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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.
| 39401. |
A tunning fork of frequency 500 Hz is sounded on a resonance tube. The first & second resonances are obtained at 17 cm and 52 cm. The velocity of sound in m/s : |
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Answer» 170 Now `l_(2) - l_(1) = LAMBDA//2"" rArr lambda = 2 (52 - 17) = 70 ` cm Now v = v `lambda = 500 xx 0.7 = 350MS^(-1)`. hence correct choice is (b) . |
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| 39402. |
A capacitor of 2.5 mu F is charged through a resistor of 4 M Omega. In how much time will potential drop across capacitor will become 3 times that of resistor (ln 20.693) |
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Answer» 13.86 s where t is any TIME during the charging. For the given circuit, as per the conditions given `V_(c )(t)=(3)/(4)th` of the voltage applied so, `(3V_(0))/(4)=V_(0)[1-e^(-t//RC)]` or, `e^(-t//RC)=(1)/(4)` `RARR t=2RC log_(e )2` `rArr t=2xx4xx10^(6)xx2.5xx10^(-6)xx0.693=13.86 s`. |
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| 39403. |
In the circuit shown in the figure X_(L)=(X_(C))/(2))=R the peck value current i_(0) is |
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Answer» An INDUCTOR of 0.103H |
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| 39404. |
a. What does the graph represent? b.What is the importance of this graph? c. What does the peak in the graph represent? |
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Answer» Solution :a. Binding energy/nucleon with atomic MASS. b. It shows the stability of atomic NUCLEUS. c. The maximum stable condition of an ELEMENT. |
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| 39405. |
The number of turns in the primary and secondary turns of a transformer are 1000 and 3000 respectively. If 80 volt A.C. is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be |
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Answer» 240 volt |
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| 39406. |
Itis now established that protonsand neutronsare themselves built out of more elementary units called quarks a proton and a neutron consist ofthree quarks eachtogether with electronsbuild up ordinary mattersuggest a possible quark composition of a protonand neutron |
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Answer» <P> SOLUTION :p,UUD, N,UDD |
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| 39407. |
A particle performings S.H.M. with amplitude 'a' has maximum velocity v, its speed at displacement a//2 will be |
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Answer» 2 v |
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| 39408. |
From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is |
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Answer» `4M^(2)` `THEREFORE` Mass of removed portion `=(9M)/(piR^(2))xxpi(R/3)^(2)=M` LET moment of INERTIA of removed portion = `I_(1)` `thereforeI_(1)=M/2(R/3)^(2)+M((2R)/3)^(2)`, by theorem of parallel or `I_(1)=(MR^(2))/2` Let `I_(2)` = Moment of inertia of the whole disc `I_(2)=(9MR^(2))/2` `therefore` Let I = Moment of inertia of remaining disc `thereforeI=I_(2)-I_(1)` or `I=(9MR^(2))/2-(MR^(2))/2=(8MR^(2))/2=4MR^(2)` or `I=4MR^2` |
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| 39409. |
The figure shows a surface XY separating two transparent media, medium - 1 and medium - 2. The lines ab and cd represent wavefronts of the light ab and cd represent wavefronts of the light wave travelling in medium-1 and incident on XY. The lines of ef and gh represent wavefronts of the light wave in medium - 2 after refraction. Answer the following question : Speed of light is: |
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Answer» the same in medium-1 and medium - 2 |
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| 39410. |
The figure shows a surface XY separating two transparent media, medium - 1 and medium - 2. The lines ab and cd represent wavefronts of the light ab and cd represent wavefronts of the light wave travelling in medium-1 and incident on XY. The lines of ef and gh represent wavefronts of the light wave in medium - 2 after refraction. Answer the following question : Light travels as a : |
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Answer» PARALLEL beam in each medium |
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| 39411. |
A weightless spring of length 60 cm and force constant 100" Nm"^(-1) is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is : |
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Answer» `(PI)/(20)s` `:. K_(p)=200+200=400" N"//"m"` `T=2pi sqrt((m)/(k_(p)))=2pi sqrt((0.25)/(400))=(2pi xx5)/(200)=(pi)/(20)` seconds. Correct choice is (a). |
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| 39412. |
A siren emitting a sound of frequency 1000 Hz moves away from you towards a cliff at a speed of 10 m/s. (a) What is the frequency of the sound you hear coming directly from the siren ? (b) What is the frequency of the sound you hear reflected off the cliff (c) What beat frequency would you hear ? Take the speed of sound in air as, 330 m/s. |
Answer» Solution :The situation is as shown in figure.  (a) FREQUENCY ofsound reaching DIRECTLY to us (by S) `f_1 (v/(v +v_s)) f = (330/(330 +10)) (1000) = 970.6 Hz` (b) Frequecny of SOUND which is reflected of from the cliff is same as from S. `f_2 = (v/(v -v_s)) f = ((330)/(330 "" 10))(1000)` ` = 1031.3 Hz` (c ) Beat FREQUNCY `=f_2-f-1=60.7 Hz` (Too high to be heared as BEATS) |
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| 39413. |
Obtain the equation for radius of illumination (or) Snell's window. |
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Answer» Solution :The radius of Snell.s window can be deduced with the illustration as shown an figure. Light is seen form a point A at a depth The Snell.s law in PRODUCT form. Equation `n_(2) sin I = n_(2) sin r` For the REFRACTION happening at the point B on the boundary between the two media its. `n_(1)sini_(c)=n_(2)sin90^(@)` `n_(1)sini_(c)=n_(2) "" thereforesin90^(@)=1` `sini, = (n_(2))/(n_(1))` From the RIGHT angle triangle `DELTA ABC`. `sini_(c)=(CR)/(AB)=(R)/(sqrt(d^(2)+R^(2))) "" ...(3)` Equating tha above two equation (3) and equation (2). `(R)/(sqrt(d^(2)+R^(2)))=(n_(2))/(n_(1))` Suquaring on both sides. `(R^(2))/(sqrt(R^(2)+d^(2)))=((n_(2))/(n_(1)))^(2)` Taking reciprocal. `(R^(2)+d^(2))/(R^(2))=((n_(1))/(n_(2)))^(2)` On further SIMPLIFYING. `1+(d^(2))/(R^(2))=((n_(1))/(n^(2)))^(2):(d^(2))/(R^(2))=((n_(1))/(n_(2)))-1:(d^(2))/(R^(2))=(n^(2))/(n_(2)^(2))-1=(n^(2)-n^(2))/(n_(2)^(2))` Again taking reciprocal and rearranging. `(R^(2))/(d^(2))=(n_(2)^(2))/(n_(1)^(2)-n_(2)^(2)): R^(2)=d^(2)((n_(2)^(2))/(n_(1)^(2)-n_(2)^(2)))` The radius of illumilation is, `R=dsqrt(n_(2)^(2)/(n_(1)^(2)-n_(2)^(2)))` If the rarer mediumoutside is air, then, `n_(2) = 1`, and we can take `n_(1) = n` `R-d((1)/sqrt(n^(2)-1))(or)R=(d)/(sqrt(n^(2)-1))` 
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| 39414. |
(a) The peak voltage of a.c. supply is 600 V. what is its rms voltage ? (b) The rms value of current in a.c. circuit is 20 A. What is its peak current ? |
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Answer» Solution :(a) Here `E_(0) = 600 V` `therefore E_(rms) = (E_(0))/(sqrt(2)) = (600)/(sqrt(2)) = 424.3` (b) Here `I_(rm) = 20 A`. We know `I_(rms) = (I_(0))/(sqrt(2))` `therefore I_(0) = sqrt(2)I_(rms)` or `I_(0) = sqrt(2) XX 20 = 1.414 xx 20 = 28.28 A` |
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| 39415. |
Two point charges , each of1 mu C,are separated from each other by a distance of 10 cm in air . Electrostatic force acting between them is ____________ |
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Answer» |
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| 39416. |
What is a hole ? Which type of doping creates a hole ? |
| Answer» Solution :A HOLE is vacant STATE in the covalent bond of a semiconductor from which an ELECTRON has been removed. When semiconductor is doped with trivalent impurity ATOMS HOLES are created. | |
| 39417. |
Name one purpose for which micro waves are used ? |
| Answer» SOLUTION :TPH and TV COMMUNICATIONS. | |
| 39418. |
A particle that carries a charge .-q. is placed at rest in uniform electric field 10N/C. It experiences a force and moves. In a certain time .t., it is observed to acquire a velocity 10 bat(i) - 10 hat(j) m/s. The given electric field intersects a surface of area 1 m^(2) in the x-z plane. Find the Electric flux through the surface. |
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Answer» SOLUTION :Force on charge `BAR(F)= q bar(E )` `therefore` particle MOVES opposite to `bar(E )" with " bar(V)` unit VECTOR in the direction of `bar(V ) " is " (bar(i))/(sqrt2) - (bar(j))/(sqrt2)` unit vector in the direction of `bar(E ) " is" (bar(i))/(sqrt2) - (bar(j))/(sqrt2)` `bar(E ) = 10 [(-i)/(sqrt2) + (j)/(sqrt2)]` ie `bar(A) = 1 xx bar(j)` Electric flux `phi = bar(E ).bar(A )= 5 sqrt2 Nm^(2)//C` |
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| 39419. |
An electric dipole consists of two opposite charges each of0.05muC separated by 30 mm. The dipole is placed in an uniform external electric field of 10^6NC^1. The maximum torque experienced by the field on the dipole is : |
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Answer» `6xx10^-3Nm` |
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| 39420. |
A sinusoidally alternating emf is applied to an LCR series circuit . (i) Define the impedance of thecircuit . (ii) Obtain theexpressions for the impedance and the phase difference between theemf andcurrent from the phasor diagram. (iii) Discuss thebehaviour of thecircuit for different relative values of thetwo reactances in thecircuit. |
Answer» Solution :Suppose a sinusoidally alternating emf E, of peak value `E_(o)` and frequency f , is APPLIED to a circuit containing a pure inductor ( inductance L), a RESISTOR ( resistance R) anda capacitor (capacitance C), all in series, Fig (a).  (i) The impedance Z of the circuit is defined as theeffective resistance offfered to analternating current and is given by` Z = V_(rms) //I_(rms)` . (ii) Let the current at any instant be`I = I_(o) sinomega t`, where `I_(o)= E_(o)//Z -=` the peak current and ` omega = 2 pi f`. Thep.d. `V_(R)` across theresistor is in phase withI and isgiven by `V_(R) = I R`. Thep.d. `V_(L)` acrose the inductor leads `90^(@)` on I, and `V_(L(rms)) = omega L I _(rms)= X_(L) I_(rms)`, where ` X_(L) = omega L` = the INDUCTIVE reactance . The p.d. `V_(C)` across the capacitor LAGS ` 90^(@)` on I, and`V_(C(rms))=(I_(rms))/(omegaC) = X_(C) I_(rms)`,where `X_(C) = 1//omegaC` =the capacitive reactance. Since the current is thesame for all the circuit elements, this is thefirst phasor drawn in thephasor diagram, Fig. (b) . Other phasors representing the instantaneous values of the sinusoidally alternating quantities arethen drawn in relation to this. In can be seen in thephasor diagram, that the p.d.s `V_(L) and V_(C)` are in antiphase so that the magnitude `E_(rms)` of thevector sum of all three p.d.s is `E_(rms) = sqrt(V_(R)^(2) +(V_(L) -V_(C))^(2))` ` = I_(rms) sqrt(R^(2) +(X_(L)-C_(C))^(2))` Hence, theimpedance of thecircuit is ` Z = sqrt(R^(2) +(X_(L)-X_(C))^(2))` ` = sqrt(R^(2)+(omegaL - 1/(omegaC))^(2))` The phase difference `phi` between the emf and current is given by ` tan phi = (V_(L) -V_(C))/V_(R) =(X_(L)-X_(C))/R = (omegaL - (1//omegaC))/R` (iii) If `X_(L) gt X_(C)`,the circuit behaves inductively and the current legs onp.d. For `X_(L) lt X_(C)`, the circuit behaves capacitively and the current leads on p.d. (`phi` negative). For ` X_(L) = X_(C), phi = 0`, the current and emf arein phase, and the circuit is purely resistive. |
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| 39421. |
A convex spherical refracting surface with radius R separates a medium having refractive index 5/2 from air. As an object is moved towards the surface from far away from the surface along the principle axis, its image |
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Answer» CHANGES from REAL to virtual when it is at a distance R from the SURFACE |
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| 39422. |
What is the de-Broglie wavelength associated with an electron,accelerated through a potential difference of 100 volts? |
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Answer» 0.01227 nm |
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| 39423. |
(a) Ratio of widths of two slits in Yougn's double-slit experiment is 4:1. evaluate the ratio of intensity at maxima and minima in the interference pattern. (b) Does the appearance of bright and dark fringes in the interference pattern violate, in any way, conservation of energy ? Explain. |
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Answer» SOLUTION :(a) Here RATIO of the widths of SLITS `r=(w_(1))/(w_(2))=(4)/(1)=4 or SQRT(r)=sqrt(4)=2`. `therefore (I_(max))/(I_("min"))=(sqrt(r)+1)^(2))/((sqrt(r)-1)^(2))=((2+1)^(2))/((2-1)^(2))=(9)/(1)impliesI_(max):I_("min")=9:1` (B) N/A |
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| 39424. |
If a shunt(1//100)^th of the coil r to moving coil galvanometer, then its sensitivity becomes _______ fold. |
| Answer» Answer :C | |
| 39425. |
A slab of material has area A, thickness L, and thermal conducitivy k. One of its surface (P ) is maintained at temperature T_(1) and the other surface (Q ) is maintained at a lower temperature T_(2). The rate of heat flow from P to Q is : |
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Answer» `(kA( T_(1) - T_(2)))/(L^(2))` |
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| 39426. |
An achromatic telescope objective is to be made by combining the lenses of flint and crown glasses. This proper choice is |
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Answer» CONVERGENT of CROWN and DIVERGENT of flint |
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| 39427. |
A stone thrown vertically up with velocity reaches three points A,B and C with velocities (V)/(2),(V)/(4) and (V)/(8) respectively.Then AB:BC is |
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Answer» `1:1` |
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| 39428. |
In the above question what should be the minimum height of plane so that it may not be hit by the shell ? (g = 10 ms^(-2) ): |
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Answer» 1 km `H=(u^(2)sin^(2)theta)/(2G)=((200sqrt(2))^(2)xx(1/sqrt(2))^(2))/(2XX10)` H=2000m=2km |
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| 39429. |
A mass hangs at theend of a massless spring and oscillates up and down at its natural frequency f. If the spring is cut at the midpoint and and mass reattached at the end, the frequency of oscillation is : |
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Answer» `sqrt(2)f` `v=(1)/(2PI) sqrt((K)/(m))` Since spring is cut into EQUAL parts `:.` spring constant becomes two times and hence frequency becomes `sqrt(2)` times So correct choice is a. |
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| 39430. |
In the circuit shown, each capacitor has a capacitance C. The emf of the cell is E. If the switch S is closed |
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Answer» POSITIVE CHARGE will flow out of the positive terminal of the CELL |
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| 39431. |
A particle moves along x-axis and its acceleration at any time t is a = 2 sin (pit), where t is in seconds and a is in m/s^2 . The initial velocity of particle (at time t = 0) is u = 0. Q. The velocity of a particle moving in the direction ofx-axis varies as v = alpha, x, where alpha is a constant. Atthe moment t = 0, the particle was located at x = 0,thenIf the average velocity and average acceleration overthe time that the particle takes to cover first s meterof the path are equal then find out value of alpha. |
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Answer» 1 |
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| 39432. |
Assertion : In a metal all the free electrons have same energy. Reason : Electrons do not obey Pauil's exclusion principle. |
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Answer» If both ASSERTION and reason are TRUE and reason is the correct explanation of assertion Assertion is correct in the sense that the free electrons occupy a single ENERGY band. But a band has a distribution of energy. One can treat assertion as correct. But the reason is false because electrons in a metal follow Pauli's exclusion preinciple. It is because of this, one different BANDS. Therefore the answer is. |
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| 39433. |
Light waves travel from optically rarer medium to optically denser medium. Its velocity decreases because of change in ....... |
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Answer» frequency |
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| 39434. |
The potential barrier of a P-N junction diode is 50 meV, When an electron having energy 400 meV travels from N to P, after crossing the junction the energy of the electron is |
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Answer» 450 me V |
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| 39435. |
A swimmer whishes to cross a river 500 m wide flowing at a rate 'u'. His speed with respect to still water is 'v'. For this, he makes an angle theta with the perpendicular as shown in the figure. For u = 5 km//hr andv=3km//hr, the swimmer : |
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Answer» Can reach B in 7.5 MIN. MINIMUM drift can't be zero. |
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| 39436. |
A circuit consists of a resistance of 10 Omega and a capacitance of 0.1 mu F. If an alternating e.m.f. of 100 V, 50 Hz is applied, calculate the current in the circuit. |
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Answer» |
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| 39437. |
In beta^(1) decay, a |
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Answer» <P>neutron converts into a proton emitting antineutrino. |
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| 39438. |
Photoelectric effect experiments are performed using three different metal plates p, and r having work functions phi_(P)=2.0eV,phi_(q)=2.5eV and wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (Take hc = 1240 eV nm) |
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Answer»
Wavelength less than `lambda_("max")` alone will cause photelectrons to be ejected `lambda_(m)=(HC)/(phi)` For `phi_(p),lambda_(m_(p))=(1240eV nm)/(2.5eV)=496 nm` For `phi_(r),lambda_(m_(r))=(1240eV nm)/(3eV)=413.3 nm` Wavelngths in the incident beam are 550 nm, 450 nm and 350 nm. 350 nm waves can generate photoelectrons from p,q and r. 450 nm is shorter than `lambda_(m)` for p and q only and 550 nm `lt` 620 nm only in this group. so it can excite on p-cell. Currept `prop` intensity. Intensity =Nhv of photoelectrons. `:.` I is maximum for p cell, one gets the maximum intensity, and next is for q cell and the r cell can give photelectrons only by 350 nm. `:.` I is minimum for r. |
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| 39439. |
An e.m.f.E=4cos(1000t)volt is applied to a series LR circuit to inductance 3mH and resisrance 4 ohms. What is the amplitude of current in the circuit ? |
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Answer» 1.0A |
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| 39440. |
A spring of constant 5 xx 1 0 ^(3)N/m is stretched initially by 5 cm from the unstreched position. Then the work required to stretch it further by another 5 cm is: |
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Answer» 6.25 N-m |
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| 39441. |
A projectile is given an initial velocity of (hati+2hatj) m//s, where hatj is along the ground and hatj is along the velocity. If g = 10 m//s^2, the equation of its trajeclory is |
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Answer» `y = x - 5x^2` |
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| 39442. |
A real point object is kept on the pricipal axis of a concave mirror (radius of curvature =20cm) at a distance 60 cm from the pole. Take pole as origin and direction of incident light as positive direction of X axis. Y axis is perpendicular to the principal axis. Select correct statement(s). |
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Answer» If velocity of object `t=1hati` and velocity of mirror `=0` then velocity of IMAGE will be `-(1)/(25)hati` `(V_(I//m))^(y)=m(V_(0//m))` `m=(f)/(f-u)` `=(-10)/(-10+60)=-(1)/(5)` |
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| 39443. |
A solid cylinder is rolling without slipping down an inclined plane. Then its angular momentum is : |
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Answer» Conserved about COM of the CYLINDER |
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| 39444. |
The anode voltage of photocell is kept fixed. The wavelength of incident light on cathode is gradually changed. The plate current I of photocell varies as : |
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Answer»
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| 39445. |
Name the EM waves which are suitable for radar system used in aircraft navigation. Write the range of frequency of these waves. |
| Answer» Solution :MICROWAVES, FREQUENCY range `10^(10) ` to ` 10^(12)` HZ. | |
| 39446. |
(A): The distance of a star from earth can be measured by parallax method. (R): The change in position of an object due to change in the point of observation iscalled parallax. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 39447. |
The plane surface of a plano convex lens of focal length f is silvered. It will behave as a |
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Answer» plane mirror |
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| 39448. |
The potential barrier of a P-N junction diode is 50 meV, When an electron having energy 400 meV travels from P to N, after crossing the junction the energy of the electron is |
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Answer» 450 meV |
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| 39449. |
An electric dipole is kept in non-uniform electric field. It generally experiences |
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Answer» A torque but not a FORCE |
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| 39450. |
A cavity of radius r is made insideasolid sphere. The volume charge density of the remaining sphere is rho. An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The centre of sphere and center of cavity are separated by a distance a. The time after which the electron again touches the sphere is |
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Answer» `SQRT((6sqrt(2)repsilon_0m)/(epa))` `A=eE//m` (in backward direction).  and `E=(pa)/(3epsilon_(0))` or `A=(epa)/(2epsilonm)` Distance traveled by electron where it HITS the wall of cavity is `S=2rcostheta=(r)/(sqrt(s)` Hence using `s=(1)/(2)At^(2)` `t=sqrt((6sqrt(2)repsilon_(0)m)/(epa))` |
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