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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.
| 5901. |
A lens behaves as converging lens in air and a diverging lens in water (n=4/3) the condition on the value of refractive index n of the material of the lens ? |
| Answer» Solution :`4/3 GT N gt 1` i.e, refractive INDEX n of the material is more than that of air (nair = 1) but LESS than that of water `(n_("water") = 4/3)` | |
| 5902. |
A battery of emf 2 V and initial resistance 1 Omega is connected across terminals A and B of the circuit shown in figure . |
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Answer» Thermal power generated in the EXTERNAL circuit will be maximum possible when `R = 16//25 Omega`. In the simplified circuit, the circuit is a balanced Wheatstone BRIDGE and a branch of `20//29 Omega and R ` is paralllel with this balanced bridge for maximum power.  `r = R_(external)` ` 1 = 1/(1/(3+6) + 1/(20//29+R)+1/(4+8))` or `R = 16/25 Omega` Maximum power developed in the external circuit is `P_(max) = i^2R = (2/(1+1))^2 xx 1 = 1W` Current through the upper branch `i_1 = i[((20/29+R)(4+8))/(9+(20/29 + R)+12)]` `i_2 = i[((20/29+R)(3+6))/(9+(20/29 + R)+12)]` Therefore, `i_1//i_2` is independent of R. |
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| 5903. |
The work of 146kJ is performed, in order to compress one kilomole of a gas adiabatically and in this process the temperature of the gas increase by 7^@C. The gas is (R=8.3Jmol^-1K^-1) |
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Answer» MIXTURE of MONOATOMIC and diatomic |
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| 5904. |
The sucepatibility of paramagnetic substance is |
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Answer» very LARGE |
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| 5905. |
Drive expression for the total resistance of a circuit in which a few resistors are connected in parallel. |
Answer» Solution :Resistance in parallel. Resistances are said to be connected in parallel if one end of each resistance is connected at one point and other end is connected to another point so that the potential difference across each resistance is the same. However, CURRENT passing through each resistance is a part of the total current.  As shown in the figure. Let `R_(1),R_(2) and R_(3)` be three resistances connected in parallel and `I_(1),I_(2) and I_(3)` be currents flowing through `R_(1),R_(2) and R_(3)` respectively. I=total current flowing between A and B. V=potential difference difference across A and B. It is clear that `I=I_(1)+I_(2)+I_(3)`. . . (1) Since the potential difference across `R_(1),R_(2) and R_(3)` is the same i.e., `V,` according to Ohm.s law. `V=I_(1)R_(1),V=I_(2)R_(2) and V=I_(3)R_(3)` `THEREFORE I_(1)=(V)/(R_(1)),I_(2)=(V)/(R_(2)) and I_(3)=(V)/(R_(3))`. . . (2) If `R_(p)` is equivalent resistance of the parallel combination, then, `therefore I=(V)/(R_(p))`. . (3) Putting (2) and (3) in (1) we get `(V)/(R_(p))=(V)/(R_(1))+(V)/(R_(2))+(V)/(R_(3))` or `(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`. . . (4) THUS we find that when a number of resistances are connected in parallel, the RECIPROCALS of the resultant resistance is EQUAL to sum of the reciprocal of individual resistances. It there are n resistances each equal to .R. and if these are in parallel, then resultant is given by : `(1)/(R_(p))=(1)/(R)+(1)/(R)+(1)/(R)+ . . .n` times `(1)/(R_(p))=(n)/(R) or R_(P)=(R)/(n)` Thus effective resitances decreases. |
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| 5906. |
A machine gun fires a bullet of mass 40 g with a velocity 1200 ms^(-1). The man holding it can exert a maximum force of 144 N on the gun. How many bullets can be fired per second at the most ? |
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Answer» One where n = no. of BULLETS in time t. `n=(Fxxt)/(m upsilon)=(144xx1)/((40)/(1000)xx1200)=3` HENCE CHOICE is (d) |
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| 5907. |
Assertion (A) : The poles of bar magnet cannot be separated by breaking it into two pieces. Reason (R) : The magnetic moment will be reduced to half when a bar magnet is broken into two equal pieces. |
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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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| 5909. |
Identify the mismatch of the following. |
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Answer» Photo diode - OPTICAL signal |
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| 5910. |
Atomic hydrogen is excited from the grounds state to the n^(th) state. The number of lines in the emission spectrum will be: |
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Answer» `(N(n+1))/(2)` |
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| 5911. |
A light ray is incident on a transparent slab of refractive index mu = sqrt2, at an angle of incidence pi//4. Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered. |
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Answer» `(sqrt3 - 1)/(SQRT6)` |
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| 5912. |
How is the capacitance of a parallel plate capacitor affected on heating the dielectric material filled between its plates? (Ignore thermal expansion and assume that dipole orientations are temperature independent). Give reasons. |
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Answer» Solution :Heating a dielectric will decrease its dielectric CONSTANT. This is because when we heat a material, the average kinetic energy per molecule increases. An electric polar molecule of a dielectric acts like an electric COMPASS needle aligning itself with the external electric field setup by the plates. But on heating the dielectric, each polar molecule will no longer be perfectly aligned with the applied electric field, but it will begin to move back and forth DUE to increase in average kinetic energy, thus effectively decreasing its CONTRIBUTION to the overall field. Dielectric constant of material decreases. As a RESULT, capacitance of the parallel plate capacitor will decrease. |
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| 5913. |
The angular speed of a body changes from omega_1 to omega_2 without applying a torque, but due to changes in M.I. the ratio of radii of gyration in the two cases is |
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Answer» `omega_2 : omega_2` |
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| 5914. |
A clock face has negative charges -q, -2q, -3q …… 12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to point charges. Time at which the hour point in the sae direction is electric field at the centre of the dial is X hours and 30 minutes, find X |
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Answer» |
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| 5915. |
चमगादड़ द्वारा परागण कहलाता है |
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Answer» आर्नीथोफिली |
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| 5916. |
A body is projected from a point with different angles of projections 20°, 35°, 45°, 60° with the horizontal but with same initial speed. Their respective horizontal ranges are R_(1), R_(2), R_(3) and R_4. Identify the correct order in which the horizontal ranges are arranged in increasing order |
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Answer» `R_(1),R_(2), R_(3)` |
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| 5917. |
सैल्विया में परागण होता है |
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Answer» जल द्वारा |
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| 5918. |
A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^(-1).Determine the amplitude of B at a point 3.0 cm from the axis between the plates. |
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Answer» Solution :Here for given point P, its distance from the AXIS of plates is r = 3 cm. But radius of plate of a CAPACITOR is R = 6 cm `rArr (r )/(R )=(3)/(6)=0.5`  Here, since the current density is same everywhere in the region between the plates, we can write, `(i_(d))/(pi r^(2))=(I_(rms))/(pi R^(2))` `therefore i_(d)=I_(rms)((r )/(R ))^(2)` Above is the displacement current enclosed by circular Amperean loop of radius r, passing through given point P. Because of this current, MAGNITUDE of magnetic field is same (equal to B) at all points on this loop. Hence, according to Ampere - Maxwell law, `int vec(B).vec(d)l=mu_(0)(i_(c )+i_(d))` `therefore int B dlcos 0^(@)=mu_(0)(0+i_(d))` (`because vec(B)"||"vec(d)l` and between the plates `i_(c )=0`) `therefore B int dl = mu_(0)i_(d) ""` (`because` B = constant) `therefore B (2pi r)=mu_(0)i_(d)` `therefore B=(mu_(0)i_(d))/(2pi r)` Since above VALUE is the rms value of induced magnetic field, `B_(rms)=(mu_(0)i_(d))/(2pi r)` `therefore B_(rms)=(mu_(0))/(2pi r)xx I_(rms)((r )/(R ))^(2)` `therefore B_(rms)=(4PI xx10^(-7))/(2pi xx 0.03)xx6.9xx10^(-6)xx(0.5)^(2)` `therefore B_(rms)=115xx10^(-13)T` `therefore (B_(0))/(sqrt(2))=115xx10^(-13)T` (Where `B_(0)` is the amplitude of induced manetic field at point P) `therefore B_(0)=sqrt(2)xx115xx10^(-13)T` `=1.414xx115xx10^(-13)T` `=162.6xx10^(-13)T` `therefore B_(0)=1.626xx10^(-11)T` Second method : From equation (3) and (4), `B=(mu_(0))/(2pi).(r )/(R^(2))I_(rms)` `therefore B_(0)=(mu_(0))/(2pi).(r )/(R^(2))I_(0) ""`....(1) (Where `B_(0)` and `i_(0)` are the amplitudes of oscillating magnetic field and current) `I_(0)=sqrt(2)I_(rms)` `=1.414xx6.9xx10^(-6)` `=9.7566xx10^(-6)A` `~~9.76 mu A` `therefore_(0)=(4pi xx10^(-7))/(2pi)xx(3xx10^(-2)xx9.76xx10^(-6))/((6xx10^(-2))^(2)) ""` [`because r=3xx10^(-2)m`and `R=6xx10^(-2)m`] `=1.6266xx10^(-11)T` `therefore B_(0)~~1.63xx10^(-11)T` |
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| 5919. |
The equation of motion of a projectile is y =sqrt(3x)-(gx^(2))/2 The angle of projection is : |
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Answer» `theta = TAN^(-1)1/SQRT3` |
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| 5920. |
State the underlying princeiple of mater bridge. Draw the circuit diagram and explain how the unknown resistance of a conductor can be determined by this method. |
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Answer» Solution :Metre bridge or slide wire bridge. It is the simplest practical application of the Wheatstone bridge that is USED to measure an unknown resistance. Principle : Its working is BASED on the principle of Wheatstone bridge. When the bridge is balanced, `(P)/(Q)=(R)/(S)` Construction : It consists of usually one metre long magnanin wire of uniform cross-section, streatched along a metre scale fixed over a wooden board and with its two ends soldered to two L-shaped thick copper STRIPS A and C. Between these two copper strips, another copper strip is fixed so as to PROVIDE two gaps ab and `a_(1)b_(1).` A source of emf `epsilon` is connected across AC. A movable jockey and a galvanometer are connected across BD, as shown in figure. After taking out suitable resistance R from the resistance box, the jockey is moved along the wire AC till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge. If P and Q are the resistances of the parts AB and BC of the wire, then for the balanced condition of the bridge, we have `(P)/(Q)=(R)/(S)`. Let total length of wire AC=100cm and AB=l cm, then BC=(100-l)cm. Since, the bridge wire is of uniform cross-section, therefore, resistance of wire `prop` length of wire `(P)/(Q)=("resistance of AB")/("resistance of BC")=(sigmal)/(sigma(100-1))=(l)/(100-l)` where `sigma` is the resistance per unit length of the wire. `"Hence, "(R)/(S)=(l)/(100-l) or, S=(R(100-l))/(l).` 
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| 5921. |
A copper wire of 10 ^(-6) m ^(2) area of cross section, carries a current of 2A. If the number of electrons per cubic meter is 8 xx 10 ^(28), calculate the current density and average drift velocity. Given, e1. 6 xx 10 ^(-19) C) |
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Answer» Solution :Current DENSITY, `J = I/A = (2)/( 10 ^(-6)) = 2 xx 10 ^(6) A //m ^(2)` `J = n E V _(d)` `or V _(d) = (J )/( n e) = ( 2 xx 10^(6))/( 8 xx 10 ^(28) xx 1. 6 xx 10 ^(-19))` `= 15.6 xx 10 ^(-5) ms ^(-1)` |
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| 5922. |
If tempreture is increased, the number of electron-hole pairs increased and is proportional to a factor (DeltaE: band gap) |
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Answer» `T^(3//2)E^(-(DELTAE)/(2kT)` |
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| 5923. |
A source of sound which emitting a sound of frequency 600 Hz is moving towards a wall with a velocity 30 m/s. Three observes A, B, C moving with velocity 20 m/s, A away from the wall and B towards the wall and Inside the source, Velocity of sound in air is 330 m/s. Find the frequency of direct sound observed by observer A |
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Answer» `600 310/360` Hz `f=f_0((vpmv_(0M))/(vpmv_(sm)))` `f=600((330-20)/(330+30))=600 310/360` Hz |
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| 5924. |
The bending of an elastic rod is described by the elastic curve passing through centres of gravity of rod's cross-sections. At small bendings the equation of this curve takes the form N(x)=EI(d^2y)/(dx^2), where N(x) is the bending moment of the elastic forces in the cross-section corresponding to the x coordinate, E is Young's modulus, I is the moment of inertia of the cross-section relative to the axis pasing through the neutral layer (I=intz^2dS, figure) Suppose one end of a steel rod of a square cross-section with side a is embedded into a wall, the protruding section being of length l (figure). Assuming the mass of the rod to be negligible, find the shape of the elastic curve and the deflection of the rod lambda, if its end A experiences (a) the bending moment of the couple N_0, , (b) a force F oriented along the y axis. |
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Answer» Solution :We use the equation given above and use the result that when y is small `1/R~~(d^2y)/(DX^2)`. Thus, `(d^2y)/(dx^2)=(N(x))/(EI)` (a) Here `N(x)=N_0` is a constant. Then integration gives, `(dy)/(dx)=(N_0x)/(EI)+C_1` But `((dy)/(dx))=0` for `x=0`, so `C_1=0`. Integrating again, `y=(N_0x^2)/(2EI)` where we have used `y=0` for `x=0` to set the constant of integration at zero. This is the equation of a parabola. The sage of the free end is `LAMBDA=y(x=l)=(N_0l^2)/(2El)` (b) In this case `N(x)=F(l-x)` because the load F at the EXTREMELY is balanced by a similar FORCE at F directed upward and they constitute a couple. Then `(d^2y)/(dx^2)=(F(l-x))/(EI)`. Integrating, `(dy)/(dx)=(F(lx-x^//2))/(EI)+C_1` As before `C_1=0`. Integrating again, using `y=0` for `x=0` `y=(F((lx^2)/(2)-(x^3)/(6)))/(EI)` here `lambda=(Fl^3)/(3El)` Here for a square cross section `I=underset(-a//2)overset(a//2)intz^2adz=a^4//12`. |
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| 5925. |
A source of sound which emitting a sound of frequency 600 Hz is moving towards a wall with a velocity 30 m/s. Three observes A, B, C moving with velocity 20 m/s, A away from the wall and B towards the wall and Inside the source, Velocity of sound in air is 330 m/s. Find the frequency of reflectedsound observed by observer B |
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Answer» `600 310/360` Hz then frequencyobserved by observer B is `f=f_"wall"{(330+20)/330}=600 330/300.350/330` `=600. 350/300`Hz |
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| 5926. |
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum. |
| Answer» SOLUTION :SPEED (or VELOCITY) | |
| 5927. |
A source of sound which emitting a sound of frequency 600 Hz is moving towards a wall with a velocity 30 m/s. Three observes A, B, C moving with velocity 20 m/s, A away from the wall and B towards the wall and Inside the source, Velocity of sound in air is 330 m/s. Frequencyof directed sound observer by observer C. |
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Answer» 450 Hz |
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| 5928. |
A TV transmission tower of antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level (ii) at a height of 25 m ? Calculate the percentage increase in area covered in case (ii) relative to case (i). |
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Answer» 2.48 If the anteenais at ground level Range, `d=sqrt(2h_(T)R)=sqrt(2xx20xx(64xx10^(5)))` `=16xx10^(3)m=16km ` Areacovered, `A=PID^(2)=(22)/(7)xx(!6)^(2)=804.6km^(2)` If the receiving ANTENNA at a height of 25m. Range `d_(1)=sqrt(2h_(T)R)_sqrt(2h_(R)R)` `=sqrt(2xx20xx(64xx10^(5)))+sqrt(2xx25xx64xx10^(5))` `=16xx10^(3)+17.9xx10^(3)=33.9xx10^(3)m=33. 9km` Area covered `A_(1)=pid_(1)^(2)=(22)/(7)xx(33.9)^(2)=3611.8km^(2)`. PERCENTAGE increase in area `=(A_(1)-A)/(A)xx100` `=((3611.8-804.6)/(804.6))xx100=348. 9%` |
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| 5929. |
Distance travelled by linear scale of screw gauge during two full rotation of circular scale is 1mm and circular scale has 50 divisions. In an experiment to measure thickness of a plate, six divisions of main scale are clearly visible and 28^(th) division of circular scale coincides with reference line. Moreover when studs touch each other, zero of circular scale lies 4 division below the reference line, thickness of the plate will be: |
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Answer» <P>3.28 mm Device has+ve ZERO error `R=6p+28(p)/(50)-4(p)/(50)=6p+(24p)/(50),3mm+(24)/(50)xx0.5=3.24mm` |
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| 5930. |
A straiht wire ABCD is bend as shown in figure. Find an expression for the magnetic field at the point P. |
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Answer» SOLUTION :`B_p = (mu_0I)/(4pi r) (sin phi_1 + sin phi_2)` (NO FIELD on the AXIS of straight CONDUCTOR) |
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| 5931. |
A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^(-1).Is the conduction current equal to the displacement current ? |
| Answer» SOLUTION :Yes, when current in the circuit changes continuously PERIODICALLY with time, we have `i_(C )=i_(d)`. (Here it should be noted that conduction current `i_(c )` flows in the external circuit only upto the plate of a capacitor whereas displacement current `i_(d)` flows in the REGION between the plates of a capacitor). | |
| 5932. |
Dust particles released by comets are continuously acted upon by sun's radiation pressure in radial outward direction. Assuming a dust particle with rho =3.5 xx10^3 kg//m^3 and reflection coefficient 0. for what value of radius r does gravitational force on dust. particle just balance the radiation force on it from sunlight ? (power radiated by sun =4xx10^(26)W , mass of sun =2xx10^(30)kg) |
| Answer» Answer :B | |
| 5933. |
Consider a monoatomic gas which undergoes a cycle as shown in P-T diagram. Match the process in column-I to description in column-II |
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Answer» PT=const. `RARR P^(2)V` =const. as , `T uarr rArr DELTAU gt 0, p DARR rArr V uarr rArr DeltaWgt 0` `rArr Delta Q =DeltaU+DeltaW rArr DeltaQ gt 0` For BC P=const. `rArr T darr DeltaU lt 0` `rArr V darr rArr DeltaW lt 0 rArr DeltaQ lt 0` For CA `T_("Isothermal process")`=const. `rArr DeltaU=0` process equation PV=const. `rArr P darr rArr V darrrArr DeltaW lt 0 rArr DeltaQ =DeltaU+DeltaWrArr DeltaQ lt 0` |
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| 5934. |
वर्गिकी का पहला चरण है |
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Answer» जीव का वर्णन |
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| 5935. |
A standing wave pattern on a string is described by y(x,t) =0.040 (sin 4pi x) (cos 4pi t). where x and y are in meters and r is in seconds For x ge 0 what is the location of the node with the (a) smallest, (b) second smallest, and (c) third smallest value of x? (d) What is the period of the oscillatory motion of any (nonnode) point? What are the (e) speed and (f) amplitude of the two traveling waves that interfere to produce this wave? For t gt 0, what are the (g) first, (h) second, and (i) third time that all points on the string have zero transverse velocity? |
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Answer» |
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| 5936. |
In a ten-wire potentiometer the first five wires are of radius r and the next five wires are of radius 2r.The wire is cnneceted to battery of steady voltage 2 V and negligible internal resistance What lenghts of this potentiometric arrangement will balacne the emf of (a) adainell cell (emf=1.0 volt),(b)a Lenhlanche cell (emf=1.5 V) (c) an unknown cell of emf 1.8 V? |
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Answer» |
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| 5937. |
A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^(-1).What is the rms value of the conduction current ? |
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Answer» SOLUTION :Here current in A.C. circuit, `I_(rms)=(V_(rms))/(|Z|)` `I_(rms)=(V_(rms))/(X_(C )) ""` (`because` Here only capacitor is CONNECTED and so `|Z|=X_(C )`) `therefore I_(rms)=(V_(rms))/(((1)/(omega C)))=V_(rms)xx omega C(because X_(C )=(1)/(omega C))` `therefore I_(rms)=(230)(300)(100xx10^(-12))` `therefore I_(rms)=6.9xx10^(-6)A` |
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| 5938. |
In the previous problem, if we connect the conductors 1 and 2, 2 and 3, 1 and 3, 1 and ground, 2 and ground, and 3 and ground, separately, find the charge flown in each case. |
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Answer» Solution :(i) If we connect the conductors 1 and 2, the potential of spheres 1 and 2 will be equal. Here `V_1 = V_2` or `K (x)/a + K (Q - x)/(2 a) + K((-4Q))/(3 a)` =`K(x)/a + K(Q - x)/(2 a) + K((-4 Q))/(3 a)` or `x = 0` The CHARGE of sphere 1 will come to sphere 2. Hence charge flown in this case is equal to `-Q`.  . (ii) If we connect the spheres 2 and 3, their potential will be same. `V_2 = V_3` or `K((-Q))/(2 a) + K(x)/(2 a) + K((-2 Q - x))/(3 a)` =`K ((-Q))/(3 a) + K(x)/(3 a) + K((2 Q - x))/(3 a)` or `x = Q` Hence, charge of amount Q will FLOW from 2 to 3.  . (iii) If we 1 and 3, their potential should be equal. `V_1 = V_3` or `K(x)/a + K(2 Q)/(2 a) + K((-5 Q - x))/(3 a)` =`K(x)/(3 a) + K(2 Q)/(3 a) + K((-5 Q - x))/(3 a)` or `(x)/a - (x)/(3 a) = (2 Q)/(3 a) - (2 Q)/(2 a)` or `(2 x)/a = -Q/(3) or x = (-Q)/(2)` Hence, `-Q//2` charge will flow from 1 to 3.  . (iv) Now sphere 1 is connected with GROUND. The potential of sphere 1 will be zero. `V_1 = 0 = K (x)/a + K(2 Q)/(2 a) + K ((-4 Q))/(3 a)` or `x = -Q + (4 Q)/(3) = Q/(3)` Hence, `Q + Q/(3) = (4 Q)/(3)` should flow from ground to sphere.  . (v) Now sphere 2 is connected with earth. The potential sphere 2 should be zero. `V_2 = 0 = K((- Q))/(2 a) + K (x)/(2 a) + K((-4 Q))/(3 a)` or `(x)/(2) = Q/(2) + (4 Q)/(3)` or `x = Q + (8 Q)/(3) = (11 Q)/(3)` hence, `(11 Q)/(3) - 2 Q = (5 Q)/(3)` charge will flow from earth sphere.  . (vi) Now sphere 3 is connected with ground. Here, `V_3 = 0 = K ((-Q))/(3 a) + K ((2 Q))/(3 a) + K((x))/(3 a)` or `x = Q - 2 Q = -Q` Hence, `- 3 Q` charge will flow from sphere to earth or `3 Q` charge will flow from earth to sphere.  .
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| 5939. |
An object 2 cm high is placed at right angles to the principal axis of a mirror of focal length 25 cm such that an erect image 6 cm high is formed. What kind of mirror is it and what is the position of the object? |
| Answer» Solution :CONCAVE, at distance `(50)/(3)`CM from the POLE | |
| 5940. |
An object 2 cm high is placed at right angles to the principal axis of a mirror of focal length 25 cm such that an erect image 0.5 cm high is formed. What kind of mirror its is and what is the position of the object? |
| Answer» SOLUTION :CONVEX, at distance 75 cm from the POLE | |
| 5941. |
The opposition offered by a capacitor to a.c. is called what ? |
| Answer» SOLUTION :CAPACITIVE REACTANCE | |
| 5942. |
STATEMENT -1 : The centre of gravity of a spherically symmetric body of uniform mass density, will lie upon the geometrical centre. and STATEMENT -2 : The location of centre of gravity depends upon the distribution of mass as well as variation of g. |
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Answer» Statement 1- TRUE, Statement -2 is True, Statement -2 is a CORRECT EXPLANATION for Statement -13 |
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| 5943. |
Two identical beam A and B of wavelength lambda fall cylindeical screen. The angle between the directions a point P on the screen at angular position phi from the beam A as shown in the figure. Find the distance between adjacent interference fringes on the screen near the point P. Asume that the distance beta between adjacent fringes is much less than the radius of the cylinder. |
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Answer» `(lambda)/(2sin ((phi)/(2))cos(theta-(phi)/(2)))` `(lambda)/(sin(theta+phir)-sin theta)` (same as in the previous PROBLEM) or`beta = (lambda)/(2sin((phi)/(2))cos(theta+(phi)/(2))` |
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| 5944. |
In the Fig. the input is across A and C and output across B and D . The output is : |
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Answer» zero |
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| 5945. |
If the amount of a radioactive substance is increased four times then how many times will the number of atoms disintegrating per unit time be increased ? |
| Answer» SOLUTION : FOUR TIMES `becauseR =-lambdaN` | |
| 5946. |
Explain, using Huygen's principle, how diffraction is produced by a narrow slit which is illuminated by a monochromatic light. Show that central maximum is twice as wide as the other maxima and the pattern becomes narrower as the width of the slit is increased. |
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Answer» Solution :When a plane wavefront of monochromatic light is incident on a narrow slit, each point of slit behaves as a secondary source of light in accordance with Huygen.s principle. These secondary sources emit wavelets which travel in DIFFERENT directions. Naturally wavelets emitted by a different PARTS of the single narrow slit superpose. as a result to superposition we obtain a broad pattern with central brigh region and on both sides there are alternate dark and BRIGHT regions. thus, intensity distribution on screen is redistributed. this phoenomenon is called diffraction of light due to a narrow slit. It can be easily shown that positions of minimas are given on the screen by `x=pm(nDlamda)/(a)`, where n is any integer a=slit WIDTH and D=distance of screen from the slit. `therefore` Width of nth maximum is twice as WIDE as the other maxima. |
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| 5947. |
इनमें से कौन-सा देश भारत से क्षेत्रफल में बड़ा नहीं है? |
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Answer» रूस |
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| 5948. |
If both the length of an antenna and the wavelength of the signal to be transmitted are doubled, the power radiated by the antenna |
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Answer» is doubled |
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| 5949. |
A magnetic needle, free to rotate in a vertical plane, orients itself vertically at a certain place on earth. What is the value of horizontal component of earth's magnetic field at this place ? |
| Answer» Solution :The place is the magnetic POLE of earth and value of HORIZONTAL component of earth.s magnetic field is ZERO at this place | |
| 5950. |
Is x = 1 the zero of x^3 − 1? |
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Answer» Yes |
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