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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.
| 38401. |
A power transmission line feedsinput potential at 2200 V to a step down transformer with its primary winding having 3000 turns. Find the number of turns in secondary to get the power output at 220 V. |
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Answer» SOLUTION :When AC voltage is applied to primary coil the resulting current produces an ALTERNATING magnetic FLUX, which links the secondary, coil. The induced emf, in the secondary coil, having N(rums, is `e_(p)=-N_(s)(dphi)/(dt)` This flux, ALSO induced an emf, called back emf, in the primary coil. `e_(p)=-N_(p)=(dphi)/(dt)` But `e_(p)=v_(P)` and `e_(s)=V_(s)` `implies(V_(s))/(V_(p))=(N_(s))/(N_(p))` Fro an ideal tranformer `I_(p)V_(p)=i_(s)v_(s)` |
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| 38402. |
In Wheat stone's bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridge is: |
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Answer» `(R_C)/(R_2) =(C_1)/(C_2)` |
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| 38403. |
If mu_r represents relative permeability and x magnetic susceptibility of a material then for a paramagnetic substance |
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Answer» `mu_r lt 1 , X lt 0` |
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| 38404. |
Two sound waves of wavelengths 0.6m and0.81m produce 8 beats per seconds in a air .what are the respective frequency of the two waves ? |
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Answer» SOLUTION :`nu=n_1lamda=n_2lamda_2` `n_2-n_1=8` `thereforen_2=n_1-8 ` `n_1xx0.8 =(n_1-8)xx0.8` SOLVING` n_1=648Hz``n_2=640Hz` |
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| 38405. |
What will be the work done in blowing a soap bubble of radius 2cm, the surface tension of the soap solution is 20 dyne per cm and temperature remaining constant? |
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Answer» `440piergs` |
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| 38406. |
Which of the following statement are correst ? |
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Answer» A PENDULUM bob suspended by a string of length l is PULLED to one side so that it is at a height `l//4`above the rest position. if the bob is now released from rest, its speed at the lowest POINT will be`gl//2`. |
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| 38407. |
Three waves A,B and C of frequencies 1500 KHz. 6MHz and 50 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communciation? |
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Answer» A is transmitted via space wave while B and C are transmitted via sky wave |
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| 38408. |
A 100(mu)F capacitor is joined to a 24 V battery through a 1.0 M(Omega) resistor.Plot qualitative graphs (a) between current and time for the first 10 minutes and (b)between charge and time for the same period . |
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Answer» SOLUTION :Time constant = rC ` = 1 XX (10^6) xx 100 xx (10^-6) ` ` 100 sec. ` ` (a) q = VC (1-e^(-t/rc))` ` I = Current = dq/dt ` ` = VC/rC .e^(-t/RC)` ` = V/re^(-t/RC)` ` = 24 xx (10^-6). E^(1/(t/100))` ` t = 10 MIN = 600 sec. ` ` q = 24 xx (10^-4) xx (1- (e^-6))` ` = 23.99 xx (10^-4) ` ` i = 24/(10^6) i/e = 5.9 xx (10^-8) amp. ` ` (b) q = VC.(1-e^(-t/rc)). |
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| 38409. |
In an isochoric process |
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Answer» Work DONE is constant `=P(0)=0`. |
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| 38410. |
A sledge and its rider, with a total mass of 85 kg, emerge fro a downhill track onto a horizontal straight track with an initial speed of 37 m/s. If a force slows them to a stop at a constant rateof 2.0 m//s^(2), (a) what magnitude F is required for the force, and (c) what work W is done on them by the force ? What work W is done on them by the force ? What are (d) F, (e) d, and (f) W if they, instead, slow at 4.0 m//s^(2) ? |
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Answer» |
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| 38411. |
In the above experiment, it is given that at time t=5min, temperature of water theta('^(@)C)=61 and temperature of water in enclosure theta_(0)('^(@)C)=30. At t=8min, if theta_(0)('^(@)C)=30, then theta^(@)(C ) will be |
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Answer» `60^(@)C` |
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| 38412. |
What had the terrible fire done? |
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Answer» BURNT LOWER HALF of the BOY's body |
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| 38413. |
Taking the open end of tube as y =0 position of pressure nodes will be . |
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Answer» `y = - 1cm y = 49CM` |
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| 38414. |
The barrier potential of a silicon diode is approximately. |
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Answer» 0.7 V |
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| 38415. |
If a proton and a deutron move in a magnetic field along the same circular path, then the ratio of their velocities is _________ . |
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Answer» `1:1` |
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| 38416. |
If in a transistor if (I_(C))/(I_(E))=alpha and (I_(C))/(I_(B))=beta , If alpha varies between (20)/(21)" and "(100)/(101), then the value of beta lies between |
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Answer» 44206 |
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| 38417. |
a.A gaint refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, What is the angular magnification of the telescope ? |
| Answer» Solution :a .`m = (f_(0))/(f_(E)) = (1500)/(L) = 1500` | |
| 38418. |
The physical quantity having the dimentiom, M^(-1) L^(-3) T^(3) A^(2)is ....... |
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Answer» resistance `sigma = (1)/(rho ) = (l)/(RA) = (Il)/(VA)= (ILC)/(VA) = (IlC)/(JA = (Il(It))/(JA) ` `therefore [sigma] = (I^(2)" lt")/(JA) " its dimention" = (A^(2) L^(1) T^(1))/(M^(1) L^(2) T^(-2) L^(2))` `M^(-1) L^(-3) T^(3) A^(2)` |
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| 38419. |
(a) A charged particle having mass m and charge q is accelerated by a potential difference V, it flies through a uniform transverse magnetic field B. The field occupies a region of space d. Find the time interval for which it remains inside the magnetic field. (b) An alpha-particle is acceleration by a potential difference of 10^(4) V. Find the change in its direction of motion if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 T. (m_(alpha) = 6.4 xx 10^(-27) kg). ( c) A 10 g bullet having a charge of 4 muC is fired at speed of 270 m//sec in a horizontal direction. A vertical magnetic field of 500 muT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations. |
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Answer» Solution :(a) `K = (1)/(2)MV^(2) - qV` `v = sqrt((2qV)/(m))` `sin theta = (d)/(R ) = (d)/(R ) = (d)/(mv// BQ) = (Bqd)/(mv)` (b) `V = 10^(4)V, q = 2e = 2 xx 1.6 xx 10^(-19)C`, `d = 0.1m, B = 0.1 T, m_(alpha) = 6.4 xx 10^(-27) kg` `K = (1)/(2)m_(alpha)v^(2) = qV = 2 eV` `v = sqrt((4 eV)/(m_(alpha)))` `R = (m_(alpha)v)/(Bq) = (m_(alpha)v)/(2eB)` `sin theta = (d)/(R ) = (d)/(m_(alpha)v//B.2e) = (2eBd)/(m_(alpha)v)` `= (2eBd)/(m_(alpha)sqrt((4eV)/(m_(alpha)))) = sqrt((E)/(m_alphaV)).Bd` `= sqrt((1.6 xx 10^(-19))/(6.4 xx 10^(-27) xx 10)) xx 0.1 xx 0.1` `= (1)/(2)` `theta = 30^(@)` ( c) `m = 10 g = 10 xx 10^(3) = 10^(-2) kg` `q = 4 muC = 4 xx 10^(-6)C` `v = 270 m//sec` `B = 500 muT = 500 xx 10^(-6) = 5 xx 10^(-4)T` `d = 100 m` `R = (mv)/(Bq) = (10^(-2) xx 270)/(5 xx 10^(-4) xx 4 xx 10^(-6)) = 13.5 xx 10^(8) m` `sin theta = (d)/(R ) = (100)/(13.5 xx 10^(8)) = (10^(-6))/(13.5)` `theta` is very small. DEFLECTION `= (d^(2))/(2R)` (as proved in previous example) `= ((100)^(2))/(2 xx 13.5 xx 10^(8)) = 3.7 xx 10^(-6) m` |
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| 38420. |
Suppose the energy of a hydrogen- like atom is given as E_(n) = (-54.4)/(n^(2))eVwhere ninN. Calculate the following: (a) Sketch the energy levels for this atom and compute its atomic number. (b) If the atom is in ground state, compute its first excitation potential and also its ionization potential. (c) When a photon with energy 42 eV and another photon with energy 56 eV are made to collide with this atom, does this atom absorb these photons? (d) Determine the radius of its first Bohr orbit. (e) Calculate the kinetic and potential energies in the ground state. |
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Answer» Solution :(a) Given that `E_(n) = -(54.4)/(n^(2)) eV` For n = 1 , the ground state energy `E_(1) = -54.4 eV` and for n = 2, `E_(2) = -13.6 eV`. Similarly, `E_(3) = -6.04 eV, E_(4) = -3.4 eV` and so on . For large value of principal quantum number - that is, `n = infty`, we get `E_(infty) = 0 eV`.  (b) For a hydrogen like ATOM, ground state energy is `E_(1) = -(13.6)/(n^(2))Z^(2)eV` where Z is the atomic number. Hence, compairing this energy with given energy , we get , `-13.6 Z^(2) = -54.4 Rightarrow Z= pm2`. Since, atomic number cannot be negative number, Z =2. (c) The first excitation energy is `E_(1) = E_(2) - E_(1) = - 13.6 eV - (-54.4 eV) = 40.8 eV` Hence, the first excitation potential is `V_(1) = (1)/(e) E_(1) = ((40.8eV))/(e) = 40.8` volt The first ionization energy is `E_(ionization) = E_(infty) - E_(1) = 0 -(-54.4 eV)` `= 54.4 eV` Hence, the first ionization potential is `V_(ionization) = (1)/(e)E_(ionization) = ((54.4 eV))/(e)` `= 54.4` volt (d) Consider two photons to be A and B. Given that photon A with energy 42 eV andphoton B with energy 51 eV. From Bohr assumption, difference in energy levels is equal to photon energy , then atom will absorb energy, otherwise, not. `E_(2) - E_(1) = 13.6 eV -(-54.4 eV)` `= 40.8 eV approx 41 eV` Similarly, `E_(3) - E_(1) = -6.04 eV - (-54.4 eV) = 48.36 eV` `E_(4) - E_(1) = -3.4 eV - (-54.4 eV) = 51 eV` `E_(3) - E_(2) = -6.04 eV - (-13.6 eV) = 7.56 eV` and so on, But note that `E_(2) - E_(1) ne 42 eV,E_(3) - E_(1) ne 42 eV, E_(4) - E_(1) ne 42 eV`and `E_(3) - E_(2) ne 42 eV`.For all possibilites, no difference in energy is an integer multiple of photon energy.Hence , photon A is not ABSORBED by this atom.But for Photon B, `E_(4) - E_(1) = 51 eV,` which means, Photon B can be absorbed by this atom. (e) Since total energy is equal to negative of kinetic energy in Bohr atom MODEL, we get `KE_(n) = - E_(n) = - (-(54.4)/(n^(2)) eV) = (54.4)/(n^(2)) eV` Potential energy is negative of twice the kinetic energy, which means, `U_(n) = -2KE_(n) = - 2((54.4)/(n^(2)) eV) = - (108.8)/(n^(2))eV` For a ground state, put n = 1, Kinetic energy is `KE_(1) = 54.4 eV` and Potential energy is `U_(1) = -108.8 eV` |
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| 38421. |
The ratio of one micron to one nanometre is |
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Answer» `10^3` |
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| 38422. |
In a semiconductor (2)/(3) rd of the total current is carried by electrons and remaining (1)/(3) rd by the holes . If at this temperature , the drift velocity of electrons is 3 times that of holes ,the ratio of number density of electrons to that of holes is |
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Answer» a) `(3)/(2)` |
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| 38423. |
If modulation frequency of an FM wave is f , then the modulation index will be directly proportional to: |
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Answer» `(1)/(F)` |
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| 38424. |
The wavelength lambda_(e) of an electron and lambda_(p) of a photon of same energy E are related by……………… |
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Answer» `gamma_ppropgamma_e` |
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| 38425. |
In quantum mechanics ,a particle … |
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Answer» can be regarded as a group of HARMONIC WAVES |
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| 38426. |
(a) Explain theprocesses ofnuclear fission and nuclear fusion by usingthe plotof bindingenergyper nucleon (BE/A) versus the mass number A. (b) A radioactive isotope has a hald -life of 10 years. Howlong will it takefor theactivity to reduceto 3.125%. |
Answer» Solution :The point of (B.F/nucleon ) verses mass number is as shown  [Note : Also accept the diagram that just shows the general shapes of the graph ] From the plot we note that (i) During nuclear Fission A heavy nucleus in the larger mass region (`A gt 200) ` breaks into two middle level nuclei , RESULTING in an INCREASE in B.E/nucleon , this results in a release of energy. (ii) During nuclear fussion Light nucleus in the LOWER mass region `(A lt 20) ` fuse to form a nucleus having HIGHER B.E/nucleon .Hence Energy gets RELEASED [Alternatively -As per the plot :During nuclear Fission as well as nuclear fussion . the final value of B.E/nucleon is more than its intial value . Hence energy gets released in both these processes.] |
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| 38427. |
In about 1915, Henry Sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. 6-35). Sincosky's mass was 79 kg. If the coefficient of static friction between hand and rafter was 0.70, what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, Sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think Sincosky's grip was remarkable, try to repeat his stunt.) |
| Answer» SOLUTION :`F_(N)=2.8xx10^(2)N` | |
| 38428. |
The ratio of the angular velocities of the earth about its own axis and the hour hand of a watch is |
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Answer» `1:2` The hour HAND completes one rotation in 12 h. Its angular velocity is `omega_2=(2pi "rad")/(12h)` `therefore` Their corresponding ratio is `omega_1/omega_2=((2pi "rad")/(24 h))/((2 pi "rad")/(12 h))=1/2` |
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| 38429. |
Non-albuminous seed is produced in- |
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Answer» Pea |
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| 38430. |
Solve the previous problem assuming the collision to take place in a colliding-beam accelerator in which the electrons and positrons meet head on with equal velocities. |
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Answer» `K=evarphi=epsilon_0^("prot")-epsilon_0^(e1)=938.3-0.511 = 937.8 MeV ~~1 GeV` Such accelerators do already exist. |
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| 38431. |
A heat source atT = 500 k is connected to another heat reservoir att = 100 kby a copper slab which is 1 m long. Given that the thermal conductivity of copper is 0.1 Wk^(-1) m^(-1) ,the energy flux through it in the steady state is: |
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Answer» `4000 Wm^(-2)`  ` (dQ)/(DT) = (KA)/(I) (500 - 100) ` Heat/time ` = (0.1A )/(1) xx 400 xx 40A` energy/flux ` = ("heat current")/(A) = 40` |
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| 38432. |
Two capacitors C_(1)=C_(2)=1/(pi^(2))xx10^(-2)F and inductor L=2xx10^(-2)H are connected in series as shown in the figure. Initially charge on each capacitors are 4sqrt(3) muC. At t=0 switch S_(1) is closed at t=1//400 sec, switch S_(2) is also closed. The maximum charge on capacitor C_(2) during LC oscillation is ksqrt(3)muC. Find k |
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Answer» `impliesT=1//50` SEC |
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| 38433. |
Two parallel infinite line charge with linear charge densities +lambda C//m and -lambda C//mare placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ? |
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Answer» `LAMBDA/(2piepsilon_(0)R)N/C` `VECE = vecE_(+lambda) + vecE_(-lambda)`  `=lambda/(2piepsilon_(0)R) + lambda/(2piepsilon_(0)R)` `=lambda/(piepsilon_(0)R) N/C` |
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| 38434. |
A TV tower has a height of 100 m. What is the maximum distance upto which the TV transmission can be received? R=8xx10^(6)m |
| Answer» Solution :`d=sqrt(2HR)=2xx100xx8xx10^(6)`=40.000 m= 40 km | |
| 38435. |
A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A= 120 with BE/A = 8.5 MeV. Calculate the released energy. |
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Answer» SOLUTION :Since the nucleus has a MASS number A = 240 and binding energy per nucleon of 7.6 MeV, its total binding energy `E_(1) = 240 xx 7.6 = 1824 MeV`. As both fragments of mass number A = 120 has a binding energy per nucleon of 8.5 MeV, total binding energy of the fragments `E_(2)=2 xx 120 xx 8.5 = 2040 MeV` `THEREFORE` Energy released during fission REACTION `E = E_(2)-E_(1) = 2040 - 1824 = 216 MeV` |
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| 38436. |
A diverging meniscus lens of radii of curvatures 25 cm and 50 cm has a refractive index 1.5. Its focal length is (in cm) |
| Answer» ANSWER :B | |
| 38437. |
A particle is moving east wards with a velocity of 5 ms^(-1). In 10S the velocity changes to 5 ms^(-1) north wards. The average acceleration in this time is |
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Answer» `(1)/(sqrt(2)ms^(-2))` TOWARDS norht east |
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| 38438. |
One of the following which does not check land degradation: |
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Answer» CHECK on overgrazing |
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| 38439. |
V - I graph of a conductor at temperature T_(1) and T_(2) are shown in the figure (T_(2)-T_(1)) is proportional to |
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Answer» Solution :Slope of LINE gives RESISTANCE So, `R_(1)=tan theta=R_(0)(1+alphaT_(1))` `R_(2)=tan (90-theta)=cot theta=R_(0)(1+alphaT_(2))`  `cot theta-tan theta=R_(0)alpha(T_(2)-T_(1))` `or (cos theta)/(sin theta)-(sin theta)/(cos theta)=R_(0)alpha(T_(2)-T_(1))` `R_(0)alpha(T_(2)-T_(1))=(cos 2 theta)/(((sin 2 theta))/(2)) or T_(2)-T_(1) alpha cot 2 theta` |
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| 38440. |
Young's doble slit experiment is carried out using microwaves of wavelength lambda = 3 cm. Distance between the slits is d=5 cm and the distance between the plane of slits and the screen is D = 100 cm. Find the distance of positions from central maximum where these maximas are formed. |
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Answer» 0 , +25 CM, -25 cm |
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| 38441. |
A neutral point is found on the prolongation of the axis of a bar magnet of length 10 cm at 10 cm from the nearer pole. How has the magnet been placed with respect to the magnetic meridian ? Calculate the magnetic moment of the magnetic given that B-vector of earth's field along the horizontal = 3.6 xx 10^(-5) tesla. Also calculate the torque required to deflect it through 30^(@) from the magnetic meridian. |
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Answer» |
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| 38442. |
In an LCR series circuit, the voltages across R, L and C at resonance are 40 V, 60 V and 60 V respectively. The applied voltage is |
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Answer» 60V |
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| 38443. |
A student measures the focal length of a convex lens by putting an object pin at a distance u from the lens and measuring the distance V of the image pin. The graph between u and V plotted by the student should look like |
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Answer»
`(1)/(v) - (1)/(u) = (1)/(f)` for a convex lens. u ha to be negative. If `u = oo, v = f and if u = oo, v = f`. A parallel beam `(u = oo)` is focussed at f and if the object is at f, the rays are parallel. The point which MEETS the curve at u = v GIVES, 2F. Therefore, v is + ve, u is negative, both are symmerical and this curve satisfies all the CONDITIONS for a convex lens. 
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| 38444. |
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related ? |
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Answer» SOLUTION :Kinetic ENERGY of the particle `K=1/2mv^(2)=(P^(2))/(2M)=(h^(2))/(2mlambda^(2))[lambda=h/P]` i.e `lambda =h/(sqrt(2mK)),lambda PROP1/(sqrtm)` `(lambda_e)/(lambda_prop)=sqrt((m_prop)/(m_e))` |
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| 38445. |
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c = Velocity of light) : |
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Answer» <P>`(E )/(c )` Momentum transferred to the surface `= p_(f)-p_(i)`  `=(E )/(c )-(-(E )/(c ))` `= (2E)/(c )` |
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| 38446. |
A solenoid hasan inductance of10 H and a resistanceR= 5 Omega . Itisconnectedta a10 V battery . How long will be ittakefor themagneticenergyto reach(1)/(4) ofitsmaximum value? |
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Answer» |
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| 38447. |
Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shownin the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces w.r.t. each other is: |
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Answer» 5g/6 |
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| 38448. |
One end of a steel rectangular girder is embedded into a wall (figure). Due to gravity it sags slightly. Find the radius of curvature of the neutral layer (see the dotted line in the figure) in the vicinity of the point O if the length of the protruding section of the girder is equal to l=6.0m and the thickness of the girder equals h=10cm. |
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Answer» Solution :A beam clamped at one end and SUPPORTING an applied load at the free end is called a cantilever. The theory of cantilevers is discussed in advanced text book on mechanics. The key result is that elastic forces in the beam GENERATE a couple, whose moment, called the moment of RESISTANCES, balances the external bending moment due to weight of the beam, load etc. The moment of resistance, also called internal bending moment (I.B.M) is given by `I.B.M. EI//R` Here R is the radius of curvature of the beam at the representative point `(x,y)`. I is called the GEOMETRICAL moment of inertia `I=intz^2dS` of the cross section relative to the AXIS passing through the netural layer which remains unstretched (figure). The section of the beam beyond P exerts the bending moment `N(x)` and we have, `(EI)/(R)=N(x)` If there is no load other than that due to the weight of the beam, then `N(x)=1/2rhog(l-x)^2bh` where `rho=` density of steel. Hence, at `x=0` `(I/R)_0=(rhogl^2bh)/(2EI)` Here b=width of the beam perpendicular to paper. Also, `I=underset(-h//2)overset(h//2)intz^2bdz=(bh^3)/(12)`. Hence, `(1/R)_0=(6rhogl^2)/(Eh^2)=(0.121km)^-1` 
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| 38449. |
Electric intensity is equal to : |
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Answer» Time RATE of change of potential |
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| 38450. |
A potentiometer wire 10m long and having 20 P Omegaresistance is connected in series with a battery and a resistance R = 2160 Omega . A daniel cell of emf 1.08 v is balanced across the P.D. of the resistance R. The emf of thermo couple is balanced across 300 cm of the wire. The emf of thermo couple is |
| Answer» Answer :C | |
