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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.
| 13251. |
What is the dimensional formula of thermal conductivity? |
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Answer» `[MLT^(-1)K^(-1)]` Substituiting DIMENSIONS for corresponding quantities in equation (i), we get `[K] = ([ML^(2)T^(-2)][L])/([L^(2)][K][T])= [MLT^(-3)K^(-1)]` |
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| 13252. |
Explain whether Archimedes' principle is applicable in the case of a freely-falling body under gravity. |
| Answer» Solution :A freely-falling body has no WEIGHT. Hence, the weight of a freely-falling body and the weight of the LIQUID displaced by it-both will be zero and so the body will not EXPERIENCE any buoyant force. It means that the APPLICATION of Archimedes. principle in this case is unnecessary. | |
| 13253. |
What is the moment of inertia of a cylinder of radius r, long its height? |
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Answer» `MR^(2)` 
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| 13254. |
A uniform wire (Young.s modulus 2xx10^(11) Nm^(-2)) is subjected to longitudinal tensile stress of 5 xx 10^(7)Nm^(-2). If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to : |
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Answer» `1.0xx10^(-4)` |
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| 13255. |
A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with gamma= 5/3 andanother gas B with gamma=7/5 at a temperature T. The gases A and B do not react with each other and assumed to be ideal. Find the number of gram moles of the gas B, if gammafor the gaseous mixture is (19/13). |
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Answer» |
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| 13256. |
A sphere of radius R of material of refractive index mu_3. Where should an object be placed so that a real image is formed equidistant from the sphere? |
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Answer» Solution :Let object be placed ata distance x from the pole `P_1` of the sphere then the ray must pass symmetrically through the sphere, as shown in the FIGURE. Applying the EQUATION at the first surface, we get `mu_2/(- infty)-mu_1/(-x)= (mu_2-mu_1)/(-R) or x=(mu_1/(mu_2-mu_1)) R` Note that the real IMAGE is formed only when the refractive index of the sphere is more than that of the surrounding i.e, `mu_2 GT mu_1`. |
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| 13257. |
As shown in figure lower portion of the manometer tube contains fluid of density rho_(1) and the upper part contains fluid of density rho_(2)(rho_(1)gtrho_(2)) . Ifpressures on the top of these two arms are P_(1)andP_(2), calculate pressure differnce (P_(1)-P_(2)). |
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Answer» Solution :Consider POINTS A and B , as SHOWN in the diagram , at EQUAL height from bottom.For these points , `P_(A)=P_(B)` `thereforeP_(1)+(h+x)rho_(2)g=xrho_(2)g+hrho_(1)g+P_(2)` `thereforeP_(1)-P_(2)=xrho_(2)g+hrho_(1)g-hrho_(2)g-xrho_(2)g` `thereforeP_(1)-P_(2)=(rho_()-rho_(2))hg` |
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| 13258. |
In the equation ((1)/( P beta) ) = (y)/( k_(B) T), when P is the pressure, y is the distance, k_(B) is Boltzmann constant and T is the temperature. Dimensions of are |
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Answer» `M^(-1) L^(1) T^(2)` |
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| 13259. |
The velocity of water in a river is 2 kmph, while width is 400 m. A boat is rowed from a point rowing always aiming opposite point at 8 kmph of still water velocity. On reaching the opposite bank the drift obtained is |
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Answer» 93 m |
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| 13260. |
An air bubble comes from the bottom to the surface of a lake of depth 2.5 m. The surface temperature of the lake is 40°C. The diameter of the bubble at the bottom and at the surface are 3.6 mm and 4 mm respectively. Find the temperature of the lake at the bottom. |
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Answer» `5.3^(@) C` |
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| 13261. |
When an object will have procession? Give one example |
| Answer» SOLUTION :The torque about the axis will ROTATE the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a precession. Example: The SPINNING TOP when it is about to COME to rest, | |
| 13262. |
Which of the following statements are correct?a) A wire is stiffer if Y is large b) A wire is stiffer if Y is small c) A wire is stronger if the breaking stress is large d) A wire is stronger if the breaking stress is small |
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Answer» B,c |
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| 13263. |
A simple pendulum is taken from the equator to the pole. Its period……….. |
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Answer» decreases |
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| 13265. |
(A) : The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. (R) : Potential energy and kinetic energy do not vary linearly with position. |
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Answer» Both 'A' and 'R' are true and R' is the correct explanation of 'A' |
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| 13266. |
A block whose mass is 1kg is fastened to a spring. The spring has a spring constant of 50 Nm^(-1). The block is pulled to a distance x=10 cm from its equilibrium position at x=0 on a frictionless surface from rest at t=0. Calculate the kinetic, potential and total energies of the block when it is 7.07 cm away from the mean position. |
| Answer» Solution :KINETIC ENERGY `=0.125J`, POTENTIAL energy `=0.125 J`, TOTAL energy `=0.25 J`. | |
| 13267. |
If vec(A) xx vec(B) = 0 and vec(A).vec(B) = - AB the angle between vec(A) and vec(B) is |
| Answer» ANSWER :D | |
| 13268. |
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^(-1),then the acceleration of the stone is |
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Answer» `12 ms^(-2)` |
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| 13269. |
A force vecF(larger than the limiting friction force ) is applied to the left to an objectmoving to the right on a rough horizontal surface . Then : |
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Answer» the OBJECT would be slowing down initially. |
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| 13270. |
A block whose mass is 1kg is fastened to a spring. The spring has a spring constant of 50 Nm^(-1). The block is pulled to a distance x=10 cm from its equilibrium position at x=0 on a frictionless surface from rest at t=0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position. |
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Answer» Solution :`m = 1kg, k = 50 Nm^(-1), A= 10 cm = 0.1 m, x=5 cm = 0.05m` Angular FREQUENCY of the BLOCK executes SHM, `omega = sqrt((k)/(m)) = sqrt((50)/(1)) =7.07 rad s^(-1)`. Its displacement at any time t is, `x(t) = A cos omega t = 0.1 cos omega t`. Now if `x= 5 cm =0.05 m` then `0.05 = 0.1 cos omega t` `therefore 0.5 = cos (omega t)` `therefore cos (omega t)= (1)/(2)` `therefore sin omega t= sin (pi)/(3) = (sqrt(3))/(2)` `sin omega t = 0.866 """........."(1)` The velocity of the block at x= 5 cm is, `x(t) = A cos omega t` `therefore v= (DX(t))/(dt) = (d(A cos omega t))/(dt)` `therefore v= -A omega sin omega t` `= -0.1 xx 7.07xx sin omega t` `= -0.1 xx7.07 xx 0.866""` [From equ. (1)] `= -0.612262 = -0.61 ms^(-1)` Kinetic energy of the block `K= (1)/(2) mv^(2)` `=(1)/(2)xx1xx(-0.61)^(2)` `=(1)/(2)xx0.3727 = 0.1864 = 0.19J` Potential energy of the block `U= (1)/(2) kx^(2)` `=(1)/(2)xx50xx(0.05)^(2) = 25xx 0.0025 = 0.0625J` Total energy of the block at distance x= 5 cm `E = K+U = 0.19+0.0625` `= 0.2525 approx 0.25J`. |
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| 13271. |
A glass plate of surface area 1m^(2) placed on a layer of turpentine of thickness 2 mm. If the coefficient of viscosity of turpentine 0.00149 Nsm^(-2). The force to be applied to move the plate horizontally with a velocity of 0.1ms^(-1) is |
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Answer» `745 N` |
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| 13272. |
When the wind blows at a high speed, the roof of a house is blown off because : |
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Answer» the pressure under the ROOF increases |
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| 13273. |
A block of mass m_(1) =1kg and another mass m_(2) = 2kg are placed together(see figure) on an inclined plane with angle of inclination theta varius value of theta are given in list 1 The coefficient of friction between the block m_(1) and the plank is always zero The coefficient of static and dynamic friction between the block m_(2) and the plank are equal to mu = 0.3 In List II experssions for the friction the block m_(2) are given Match the correct experssions of the frictionless in List II with the angle given in list 1 and choice the correct option The acceleration due too gravity detented by g [Useful information tan (5.5^(@)) = 0.1 tan (11.5^(@)) = 0.2 tan (16.5^(@)) = 0.3] List I P.theta = 5^(@) Q. theta = 10^(@) R. theta = 15^(@)S. theta = 20^(@) List 2 1.m_(2)g sin theta 2.(m_(1) + m_(2))g sin theta 3.mu m_(2)g cos theta 4. mu (m_(1) + m_(2))g cos theta |
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Answer» `P-1, Q-1,R-1,S-3`  `f_(max) = (m_(1) + m_(2))g sin THETA` `mu N gt (m_(1) + m_(2))g sin theta` `0.3m_(2)g costheta ge (1+2)10 sin theta` `0.3 xx 2 xx 10 cos theta ge (1+2)10 sin theta` `6 ge 30 tan theta` `1//5 TANTHETA rArr 0.2 ge tan theta` Now it is clear in case`(P)(Q)` it will slip friction is `(m_(1)+ m_(2))g sin theta` `rArr F = (m_(1)+ m_(2))g sin theta` in the case `(R) (S)` of they slipso friction should be kinetic i.e. `mu m_(2) g cos theta` |
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| 13274. |
Three balls of masses m, n, and m, are lying on a straight line. The first ball is moved with a certain velocity so that it strikes the second ball directly and itself comes to rest. The second ball collides with the third and is itself reduced to rest. If .e. be the coefficient of restitution for each ball, then what will be the relation between m_(1), m_(2) and m _(3)? |
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Answer» |
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| 13275. |
A spring of force constant 1200 Mm^(-1) Is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum acceleration of the mass. |
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Answer» Solution :Here , `k= 1200NM^(-1), m= 3.0 kg, A= 2.0 CM = 0.02m` ACCELERATION `a= omega^(2)y= (k)/(m)y` Acceleration will be maximum when y is maximum i.e. y=A `:. "Max. acceleration" , a_("max")= (kA)/(m)= (1200xx 0.02)/(3)= 8ms^(-2)` |
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| 13276. |
A spring of force constant 1200 Mm^(-1) Is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum speed of the mass. |
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Answer» SOLUTION :Here , `k= 1200Nm^(-1), m= 3.0 KG, A= 2.0 cm = 0.02m` Max. SPEED of the MASS will be when it is passing through the MEAN position, which is given by `V_("max")= Aomega=A sqrt((k)/(m))= 0.02 xx sqrt((1200)/(3))= 0.4 ms^(-1)` |
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| 13277. |
The pressure of air increases by 100mm of Hg and the temperature decreases by 1^(@)C. The change in the speed of sound in air at STP is (V_(0)=333m//s) |
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Answer» `61ms^(-1)` |
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| 13278. |
Temperature of plate is increased by Delta theta then find new |
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Answer» inner radius |
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| 13279. |
A spring of force constant 1200 Mm^(-1) Is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the frequency of oscillation of the mass |
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Answer» Solution :Here , `K= 1200Nm^(-1), m= 3.0 kg, A= 2.0 cm = 0.02m` FREQUENCY , `f= (1)/(T)= (1)/(2A)SQRT((k)/(m))= (1)/(2 XX 3.14)sqrt((1200)/(3))= 3.2 s^(-2)` |
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| 13280. |
(A) : If bomb is dropped from an aeroplane moving horizontally with constant velocity then the bomb appears to move along a vertical straight line for the pilot of the plane. ( R) : If a bomb is dropped from an aeroplane moving horizontally with constant velocity then horizontal component of velocity of the bomb remains constant and same as the velocity of the plane during the motion under gravity. |
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Answer» Both (A) and ( R) are TURE and ( R) is the CORRECT EXPLANATION of (A) |
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| 13281. |
The motion of satellite around the Earth is an example for |
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Answer» CIRCULAR MOTION |
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| 13282. |
Statements :(a) In translatory motion each point of the body experiences the same displacement as any other point as time goes on, so that the motion of any particle epresents the motion of the whole body.(b) In rotatory motion, all the particles in the rigid body moves in concentric circles about the axis of rotation. |
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Answer» a is TRUE, B is FALSE |
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| 13283. |
A homogeneous rod of length L=1.8 m and mass M is pivoted at the centre O in such a way that it can rotate freely in the vertical plane as shown fig. The rod is initially in the horizontal position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the point O & B. Immediately after falling the insect moves towards the end B such that the rod rotates with constant angular velocity omega |
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Answer» The VELOCITY `omega` in terms of V & L `=(12V)/(7L)` |
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| 13284. |
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1 m, what is the speed with which the bob reaches the lower most point, given that it dissipates 10% of the initial energy against air resistance ? |
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Answer» Solution :When the BOB is the horizontal POSITION its POTENTIAL energy = MGH. When it reaches the lowest position, its kinetic energy `= (1)/(2) mv^(2)`. As part of the INITIAL energy is dissipated against air resistance, by the principle of conservation of energy. Potential energy in the horizontal position = Kinetic energy at the lower most point + Energy dissipated `mgh = (1)/(2) mv^(2) + ((10)/(100))` initial energy `mgh = (1)/(2) mv^(2) + 0.1` mgh 0.9 mgh `= (1)/(2) mv^(2) rArr v = sqrt(2 xx 0.9 xx g xx h)` `= sqrt(2 xx 0.9 xx 9.8 xx 1) (because h = l) = 4.2 ms^(-1)` |
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| 13285. |
A balloon of total mass 1000 Kg float motionless over the earth's surface. If 100 kg of sand ballast are thrown over board, the balloon starts to rise with an acceleration of |
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Answer» `10 m//s^(2)` |
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| 13286. |
A: A body is released from a height. As it is falling vertically downwards, at some position, it explodes into fragments under purely internal force. Centre of mass of the system of fragments will keep moving along the original vertical line and also accelerate downwards with an acceleration g. R: Whenever linear momentum of a system is conserved, its centre of mass always remains at rest. |
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Answer» If both A and B are true and R is the CORRECT explanation of A |
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| 13287. |
A flywheel starts rotating from rest when a torque is applied on it, with a uniform angular acceleration of 5 "rad"//s^(2). What is the angular displacement of the flywheel in the 5th Second ? |
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Answer» Solution :`omega_(0) = 0, ALPHA = 5 "rad"//s^(2), N = 5 s` Angular displacement in the 5th SECOND `=theta_("5th")`= ? `theta_("5th") = omega_(0) + alpha/2 (2n-1) = 0 +5/2 (2 xx 5-1) = 22.5` rad. |
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| 13288. |
Find the change in internal energy in joulewhen 20gm of a gas is heated from 20^(@)C to30^(@)C (G = 0.18 kcal/kg K, J = 4200J/kcal) |
| Answer» ANSWER :B | |
| 13289. |
Mass of moon is almost 10 % lesser than that of earth. Then, the gravitational force due to moon on the earth is how much than that on moon due to earth ? |
| Answer» SOLUTION :According to `F alpham_(1) m_(2)`, gravitational force depend on the MULTIPLICATION of masses not on the mass of one body and HENCE both moon and EARTH exerts same force. | |
| 13290. |
Air coming out form a burst bicycle or motorcar tube appears to be cold. Why? |
| Answer» SOLUTION :DUE to ADIABATIC EXPANSION | |
| 13291. |
A train moving at a constant velocity of 54 km/hr moves east wards for 30 minutes, then due north with the same speed for 40 minutes. What is the average velocity of the train during this run? (in km/hr) |
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Answer» 30 |
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| 13292. |
The rotational analogue of force in translatory motion is |
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Answer» torque |
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| 13293. |
Two bodies of masses, M_(1) and M_(2) are placed at a distance .r.. The gravitational potential at the point where the gravitational field is zero |
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Answer» `-(G)/(r)(M_(1)+2sqrt(M_(1)M_(2))+M_(2))` |
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| 13294. |
Figure 10.24(a) shows a thin liquid film supporting a small weight =4.5 times 10^-2 N What is the weight supported by a film of the same liquid at the same temperature in Fig.(b) and ( c) ? Explain your answer physically. |
| Answer» SOLUTION :`4.5 TIMES 10^-2 N` for (B) and (C ) the same as in (a). | |
| 13295. |
A large force acting for a short interval of time is called impulsive force. What is the SI unit of impulse ? |
| Answer» SOLUTION :NS or `KG MS^(-1)` | |
| 13296. |
The potential energy U [in J] of a particle is give (ax+by), where a and b are constants. The mass of the particle is 1kg and x- and y-are the co-ordinates of the particle in metre. The particle at t=0 is rest at (4a,2b).Then Find the speed of the particle when it crosses y-axis |
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Answer» `4sqrt(a^(2)+B^(2))` |
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| 13297. |
Suppose the bob in a simple pendulum is a hollow sphere with a small hole at the bottom. The sphere is filled with water and is made to oscillate. What happens to the period ? |
| Answer» Solution :T depends on length of the simple pendulum which is the distance between CENTRE of gravity of thebob and the point of suspension. As the water flows out the the C.G. shifts DOWNWARDS. So the effective length increases and period increases. When the whole water has flown out the C.G again coincides with the centre of the bob and the time period BECOMES same as in the beginning. THUS the period first increases and then decreases to the original value. | |
| 13298. |
If vec(A) = 7hat(i) + 2hat(j)- 3hat(k) and vec(B) = hat(i) + hat(j) + 2chat(k) are perpendicular vectors, the value of c is |
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Answer» `-1.5` |
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| 13299. |
The potential energy U [in J] of a particle is give (ax+by), where a and b are constants. The mass of the particle is 1kg and x- and y-are the co-ordinates of the particle in metre. The particle at t=0 is rest at (4a,2b).Then Find the co-ordinates of the particle at t=1 sec |
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Answer» (3.5a,1.5b) |
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| 13300. |
Define phase of the wave. |
| Answer» Solution :The FRACTION of the time elapsed since the wave had crossed the EQUILIBRIUM POSITION is KNOWN as the PHASE. | |