius of curvature of R metres if the mirror at the centre of curvature of the screen? |
| Answer» ANSWER :A | |
| 14208. |
A flywheel of M.I. 6xx10^(-2)kgm^(2) is rotating with an angular velocity of 20rads^(-1) . The torque required to bring it to rest in 4s is |
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Answer» `1.6`NM |
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| 14209. |
What force is required to stretch a steel wire to double its length when its area of cross section 1 cm2 and Young.s modulus 2xx10^(11)N//m^(2). |
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| 14210. |
Match the following columns, {:("Column1","Coloumn2"),("a Specific heat","p [MLT^(-3)K^(-1)]"),("b Coefficient of thermal conductivity","q [MT^(-3)K^(-4)]"),("c Boltzmann constant","r [L^(2)T^(-2)K^(-1)]"),("d Stefan's constat","s [ML^(2)T^(-2)K^(-1)]"):} |
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| 14211. |
A mass m moving horizontally (along the X-axis) with velocity v collides and sticks to a mass of 3 m moving vertically upward (alongthe Y - axis) with velocity2v. The final velocity of the combination is ………. |
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Answer» `1/4 v HAT(i) +3/2 v hat(j)`  Both STICKS in collision .hence this collision is INELASTIC collision and only momentum conserved . ` :. Mv hat(i) +3m (2V) hat(j) = 4m vec(v)` The velocity of the both the stick body is `vec(v)` . ` :. vec(v) =1/4 v hat(i) +3/2 v hat(j)` |
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| 14212. |
When a body is suspended from a spring its time period 'T'. If the spring is cut into two equal parts and the body is suspended from one part, its time period will be |
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Answer» `T//SQRT(2)` |
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| 14213. |
A thin equi-convex lens having radius of curvature 10cm is placed as shown in figure . Calculate the focal length of the lens, if parallel rays are incident as shown |
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Answer» Solution :Refraction at 1st SURFACE `u=infty, mu_1=5/4 mu_2=5/3` `mu_2/v-mu_1/u=(mu_2-mu_1)/R_1 implies 5/(3v)=(5/3-5/4)/10 implies v=40cm` Refraction at SECOND surface `mu_1=5/3,mu_2=3/2 mu=+40 v=? R=-10` `3/(2v)-5/(3 times 40)=(3/2-5/3)/(-10)` After solving we get `V=180/7 cm` So FOCAL LENGTH `=180/7 cm` |
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| 14214. |
vecA*vecA=? |
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Answer» 0 |
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| 14215. |
An open glass tube is immersed in mercury in such a way that the length 8 cm extends above the mercury level. Now the open end of the tube is closed by a finger and raised by further by 44 cm. What will be the length of air column above mercury in the tube. Take atmospheric pressure to be 76 cm of mercury. Neglect capillary effect. |
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Answer» Solution :Figure- (a) SHOWS the tube in initial state then it is closed at top and raised up so that its length above mercury becomes 8 + 44 = 52 cm as shown figure- (b). As initially tube is open to atmosphere, pressure of AIR inside is 76cm of Hg and its volume is 8A if A is the cross-sectional area of the tube. When its upper end is closed and raised up, situation in shown figure- (b). Let the air column be of length x then its final volume will be xA. pressure at the mercury level in the container is equal to atmospheric pressure 76 cm of Hg and the air column is separated by another mercury column of height 52-x above this level which will oppose the atmospheric pressure inside the column will be 76-(52-x) = (24+x)cm of mercury. As during the process temperature of system remains constant, thus we can USE Boyle.s law for the initial and final states as `P_(1)V_(1)=P_(2)V_(2),76 TIMES 8A=(24+x) times xA" or "608=24x+x^(2)" or "x^(2)+24x-608=0` On solving we get x = 15.4 cm or x = -39.4 cm Thus the ACCEPTABLE value of final air column is 15.4 cm 
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| 14216. |
A block of mass .m. is connected to one end of a spring of .spring constant. k. The other end of the spring is fixed to a rigid support. The mass is released slowly so that the total energy of the system is then constituted by only the potentiasl energy, thn .d. is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g - acceleration due to gravity) |
| Answer» Answer :C | |
| 14217. |
Value ofN divison on a vernier scale is equal to (N-1) scale on main scale of vernier. If value of 1 scale on main scale is 1 mm then leas count will be.... |
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Answer» SOLUTION :Least count= {value of division on main SCALE}-{value of 1 division on vernier scale} `:.L*C=1MSD-1VSD` `=1MSD-((N-1)/(N))MSD` `[{:( :. N VSD=(N-1)MSD),( :. 1VSD=((N-1)/(N))MSD):}]` `L*C=((N-N+1)/(N))MSD` `=(1)/(N) MSD` `=(1)/(N)x1mm` `=(1)/(N)xx(1)/(10) cm` `=(1)/(10N) cm` |
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| 14218. |
Which physical quantities are expressed by the following? (i) moment of linear momentum. (ii) rate of change of angular momentum. |
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Answer» SOLUTION :(i) ANGULAR MOMENTUM. (II) TORQUE. |
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| 14219. |
A girl is standing at the centre of a rotating horizontal platform with her hands drawn inwards. What will happen if she stretches her hands horizontally? |
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| 14220. |
The rotational kinetic energy of a body is given byE=1/2I omega^(2). Use this equation to get the dimensional formula of I? |
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Answer» Solution :`E=1/2Iomega^(2),I=(2E)/(omega^(2))` Dimensional formula of I= Dimensional formula of `[E/(omega^(2))]` `[I]=([ML^(2)T^(-2)])/([T^(-1)]^(2)=[ML^(2)]` |
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| 14221. |
Two moles of helium gas (gamma=(5)/(3)) at 27^(@)Cisexpanded at constant pressure until its volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. The work done during adiabatic process is ________ (universal gas constant = 8.3 "J mol"^(-1)K^(-1)) |
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Answer» 7470 J |
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| 14222. |
A small piece of wire 4 cm long is floating on the surface of water. If a force of 560 dnes in excess of its weight is required to pull it up from the surface, find surface tension of water. |
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Answer» SOLUTION :Length of WIRE l=4cm CONTACT length of solid with liquid surface `L=2l =8cm` Surface TENSION `S=F/LimpliesS=560/8=70` DYNE /cm Surface tension of water `=70"dyne"cm^(-1)=0.07Nm^(-1)` |
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| 14223. |
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed upsilon the net force on the particle (directed towards the centre) is |
| Answer» Answer :A | |
| 14224. |
If momentum of object becomes twice, then its kinetic energy would bexome …... times . |
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Answer» `Kpropp^2` `:. K_2/K_1=(p_2/p_1)^2=((2p_1)/p_1)^2=4` |
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| 14225. |
The gravitational force between two bodies is 1 N. If the distance of these two bodies is doubled, what will be the force of attraction between them ? |
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Answer» SOLUTION :`IMPLIES F_(1) prop 1/(r_1^2) and F_(2) prop 1/r_2^2` `:. F_2/F_1 = (r_1/r_2)^2 =(R/(2R))^2=1/4` `:. F_(2) = F_(1)/4 =1/4 = 0.25 N` |
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| 14226. |
A particle execute S.H.M from the mean position. Its amplitude is A, its time period is 'T'. At what displacement, its speed is half of its maximum speed. |
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Answer» `(SQRT(3)A)/(2)` |
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| 14227. |
A satellite of mass 400 kg is in a circular orbit of radius 2R about the earth where R is radius of the earth. How much energy is required to transfer it to a circular orbit of radius 4R ? Find the change in the kinetic and potential energies ? (R= 6.37 xx10^(6)m) |
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Answer» Solution :INITIAL total energy is - `(GMm)/(2r_(2)) =-(GMm)/(8R) =E_2` Final total energy is `-(GMm)/(2r_(2))=-(GMm)/(8R)=E_2` The change in the total energy is `DeltaE=E_(2)-E_(1)=(GMm)/(8R) implies DeltaE=((GM)/R^2) (mR)/8` `DeltaE=(gmR)/(8)=(9.8xx400xx6.37xx10^(6))/8 = 3.13 xx10^(9)J` Change in KINETIC energy `=K_(2)-K_(1) =-3.13 xx10^(9)J` Change in potential energy `=U_(2)-U_(1) =-6.25 xx10^(9)J` |
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| 14228. |
A bullet is fired from a rifle. The rifle is free to recoil. Compare the kinetic energy of the rifle with that of the bullet . |
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Answer» SOLUTION :`K.E , K =1/2 mv^2 =1/2 ((mv)^2)/(m) =(p^2)/(2m)` [ p=momentum ] Both the rifle and the bullet were at REST before firing . After firing , the momenta of the rifle and the bullet MUST be equal in magnitude and OPPOSITE in direcition in order to conserve linear momentum. As p is thesame for both `Kprop 1/m`. As the rifle is HEAVIER, its K.E is less than that of the bullet. |
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| 14229. |
What is thedisplacementof thepoint ofa wheelinitiallyin contactwiththe groundwhenthe wheelrollsforward halfa revolution? Takethe radiusofthewheelas Rand thethe x - axisas theforwarddirection ? |
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Answer» `R sqrt(PI^(2) + 9), Tan^(-1) ((3)/(pi))` with x-axis |
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| 14230. |
A man with a wrist watch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ? |
| Answer» SOLUTION :YES. The working of the wrist WATCH is based on the energy storted in the spring which is independent of acceleration due to GRAVITY. | |
| 14231. |
The springs in figure. A and B are identical but length in A is three times each of that in B. (A) , (B) |
| Answer» SOLUTION :FIGURE | |
| 14232. |
In the proscess Tprop W, pressure of the gas increase from p_(o) to t_(o) Match the columns. {:(,"ColumnI",, "ColumnII"),((A),"Temperature of teh gas",(p),"Pssitive"),((B),"Volume of the gas",(q),"Nagative"),((C),"work done by the gas",(r),"Two times"),((D),"Heat supplies to the gas",(s),"Cannot say anything"):} |
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Answer» p has become four times. Therefore. T has BECOMES two times or V will have become `1/2` times. Since V is decreasing, WORK done by the gas is negative. Further. T is increasing, hence `DeltaU` will be CONNOT SAY abour `DELTAQ.` negative and `DeltaU` is positive Hence, ew cannot say about `DeltaQ` |
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| 14233. |
An object A is dropped from rest the top of a 30m high building and at the same moment another object B is projected vertically upwards with an initial speed of 10 m/s from the base of the building. Mass of the object A is 2 kg while mass of the object B is 4 kg. Find the maximum height above the ground level attained by the centre of mass of the A and B system (take g= 10 m//s^2) |
Answer» Solution :`m_(1) = 4 kg, m_(2) = 2kg`  Initially 4 kg is on the GROUND `therefore x_(1)=0` 2 kg is on top of the building `therefore x_(2) = 30 m` `x_(cm) =(m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)) =(0 + 2 xx 30)/(4+2) = 10 m` `therefore` Initial HEIGHT of CM = 10m, Initial velocity of CM, `u_(cm) =(m_(1)u_(1) + m_(2)u_(2))/(m_(1) + m_(2))` `u_(cm) =(4 xx 15 + 0)/(4+2) = 10` m/s upward. Acceleration of CM, `a_(cm) = g = 10 m//s^(2)` downwards. `therefore` Maximum height attained by CM from initial position `h_(CM) =(u_(cm)^(2))/(2g) = 10^(2)/20 = 5m` `therefore` Maximum height attained by CM of 4 kg and 2 kg from the ground `=10 + 5= 15 m` |
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| 14234. |
A quantity X is given by epsilon_(0)L(DeltaV)/(Deltat)where epsilon_(0) is the permittivity of free space, L is a length, DeltaV is a potential difference and Deltat is a time interval. The dimensional formula for X is the same as that of |
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Answer» resistance or `epsilon_(0)= ((Deltaq)L)/(A(DeltaV))` `:. X= epsilon_(0)L((DeltaV))/(Deltat)= ((Deltaq)L)/(A(DeltaV))L((DeltaV))/(Deltat)` (using (i)) but `[A]= [L]^(2)` `:. X= (Deltaq)/(Deltat)= "current"` |
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| 14235. |
(A) : Reversible systems are difficult to find in real world. (R ): Most processes are dissipative in nature |
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Answer» Both (A) and(R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 14236. |
It is better to wash clothes in hot water than in cold water .Explain . |
| Answer» SOLUTION :The SURFACE tension of hot water is LESS so it wets more to the clothes .Hence clothes wash better . | |
| 14237. |
Bernoulli's equation is applicable in the case of ………… . |
| Answer» Answer :A | |
| 14238. |
If .g. on the surface of the earth is 9.8 ms^(-2) , find its value at a depth of 3200km (radius of the earth = 6400 km )D |
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Answer» Solution :epth , d= 3200km The EXPRESSION for the ACCELERATION due to gravity at depth .d. is `g_d = G |1 - d/R| rArr g_d = 9.8 |1- 3200/6400| rArr g_d = 4.9 ms^(-2)` |
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| 14239. |
The magnetic susceptibililty of a paramagnetic substance changes with absolute temperature as chi=c/(T=223^(@)) when Tgt223 and c is a constant. Derive an expression for the Celsius temperature t baed on this property and establish the relation betwen t and T. What is the value of t corresponding to T=423K? Take ice point =273K |
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| 14240. |
A runner moves along the road at 2.0 ms^(1)in still air that is at a temperature of 29.0^(0)C. His surface area is 1.4 m^(2), of which approximately 85% is exposed to the air. Find the rate of convective heat loss from his skin at a temperature 35.0^(0) C to the outside air? Coefficient of convection for dry air and bare skin at wind speed 2.0 ms^(-1)' is 22 W/m2""^(@) C . |
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Answer» Solution :FromEq. `(Q)/(t) = lambda A Delta T ` Given `lambda ` = Coefficientof conduction ` = 22 W // m^(2 0) C, Delta T= 35.0^(@)C - 29.0^(@)C = 6.0^(@)C` A =85% of the surfacearea of the runner `1.4m^(2) = 1.2 m^(2)` UPON substituting the value `(Q)/(t) = 22W // m^(20)C xx 1.2 m^(2) xx 6.0^(@)C = 160 W ` |
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| 14241. |
Whycopperis used at the bottom of cooking untensilsand why they are provided withwooden handles ? |
| Answer» Solution :Copperis a goodcondutorheat ,So copperis usedat thebottomof cookingutensils . Woods has low valueofthermalconducitivity K = 0.13 SI unithencea BD condutorof HEATTO provieditashandles. | |
| 14242. |
Describe galileo experiments concerning motion of object on inclined planes galileo 's experiment with the second plane(a) at same inclination angle as the first (b) with increased smoothness (c ) with reducedangle ofinclination (d) with zero angle of inclination |
| Answer» SOLUTION :Whena ballrollsfrom thetop of an inclinedplaneto itsbottom , it movesomedistanceafterreachingthe groungcontinuesto moveon toanotherinclinedplaneof sameangleofinclination. It isshownin the figure(a) . Theballreachesalmostthe sameheight(h) fromwhereis was released (L1)in thesecond PLANE(L2) it isshownin the figure(b) byincreasingthe MOTIONOF the ballis thenobserved byvaryingthe angleof inclinationof thesecondplanekeepingthe samesmoothnese . If theangle ofinclinations is madezero,thentheballmovesforeversin the horizontalideaweretrue, the ballwouldnot havemovedin thesecondplaneevenif itssmoothnessis mademaximumsinceno forceactedon ITIN thehorizontaldirection. | |
| 14243. |
Given samples of 1 cm^3 of Hydrogen and 1 cm^3 of Oxygen both at N.T.P., which samples has larger number of molecules |
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Answer» Hydrogen |
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| 14244. |
APPLICATION OF SURFACE TENSION |
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Answer» Solution :Applications are:(i) Mosquitoes LAY their eggs on the surface of water. To reduce the surface tension of water, a small amount of oil is poured. This BREAKS the elastic film of water surface and eggs are killed by drowing. (ii) Chemical engineers must finely adjust the surface tension of the liquid, so it forms droplets of designed SIZE and so it adheres to the surface without smearing. This is used in desktop printing, to paint automobiles and decorative items. (III) Specks of dirt get removed when detergents are added to hot wate while wasing clothes because tension is reduced. (iv) A fabric can be made waterproof, by adding SUITABLE waterproof materia (wax) to the fabric. This increases the angle of contact. |
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| 14245. |
(A) : The time period of geostationary satellite is 24 hours. (R) : Geostationary satellite must have the same time period as the time taken by the earth to complete one revolution about its axis. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 14246. |
A uniform rod of mass 2M is bent into four adjacent semicircles each of radius r, all lying in the same plane. The moment of inertia of the bent rod about an axis through one end A and perpendicular to plane of rod is |
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Answer» `22Mr^(2)` |
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| 14247. |
Keeping the mass of earth as constant, if its radius is reduced to 1//4^(th) of its initial value, then the period of revolution of earth about its own axis and passing through the centre, in hours, in is (Assure earth to be a solid sphere and its initial period of rotation as 24 hrs.) |
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Answer» `1.5` |
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| 14248. |
The position of a particle moving along a. straight line is given by x = 2 - 5t + t ^(3)The acceleration of the particle at t = 2 sec. is ...... Here x is in meter. |
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Answer» `12m/s ^(2)` `thereforex = t ^(3) - 5t +2` So, velcity` v = (dx)/(dt) = (d)/(dt) (t ^(3) - 5t +2)` Thus, acceleration ` = (dv)/(dt) = (d)/(dt) (3T ^(2) -5)` `therefore a = 6t` Taking `t =2s` in above FORMULA, we get `a =12 m//s^(2)` |
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| 14249. |
Derive the expression for centripetal acceleration. |
Answer» Solution :In uniform circular motion the velocity vector turns continuously WITHOUT changing its magnitude (speed), as shown in figure.  Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is CALLED centripetal acceleration. It always points towards the center of the circle. This is shown in the figure The centripetal acceleration is derived from a simple geometrical RELATIONSHIP between position and velocity vectors  Let the directions of position and velocity vectors shift through the same angle `theta` in a small interval of time `Deltat`, as shown in figure. For uniform circular motion, `r=|vecr_(1)|=|vecr_(2)|andv|vecv_(1)|=|vecv_(2)|`.If the particle moves from position vector `vecr_(1)` to `vecv_(2)`, the displacement is given by `Deltavecv=vecr_(2)-vecr_(1)`and the change in velocity from `vecv_(1)` to `vec_(2)` is given by `Deltavecv=vecv_(2)=vecv_(1)`. The magnitudes of the displacement `Deltar` and `Deltav` satisfy the following relation `(Deltar)/(r)=(-Deltav)/(V)=theta`  Here the negative sign implies that Av points radially inward, towards the center of the circle. `Deltav=v((Deltar)/(r))` Then, `a=(Deltav)/(Deltat)=(v)/(r)((Deltar)/(Deltat))=-(v^(2))/(r)` For uniform circular motion `v=omegar`, where `omega` the angular velocity of the particle about the center. Then the centripetal acceleration can be written as `a=-omega^(2)r` |
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