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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.
| 38251. |
Find the position and nature of the image of an object of height 3 cm when placed 60 cm from a mirror of focal length 15cm, when the mirror isconcave, |
| Answer» SOLUTION :Real, INVERTED, of size 1 CM, at a distance of 20 cm in front of the mirror | |
| 38252. |
Three positive charges of equal value q are placed at vertices of an equilateral triangle. The resulting lines of force should be sketched as in |
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Answer»
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| 38253. |
What is displacement current? |
| Answer» SOLUTION :The displacement CURRENT can be defined as the current which comes into play in the region is which is the ELECTRIC field and the electric are changing with time. | |
| 38254. |
The fuel in a rocket is burnt at the rate of 50 kg/s and the gases are ejected at the rate of 6 xx 10^4m/s with respect to the rocket, What is the thrust developed? |
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Answer» SOLUTION :Thrust = `V(dm)/dt`, V is the EXHAUST velocity of the EJECTED gases w.r.t the rocket = `6xx10^4xx50 = 3xx10^6N`. |
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| 38255. |
Statement-I : In simple harmonic motion, the motion is 'to and fro' and periodic. Statement-II : Velocity of particle is v=w sqrt(r^(2)-x^(2)) where x is displacement and r is amplitude. |
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Answer» Statement-I is TRUE, Statement-II is true and So correct CHOICE is (b). |
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| 38256. |
A long straight wire AB carries a current of 4 A. A proton P travels at 4 xx 10^(6) ms^(-1) parallel to the wire 0.2 m from it and in a direction opposite to the current as shown in the fig. Calculate the force which the magnetic field due to the current carrying wire exerts on the proton. Also specify its direction. |
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Answer» |
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| 38257. |
Assume that silicon diode in the circuit Fig.requires minimum current of 1 mA to be above the knee point (0.7 V) of its I-V characteristics. Also assume that the voltage across the diode is independent of current above the knee point. If V_(B)=5V, what should be the minimum value of R so that the voltage is above the knee point? If V_(B)=5V, what should be the value of R to establish the current of 5mA in the circuit? What is the power dissipated in the resistance R and in the diode, when a current of 5mA flows in the circuit at V_(B)=6V. If R=1kOmega, what is the minimum voltage V_(B) required to keep the diode above the knee point? |
Answer» Solution :Here the DIODE is non-ideal diode. The equivalent circuit for it is shown in Fig. Here the knee point VOLTAGE has been replaced by a CELL connected in reverse bias.  Given, minimum current, `I_(min)=1mA=10^(-3)A`. If `R_(max)` is the maximum value of resistance used, then `R_(max)I_(min)=5V-0.7=4.3V` or `R_(max)=4.3/I_(min)=4.3/10^(-3)=4.3xx10^(3)OMEGA` Here, `V_(B)=5V`, knee voltage, `V_(k)=0.7V, I=5mA=5XX10^(-3)A`. From the equivalent circuit as given above `IR=V_(B)-0.7=5-0.7=4.3V` `R=4.3/I=4.3/(5xx10^(-3))=860Omega` Here, `V_=6V, I=5xx10^(-3)A` `R=(V_B-0.7)/I=(6-0.7)/(5xx10^(-3))=1060Omega` Power dissipated in resistance R is `=I^(2)R=(5xx10^(-3))^(2)xx1060` `=0.0265W=26.5mW` Power dissipated in diode is `=IV_(k)=(5xx10^(-3))xx0.7=3.5xx10^(-3)W` `=3.5mW` Here, `R=1kOmega=1xx10^(3)Omega`, `I_(min)=1mA=10^(-3)A` `V_(B)-0.7= RI_(min) or V_(B)=0.7+RI_(min)` `:. V_(B)=0.7+10^(3)xx10^(-3)=1.7V` |
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| 38258. |
In the Geiger-Marsden scattering experiment find the distance of closest approach to the nuclcus of a 7.7 Mev alpha-particle before it comes momentaily to rest and reverses its direction. (Z for gold nucleus = 79 ) |
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Answer» Solution :The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an a-particle and a gold nucleus is conserved. The system’s initial mechanical energy is `E_i`, before the particle and nucleus interact, and it is equal to its mechanical energy `E_f` when the a-particle momentarily STOPS. The initial energy `E_i` is just the KINETIC energy K of the incoming a- particle. The final energy `E_f` is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. Let d be the centre-to- centre distance between the a-particle and the gold nucleus when the a- article is at its stopping point. Then we can WRITE the conservation of energy `E_i = E_f` as `K = 1/(4 pi epsilon_0) ((2e)(Ze))/(d) = (2 Ze^2)/(4 pi epsilon_0 d)` Thus the distance of closest approach d is given by `d = (2 Z e^2)/(4 pi epsilon_0 K)` The miximum kinetic energy found in `alpha`- PARTICLES of natural origin is `7.7 MeV` or `1.2 xx 10^(-12) J`. Since `1//4 pi epsilon_0 = 9.0 xx 10^(9) N m^2//C^2`. Therefore with `e = 1.6 xx 10^(-19) C`, we have `d = ((2)(9.0 xx 10^(9) Nm^2//C^2)(1.6 xx 10^(-10)C)^2 Z)/(1.2 xx 10^(-12) J)` `= 3.84 xx 10^(-16) Z m`. The atomic number of foil material gold is `Z = 79`, so that `d (Au) = 3.0 xx 10^(-14) m = 30 fm`. (1 fm (i.e. fermi) = `10^(-15) m`.) The radius of gold nucleus is, therefore, less than `3.0 × 10^(-14)` m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the `alpha`-particle reverses its motion without ever ACTUALLY touching the gold nucleus. |
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| 38259. |
The solution of equations x-y=2 and x+y=4 is:- |
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Answer» 3 and 5 |
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| 38260. |
A gas rotates in a centrifuge. Taking into account that the field of centrifugal forces of inertia is equivalent to a gravitational field, write the expression for the barometric distribution in the centrifuge. |
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Answer» `U = -W = -1/2 FL = -1/2I_(ct) t = -1/2 m omega^2 r^2`. Substituting it into FORMULA (26.26), we obtain the result sought. |
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| 38261. |
A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases its speed. The angular acceleration of particle at time (t) is proportional |
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Answer» `(1)/(sqrtt)` |
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| 38262. |
A small spherical body of radius r and density rhomoves with the terminal velocity v in a fluid of coefficient of viscosity etaand density sigma . What will be the net force on the body? |
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Answer» `(4pi )/(3) r^3 (rho - sigma) g ` Now, when the spherical body moves DOWNWARDS in the fluid medium, with terminal velocity (say, `V_("terminal")`) then, sum of the viscous drag and upthrust will be EQUAL to the weight of the spherical body. ` therefore F_v + F_T -W` ` rArr 6pi eta r v_("terminal") + 4/3 pi r^3 sigma g = 4/3 pi r^3 rho g ` ` rArr 6pi eta r v_("terminal") = 4/3 pi r^3 (rho - sigma) g ` ` rArr v_("terminal") = (2r^2 (rho - sigma))/(9 eta) g ` `rArr F_("net") = 0` |
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| 38263. |
Solve the above problem if the planes of coils make theta angle with each other. |
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Answer» SOLUTION :If i current flows in the LARGER coil, magnetic field PRODUCED at the centre will be perpendicular to the plane of larger coil. Now the area VECTOR of smaller coil which is perpendicular to the plane of smaller coil will make an angle `theta` with the magnetic field. Thus flux `=vecB,.vecA=(mu_(0)i)/(2a_(2)).pia_(1)^(2)costheta` or `M=(mu_(0)pia_(1)^(2)costheta_(1))/(2a_(2))` |
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| 38264. |
An electron is accelerated through a potential difference of 10,000 V.Its de-Broglie wavelength is ,(nearly)(m_(e)=9xx10^(-31)kg) |
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Answer» 12.2 m `THEREFORE lambda=(6.6xx10^(-34))/(sqrt(2xx9)xx10^(-31)xx1.6xx10^(-19)xx10^(4))` `=(6.6xx10^(-34+23))/(sqrt(28.8))=(6.6xx10^(-11))/(5.366)` `=1.2299xx10^(-11)` `therefore lambda~~12.2xx10^(-12m)` |
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| 38265. |
Considerthe equationT=2pi sqrt((l)/(g)) and check wheter it is correct or not . |
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Answer» SOLUTION :We know dimensional formulae of T `L` and g are [T], [L], `[LT^(-2)]` respectively . `:. ` Dimensional formula of `sqrt((l)/(g))=sqrt((L)/(LT^(-2)))=sqrt(T^(2))=|T|` Dimensional formula of left side is also [T]. As 2t is a dimensionless quantity, the dimensional formula of lefi hand side of the equation is EQUAL to dimensional formula of right hand side. HENCE we conclude that thegiven equation is a correct ONE. |
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| 38266. |
In radial magnetic fieldlines of force are |
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Answer» PARALLEL |
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| 38267. |
Two sinusoidal waves with the same amplitude and wavelength travel along a string that is stretched along an x axis, the resultant wave due to their superposition was recorded on video tape. The curves in Fig. 16-31 represent the resultant wave in four freeze frames in the sequence of a, b, c, and d, with 1.0 ms elapsing between curves a and d. The grid lines along the x axis are 1.0 cm apart, and the string elements oscillated perpendicular to the r axis by 16.0 mm (between the extreme displacements shown by curves a and d) as the resultant wave passed through them. Write equations for the two superposition waves and for their resultant wave. |
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Answer» Solution :FIGURE 16-31, shows a standing wave and so the two waves must be travelling in opposite directions. The two waves have the forms given by `y_(1)(x,t) =y_(m) sin (kx-omegat)` and `y_(2) (x,t) =y_(m) sin (kx+omega t)` Hence, we can write the RESULTANT wave as `y. (x,t)=(2y_(m), sin kx) cos omega t` We need to find the common values of `y_m, k,` and a in the three equations. Calculations: To find `y_m` we use the key idea that the amplitude of a standing wave is a maximum at an antinode, and there the amplitude is `2y_m.` Here the maximum amplitude of the standing wave is 8.0 mm, so `y_(m)=4.0mm` To find the angular wave number k, we use two more key ideas. First, the common angular wave number k in Eqs. 16-71, 16-72 and 16-73 is related to the common wavelength `lambda` by equation `k= 2pi//lambda.` Second, the nodes in a standing wave arc separated by 0.502. Because the antinodes in Fig. 16-31 are separated by 2.0 cm, we have `lambda=2 xx 2.0 cm =4.0cm` .Then `k=(2pi)/(lambda) =(2pi)/(0.040 m)=157 m^(-1) APPROX 160m^(-1)` To find the common angular frequency `omega` we again use two more key ideas: We can relate the common value of `omega` in Eqs. 16-71, 16-72, and 16-73 to the period T of the standing wave with equation o = 24/7. A string element at an antinode TAKES time T to move through a full oscillation from one extreme and back to the first extreme. According to the given data, the string element at the dot on curve a in Fig. 16-31 takes 10 ms to move to the POSITION of the dot on curve d, which is one half a full oscilations. Thus, T=2.0 ms, and `omega=(2pi)/(T)=(2pi)/(0.0020 s)=3142 s^(-1) approx 3100 s^(-1)` Now we can write Eqs. 16-71 and 16-72 for the interfering waves as `y_(1) (x,t) =(4.0 mm) sin (160x-3100t)` `and y_(2) (x,t)=(4.0mm) sin (160x-3100t)` with x in meters and t in seconds. We can also write the standing wave of Eq. 16-73 as |
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| 38268. |
Electric current through a conductor varies with time as I(t)=50sin(100pi t). Here I is in amperes and t in seconds. Total charge that passes any point from t=0 tot= (1)/(200)s is |
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Answer» 1.2 C |
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| 38269. |
In single slit diffraction a= 0.14mm, D= 2m and distance of second dark band from central maxima is 1.6 cm. The wavelength of light is |
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Answer» `6500 A^(0)` |
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| 38270. |
(a) Write the expression for the speed of light in a material medium of relative permittivity & and relative magnetic permeability mu_(r). (b) Write the wavelength range and name of the electromagnetic waves which are used in (i) radar systems for aircraft navigation, and (a) earth satellites to observe the growth of the crops. |
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Answer» Solution :(a) Speed of LIGHT in a material medium of relative permittivity E, and relative magnetic permeability `mu,` is given as `v=(1)/sqrt(in_(0) in_(r) mu_(0) mu_(r))=e/sqrt(in_(r) mu_(r))` (b) (i) Electromagnetic waves used in radar systems are microwaves whose wavelength varies from 1 MM to 10 cm. (ii) Infrared waves are used in remote sensing EARTH satellites and their wavelength range is from 700 nm to 1 mm. |
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| 38271. |
(a) Assume that the light of wavelength 6000 Å is coming from a star. Find the limit of resolution of a telescope whose objective has a diameter of 250 cm. (b) Two slits are made 1 mm apart and the screen is placed 1m away. What should be the width of each slit to obtain 10 maxima of the double-slit patterns within the central maximum of the single-slit pattern? |
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Answer» Solution :(a) ` lambda = 6000 Å` `D = 250 cm = 250 xx 10^(-2) m` Limit of resolution = `(1.22 lambda)/D = (1.22 xx 6000 xx 10^(-10))/(250 xx 10^(-2))` `= 29.28 xx 10^(-8) m`. (B) `d = 1 MM = 1 xx 10^(-8) m` `D = 1m` Width of 10 MAXIMA of double -slit pattern `= (10 lambda D)/(d)` The width of central maxima of single-slit pattern `=(2 lambda D)/a` a is aperture of slit `:. (10 lambda D)/d = (2 lambda D)/(a) a = (2 lambda d)/(10 lambda D)` `a = (2D)/(10 D)` `= (2 xx 10^(-3))/(10)` `= 2 xx 10^(-4) m`. |
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| 38272. |
A : No interference pattern is detected when two coherent sources are very close to each other. (i.e separation almost zero) R : The fringe width of interference pattern is inversely proportional to the distance between the two slits. |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION of A |
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| 38273. |
The most common kind of iron nucleus has a mass number of 56. Find the approximate density of the nucleus. |
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Answer» `2.29xx10^16 "KG m"^(-3)` |
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| 38274. |
Projection of U.C.M, along any diameter is |
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Answer» LINEAR S.H.M. |
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| 38275. |
….. NAND gate is required for OR gate. |
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Answer» 1 |
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| 38276. |
A ray of light passes through an equilateral prism withmu=1.5. The angle of deviation is |
| Answer» Answer :D | |
| 38277. |
A deuteron and an alpha particl are accelerated with the same potential. Which one of the two hasgreater value of de Broglie wavelength associated with it and i |
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Answer» Solution :(i) Using de-Broglie wavelength formula, the dueteron and alpha particle are accelerated with same potential. So, boh their velocities are same. `lambda = (h)/(sqrt(2 mV_(0) Q)), lambda prop (1)/(sqrt(mq))` So, `lambda_(due) prop (1)/(sqrt(2m_(d) q_(d))) and lambda_(alpha) prop (1)/(sqrt(8m_(alpha)q_(alpha)))` `(lambda_(alpha))/(lambda_(d)) = (sqrt(2m_(d)q_(q)))/(sqrt(8m_(alpha) q_(alpha))) = sqrt((2)/(8)) = sqrt((1)/(4)) = (1)/(2)` `lambda_(d) = 2 lambda_(alpha)` (ii) For same potential of ACCELERATION, KE is directly proportional to the .q. Charge of duetron is +e Charge of alpha is +2e So, `K_(d) = (K_(alpha))/(2)` Charge of alpha particle is more than the duetron. |
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| 38278. |
Huygen's wave theory of light could not explain |
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Answer» reflection |
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| 38279. |
The frequency of an alternating voltage is 50 cycles/s and its amplitude is 120 V. Then what is the rms value of voltage ? |
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Answer» 101.3 V `THEREFORE V_(rms) APPROX` 84.9 V |
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| 38280. |
It is safe to be inside a vehicle rather than outside, even when there is lightning and thunder. Comment on this. |
| Answer» SOLUTION :Inside a SPHERICAL SHELL electrical field is zero. This is called electrostatic shielding. HENCE it is SAFE to be inside a vehical rather than outside | |
| 38281. |
Mention the three types of electron emission. |
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Answer» SOLUTION : Depending UPON the sources of EXTERNAL energy, electron EMISSION is classified intofour groups. 1. Thermionic emission 2. Photoelectric emission 3. FIELD emission 4. Secondary emission. |
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| 38282. |
Plane-polarized light of wavelength 589nm propagates along the axis of a concentration 500 g//l. Viewing from the side, one can see a system of helical fringes, with 50 cm between neighbouring dark fringes along the axis. Explain the emergence of the fringes and determine the specific rotation consatnt of the solution. |
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Answer» Solution :Two effects are involved here : rotation of PLANE of POLARIZATION by sugar solution and the effect of that rotation on the scattering of light in the transverse DIRECTION. The latter is shown in the figure given below. It is easy to see from the figure that there will no scattering of light in this transverse direction if the incident light has its electric vector parallel to the line of sight. In such a situation, we expect fringes to occur in the given experiment. From the given data see that in a disatnce of `150 cm`, the rotation of plane of polarization must be `180^(@)`. Thus the specific rotation constant of suger `= ("rotation constant")/("CONCENTRATION")` `= (180//50)/(500(g)/(l))ANG//deg//cm = (180)/(5.0dm xx (.500 gm//c c))` `= 72^(@) ang deg// (dm gm//c c) (1dm = 10 cm)` 
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| 38283. |
The V - I character for a p - n junction diode is plotted as shown in the figure. From the plot, we can conclude that [V_(b) to breakdown voltage V_(k) to knee voltage ] |
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Answer» the FORWARD BIAS RESISTANCE of diode is very high almost infinity for small values of V and after a certain VALUE it become very low |
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| 38284. |
Consider two identical inicroscope slides in air illumi- nated with monochromatic light. The bottom slide is rotated (counterclockwise about the point of contact in the side view) so that the wedge angle gets a bit smaller. What happens to the fringes? |
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Answer» They are SPACED FARTHER apart |
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| 38285. |
The wavefront coming from a point source of light is a ______wavefront. |
| Answer» SOLUTION :SPHERICAL | |
| 38286. |
There is a point object at the center of a glass sphere of diameter 12cm and refractive index 1.5 . The distance of the virtual from the surface of the sphere is |
| Answer» ANSWER :B | |
| 38287. |
If n(A) = 3 and n(B) = 2 thennumber of subsets (AxxB) has- |
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Answer» 0 |
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| 38288. |
The number of capital latters such as A,B,C,D"….." which are not latterally inverted by a plane mirror ? |
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Answer» 6 |
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| 38289. |
Find the number of half lives elapsed , before which , 93.75% of a radioactive sample has decayed . |
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Answer» Solution :Fraction of ATOMS left undecayed is 100 - 93.75% i.e `N/N_(0)=(6.25)/100 = 1/(16)` or , `N/N_(0) = (1/2)^(4) ` or `N = N_(0) (1/2)^(4)` Comparing with `N = N_(0) (1/2)^(n) , ` yields n = 4 . |
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| 38290. |
The magnetic flux through a circuit of resistance R changes by an amount Delta phi in a time Delta t. Then the total quantity of electric charge Q that passes any point in the circuit during the time Delta t is represented by |
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Answer» `Q=(1)/(R ).(DELTA phi)/(Delta t)` `rArr` current `(i)=(Q)/(Delta t)=(Delta phi)/(Delta t)xx(1)/(R ) ""` [where Q is total charge in time `Delta t`] `rArr Q=(Delta phi)/(R )` |
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| 38291. |
A rock is dropped from an 80- meter cliff. How long does it take to reach the ground ? |
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Answer» Solution :SINCE the rock's displacement is down, we call down the positive direction, so a = +G. We're GIVEN `v_(0) s`, and a , and asked for t. So v is missing, it isn't given and it isn't asked for, and we use Big FIVE #2. `Delta s= v_(0) t+(1)/(2) at^(2) implies Delta s = (1)/(2) a t^(2) (" since" v_(0)=0)` `t= sqrt((2Delta s)/(a))` `sqrt((2Delta s)/(+g))=sqrt((2(+80m))/(+10 m//s^(2)))=4.0s` |
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| 38292. |
A particle of charge 'q' and mass 'm' moves in a circular orbit of radius 'r' with angular speed 'omega '. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on |
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Answer» `OMEGA ` and q |
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| 38293. |
A circuit using a potentiometer and battery of negligible internal resistance is set up as shown in Fig. to develop a constant potential gradient along the wire AB. Two cells of emfs epsi_1 and epsi_2are connected in series as shown in combinations (1) and (2). The balance points are obtained respectively at 400 cm and 240 cm from the point A. Find(i) epsi_1/epsi_2 , (ii) balancing length for the cell epsi_1 only . |
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Answer» Solution : If k be the potential gradient along the potentiometer wire, then as per data given we have : `epsi_1 + epsi_2 = k xx 400 CM `...(i) and `epsi_2- epsi_1= k xx 250 cm`...(ii) `rArr epsi_1 = 75k`and`epsi_2 = 325 k` ` therefore epsi_1/epsi_2 = (75k)/(325k) = 3/13` If BALANCING length for the cell `epsi_1`only be L cm, then `epsi_1= lk `. but `epsi_1 = 75k rArr l = 75 cm ` |
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| 38294. |
A system of capacitors , connected as shown , has a total energy of 160 mJ stored in it. Obtain the value of the equivalent capacitance of this system and the value of x . |
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Answer» Solution :In the arrangement of capacitors shown , capacitors of capacitances `C_2` and `C_3` are JOINED in parallel and their equivalent capacitance `C_(23) = 7 + 3 = 10 mu F ` Now `C_(1) , C_(23)` and `C_(4)` are joined in series , hence their equivalent capacitance C. is given by `(1)/(C.) = (1)/(C_(1)) + (1)/(C_(23)) + (1)/(C_(4)) = (1)/(10) + (1)/(10) + (1)/(15) = (8)/(30)` `IMPLIES C. = (30)/(8) mu F = 3.75 mu F ` As `C_(5)` is in parallel to C. , hence equivalent capacitance `C_(eq)` of the arrangement is `C_(eq) = C. + z = (3.75 + z) mu F ` `therefore` Energy of the system `U = (1)/(2) C_(eq) , V^(2)` , where U = 160 mJ = 0.16 J and V = 200 V `therefore 0.16 = (1)/(2) XX C_(eq) (200)^(2) implies C_(eq) = (0.16 xx 2)/((200)^(2)) = 8 xx 10^(-6) F = 8 mu F ` As `C_(eq) = (3.75 + z) = mu F = 8 mu F ` , hence we have z = `4.25 mu F ` |
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| 38295. |
A uniform ring of mass M and radius R carries a current I (see figure). The ring is suspended using two identical strings OA and OB. There exists a uniform horizontal magnetic field B_(0) parallel to the diameter AB of the ring. Calculate tension in the two strings. [Given theta = 60^(@)] |
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Answer» `T_(BO)=Mg-piIRB_(0)` |
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| 38296. |
A steady current is set up in a metalic wire of non- uniform cross-section. How is the rate of flow free electrons related |
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Answer» `R PROP A^(-1)` |
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| 38298. |
The amplitude of the sinusoidally oscillating electric field of a plane wave is 60 V/m. Then the amplitude of magnetic field is |
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Answer» `2 XX 10^7 T` |
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| 38299. |
How capacitive reactance depends on the frequency of AC ? |
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Answer» SOLUTION :It is INVERSELY PROPORTIONAL to the frequency of AC ` X_(C) = (1)/(omega C)` |
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| 38300. |
A voltmeter of variable ranges 3V, 15V, 150 V is to designed by connecting resistances R_(1), R_(2), R_(3) in series with a galvanometer of resistance G=20Omega, as shown in the figure. The galvanometer gives full pass through its coil. Then, the resistances R_(1), R_(2) and R_(3) (in kilo ohms) should be, respectively |
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Answer» 3, 12, 135 |
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