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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.
| 7051. |
Using Bohr's formula for energy quantization determine the longest wavelength in Lyman series. |
| Answer» SOLUTION :As `1/x = R(1/n_f - 1/n_c)`, for LYMAN SERIES , `1/x= R(1/1_2 - 1/2_2), LAMBDA = 1216 A^@`. | |
| 7052. |
Does the law of conservation of energy hold good at the point of destructive interference? |
| Answer» SOLUTION :YES, it does. | |
| 7053. |
Optical density of a transparent medium is a measure of ? |
| Answer» SOLUTION :REFRACTIVE INDEX | |
| 7054. |
A well collimated parallel pencil of cathode rays falls through a potential difference 3kV & enters the spacing between two parallel metallic plates, parallel to their length the spacing between the plates being 0.5cm. The pencil strikes a fluorescent screen, mounted perpendicular to the length of the plates at the farther end of the plates & produces fluorescent spot. if now a potential difference of 3V is applied across the two plates, calculate the linear deflection of the spot on the screen. Given the length of the plates is 10 cm. |
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| 7055. |
A hollow cylinder of specific resistance rho, inner radius R, outer radius 2R and length l is as shown in figure. What is the net resistance between the inner and outer surface? |
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Answer» Solution :Consider a ring of width .dr. and RADIUS .r.. RESISTANCE across the ring is `dR=(rhodr)/(dA)=(rhodr)/(2PI rl)`  Net resistance `=int_(R)^(2R)(rho(dr))/((2pirl))=((rho)/(2pil))ln(2)` |
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| 7056. |
(i) Use: (i) the Ampere's law for H and (ii) continuity of lines of B, to conclude that inside a bar magnet, (a) lines of overset(to)(H) run from the N pole to S pole, while b) lines of overset(to)(B) must run from the S pole to N pole. |
Answer» Solution :Let us consider a magnetic field line of `overset(to)(B)` through the BAR magnet as given in the figure below.  It must be a closed loop as SHOWN in figure. Let C be the amperian loop, then `int_(Q)^(P) overset(to)(H). overset(to(d) l = int_(Q)^(P) ( overset(to)(B) )/(mu_(0) ). overset(to)(d) l ""[because B= mu_(0) H]` The angle between `overset(to)(B) and overset(to)(d) l` is less than `90^@` inside the bar magnet, so it is positive hence `cos theta GT 1` `int_(Q)^(P) overset(to)(H). overset(to)(d) l = int_(Q)^(P) ( overset(to)(B))/(mu_0) . overset(to)(d) l lt 0` Hence, the line of `overset(to)(B)` must run from south pole (S) to north pole (N) inside the bar magnet. According to Ampere.s law, `oint_("PQP") overset(to)(H). overset(to)(d) l = 0` `oint_("PQP") overset(to)(H). overset(to)(d)l = int_(P)^(Q) overset(to)(H). overset(to)(d)l + int_(Q)^(P) overset(to)(H). overset(to)(d)l=0` but `int_(Q)^(P) overset(to)(H). overset(to)(d) LGT 0 ""` (outside magnet) Hence `int_(P)^(Q) overset(to)(H). overset(to)(d) l lt 0""` (Inside magnet) If angle between `overset(to)(H) and overset(to)(d) l` is more than `90^@`, sothat `cos theta` is negative. It means the line of `overset(to)(H)` must run from N pole to S pole inside the bar magnet. |
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| 7057. |
For an inductor having self inductance 4.6 H, emf induced in it for time interval from 5 ms to 6 ms is …… |
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Answer» ZERO `=(0-5)/(1xx10^(-3))=-5xx10^3 A/s` Now , self induced emf is, `epsilon=-L (dI)/(dt)` =-(4.6)`(-5xx10^3)` `THEREFORE epsilon=23xx10^3` V |
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| 7058. |
Name the electromagnetic radiations used for (a) water purification, and (b) LASIC eye surgery. |
| Answer» Solution :Ultraviolet RAYS are USED for water PURIFIERS as WELL as in LASIC eye SURGERY. | |
| 7059. |
A generator at one end of a very long string creates a wave given by y=(6.0 cm) cos ""pi/2 [(2.00 m^(-1)) x+(6.00 s^(-1))t], and a generator at the other end creates the wave y=(6.0 cm) cos""pi/2 [(2.00 m^(-1)) x-(6.00 s^(-1)) t] . Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For x ge 0, what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of x? For x ge , what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of x? |
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| 7060. |
The focal length of the objective of an astronomical telescope is 75 cm and that of the eyepiece is 5 cm . If the final image is formed at the least distance of distanct of distinct vision from the eye, calculate the magnifying power of the telescope. |
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Answer» SOLUTION :DATA supplied , `f_(0) = 75 cm , "" f_(e) = 25 cm , "" D = 25 ` cm m = `- (f_(0))/(f_(e)) ( 1 + (f_(e))/(D) ) = (-75)/(5) (1 + (5)/(25)) = -15 xx 1.2 = - 18` |
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| 7061. |
The displacemnet of particle executing simple harmonic motion is given by y=5/pi sin(20pi t)m, then what is its amplitude ? |
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Answer» Solution :`y=a SIN omegat _____(i)` `y=5/pi sin(20pit) ____(ii)` by COMPARING EQUATION (ii) with standard EQUTION(i) we get `a=5/pi` |
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| 7062. |
The potential in certain region is given as V = 2x^(2), then the charge density of that region is |
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Answer» `-(4x)/(epsilon_(0))` |
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| 7063. |
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on atall that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration veca and the normal force vecF_(N) on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride? (c) How does the magnitude of the acceleration at the highest point compare with that at the lowest point? (d) How do the magnitude of the normal force compare at those two points? |
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| 7064. |
Obtain the equation of power for combination of lenses. |
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Answer» Solution :As power of lens is `P = 1/f`, equivalent power of COMBINATION, `P = P_1+ P_2 + P_3 + ...` Note that it is an algebraic sum of individual powers. So sum of the terms on the right side may be positive for convex lenses and negative for concave lenses. Combination of lenses helps to obtain DIVERGING or converging lenses of desired magnification. It also enhances sharpness of the image. Combination of lenses is used in camera, MICROSCOPE, telescope and other optical instruments. |
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| 7065. |
A stone is thrown upwards with a speed u from the top of a tower reaches the ground with a velocity 3 u . The height of the tower is |
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Answer» `(3a^2)/G` |
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| 7066. |
A point charge causes an electric field of -1.0xx10^3Nm^2//C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge The charge inside of surface. |
| Answer» Solution :By Gauss’s THEOREM Total ELECTRIC FLUX `=phi=q/epsilon_0q=epsilon_0xxphi=8.85xx10^-12xx(-1.0xx10^3)=-8.85xx10^-9C.` | |
| 7067. |
A current carrying wire is placed below a coil in its plane, with current flowing as shown. If the current increases. |
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Answer» No current will be induced in the COIL |
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| 7068. |
Which of the following graph represents uniform motion . |
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Answer»
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| 7069. |
The figure shown is part of the circuit at steady state The charge on 2mu F capacitor is : |
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Answer» SOLUTION :`V_P=x` `(3-x)/1=(x-4)/2+(x+10)/5` SOLVE `q=2xx4=8muc` |
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| 7070. |
Two longparallelwiresof negligibleresistance are connectedat one endto a reasitanceR and at the otherend to adcvoltagesource. The distancebetweenthe axesof the wires is eta = 20 timesgreaterthan the cross-sectionalradiusof each wire. At whatvalueof resistanceR doesthe resultantforceof interaction betweenthe wiresturn into zero ? |
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Answer» Solution :There are EXCESS surface chargeson eachwire (irrespectiveof whether the current is flowingthroughthen or net). Hencein addtiionto the magneticforce `vec(F)_(m)`, we must takeinto accountthe ELECTRICFORCE `vec(F)_(e)`. Suppose that an excesschagre `lambda` correspondsto a unitlength of the wire, then electricforceexertedper unitlength of the wireby otherwire can befound with the help of Gauss's theorem. `F_(e) = lambda E = lambda (1)/(4pi epsilon_(0)) (2 lambda)/(l) = (2 lambda^(2))/(4pi epsilon_(0) l)` ....(1) where `l` is the distance between the area of the wires. The magentic forceacting per unitlength of the wirecan be found with the helpof the theroem on circularion of vector `vec(B)` `F_(m) = (mu_(0))/(4pi) (2i^(2))/(l)`, where `i` is the current in the wire. ..(2) Now, from the relation, `lambda = C varphi`, where `C` is the capacitanceof the wires per unit lengthsand is given in problem3.108 and `varphi = iR` `lambda = (pi epsilon_(0))'/(ln ETA) iR` or, `(i)/(lambda) (ln eta)/(pi epsilon_(0)R)` .....(3) Dividing(2) by (1) and then substuting the valuesof `(i)/(lambda)` from (3) we get, `(F_(m))/(F_(e)) = (mu_(0))/(epsilon_(0)) ((ln eta)^(2))/(pi^(2) R^(2))` The resultantforceof interaction vanisheswhen this ratiounity. This is possible when `R = R_(0)`, where `R_(0) = SQRT((mu_(0))/(epsilon_(0))) (ln eta)/(pi) = 0.36 k OMEGA` 
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| 7071. |
Calculate the length of a nichrome spiral for an electric hot plate capable of healing 21 of water to the boiling point in 8 min. The initial temperature of water is 20^@C, the efficiency is 60%, the diameter of the wire is 0.8 mm, the voltage is 220 V. the resistivity of nichrome is 10^(-4) ohm-m. Neglect the heat required to heal the kettle. |
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| 7072. |
The time varying electric and magnetic fields in space |
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Answer» produce EM wave which is PROPAGATED with a velocity LESS than velocity of light |
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| 7073. |
Prove that in similar conditions a cylindrical container will withstand a pressure half as great. |
Answer» Solution :Imagine a small area along a generatrix to be cut out of the cylindrical surface (Fig.). Four elastic FORCES are seen to act on the area. Two of them are parallel to the generatrix, their normal components are zero and so they offer no resistance to the gas pressure and should not be taken into account. The remaining two are PERPENDICULAR to the GENERATRICES, the sum of their normal components is the normal force `T_n = 2 sigma ld sin alpha`. The force of pressure is `F = pS = 2pal - 2plR sin alpha` .  It follows from the balance of the forces that `p = sigma d//R`which was to be proved. |
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| 7074. |
Water is flowing through a horizontal pipe of non -uniform cross section. The speed of water is 30 cm/s at a place where pressure is 10 cm (of water). Calculate the speed of water at the other place where the pressure is half of that of the first place |
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Answer» 100.4 cm/s |
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| 7075. |
The TV transmission tower at a particular place has a height of 160 m. What is the coverage range? By how much should the height be increased to double it's coverage range? Given that the radius of earth=6400 km. |
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Answer» SOLUTION :d =`sqrt(2Rh)`, Given R = 6400 KM `= 6400xx10^3m, H = 160 m` `d=sqrt(2xx6400xx10^3xx160` `thereforeh_2/h_1=d_2^2/d_1^2=(2)^2=4` or `h_2=4h_1=4xx160=640m` |
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| 7076. |
A particle is acted upon by a constant force always normal to the direction of motion of the particle. It is therefore inferred that- (a) Its velocity is constant (b) It moves in a straight line (c) Its speed is constant (d) It moves in circular path |
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Answer» a, d |
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| 7077. |
In the initial state of the circuit shown in the figure, the capacitors are not charged. At the time t=0 key is closed. Find the charge accumulated on the capacitor if R=10kOmega,C=10muF, and battery internal resistance r=10Omega. Volt-ampere characteristic of a diode is show in the chart, with U_(0)=0.7V |
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| 7078. |
In the case of diffraction bands due to a straight edge, as we move away from the edge |
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Answer» INTENSITY of the BRIGHT BAND INCREASES and that of the dark band DECREASES |
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| 7079. |
Linear simple harmonic motion. |
| Answer» Solution :A particleis said to perform linear simple harmonic MOTION (SHM) if it OSCILLATED about a point of stable equilibrium , subject to a FORCE always DIRECTED towards that point and WHOSE magnitude is proportional to the displacement of the particlefrom the point . | |
| 7080. |
Three resistors 1Omega , 2Omega " and " 3Omegaare combined in series. What is the total resistance of thecombination ?(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance,obtain the potential drop across each resistor. |
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Answer» Solution :(a) Here `R_1 = 1Omega , R_2 = 2 Omega` and `R_3 = 3Omega ` TOTAL resistance in series combination`R = R_1 + R_2 + R_3 = 1 + 2 + 3 = 6Omega` (b) As `epsi = 12 V, r = 0 `, hence current `I = (epsi)/(R+r ) = (12)/(6+0) = 12/6= 2.0 A` POTENTIAL drop across the three RESISTORS are respectively `V_1 = IR_1 = 2 xx 1 = 2V, V_2 = IR_2 = 2 xx 2 = 4V " and " V_3 = IR_3 = 2 xx 3 = 6V` |
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| 7081. |
Figure shows a point object placed in front of a transparent sphere of radius 20 cm and refractive index 1.5. The sphere forms an image due to refraction at surface I followed by refraction at surface II. The sphere is kept in air and the object is located at distance x from P_(1). It is found that for x gt x_(1), the image formed due to refraction at surface I is real, the final image formed by the sphere is also real and forms to the right of P_(2). For x_(2) lt x lt x_(1), the image formed by surface I is virtual while the final image formed by the sphere is real and forms to the right of P_(2). However for x lt x_(2), the image formed by surface I and also the final image formed by the sphere are virtual. It is also found that the final image is virtual if the distance of image formed by surface I from P_(2) is less than y. The object is now moved away from the sphere so that x becomes large and say, x to oo. In the given sphere is replaced by another one of the same radius but of refractive index 2.5, the final image is formed at distance y_(2) from P_(2). Distance x_(1) such that for x gt x_(1), the image formed by surface I and also the final image are real, is |
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Answer» 60 cm |
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| 7082. |
The focal lengths of the objective and eye lenses of a telescope are respectively 200 cm and 5 cm. The maximum magnifying power of the telescope will be : |
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Answer» -40 |
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| 7083. |
Figure shows a point object placed in front of a transparent sphere of radius 20 cm and refractive index 1.5. The sphere forms an image due to refraction at surface I followed by refraction at surface II. The sphere is kept in air and the object is located at distance x from P_(1). It is found that for x gt x_(1), the image formed due to refraction at surface I is real, the final image formed by the sphere is also real and forms to the right of P_(2). For x_(2) lt x lt x_(1), the image formed by surface I is virtual while the final image formed by the sphere is real and forms to the right of P_(2). However for x lt x_(2), the image formed by surface I and also the final image formed by the sphere are virtual. It is also found that the final image is virtual if the distance of image formed by surface I from P_(2) is less than y. The object is now moved away from the sphere so that x becomes large and say, x to oo. In the given sphere is replaced by another one of the same radius but of refractive index 2.5, the final image is formed at distance y_(2) from P_(2). Distance x_(2) such that for x_(2) lt x lt_(1), the image formed by surface I is virtual but the final image is virtual. Then, y will be |
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Answer» 10 cm |
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| 7084. |
(a) Distinguish betweenunpolarized light and linearly polarizedlight. How does one getlinearly polarisedlight with thehelp of a plaroid ? (b) A narrowbeam of unpolarised light of intensityI_(0) is incidenton a polaroid P_(1).The light transmittedby it is then incidentona secondpolaroidP_(2) withits pass axismakingangleof 60^(@)relative to the pass axis of P_(1). Find the intensityof the lighttransmittedby P_(2). |
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Answer» Solution :Distinction between unpolarised and linearly polarized light obtainigng linearly polarized Light 1 In an unpolarised th eoscillations, of the electric field, arc in random directions, in planes perpendicular to the direction of propagation. For a polarized light, the oscillations are aligned along one particular direction. Alternatively. Polarized light can be DISTINGUISHED, form unpolarised light when it is allowed to pass through a POLAROID. Polaroid light does can sho CHANGE in its intensity remains same in CASE of unpolarised light. When unpolarised light WAVE is incident on a polaroid, then the electric vectors along teh direction of its aligned molecules, get absorbed , the electric vector, oscillation alogn a direction perpendicular to thealigned molecules, pass thorugh. This light is called linearly polarized light. |
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| 7085. |
Figure shows a point object placed in front of a transparent sphere of radius 20 cm and refractive index 1.5. The sphere forms an image due to refraction at surface I followed by refraction at surface II. The sphere is kept in air and the object is located at distance x from P_(1). It is found that for x gt x_(1), the image formed due to refraction at surface I is real, the final image formed by the sphere is also real and forms to the right of P_(2). For x_(2) lt x lt x_(1), the image formed by surface I is virtual while the final image formed by the sphere is real and forms to the right of P_(2). However for x lt x_(2), the image formed by surface I and also the final image formed by the sphere are virtual. It is also found that the final image is virtual if the distance of image formed by surface I from P_(2) is less than y. The object is now moved away from the sphere so that x becomes large and say, x to oo. In the given sphere is replaced by another one of the same radius but of refractive index 2.5, the final image is formed at distance y_(2) from P_(2). As stated above, if the distance of image formed by Surface 1 from P_(2) is less than y, the final image is virtual. Then, y will be |
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Answer» 40 cm |
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| 7086. |
Define half life of a radiocative sample. |
| Answer» Solution :Time during which the number of radioactive atoms will reduce to HALF the ORIGINAL number is known as half - LIFE of radioactive element. | |
| 7087. |
Figure shows a point object placed in front of a transparent sphere of radius 20 cm and refractive index 1.5. The sphere forms an image due to refraction at surface I followed by refraction at surface II. The sphere is kept in air and the object is located at distance x from P_(1). It is found that for x gt x_(1), the image formed due to refraction at surface I is real, the final image formed by the sphere is also real and forms to the right of P_(2). For x_(2) lt x lt x_(1), the image formed by surface I is virtual while the final image formed by the sphere is real and forms to the right of P_(2). However for x lt x_(2), the image formed by surface I and also the final image formed by the sphere are virtual. It is also found that the final image is virtual if the distance of image formed by surface I from P_(2) is less than y. The object is now moved away from the sphere so that x becomes large and say, x to oo. In the given sphere is replaced by another one of the same radius but of refractive index 2.5, the final image is formed at distance y_(2) from P_(2). For the object placed at x to oo, the given sphere of refractive index 1.5 forms an image at distance y_(1) from P_(2). |
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Answer» this image is real and `y_(1)=12.5 cm " RIGHT of " P_(2)` |
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| 7088. |
Two point - like chargesQ_(1) and Q_(2) are positioned at point 1 and 2 respectively. The field intensity to the right of the charge Q_(3) on the line that passes through the two charges varies according to a law that is represented schematically in the figure. The field intensity is assumed to be positive if its direction with the positive direction on the x-axis. The distance between the charges is l, then which of the folllowing is true |
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Answer» `Q_(1)` is positive charge while `Q_(2)` is negative charge |
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| 7089. |
The slope of the graph drawn between stopping potential and frequency of incident radiation will be |
| Answer» ANSWER :A | |
| 7090. |
Show that the superposition of the waves originating from two coherent sources s_(1) and s_(2) having displacement y_(1)=acosomegat and y_(2)=acos(omegat+phi) at a point produce a result intensity I_(R)=4a^(2)"cos"^(2)(phi)/(2). Hence, write the conditions for the appearance of dark and bright fringes. |
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Answer» Solution :If two waves are being REPRESENTED as: `y_(1)=acosomegat and y_(2)=acos(omegat+theta)`, then the displacement of resultant wave formed the to their superposition is given by `y=y_(1)+y_(2)=acosomegat+acos(omegat+phi)=2acos((phi)/(2)).cos(omegat+(phi)/(2))` Obviously the AMPLITUDE of the resultant wave is `A=2acos((phi)/(2))` and therefore the resultant intensity at a point will be given by `I_(R)=KA^(2)=k.4a^(2)cos^(2)((phi)/(2))` But `ka^(2)=I=`Intensity of light due to any one superposing wave. so, we have `I_(R)=4Icos^(2)((phi)/(2))`. Condition for bright: If `theta=2npi` (where n=0,1,2,3. . ), then `I_(R)=4Icos^(2)(npi)=4I=`maximum and we obtain bright fringe. Condition for dark fringes: If `phi=(2n-1)pi` (where n=1,2,3. . ), then `I_(R)=4picos^(2)((2n-1)/(2)pi)=0` and we obtain dark fringe. |
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| 7091. |
In circuit shown in figure, terminal X is brought from point A to B. Then reading of ammeter . |
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Answer» increases Raeading of ammeter `I_(0) = epsilon/R` `R = x+y` or `I_(0 )= epsilon/(x+y)`  ,  When sliding key is at any other position `I_(1) = epsilon/((R_0x)/(R_0+x) +y)` `Reading of ammeter `I = I_(1) (R_0)/(R_0+x)` `= (epsilon(R_0+x))/([R_0x+(R_0+y)] (R_0+x))` `= (epsilonR_0)/(R_0(x+y)+xy) = (epsilon)/((x+y)+(xy)/(R_0))` ,BRGT Hence `I' ltI_0` Hence the reading of ammeter first decreases then increases. |
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| 7092. |
Graph showing the variation of current versus voltagefor a material Ga As (a semiconductor) is shown in the Fig. Identify the region of (i) negative resistance,(ii) where Ohm's law is obeyed. |
| Answer» SOLUTION :(i) The RESISTANCE of MATERIAL is negative in the region BC.(ii) The material OBEYS Ohm.s law with in the region OA. | |
| 7093. |
What are unit of energy of an electron ? What is electron volt ? |
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Answer» Solution :ELECTRON volt: It is the energy required by an electron when it is accelerated through potential difference of 1 volt and is denoted by eV. 1eV=`1.602xx10^(-19) J` It is a convenient UNIT of energy used commonly in ATOMIC physics. |
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| 7094. |
A potentiometer wire is 100 cm long and a constant potential differenceis maintained across it. Two cells are connected inseries first to support one another and then in opposite direction. The balance points are obtained at 64 cm and 32 cm from the positive end of the wire in the two cases. the ratio of emf's is |
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Answer» `5: 4` In helping CONDITION `epsilon_(1) + epsilon_(2) prop l_(3)` In OPPOSING condition `epsilon_(1) - epsilon_(2) prop l_(4) "" [ because epsilon_(1) gt epsilon_(2) ] ` `therefore (epsilon_(1) + epsilon_(2))/(epsilon_(1) - epsilon_(2)) = (l_(3))/(l_(4)) = (50)/(10) = (5)/(1)` ` (2 epsilon_(1))/(2 epsilon_(2)) = (5 + 1)/(5 - 1) ` `therefore (epsilon_(1))/(epsilon_(2)) = (6)/(4) "" therefore (epsilon_(1))/(epsilon_(2))= (3)/(2)` |
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| 7095. |
Is Huygen's principle valid for longitudunal sound waves ? |
| Answer» SOLUTION :YES, Huygen.s principle is true and valid for all types of waves. | |
| 7096. |
Figure shows a point object placed in front of a transparent sphere of radius 20 cm and refractive index 1.5. The sphere forms an image due to refraction at surface I followed by refraction at surface II. The sphere is kept in air and the object is located at distance x from P_(1). It is found that for x gt x_(1), the image formed due to refraction at surface I is real, the final image formed by the sphere is also real and forms to the right of P_(2). For x_(2) lt x lt x_(1), the image formed by surface I is virtual while the final image formed by the sphere is real and forms to the right of P_(2). However for x lt x_(2), the image formed by surface I and also the final image formed by the sphere are virtual. It is also found that the final image is virtual if the distance of image formed by surface I from P_(2) is less than y. The object is now moved away from the sphere so that x becomes large and say, x to oo. In the given sphere is replaced by another one of the same radius but of refractive index 2.5, the final image is formed at distance y_(2) from P_(2). For the object placed at x to oo, a sphere of the same radius but of refractive index 2.5 forms an image at distance y_(2) from P_(2) |
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Answer» <P>this image is REAL and `y_(2)=2.5 cm " left of " P_(2)` |
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| 7097. |
A string 120 cmin length sustains a standing wave, with the points ofstring at which the displacement amplitude is equal to sqrt2mm being separated by 15.0 cm, Find the maximum displacement amplitude. Also find the harmonic corresponding to this wave. |
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Answer» Solution :From figure. Point A, B, C, D and E are having equal displacement amplitude. Further, `x_E - x_A = lambda = 4 xx15 = 60 cm`  As `lambda =(21)/n = (2 xx 120)/n =60` `:.n = (2 xx 120)/60 =4` So, it corresponds to `4^(th)` harmonic. Also,distance of node from A is 7.5 cm and no node is between the.Taking node at ORIGIN, the amplitude of STATIONARY wave can be written as, a = A sin kx Here `a = sqrt2 mm, k = (2pi)/lambda =(2pi)/60 and x =7.5 cm` `:. sqrt2 = A sin ((2pi)/(60) xx 7.5)= A sin ""(pi)/4` Hence, A = 2 mm |
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| 7098. |
A geostatationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time peiod of another satellite in hours at a height of 2R from the surface of the earth is : |
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Answer» `(6)/(sqrt(2))` `=(1)/(2sqrt(2))` `THEREFORE T_(2)=(1)/(2sqrt(2))xx24=6sqrt(2)` Thus correct choice is (d). |
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| 7099. |
Interference of light was first experimentally observed by : |
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Answer» Newton |
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| 7100. |
A steady current is set up in a metallic wire of non uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A? |
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Answer» K is INDEPENDENT of A |
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