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InterviewSolution
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.
| 901. |
In the photoelectric phenomenon if the ratio of the frequency of incident radiationincident on a photosensitive surface is 1:2:3 the ratio of the photoelectric current is |
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Answer» `1:2:3` |
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| 902. |
In a conductor, the forbidden energy gap is of the order of |
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Answer» 1.1eV |
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| 903. |
(A): For a practical choke coil the power factor is very small (R): In a practical choke coil the power dissipation reduces if frequency of the a.c. is increased. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct explanation of 'A'. |
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| 904. |
A parallel beam of light of intensity I_0 is incident an a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected by lower surface. The ratio of maximum to minimum intensity is interference region of reflected rays is (Assume that cohernt light reflected from lower surface I refects from upper surface it entirely come out). |
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Answer» Solution :`I_(1)= IXX (25)/(100)= (I)/(4)` The intensity of transmitted LIGHT is `I.= I-(I)/(4)= (3I)/(4)` The intensity of light reflected fro lower surface is `I_(2)= (3I)/(4)xx(50)/(100)= (3I)/(8)` `(I"MAX")/(I"min")= ((SQRT(I_1)+sqrt(I_2))/(sqrt(I_1)+sqrt(I_2)))^(2)= ((sqrt(I/4)+sqrt((3I)/(8)))^2)/((sqrt(I/4)-sqrt((3I)/(8)))^2)=(((1)/(2)+sqrt((3)/(8)))^2)/((1)/(2)-sqrt((3)/(8)))^(2)`. 
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| 905. |
For a CE- transistor amplifier , the audio signal voltage across the collected resistance of 2 k Omega is 2 V . Suppose the current amplification factor of the transistor is 100 , find the input signal voltage and base current , if the base is 1 k Omega. |
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Answer» Solution :`V_(o) - 2V , R_(C) =2 K OMEGA = 2000 Omega , I_(B) = ? , V_(i) = ? , r_(i) = 1 k Omegabeta = 100` `V_(o)= I_(C) xx R_(C) , I_(C) = (V_(B))/(R_(C)) = (2)/(2000) = 10^(-3) A = 1 m A` `A_(V) = (V_(o))/(V_(C)) = BETA ((R_(C))/(r_(i)))` `V_(i) = V_(o) , (r_(i))/(beta R_(C)) = (2 xx 1)/(100 xx 2) = 0.01 V ` `beta = (I_(C))/(I_(B)) , I_(B) = (I_(C))/(beta)= (1)/(100) = 10^(-2) mA` |
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| 906. |
The organelle that differentiates plant cell and animal cell |
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Answer» Vacuoles |
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| 907. |
The proteins and lipids, essential for building the cell membrane, are manufactured by |
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Answer» ENDOPLASMIC reticulum |
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| 908. |
What is image ? Explain its types. |
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Answer» Solution :If rays EMANATING from a point actually meet at ANOTHER point after reflection and/or refraction, that point is called the IMAGE of the FIRST point. The image is real if the rays actually converge to the point, it is virtual if the rays do not actually meet but appear to diverge from the point when produced backwards. |
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| 909. |
Give analytical solution to a series RLC circuit connected to ac. |
Answer» Solution : The voltage equation for the CIRCUIT is `L(di)/(dt)+Ri+(q)/(c)=v=v_(m)SINOMEGAT` The solution to the equation will be `q=q_(m)SIN(omegat+theta)` `therefore q_(m)OMEGA[Rcos(omegat+theta)+(X_(C)-X_(L))sin(omegat+theta)]=v_(m)sinomegat` where, `X_(C)=(1)/(Comega)andX_(L)=Lomega` Also, `q_(m)omegaZ[(R)/(Z)cos(omegat+theta)+(X_(C)-X_(L))/(Z)sin(omegat+theta)]=v_(m)sinomegat`  Put, `(R)/(Z)=cosphiand((X_(C)-X_(L))/(Z))=sinphi""...(1)` Hence, `q_(m)omegaZ(cos(omegat+theta-phi))=v_(m)sinomegat` i.e., `[q_(m)omega(cos(omegat+theta-phi)]Z=v_(m)sinomegat` If `theta-phi=-pi//2`, then `cos(omegat+theta-phi)=cos(pi//2-omegat)=sin(omegat)` `therefore v_(m)=q_(m)omegaZ,i_(m)=q_(m)omegaandZ=(v_(m))/(i_(m))` Since, `i=(dq)/(dt)=(d)/(dt)(q_(m)sin(omegat+theta)=q_(m)omegacos(omegat+theta)` i.e., `i=i_(m)cos(omegat-(pi)/(2)+phi)=i_(m)cos((pi)/(2)-(omegat+phi))` i.e., `i=i_(m)sin(omegat+phi)` By using (1), `tanphi=(X_(C)-X_(L))/(Z)` graphically we represent it as in fig (i) and fig (ii) Hence, `v^(2)=v_(R)^(2)+(V_(C)-V_(L))^(2)` using fig (ii) or `Z=((v_(m))/(i_(m)))=sqrt(R^(2)+(X_(C)-X_(L))^(2))` 
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| 910. |
A short bar magnet is placed horizontallyin the magnetic meridian with its north pole pointing north and the neutral points are found 0.15m from the centre of the magnet. Find the distance between the neutral points if the magnet is reversed. |
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| 911. |
A conducting rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends.An external magnetic field of 2T is applied normal to the rod. Now, the current to be passed throughthe rod so as to make the tension in the wires zero is : (Take g = 10 ms^(-2)) : |
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| 912. |
Using Kirchhoff's rules, calculate the potential difference between B and D in the circuit diagram as shown in Fig. |
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Answer» SOLUTION : LET currents flowing through different branches of the circuit beas shown in Fig. Then, as per Kirchhoff.s first rule `I = I_1 + I_2` Applying Kirchhoff.s second rule to MESH BADB, we have `I_1. 2 - I_1.1+ I_2.2= 2 - 1 ` ` RARR I_1 + 2I_2 = 1` Similarly for mesh DCBD, we have `I.3 - I.1 - I_2.2 = 3 - 1 rArr 4I - 2I_2 = 2` `4(I_1 + I_2) - 2I_2 = 2 " or4I_1 + 2I_2 = 2`...(ii) Solving (i) and (ii), we get`I_1 = I_2 = 1/3A` ` therefore `Potential difference between B and D, `V_(BD) = I_2 xx2= 5 xx2=+ 2/3 ` |
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| 913. |
In one refrigeration cycle, the energy absorbed as heat by the refrigerant from the cold reservoir is Q_c and the energy discharged as heat to the hot reservoir is Q_h. What is the coefficient of performance of the refrigeration? |
| Answer» SOLUTION :The COEFFICIENT of PERFORMANCE of a REFRIGERATION, `alpha=Q_c/(Q_h-Q_c)`. | |
| 914. |
Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is |
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Answer» `(Wtantheta)/(mu)` Weight of block A = W. T = tension in the string Normal on block B, N= W FRICTION force on block B,f= μN= μW  System will be in EQUILIBRIUM, if `Tcostheta=f=muW` ...(i) `Tsintheta=W.` ...(ii) Divide eqn (ii) by eqn (i), we get `(sintheta)/(COSTHETA)=(W.)/(muW),thereforeW.=muWtantheta` |
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| 915. |
A monochronicgreen light of wavelength 5xx10 m illuminates |
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Answer» 0.25 mm |
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| 916. |
Explain amplitude modulation ? |
| Answer» Solution :When the modulation wave is superimposed on high frequency carrier wave in a manner that the frequency of the modulated wave is the same as that of the carrier wave but its AMPLITUDE is modified in ACCORDANCE with that of the MODULATING wave, the PROCESS is called amplitude modulation. | |
| 917. |
Which electromagnetic waves can be used to obtain electrical energy ? |
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Answer» RADIO waves |
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| 918. |
The equation of A.C. voltage is given by V = 158 sin 200pit. The value of voltage at time t =1/400sec is …... V. |
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Answer» `-79` `THEREFORE V=158 sin 200pixx1/400` `therefore V=158 sin (pi/2)` `therefore` V=158 Volt [`because "sin" pi/2 =1`] |
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| 919. |
Inverse square law for illuminance is valid for |
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Answer» Isotropic POINT SOURCE |
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| 920. |
A particle of rest mass m_0 starts moving at a moment t=0 due to a constant force F. Find the time dependence of the particle's velocity and of the distance covered. |
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Answer» SOLUTION :The equation of MOTION is `(d)/(DT)((m_0v)/(sqrt(1-v^2/c^2)))=F` Integrating `=(v//c)/(sqrt(1-v^2/c^2))=(beta)/(sqrt(1-beta^2))=(F_t)/(m_0c)` using `v=0` for `t=0` `(beta^2)/(1-beta^2)=((Ft)/(m_0c))^2` or, `beta^2=((Ft)^2)/((Ft)^2+(m_0c)^2)` or, `v=(Fct)/(sqrt((m_0c)^2+(Ft)^2))` or `x=int (Fctdt)/(sqrt(F^2t^2+m_0^2c^2))=c/Fint(xidxi)/(sqrt(xi^2(m_0c)^2))=c/Fsqrt(F^2t^2+m_0^2c^2)+const ant` or using `x=0` at `t=0`, we get, `x=sqrt(c^2t^2+((m_0c^2)/(F))^2)-(m_0c^2)/(F)` |
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| 921. |
The ground state energy of hydrogen atom is - 13.6 eV. What is the potential energy of the electron in this state? |
| Answer» Solution :`U = 2E = 2 (- 13.6eV) = - 27.2 eV` | |
| 922. |
As the temperature rises,,the resistance offered by metals_____. |
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| 923. |
Poet talks about the damage that has been done to the.................. |
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Answer» Environment |
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| 924. |
In a n-p-n transistor, the collector current is 10 mA . If 90%the electrons emitted reach the collector. |
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Answer» the emitter current will be9mA `=(90)/100 I_e=9/10 I_e:. I_e=10/9I_c` . `=10/9xx10=11.1 mA` Now `I_b=I_e -I_c = 11.1 -10=0.1mA` |
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| 925. |
The distance of geostationary satellite from centre of earth is nearest to : |
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Answer» Solution :Height of GEOSTATIONARY satellite from surface of earth is 36000 km and height from centre of earth is 42400 km which is nearly equal to 7R. Thus CORRECT choice is (c ). |
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| 926. |
A nonconducting of radius r has charge Q. A magnetic field perpendicular to the plane of ring changes at the rate (dB)/(dt). The torque experienced by the ring is |
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Answer» zero |
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| 927. |
What are electromagnetic waves ? Write the expression for the velocity of electromagnetic waves in terms of permittivity and magnetic permeability of free space. |
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Answer» Solution :Waves radiated by accelerated charges and consist of time varying, TRANSVERSE electricand magnetic fields are CALLED electromagnetic waves. `C=(1)/( sqrt(mu_0 epis_0))` Where .C.is VELOCITY of light `mu_0` permeability of free space `epsi_0`PERMITTIVITY of free space |
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| 928. |
An a.c. voltage represented be e = 310 sin 314 t is connected in series to a 24Omega resistor, 0.1 Hinductor and a 25 muFcapacitor. Find the value of the peak voltage rms voltage frequency rectus of the circuit impedance of the circuit and phase angle of the current. Data : R = 24 Omega, L = 0.1 H, C = 25 xx 10^(-6)F |
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Answer» Solution :E = 310 sin 314 t …(i) and `r = E_(0) sin omega t` …(ii) comparing equations (i) & (ii) `E_(0) = 310 V` `E_(rms) = (E_(0))/(sqrt(2)) = (310)/(sqrt(2)) = 219.2 V` `omegat = 314 t` `2pi v = 314` `v = (314)/(2 xx 3.14) = 50 Hz` Reactance `= X_(L) - X_(C) = Lomega - (1)/(C omega` `= L.2pi v - (1)/(C.2pi v)` `= 0.1 xx 2 pi xx 50 - (1)/(25 xx 10^(-6) xx 2pi xx 50)` `= 3.14 - 127.4` `= -96 Omega` `X_(L) - X_(C) = -96 Omega` `:. X_(C) - X_(L) = 96 Omega` `Z = sqrt(R^(2) + (X_(C) - X_(L))^(2))` `= sqrt(24^(2) + 96^(2))` `= sqrt(576 + 9216)` `= 98.9 Omega` `tan phi = (X_(C) - X_(L))/(R )` `= ((127.4 - 31.4)/(24))` `tan phi = (96)/(24) = 4` `phi = 76^(@)` Predominance of capacitive reactance signify that current leads the EMF by `76^(@)`. |
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| 929. |
The root mean square velocity of the molecules of a gas is 1260 m/s. The average speed of the molecules is |
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Answer» 1029 m/s |
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| 930. |
A stationary wave is produced if the individual wave is |
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Answer» PROGRESSIVE |
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| 931. |
Define electric intensity at a point in an electric field. |
| Answer» SOLUTION :It is DEFINED as the electrostatic force per unit POSITIVE charge placed at that POINT. | |
| 932. |
The figure shows four situations in which light reflects perpendicularly from a thin film of thickness L, with indexes of refraction as given. (a) For which situations does reflection at the film interfaces cause a zero phase difference for the two reflected rays? (b) For which situations will the film be dark if the path length difference 2L causes a phase difference of 0.5 wavelength? |
| Answer» SOLUTION :(a) 1 and 4, (B) 1 and 4 | |
| 933. |
Mention the value of power factor of a pure capacitor. |
| Answer» SOLUTION :The POWER FACTOR of a PURE capacitor is ZERO. | |
| 934. |
Light of wavelength 500 nm Pass through a slit of 0.2 mm wide. The diffraction pattern is formed on a screen 60cm away. Determine the. (i) angular spread of centeral maximum (ii) the distance between the central maximum and the second minimum. = |
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Answer» Solution :`lambda=500nmxx500xx10^(-9)m,a=0.2xx10^(-3)m,D=60cm=60xx10^(-2)m` (i) Equation for diffraction minimumis, `a sin 0 = n lambda` The central maximum is spread up to the first minimum. Hence, n = 1 Rewriting, `sin 0 (lambda)/(a) (or) = sin^(-1) ((lambda)/(a))`  `theta = 0.0025 rad` (ii) To find the value of `y_(1)` for maximum, which is spread up to first minimum with `(n = 1) is, a sin theta = lambda`  As `theta` is very small, `sinthetaapproxtantheta=(y_(1))/(D)` `a(y_(1))/(D)=lambda "rewriting", lambda_(1)=(yD)/(a)` Substituting, `y_(1)=(500xx10^(-9)xx60xx10^(-2))/(0.2xx10^(-3))=1.5xx10^(-3)=1.5 mm` To find the value of `y_(2)` for SECOND minimum with `(n=2)is,asintheta=2lambda` `a(y_(2))/(D)=2lambda "rewriting", y_(2)=(2lambdaD)/(a)` Substituting, `y_(2)=(2xx500xx10^(-9)xx60xx10^(-2))/(0.2xx10^(-3))=3xx10^(-3)=3mm` The distance between the central maximum and second minimum is, `y_(2) - y_(1) y_(2) - y_(1) = 3mm - 1.5 mm` NOTE: The above calculation SHOWS that in the differactionpatten caused by single slit, the width of each maximum is equalwith centralmaximum as the double that of others. But the bright and dark FRINGES are not of equal width. |
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| 935. |
What is polarisation of light? Name any one method of producing plane polarised light. |
| Answer» Solution :The phenomenon of confining the vibrations of the ELECTRIC vector in a specific direction perpendicular to the direction of WAVE MOTION is called polarisation, The method of producing plane POLARISED light is reflection The other methods are by refraction, double refraction, scattering and selective ABSORPTION. | |
| 936. |
Give any one defination of power factor. |
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Answer» Solution :The POWER FACTOR is DEFINED as the RATIO of true power to the apparent power of an a.c. circuit. It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit. `COS phi=(True power )/(Apparent power)=(P_(av))/(V_(rms)I_(rms))` |
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| 937. |
A pair of physical quantities having same dimensional formula is |
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Answer» FORCE and TORQUE |
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| 938. |
The emf of the driver cell in the potentiometer experiment should be greater than the emf of the cell to be determined. Why? |
| Answer» SOLUTION :Ifemfofa drivercell is LESS,then null POINT will notbe obtained onthe POTENTIOMETER wire. | |
| 939. |
A particle executing S.H.M. has total energy of 20 J. If the potential energy of the particle midway between mean and extreme position is 5 J, then the average total energy will be |
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Answer» 10 J |
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| 940. |
The atom ""_8O^(16) has 8 protons 8 neutrons and 8 electrons while atom ""_4Be^8 has 4 proton, 4neutrons and 4 electrons, yet the ratio of their atomic masses is not exactly 2. Why ? |
| Answer» Solution : DUE to MASS defect or DIFFERENT binding energies | |
| 941. |
In the givencircuit , calculate the (i) effective resistance between A and B (ii) current through the circuit and (iii) current through 3 Omega resistor. |
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Answer» SOLUTION :`R_(1) = 12 Omega ""R_(3) = 4 Omega` `R_(2) =8Omega ""R _(4) =6 Omega` ` R_(5) =3 Omega ""R_(6) = 2 Omega` Now ` R _(1) || R_(2)` `R _(n et1 ) = (R_(1) R_(2))/(R_(1) + R _(2)) = (12xx 8)/( 12 +8)` Again `R _(n et1) = 4.8 Omega` `R _(3) || R_(4)` ` R _( n et2) = ( R _(3) R_(4))/(R_(3) + R_(4)) = ( 4xx 6)/( 4 + 6)` `R _(n et2) =2.4 Omega` Now `R_(n et1)` is series to `R _( n et2)` `R _( n et3) = R _(n et1 ) + R _(n et2)` `R _(n et3) = 7.2Omega` Here `R _(n et3) || R_(5) ||R_(6)` `(I)/( R _(eff)) = (1)/(7.2) + 1/3 + 1/2 = (70)/(72)` `R _(eff) = (72)/(70 ) = (36)/(35)` Now `I = V/R = (35)/(36) xx 12` `I = ( 12 xx 35)/(36)` `I = 11. 66 A` Current through `3 Omega` resistor is `I = V/R = (12)/(3), I = 4 A.` |
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| 942. |
A dust particle of mass 10^(-6) gm moves between two walls separated by 1.0 mm. If speed of the particle is 10^(-4)ms^(-1) then nth state of the particle will be : |
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Answer» `3 xx 10^(10)` `n^(2)=(4M^(2) V^(2) l^(2))/(h^(2)) RARR n=(2mvl)/(h)` `rArr n=3 xx 10^(17)` |
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| 943. |
Uniform rope of length Tlies on a table with coefficient of friction between the rope and table being u. What is the maximum length of the rope which can over hang from the edge of the table without sliding down? |
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Answer» `(1)(mu)` `m (l-x) gmu = mx.g or mul-mu x = x` or`x=(mul)/(mu+1)` Hence correct CHOICE is (c). 37. The block will move if mg sin is greater than or equal to force of friction. |
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| 944. |
A source .S. emitting a sound of frequency 300 Hz is fixed on block .A. while detector is fixed on block .B. detects the sound. The block .A. and .B. are simultaneously displaced towards each other through a distance of 1 m and then left to vibrate. If the velocity of sound in air is 340ms^(-1)and S , S_(A), S_(B)are identical the frequency of block A is 2Hz. Then |
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Answer» Maximum velocity of DETECTOR is `4pi ms^(-1)` |
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| 945. |
Classify the following six nuclides into (i) isotones, (ii) isotopes, and (iii) isobars : " "_(6)^(12)C, " "_(2)^(3)He, " "_(80)^(198)Hg, " "_(1)^(3)H, " "_(79)^(197)Au, " "_(6)^(14)C |
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Answer» SOLUTION :(i) `" "_(80)^(198)HG` and `" "_(79)^(197)Au` are isotones because both nuclides have 118 neutrons. (ii) `" "_(6)^(12)C` and `" "_(6)^(14)C` are isotopes because their atomic numbers are same but mass numbers are different. (iii) 3He and 3 H are isobars because their mass numbers are same but atomic numbers are different. |
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| 946. |
You are given 10 resistors each of resistance 2 Omega. First they are connected to obtain possible minimum resistance. Then they are connected to obtain possible maximum resistance. The ratio of maximum and minimum resistance is .... . |
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Answer» Solution :100 Minimum RESISTANCE : `R_("max")= 10 xx 2 = 20 Omega` Minimum resistance : `R_("MIN") = (2)/(10)= 0.2 Omega` `THEREFORE (R_("max"))/(R_("min")) = (20)/(0.2) = 100` |
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| 947. |
A smooth horizontal disc rotates with a constant angular velocity omega about a stationary vertical axis passing through its centre, the point O. At a moment t=0 a disc is set in motion from that: point with velocity v_0. Find the angular momentum M(t) of the disc relative to the point O in the reference frame fixed to the disc. Make sure that this angular momentum is caused by the Coriolis force. |
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Answer» Solution :The Coriolis force is `(2moverset(rarr')vxxvecomega)`. Here `vecomega` is along the z-axis(VERTICAL). The MOVING disc is moving with velocity `v_0` which is constant. The motion is along the x-axis SAY. Then the Coriolis force is along y-axis and has the magnitude `2mv_0omega`. At time t, the DISTANCE of the centre of moving disc from O is `v_0t` (along x-axis). Thus the TORQUE N due to the coriolis force is `N=2mv_0omega*v_0t` along the z-axis. Hence equating this to `(dM)/(dt)` `(dM)/(dt)=2mv_0^2omegat` or `M=mv_0^2omegat^2`+constant. The constant is irrelevant and may be put equal to zero if the disc is originally set in motion from the point O. This discussion is approximate. The Coriolis force will cause the disc to swerve from straight line motion and thus cause deviation from the above formula which will be substantial for large t. |
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| 948. |
If the coefficient of friction is sqrt3, the angle of friction is |
| Answer» Answer :B | |
| 949. |
What are the disadvantages of using a reflecting telescope? |
| Answer» Solution :Disadvantages : The objective mirror WOULD FOCUS the light inside the TELESCOPE TUBE. (ii) ONE must have an eye piece inside obstructing some light. This problem could also be overcome by introducing a secondary mirror which would take the light outside the tube for view. | |
| 950. |
If the length of the tube of a compound microscope is increased , the magnification increases -Is this statement trueof false ? |
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