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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.
| 3151. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is 10Vm^-1. The frequency of wave is 5xx10^(14)Hz . the wave is propagating along z-axis. If mu_0,mu_r, in_0 and in_r as the absolute permeabilty, relative permeability , absolute permitivity and relative permitivity of the medium, then the velocity of em wave in a medium is |
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Answer» `1/(SQRT(mu_0in_0))` `v=1/(sqrt(MU in))=1/(sqrt(mu_0 in_0 mu_0 in_r))` |
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| 3152. |
A square loop of side 10 cm and resistance 0.5Omega is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval. |
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Answer» Solution :The angle `theta` MADE by the area vector of the loop with the magnetic field is `45^(@)`. The initial magnetic flux is, `phi=BAcostheta=(0.1xx10^(-2))/(SQRT(2))Wb` Final flux, `phi_("min")=0` The change in flux is brought about in 0.70s. The magnitude of the induced emf is GIVEN by `epsilon=(|Deltaphi_(B)|)/(Deltat)=(|(phi-0)|)/(Deltat)=(10^(-3))/(sqrt(2)xx0.7)=1.0mV` And the magnitude of the current is `I=(epsilon)/(R)=(10^(-3)V)/(0.5Omega)=2mA` |
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| 3153. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is 10Vm^-1. The frequency of wave is 5xx10^(14)Hz . the wave is propagating along z-axis. In em wave, the average energy density due to magnetic field is |
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Answer» `8.85xx10^(-10)Jm^-3` em wave is `u_B=1/2 (B_(RMS)^2)/(mu_0) =1/(2mu_0) ((B_0)/(sqrt2))^2=1/4 (B_0^2)/(mu_0)` `=1/4 ((E_0//c^2)/(1//mu_0epsilon_0))=1/4 in_0 E_0^2` `=1/4xx(8.85xx10^(-12))xx10^2=2.21xx10^(-10)Jm^-3` |
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| 3154. |
Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefront |
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Answer» y = CONSTANT |
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| 3155. |
From source of power 4 kW,radiation emitted are such that photon are emitted at rate of 10^(20) photon/sec.This radiation will be in …..region of electromagnetic waves… []h=6.6xx10^(-34)Js |
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Answer» gamma rays `therefore F=(p)/((n)/(t)h)=(4000)/(10^(20)xx6.6xx10^(-34))` `=6.06xx10^(16)HZ` The spectrum containing frequency are of X-region. |
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| 3156. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is 10Vm^-1. The frequency of wave is 5xx10^(14)Hz . the wave is propagating along z-axis. Sun also sends em wave to earth. which one of em wave out of the visible portion, from sun will be reaching the surface of earth earlier than others: |
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Answer» VIOLET waves As `lambda_r GT lambda_v` so `mu_r LT mu_v`, THEREFORE `v_r gt v_v` |
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| 3157. |
Figure shows a torch producing a straight light beam falling on a plane mirror at an angle 60^@. The reflected beam makes a spot P on the screen along y-axis. If at t=0, the mirror starts rotating about the hinge A with an angular velocity omega = 1^@per second clockwise, find the speed of the spot on screen after time t = 15 s. |
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Answer» `(pi)/(15) m//s ` |
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| 3158. |
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red and end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used: lamda_(1)=3650Å, lamda_(2)=4047Å,lamda_(3)=4358Å,lamda_(4)=5461Å,lamda_(5)=6907Å The stopping voltages, respectively, were measured to be: V_(01)=1.28V,V_(02)=0.95V,V_(03)=0.74V,V_(04)=0.16V,V_(05)=0V Determine the value of Planck's constant h, the threshold frequency and work function for the material. |
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Answer» Solution :Frequencies corresponding to different wavelengths of spectral lines of mercury are `v_(1)=(c)/(lamda_(1))=(3xx10^(8))/(3650xx10^(-10))=8.219xx10^(14)Hz`  `v_(2)=(c)/(lamda_(2))=(3xx10^(8))/(4047xx10^(-10))=7.412xx10^(14)Hz` `v_(3)=(c)/(lamda_(3))=(3xx10^(8))/(4358xx10^(-10))=6.883xx10^(14)Hz` `v_(4)=(c)/(lamda_(4))=(3xx10^(8))/(5461xx10^(-10))=5.493xx10^(14)Hz` and `v_(5)=(c)/(lamda_(5))=(3xx10^(8))/(6907xx10^(-10))=4.343xx10^(14)Hz` From the given data, we plot the graph, which comes out to be as shown in figure. first FOUR points lie on a straight line but the FIFTH point does not lie on straight line. it means that `v_(5) lt v_(0)`. (a) A careful measurement shows that slope of straight line graph is `4.15xx10^(-15)VS` `therefore (h)/(e)=4.15xx10^(-15)=exx4.15xx10^(-15)=1.6xx10^(-19)xx4.15xx10^(-15)=6.64xx10^(-34)Js` (b) From graph intercept along v-axis is found to be `5xx10^(14)Hz`. `therefore`Threshold frequency `v_(0)=5xx10^(14)Hz` `therefore`Work function `phi_(0)=6.64xx10^(-34)xx5xx10^(14)J=(6.64xx10^(-34)xx5xx10^(14))/(1.6xx10^(-19))eV` `=2.075eV=2.1eV`. |
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| 3159. |
V_("BB") can vary between 0 and 5 V. Find the minimum value of base current and V_("BB"), so that transistor works in saturation mode. ["Given "beta =200 and V_("BE")=1V] |
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Answer» `20muA, 2.8V` |
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| 3160. |
The combination of gates shown below yields : |
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Answer» NAND gate |
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| 3161. |
Cell wall is made up of polysaccharide and amino acid in most the members of? |
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Answer» Monera |
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| 3162. |
Column I contains a list of processes involving expansion of an ideal gas. Match this with column II describing the thermodynamic change during this process. |
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Answer» (A) |
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| 3163. |
A parallel plate capacitor with air between the plates has a capacitor of 8pF ("1pF="10^(-12)F). What will be the capacitance if the distance if the distance between the plates is reduced by, half and the space between them is filled with a substance of dielectric constant 5? |
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Answer» SOLUTION :`C=8pF, C=(epsi_(0)A)/(d)` `C.=(epsi_(0) epsi_(0)A)/(d/2)=2epsi_(r) C THEREFORE C.=2 xx 6 xx 8 xx 10^(-12)F=96pF` |
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| 3164. |
There is an air bubble of radius R inside a drop of water of radius 3R. Find the ratio of gauge pressure at point B to that at point A. |
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Answer» `(1)/(2)` |
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| 3165. |
At t = 0 a travelling wave pulse on a string is described by the function y = (10)/(x^2 + 2) , here x and y are in meter and t in second. What will be the wave function representing the pulse at time t, if the pulse is propagating along the positive x-axis with speed 2m/s ? |
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Answer» `y = (10)/((X^2 + 2Y)^2 + 2)` |
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| 3166. |
An ammeter has a resistance of 50Omega and a full scale deflection current 50muA. It can be used as a voltmeteror as a higher range ammeter provided that a resistance is added to it. Which of the following is/are true? |
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Answer» 10 V range with APPROXIMATELY `200kOmega` resistance in series |
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| 3167. |
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. In which direction will it move and why? |
| Answer» Solution :A SUPERCONDUCTOR BEHAVES as an ideal diamagnetic substance. Hence, it MOVES away from the bar magnet when it is PLACED NEAR the bar magnet. | |
| 3168. |
When the incident wavelengths are lamda and lamda//2 the kinetic energies of the emitted photo electrons are E and 2E. The work function of the metal is |
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Answer» E/4 |
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| 3169. |
Figure gives alpha versus the sine of the angle theta in a single-slit diffraction experiment using light of wavelength 610 nm. The vertical axis scale is set by alpha_(s) = 12 rad. What are (a) the slit width, (b) the total number of diffraction minima in the pattern (count them on both sides of the center of the diffraction pattern), (e) the least angle for a minimum, and (d) the greatest angle for a minimum? |
| Answer» SOLUTION :(a) 2.33 `MU m`, (B) 3.82, (C) `15.2^(@)`, (d) `51.8^(@)` | |
| 3170. |
When only carrier of transmitted antenna current observed is 8A. When it is modulated with 500 Hz sine wave antenna current becomes 9.6 A. Find % modulation. |
| Answer» ANSWER :C | |
| 3171. |
Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of Magnetic flux |
Answer» Solution :Plot of VARIATION of MAGNETIC FLUX with time, 
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| 3172. |
A variable frequency ac source is connected to a capacitor. |
| Answer» SOLUTION :CAPACITIVE RECTANCE `X_c = 1/(comega)`. | |
| 3173. |
A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 xx 10^(-3) Wb. The self-inductance of the solenoid is |
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Answer» 3H `=(1000xx4xx10^(-3))/4` `THEREFORE` L=1H |
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| 3174. |
(a) Draw a schematic diagram of an AC generator. Explain is working and obtain the expression for the instantaneous value of the emf in terms of the magnetic field B, number of turns N of the coil of area A rotating with angular frequency omega. Show how an alternating emf is generated by loop of wire rotating in a magnetic field. (b) A circular coil of radius 10 cm and 20 turns is rotated about its vertical diameter with angular speed of 50 rad s^(-1). in a uniform horizontal magnetic field of 3.0xx10^(-2)T. (i) Calculate the maximum and average emf induced in the coil. (ii) If the coil forms a closed loop of resistance 10 Omega, calculate the maximum current in the coil and the average power loss due to Joule heating. |
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Answer» SOLUTION : PLOT of variation of ALTERNATING emf with time. `e=NABomega SIN m ea=sin omegat ` 
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| 3175. |
The angle of incidence and emergence of a ray of light through an educational prism are 50^@ and 40^@. Find the angle of deviation. |
| Answer» SOLUTION :d = (e+i)-A = ( | |
| 3176. |
In Young's double slit experiment if width of the slit is made (1)/(3) th times width of fringe becomes n times then n ...... |
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Answer» 3 `:. Beta prop (1)/(d)` `:. (beta_(2))/(beta_(1))=(d_(1))/(d_(2))` `:. (beta_(2))/(beta_(1))=(d)/((d)/(3))=3` `:. (n beta_(1))/(beta_(1))=3 "" :. n=3` |
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| 3177. |
At a certain location the horizontal and the vertical components of the earth's magnetic field are equal. The value of dip at that place is |
| Answer» ANSWER :D | |
| 3178. |
(a) Three resistor 1Omega ,2Omega, and 3Omega are combined in series. What is the total resistance of the ombination ? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor. |
Answer» Solution : CURRENT PASSING through battery in above loop, `I =(epsilon)/(R_(s)+ R) = (epsilon)/(R_(s)) "" (because" Here, r = " 0 )` `= (12)/(6)` `therefore I = 2 A ` Voltages across `R_(1), R_(2) , R_(3)` are respectively, `therefore V_(1) = IR_(1) = 2 xx 1= 2 ` V `therefore V_(2) = IR_(2) = 2 xx 2 = 4 `V `therefore V_(3) = IR_(3 ) = 2 xx 3 = 6` V |
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| 3179. |
A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight during each full rotation of the rod. The strobe rate is 2000 flashes per second, the bolt has mass 33 g and is at radius 4.0 cm. What is the magnitude of the force on the bolt from the rod? |
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Answer» |
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| 3180. |
Give the expression for the magnetic force on a moving charge in an uniform magnetic field. What will be the maximum magnetic force on the moving charge? |
| Answer» SOLUTION :The magnetic force on a moving CHARGE be maximum if VELOCITY and magnetic field are PERPENDICULAR to each other. | |
| 3181. |
A laser emit monochromatic light of 6.0xx10^(14)Hz frequency and produce power of 2xx10^(-3) W.No of photon produced per second will be….. |
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Answer» <P>`5xx10^(16)` `impliesn=(P)/(hf)=(2xx10^(-3))/(6.6xx10^(-34)xx6xx10^(14))=5xx10^(15)` |
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| 3182. |
State the necessary conditions for the phenomenon of total internal reflection to occur. |
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Answer» Solution : (i) The ray of light should travel in the OPTICALLY denser medium towards the rarer medium. (ii) The angle of incidence at the REFRACTIVE SURFACE should be greater than the critical angle for given PAIR of media. |
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| 3183. |
Two independent light sources are always incoherent. This is because phase of light emitted is random and phase difference, at any point, changes very rapidly with time. So as to obtain observable interference, we require coherent sources which emit light of same wavelength such that phase difference at any point does not change with time. In different interference experiments, the method to obtain two coherent sources could be different. In Fresnel's biprism experiment, two virtual images of a real source, which are formed due to refraction at the biprism, act as coherent source. In Lloyd’s mirror experiment, a real source and its virtual image formed due to reflection at a plane mirror act as coherent source. Two images of a source or a source and its image are coherent because of their phase correlation. It is also possible to obtain coherent sources by forming two images of a point source in the following manner.As shown in the figure, S is a monochromatic point source. A thin circular lens is cut into two identical halves L, and L_(2) by a plane passing through a diameter. L_(1) is kept above the axis and L_(2) below it is in a symmetrical manner such that the gap between the two is 0.1 mm. Each forms an image Qf the source and the two images, say, Si and S_(2) , act as coherent sources. Superposition of waves from these coherent sources will result in observable interference and a pattern can be obtained on a screen.The lens which has been cut to from L, and L_(2) is made of a material of refractive index 1.5 for the wavelength emitted by the source S. Radius of curvature of each surface of the lens is 18 cm. Source S is 30 cm along the axis fro Li and L_(2) . It is found that smallest distance from O along the screen at which intensity is half of the maximum intensity is 0.1 mm. It is also observed that the highest order of maximum intensity in the interference patter is 416. Q Fringe width is nearly |
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Answer» 0.6 MM |
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| 3184. |
Answer the following questions : (a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? (b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (c ) When a tiny circular obstacle is placed in the path of light from a distant source,a bright spot is seen at the centre of the shadow of the obstacle. Explain why? (d) Two students are separated by a 7 m partition wall in a room 10 m high. If both ligth and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily ? (e ) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of locaiton and several other properties of images in optical instruments. What is the justification? |
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Answer» Solution :(a) The angular size of central diffraction band reduces by half as if angular width of central maximum is given by `theta = lambda//d`. The overall AREA of central maximum thus reduces by `1//4^(th)`. Therefore, the intensity increases to four fold. (b) When the width of each slit is of the order of incident wavelength, the diffraction effects appear. Diffraction pattern is formed by each slit and then these two diffraction patterns are superimposed. The interference pattern in the double-slit experiment is modified by the diffraction pattern obtained from each of the two slit. (c ) WAVES diffracted from the edges of a tiny circular obstacle interfere constructively at the cantre of the shadow, thereby producing a bright spot at the centre. (d) For diffraction (bending of waves around an obstacle), the size of the obstacle (a) should be comparable to wavelength `(lambda)` of wave. If the size of the obstacle is too large compared to wavelength, diffraction observed is only by a small angle according to `sin theta = lambda//a`. Here, the size of wall is of the order of a few meters , i.e., 10 m. The wavelength of light is about `5 xx 10^(-7) m`. Thus, `sin theta = 5 xx 10^(-7)//10 = 10^(-8)m` `implies theta to 0`, which results in almost zero bending. Sound waves on the other hand of, say 1 kHz frequency, have wavelength of about 0.3 m. Thus, `sin theta = 0.3//10 = 0.03` Which gives a definite value of `theta`. Therefore, sound waves can BEND around the partition while light waves cannot, enabling students to converse without seeing each other. (e ) Typical sizes of apertures involved in ordinary optical instruments are much LARGER than the wavelength. Diffraction or bending of light is not significant in these instruments. |
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| 3185. |
In a Young's double slit experiment. Let A and B be the two slits. A thin film of thickness t and refractive index mu is placed in front of A. Let beta = frine width. The central maximum will shift : |
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Answer» towards A |
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| 3186. |
Two independent light sources are always incoherent. This is because phase of light emitted is random and phase difference, at any point, changes very rapidly with time. So as to obtain observable interference, we require coherent sources which emit light of same wavelength such that phase difference at any point does not change with time. In different interference experiments, the method to obtain two coherent sources could be different. In Fresnel's biprism experiment, two virtual images of a real source, which are formed due to refraction at the biprism, act as coherent source. In Lloyd’s mirror experiment, a real source and its virtual image formed due to reflection at a plane mirror act as coherent source. Two images of a source or a source and its image are coherent because of their phase correlation. It is also possible to obtain coherent sources by forming two images of a point source in the following manner.As shown in the figure, S is a monochromatic point source. A thin circular lens is cut into two identical halves L, and L_(2) by a plane passing through a diameter. L_(1) is kept above the axis and L_(2) below it is in a symmetrical manner such that the gap between the two is 0.1 mm. Each forms an image Qf the source and the two images, say, Si and S_(2) , act as coherent sources. Superposition of waves from these coherent sources will result in observable interference and a pattern can be obtained on a screen.The lens which has been cut to from L, and L_(2) is made of a material of refractive index 1.5 for the wavelength emitted by the source S. Radius of curvature of each surface of the lens is 18 cm. Source S is 30 cm along the axis fro Li and L_(2) . It is found that smallest distance from O along the screen at which intensity is half of the maximum intensity is 0.1 mm. It is also observed that the highest order of maximum intensity in the interference patter is 416. Q Distance of a point from O, at which the sixth order minimum is formed, will be |
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Answer» 1.4 mm |
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| 3187. |
The activity of a freshly prepared radioactive sample is 10^10 disintegrations per second, whose mean life is 10's. The mass of an atom of this radioisotope is 10^(-25) kg. The mass (in mg) of the radioactive sample is |
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Answer» 1 `N=((d N)/(dt))/(lambda) =10^(10) XX 10^(9) =10^(19)"atoms"` `m_("sample")=10^(-25) xx 10^(19)kg =1mg` |
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| 3188. |
What is the capacitance of an oscillating LC circuit if the maximum charge on the capacitor is 1.60 muC and the total energy is 140 muJ? |
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Answer» |
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| 3189. |
Electromagnetic waves consists of what type of varation of electric and magentic fields at right angles to each other and to the direction of propagation of wave. |
| Answer» SOLUTION :SINUSOIDAL | |
| 3190. |
What is a series resonance circuit ? |
| Answer» Solution :A CIRCUIT in which resistance , inductor and capacitor are CONNECTED in series and the MAXIMUM current flows through the circuit for a certain FREQUENCY is KNOWN as resonance frequency. | |
| 3191. |
A beam of light travelling in water falls on a glass plate immersed in water. When the incident angle is 51^(@), the reflected beam of light is found to be completely plane polarised. Determine therefractive index of glass. Given refractive idex of water =4/3. |
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Answer» |
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| 3192. |
How much work must be done to increase the speed of an electron (a) from 0.18c to 0.20c and (b) from 0.97c to 0.99c? Note that the speed increase is 0.01c in both cases. |
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Answer» |
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| 3193. |
In a hydrogen atom , the electron and proton are bound at a distance of about 0.53Å. (a) Estimate the potential energy of the system in e V, taking the zero of the potential energy at infiniate separation of the electron from proton. (b) What is the minimum work required to free the electron, given that is kinetic energy in the orbit is half the magnitude of potenital energy obtained in (a) ? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06Å separation ? |
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Answer» Solution :Here `q_(1) = - 1.6 xx 10^(-19) C`, `q_(2) = + 1.6xx 10^(-19) C`. `r = 0.53 lambda = 0.53 xx 10^(-10) m` Potential energy = P.E at `prop` - P. E at r `= 0 - (q_(1) q_(2))/(4 pi r epsilon_(0) r) = (- 9 xx 10^(9) (1.6 xx 10^(-19))^(2))/(0.53 xx 10^(-10))` `= - 43.47 xx 10^(-19)` Joule `= (-43.47 xx 10^(-19))/(1.6 xx 10^(-19)) eV = - 27.16 eV`. (b) K.E in the orbit `= (1)/(2) (27.16) eV` Total energy = K.E + P.E = 13.58 - 27.16 = - 13.58 eV WORK required to free the electron = 13.58 eV Potential energy at a seperation of `r_(1) (= 1.06 Å)` is `= (q_(1) q_(2))/(4 pi epsilon_(0) r_(1)) = (9 xx 10^(9) (1.6 xx 10^(-19)))/(1.06 xx 10^(-10))` `= 21.73 xx 10^(-19) J = 13.58 eV` Potential energy of the system, when zero of P.E is taken at `r_(1) = 1.06Å` is = P.E at `r_(1)` - P.E at r = 13.58 - 27.16 = - 13.58 eV By shifting the zero of petential energy work required to free the electron is not affected. It CONTINUES to be the same, being equal to + 13.58 eV. |
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| 3194. |
A man standing between two parallel hills fires a gun and hears two echoes, one 2.5 s and the other 3.5 s after the firing. If the velocity of sound is 330 ms^(-1), how long will it take him to hear the third echo? |
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Answer» 4s |
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| 3195. |
A block of mass 2 kg is dropped from a height of 0-4 mon a spring whose force constant is 1960 N/m. What will be the maximum distance x of the compression of the spring ? |
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Answer» 0.5 m `1/2 kx^(2)=mg(h+X)` `2(0.4+x)9.8=(1)/(2)1960xxx^(2)` `19.6(0.4+x)=980 x^(2)` Which gives on SOLVING, the quadratic `x=(1)/(10)=0.1 m` |
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| 3196. |
Equipotential surfaces, |
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Answer» Arecloserin regionsof largeelelctricfieldscomparedto regionsof lower electricfields Electric field and electric potential is RELATED to, `E=- (dV)/(dr)` `:. E prop (1)/(dr)` means the electric field INTENSITY E is inversely proportional to the separation between two equipotential surfaces. So equipotential surfaces are closer in regions of large electric fields. And the electric field is larger near the sharp edge due to larger charge density as A is very small surface charge density. `:. sigma= (V)/(A)` and according to E `= (kq)/(r^(2))` as `sigma` increases q increases and E increase . Hence, as the dimension of conductor increases electric potential and electric field DECREASES. Hence, near sharp edges of charged conductor both electric potential and electric field Increases and so equipotential surfaces are crowded. |
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| 3197. |
Give an expression for the electric potential at a point due toa point charge. |
| Answer» Solution :Electric POTENTIAL at any point due to a point CHARGE is defined as the amount of WORKDONE in bringing a unit POSITIVE charge (without acceleration) from infinity to that point against the electric field. | |
| 3198. |
Why are inductance coils made of copper? |
| Answer» Solution :It is DESIRABLE that an inductor should have as small RESISTANCE as possible. Since the RESISTIVITY of copper is very small, INDUCTANCE COILS are made up of copper. | |
| 3199. |
Marke the correct statements : |
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Answer» a given conducting SPHERE can be charged to any extent |
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| 3200. |
The mean lives of a radioactive substance are 1620 year and 405 year for alpha- emission respectively . Find the time during which three - fourth of a sample will decay if it is decaying both by alpha - emission and beta - emission simultaneously . |
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Answer» Solution :The decay constant `lamda` is the reciprocal of the mean life `TAU ` THUS, `lamda_(alpha)=1/(1620)` per year and `lamda_(beta) = 1/(405) ` per year `:.` TOTAL decay constant , `lamda=lamda_(alpha)+lamda_(beta)` (or) `lamda=-(1)/(1620)+1/(405)=1/324` per year . We know that `N=N_(0)e^(-lamdat)` When `3/4` th PART of the sample has disintegrated , `N = N_(0)//4` `:.N_(0)/4=N_(0)e^(-lamdat)("or")e^(lamdat)=4` Taking logarithm of both sides , we get `lamdat = log_(e)4` (or) `t=(1)/lamdalog_(e)2^(2) =2/lamdalog_(e) 2 = 449` year |
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