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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.
| 40951. |
At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10ms^(-1). The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g=10ms^(-2)) |
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Answer» 1s |
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| 40952. |
A liquid drop always tends to have spherical shape because - |
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Answer» Sphere has minimum volume |
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| 40953. |
A: Resolution of telescope is high if diameter of objective is large.R: More light rays are converged by objective of large diameter. |
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Answer» Both ASSERTION and REASON are TRUE and the reason is CORRECT EXPLANATION of the assertion. |
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| 40954. |
Try to derive the barometrical formula for an atmosphere in which the temperature decreases linearly with the altitude. |
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Answer» `T = T_(0) (1 - alpha h)` we obtain `dT = -alpha T_0 dh`. Hence `(dp)/(p) = (Mg)/(alpha RT_0) . (dT)/(T)` Integrating , we obtain In `p = (Mg)/(alpha RT_0) ln T` + const. Since the equation is valid at any point of the gravitaional field, we have on the planet.s surface `ln p_0 = (Mg)/(alpha RT_0) ln T_0 + "const"` SUBTRACTING this force the preceding equation, we eliminate the integration constant to obtain `ln p - ln p_0 = (Mg)/(alpha RT_0) (ln T - ln T_0)` Whence the formula sought for the barometric destribution. |
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| 40955. |
In A young.sdouble - slitexperiment ,let A and Bbe twoslits , A thin filmof thickness t and refractive indexmuis placed in front of A .Let beta= fringewidth .The central maximum will shift . |
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Answer» TOWARDS A |
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| 40956. |
An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? |
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Answer» Solution :For CONCAVE lens, Here object height h =3 cm Object distance U = - 14 cm Focal length f = - 21 cm Image distance v = ?  `rArr` Lens formula, `1/f=1/v - 1/u ` `therefore1/v = 1/f + 1/u = 1/(-21) + (1)/(-14)` `therefore 1/(v) = -(1)/(21) - (1)/(14) = (-2-3)/(42) = (-5)/(42)` `therefore v = - (42)/(5) cm = - 8.4 cm ` Negative sigh represent that image is virtual erect and siminised and toward object Maginification `m = v/u ` `therefore h./h = (-8.4)/(-14)` `thereforeh. = 0.6 xx h = 0.6 xx 3` `therefore` Height of image = 1.8 cm As the object GOES away from lens , the vritual image goes toward the optica centre oflens (but it is never obtained away from optical centre) but it is obtained of dereasing size. |
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| 40958. |
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Omega per km. The town gets power from the line through a 4000 - 220 V step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage ? (c) Characterise the step-up transformer at the plant. |
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Answer» Solution :(a) Here power required by the town, P = 800 kW `=800 xx 10^(3) W = 8 xx 10^(5)` W Total resistance of 2 wire of line of 15 km at 0.5 `Omega` PER km, `R = 2 xx 15 xx 0.5 = 15 Omega` s supply voltage is through 4000 - 220 V line, the transmission line transmits power at 4000 V `therefore` Line current `I = P/V = (8 xx 10^(5))/(4000) = 200 A` `therefore` LIne power loss in the form of heat `= I^(2)R = (200)^(2) xx 15 = 6 xx 10^(5) W = 600 kW` (b) (b) If there is no power loss due to leakage then the total power to be SUPPLIED by power plant= Power NEEDED by town + Power loss during transmission = 800 kW + 600 kW = 1400 kW ( C) Voltage drop along the transmission line = I R = 200 x 15 = 3000 V As the transmission line supplies power to sub-station at 4000 V, hence, voltage at generating station side of line = 4000 + 3000 = 7000 V Hence, a step up transformer of 440 V - 7000 V be used at the generator plant. |
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| 40959. |
A surface of a prism having refractive index 1.5 is covered with a liquid of refractive index (3sqrt(2))/(4) . When be the minimum angle of incidence of an incident ray so that on the other surface of the prism the the ray will be totally reflected from the surface covered with liquid? The refracting angle of the prism75^(@) [sin48^(@)36' = 0.75]. |
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Answer» Solution :Let the critical angle between the prism and the LIQUID be `theta_(c)`. If the ray of light is totally REFLECTED from the surface covered with liquid then the angle of incidence of the ray on the surface is `theta_(c)` [Fig 2.63]. `1.5 sintheta_(c) = (3sqrt(2))/(4) sin90^(@) = (3sqrt(2))/(4)` `sintheta_c = (3sqrt(2))/(4) xx (10)/(15) = (1)/(sqrt(2))` `theta_(c) = 45^(@)` `"Again", r_(1) + theta_(c)= A or, r_(1) + 45^(@) = 75^(@) or, r_(1) = 30^(@)` `"Now" " " sini_(1) = musinr_(1) = 1.5 sin30^(@) = 0.75 or, i_(1) = 48^(@)36.` 
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| 40960. |
Calculate the a. momentum, and b. de Broglie wavelength of the electrons acceleratedthrough a potential difference of 56V. |
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Answer» SOLUTION :a. `v=56V` `p=mv=sqrt(2mK_("max"))=sqrt(2meV)=sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx56)` `=40.38xx10^(-25)KG ms^(-1)=4.038xx10^(-24) kg ms^(-1)` b. `lambda=(h)/(p)=(6.6xx10^(-34))/(4.038xx10^(-24))=0.163nm` |
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| 40961. |
What is the difference between n-type and p-type semiconductor? |
Answer» SOLUTION :
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| 40962. |
What is a proton ?What's it's mass and charge? |
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Answer» SOLUTION :A proton is a+vely charged PARTICLE and has a CHARGE equal to that of the electron.But,mass of proton is about 1840 TIMES that of electrons Charge of proton=`+1.6xx10^(-19)C` Mass of proton =`1.6726xx10^(-27)KG` |
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| 40963. |
Intensity level of sound whse intensity is 10^(-8) Wm^(-2) is ..... dB |
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Answer» 8 L = 10` LOG" " (I)/(I_(0))` = 10 ` LOT (10^(-8))/(10^(-12)) = 10 log 10^(4)` = 40 dB. so correct choice is (c) . |
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| 40964. |
An alpha particle of 10 MeV is moving forward fo a head on collision. What will be the distance of closest approach from the nucleus o atomic number Z = 29 ? |
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Answer» `8.4xx10^(-15)cm` Potential energy of `alpha`-particle at distance d=kinetic energy of `alpha`- particle at large distance `10xx10^(6)xx1.6xx10^(-19)=(K(2qe)(2e))/(r_(0))` `:.r_(0)=(9xx10^(9)xx29xx2xx(1.6xx10^(-19))^(2))/(16xx10^(-13))` `:.r_(0)=83.52xx10^(-16)` `:.r_(0)=8.4xx10^(-15)m` |
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| 40965. |
Which of the followingcarbidesgivespropyne on hydrolysis? |
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Answer» `CaC_(2)` |
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| 40966. |
Explain, with the help of diagram, how plane polarised light is obtained by scattering. |
Answer» SOLUTION : The sunlight is ordinary unpolarised LIGHT. However, when the sunlight is scattered on encountering the molecules of earth.s atmosphere, the scattered light as seen by an observer LOOKING at `90^(@)` to the direction of Sun is found to be plane polarised as SHOWN in figure. |
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| 40967. |
अन्तः केंद्रित यूनिट सेल में अवयवी कणों की संख्या होगी - |
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Answer» 1 |
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| 40968. |
The resistance R=V/IWhere V = (100 p5) V and I=(10 pm0.2) . The percentage error in R is |
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Answer» SOLUTION :`R=V/I` `(DeltaR)/Rxx100=(DeltaV)/Vxx100+(DeltaI)/Ixx100` = `5%+2%=7%` |
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| 40969. |
The distance beween the toothed wheel with 720 teeth and the mirror in an experiment based on Fizeau's method is 7 km. The two consecutive speeds of rotation of the wheel for which light disappeared were 283 rps and 313 rps. Find the velocity of light. |
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Answer» |
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| 40970. |
A resistance coil of 60Omega is immersed in 42 kg of water. A current of 7A is passed through it. The rise in temperature of water per minute is |
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Answer» `4^(@)C` |
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| 40971. |
In a straight conductor of uniform cross-section charge qis flowing for time t. Lets be the specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is |
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Answer» `(q)/(TS)` |
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| 40972. |
N small conducting liquid droplets, each of radius r, are charged to a potential V each. These droplets coalesce to form a single large drop without any charge leakage. Find the potential of the large drop. |
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Answer» SOLUTION :Let N small conducting liquid droplets each of radius r. CARRY a charge q so as to have a potential V) then `V=q/c=(q)/(4PI epsi_(0) r)` If the radius of single bigger drop formed due to coalescence of these droplets be r., then, `4/3 pi Nr^(3) =N xx 4/3 pi r^(3) rArr r=(N)^(1/3).r` and charge on this drop q.=Nq Potential of single drop `V.=(q.)/(C.)=(q.)/(4pi epsi_(0) r.) =(Nq)/(4pi epsi_(0) (N)^(1/3) .r) =(N)^(2/3) . (q)/(4pi epsi_(0) r)=N^(2/3).V` |
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| 40973. |
A detector is release from rest over a source of sound of frequency f_(0)= 10^(3) Hz. The frequency observed by the detector at time t is plotted in the graph. The speed of sound in air is: (g =10m/s) |
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| 40974. |
The revolution of the electron in the first Bohr orbit of a hydrogen atom constitutes a current loop of area 8.8 xx 10^(-21)m^(2). If the frequency of revolution is 6.6 xx 10^(15) Hz, calculate the equivalent magnetic moment and the orbital angular momentum. [y_(0) = 8.975 xx 10^(10)C//kg] |
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Answer» Solution :Data: `A=8.8 xx 10^(-21) m^(2), f=6.6 xx 10^(15)Hz, y_(0)= 8.975 xx 10^(10) C//Kg, e=1.6 xx 10^(-19) C` The equivalent magnetic moment, `M=IA = efA = (1.6 xx 10^(19))(6.6 xx 10^(15))(8.8 xx 10^(-21))= 1.6 xx 48 xx 1.21= 9.293xx 10^(24)A.m^(2)` GYROMAGNETIC ratio, `y_(0)=(M_(0))/(L_(0))` `THEREFORE` The orbital angular momentum, `I= (M_(0))/y_(0) = (9.293 xx 10^(-24))/(8.795 xx 10^(10)) = 1.056 xx 10^(-34) kg.m^(2)//s` |
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| 40975. |
What is optical fibres? What principle is used in it? |
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Answer» reflection |
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| 40976. |
Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident on the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere will be a |
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Answer» PARALLEL beam |
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| 40977. |
A vector sqrt(3) hati+hatj rotates about its fail through an angle 300 in clock wise direction then the new vector is |
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Answer» Solution :The magnitude of `SQRT(3)hati+hatj` is `sqrt(3+1)=2` the ANGLE made by the vectorwith x-axis is `"tan" theta=(A_(y))/(A_(x))=(1)/(sqrt(3))` `:. theta=30^(@)`  When the given vector rotates 30 in clock wise its direction changes ALONG x - axis but its magnitude does not change. `:.` The new VETOR is `2hati` 
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| 40978. |
A wide copper strip of width l is bent into piece of slendder tubing of radius R with two plane extension as shown (Fig. 1.12). A current I flows through the strip, distributed uniformly over its width. Ib this way, one-turn solenoid has been formed (a) Derive an expression for the magnetic field in the tubular part. |
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Answer» |
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| 40979. |
In Fig. 8-15b, an 8.0 kg block slides along a frictionless floor as a force acts on it, starting at x_(1)=0 and ending atx_(3) = 6.5m. As the block moves, the magnitude and direction of the force varies according to the graph shown in Fig. 8-15a. Forexample, from x = 0 to x = 1 m, the force is positive (in the positive direction of the x axis) and increase in magnitude from 0 to 40 N. And from x = 4 m to x = 5, the force is negative and increases in magnitude from 0 to 20 N. (Note that this latter value is displayed as -20 N.) The block's kinetic energy atx_(1) is K_(1)=280 J. What is the block's speed at x_(1) = 0, x_(2)=4.0m, and x_(3) = 6.5 m ? |
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Answer» Solution :KEY IDEAS (1) At any POINT, we can relate the speed of the block to its kinetic energy with EQ. 8-1 `(K=1//2mv^(2))`. (2) We can relate the kinetic energy `K_(f)` at a later point to the initial kinetic `K_(i)`and the work W done on the block by using work-kinetic energy theorem of Eq. 8-10 `(K_(f)-K_(i)=2)`. (3) We can calculate the workk W done by a variable force F(X) by integrating the force versus position x.Equation 8-32 tells us that `W = int_(x_(i))^(x_(f)) f (x)dx`. We do not have a functon F(x) to carry out this INTEGRATION, but we do have a graph of F(x) where we an integrate by finding the area between the plotted line  Figure 8-15 (a) A graph indicating the magnitude and direction of a variable force that acts on a block as it moves along an x axis on a floor. (b) The location of the block at several times. and the x axis. Where the plot is obove the axis, the work (which is equal to the area) is positive. Where it is below the axis, the work is negative. Calculations: The requested speed at x=0 is easy because we already Know the kinetic energy. So, we just plug the kinetic energy into the fromula for kinetic energy: `K_(1) = 1/2 mv_(1)^(2), 280 J = 1/2 (8.0 kg) v_(1)^(2)`, and then `v_(1)= 8.37 m//s ~~ 8.4 m//s`. As the block moves from x = 0 to x = 4.0 m, the plot in Figure 8-15a is above the x axis, which means that positive work is being done on the block. We split the area under the plot into a triangle at the left, a rectangle in the center, and a triangle at the right. Their total area is `1/2 (40 N) (1m) + (40 N)(2m) +1/2 (40 N) (1m) = 120 N*m` `=120J`. This means that between x = 0 and x = 4.0 m, the force does 120 J of work on the block, increasing the block reaches x = 4.0 m, the work-kinetic energy theroem tells us that the kinetic energy is `K_(2) = K_(1) +W = 280 J + 120 J = 400 J`. Again using the definition of kinetic energy, we find `K_(2) = 1/2 mv_(2)^(2), 400 J = 1/2 (8.0 kg)v_(2)^(2)`, and then `v_(2) = 10m//s` This is the block.s greatest speed because from x = 4.0 m to x = 6.5 m the force is negative, meaning that it opposes the block.s motion, doing negative work on the block and thus decrasing the kinetic energy and speed. In that range, the area between the plot and the x axis is `1/2(20N) (1m) +(20N)(1m) + 1/2 (20 N ) (0.5 m) = 35 N*m` `= 35J`. This means that the work done by the forcein that range is -35 J. At x =4.0 m, the block .s K =400 J. At x =6.5 m, the work-kinetic energy theroem tells us that its kinetic energy is `K_(3) = K_(2) +W = 400 J - 35 J = 365 J`. Again using the definition of kinetic, we find `K_(3) = 1/2 mv_(3)^(2), 365 J = 1/2 (8.0 kg)v_(3)^(3)`, and then `v_(3) = 9.55 m//s ~~ 9.6 m//s`. The block is still moving in the positive direction of the x axis, a bitfaster than initally. |
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| 40980. |
Assertion (A) : The images formed by total internal reflections are much brighter than images formed by mirrors or lenses. Reason (R ) : There is no loss of intensity in total internal reflection. |
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Answer» If both ASSERTION and REASON are true and the reason is the CORRECT explanation of the assertion. |
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| 40981. |
Which physical quantity has unit newton (coulomb)^(-1)? Is it vector or scalar quantity? |
| Answer» SOLUTION :ELECTRIC FIELD, which is a VECTOR QUANTITY. | |
| 40982. |
Fig. shows two identical capacitorsC_(1) and C_(2) each of 1 muF capacitance, connected to a batteryof 6VInitially,swich S is closed. Aftersometime,S is leftopen and dielectricslabs of dielectricconstant K = 3 are insteredto fillcompelelty the space betweenthe plates of two capacitors. How will the (i) charge and (ii) potential difference between the platesof the capacitors be affectedafter teh slabs are inserted ? |
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Answer» Solution :In Fig, when switch S is closed, both`C_(1) and C_(2)` are chargedto 6Vpotential, i.e, `V_(1) = 6V and V_(2) = 6V` when S is left open, `C_(1)` is still connectedto battery. On introducingslabof DIELECTRICCONSTANT. K = 3. `C'_(1) = KC_(1) = 3mu F= 3xx10^(-6) F`, `V'_(1) = V_(1) = 6V` `Q_(1) = C'_(1) V'_(1) = 3xx10^(-6) xx6 = 18xx10^(-6)C`. However `C_(2)` is disconnecedfrom battery now, when S is left open. Due to introducedof slab, `C'_(2) = KC_(2) = 3xx10^(-6)F` `Q'_(2) = Q_(2) = C_(2) V_(2) = 1XX10^(-6)xx6 = 6xx10^(-6) C` `:. V_(2) = (Q_(2))/(C_(2)) = (6xx10^(6))/(3xx10^(-6)) = 2V` |
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| 40983. |
A force vecF = 6t^(2)hat i +4thatj is acting on a particle of mass 3 kg, then what will be velocity of particel at t = 3 sec if at t = 0, particle is at rest : |
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Answer» `181hati+6hatj` |
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| 40984. |
Two coherent sources are placed 0.9 mm apart and the fringes are observed one away .If it produces the second dark fringe at a distance of 10 mm from the central fringe . The wavelength of monochromatic light is x xx 10^(-4) cm , What is the value of x ? |
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Answer» |
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| 40985. |
To lift a load by a metallic rope, its radius of cross-section should be |
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Answer» `GE1 mm` |
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| 40986. |
A flexible steel cable of total length .L. and mass per unit length mu hangs vertically (under it.s own weight) from a support at upper end. If transverse pulse starts to move down in the wire from its support. The ratio of acceleration of the pulse at distance L/4 from the support end to the acceleration of the of the pulse at distance L/2 from the support end K : 1 then the value of K is |
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Answer» |
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| 40987. |
The density of solid ball is to be determined in an experiment the diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the relative error of 2%, the relative percentage error error in the density is |
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Answer» 0.009 |
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| 40988. |
Two satellites of earth, S_(1) and S_(2) are moving in the same orbit. The mass of S_(1) is four times the mass of S_(2). Which of the following statements is true? |
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Answer» The potential energies of earth and satellite in the two cases are EQUAL. So, `S_(1) and S_(2)` are moving with same speed. Thus correct CHOICE is (b). |
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| 40989. |
A combinationarrangement of the capacitors is shown in the figure (ii) If a potentialdifference of 48 V is appliedacross points a and b, then charge on the capacitor C_(3) at steady state condition will be : |
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Answer» `8 mu C` |
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| 40990. |
There are tow identical spring each of spring constant k. here springs, pulley and rods are massless and the block has mass m . What is the extension of each spring at equilibrium? |
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Answer» `(2 MG)/K` |
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| 40991. |
A capacitor of 1 muF and resistance 82 kOmegaare connected is series with a d.c. source of 100 volt. Calculate the magnitude of energy and the time in which energy stored in the capacitor will reach half of its maximum value. |
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Answer» Solution :i) Maximum energy stored `=(CV_0^2)/(2)` Energy stored `ALPHA V_0^2` Half of maximum will be stored when voltage across capacitor is `V=(100//sqrt2)` volt = 70.7 Volt Energy stored `=1/2 CV^2 = 1/2 xx(1xx10^(-6)) xx (70.7)^2 = 0.0025 J` II) `(V_0)/(sqrt2) = V_0 (1- E^(-t//RC))` or , `70.7 =100 [1-e^(-t//(82xx10^3 xx 10^(-6)))]` or , `70.7 = 100[1-e^(t//0.082)]` or, `e^(-t//0.082) = 1-(70.7)/(100) = 0.293` or, `e^(t//0.082) = 3.413` `therefore t= 0.082 log_e = 3.413 = 0 ` SEC. |
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| 40992. |
To which part of the electromagnetic spectrum does the wave of frequency (i) 3xx10^(13)Hz(ii) 5xx10^(11)Hz belong? |
| Answer» SOLUTION :(i) INFRARED RAYS, (II) MICROWAVES. | |
| 40993. |
Figure shows the position-time graph of a particle of mass 4 kg, then a) force on the particle for t lt 0, t gt 4 s, 0 lt t lt 4 s?? b) impulse at t = 0 and t = 4 s. (consider one-dimensional motion only) |
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Answer» Solution : i) For `t lt 0`in the position time graph is at O which means displacement of the particle is zero. i.e. particle is at rest at the origin. Hence force on the particle must be zero II) For 0 < t < 4 s, the position time graph OA has a constant slope. THEREFORE velocity of the particle is constant in this interval i.e. particle has zero acceleration. Hence force on the particle must be zero. iii) For t > 4 s, the position time graph AB is parallel to time AXIS. Therefore, the particle remains at a distance 3M from the origin, i.e. it is at rest. Hence force on the particle is zero. B)i) impulse at t = 0 We know, Impulse = change in linear momentum, Before t = 0 particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity `v=3/4 = 0.75 m//s` `therefore` Impulse `=m(v-u) =m(0.75 -0) = 3` kgm//s (ii) `therefore`Impulse at t = 4 s, Before t = 4s, particle has a constant velocity u = 0.75 m/s After t = 4s, particle is at rest i.e. v = 0 Impulse = `m(v - u) = 4(0 - 0.75 - - 3` kg m/s |
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| 40994. |
In an oscillating L-C circuit the maximum charge on the capacitor is Q. What will be the charge on the plate of the capacitor, when energy stored in magnetic field and electric field are equal ? |
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Answer» `Q/3` `U=Q^2/(2C)` When the energy in capacitor and in inductor becomes EQUAL then energy `Q^2/(4C)` is same on both these components and suppose at this time CHARGE on capacitor is Q.. `therefore Q^2/(4C)=(Q.)^2/(2C)` `therefore (Q.)^2=Q^2/2` `therefore Q.=Q/sqrt2` |
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| 40995. |
A charged particle is held a the centre foasperical shell. The following figure gives the magnitude E of the electric field versus radial distance r. what is the net charge on the shell ? |
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Answer» |
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| 40996. |
Why did people discourage (हतोत्साहित ) Evelyn to pursue music? |
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Answer» it was not CONSIDERED good |
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| 40997. |
C_(V)andC_(P) denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then |
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Answer» `C_(P)-C_(V)` is larger for a DIATOMIC ideal gas than for a monoatomic ideal gas `C_(P)-C_(V)=R` (for all GASES). `C_(P)+C_(V)=4R` (for monoatomic gases), `C_(P)+C_(V)=6R` (for diatomic gases) `(C_(P))/(C_(V))=(5)/(3)=1.67` (for monoatomic gases), `(C_(P))/(C_(V))=(7)/(5)=1.4` (for diatomic gases) `C_(P)xxC_(V)=(15)/(4)R^(2)` (for monoatomic gases) `C_(P)xxC_(V)=(35)/(4)R^(2)` (for diatomic gases) |
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| 40998. |
For the current in LCR circuit to be maximum |
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Answer» `X_L=0` |
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| 40999. |
Whatis the net flux of the uniformelectric field ofexercisethrough a cube of side 20cm oriented so that its faces are parallel to the coordinate planers |
| Answer» Solution :(a) 30 `Nm^(2)//C`(B) 15 `Nm^(2)//C` | |
| 41000. |
Explain the production of x-rays. |
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Answer» Solution :(i) X-rays are produced in x-ray tube which is ESSENTIALLY a discharge tube. (ii) A tungsten filament F is heated WO incandescence by a battery. As a result, electrons are emitted from it by thermionic emission.  (iii) The electrons are accelerated to high SPEEDS by the voltage applied between the filament F and the anode. (iv) The target materials like tungsten, molybdenum are embedded in the face of the solid copper anode. (v) The face of the target is inclined at an angle with respect to the electron beam so that x-rays can leave the tube through its SIDE. (vi) When high-speed electrons strike the target, they are decelerated suddenly and lose their kinetic energy. (vii) As a result, x-ray photons are produced. Since most of the kinetic energy of the bombarding electrons gets converted into heat, targets made of high-melting-point metals and a cooling system are usually employed. |
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