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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.
| 4301. |
A boat is travelling on a river with a speed of 3m/sec. The force on the boat by water is 500N. The power delivered by the engine of the boat is : |
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Answer» 1.5 kW |
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| 4302. |
It it is not aligning horizontally , what does it mean ? |
| Answer» SOLUTION :The NET MAGNETIC FIELD is not HORIZONTAL. | |
| 4303. |
An electric bulb is designed to operate at 12 V.D.C. if it is connected to A.C. it gives the same brightness. Then the peak value of the A.C. voltage is : |
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Answer» 12 V |
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| 4304. |
Three large indentical conducting plates of area A are closely placed parallel to each other as shown (the area A is perpendicular to plane of diagram). The net charge on left. Middle and rightplates are Q_(L) , Q_(M) and Q_(s)respectively. Threeinfinitely large parallel surfaceS_(c ), S_(m) and S_(R) are drawn passing through middle of each plate such that surfaces are perpendicular to plane of diagram as shown. Then pick up the correct option(s). |
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Answer» The net charge on the LEFT side of SURFACE `S_(L)` is equal to net charge on RIGHT side of surface `S_(R)`. |
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| 4305. |
Relation between the magnetic field vectorvec(B) and magnetic intensityvec(H) at a point in a magnetic field vec(B)=muvec(H) wheremuis the magnetic permeability of the medium in which the point is situated . Magnetic permeability of vacuum mu_(0)=4pixx10^(-7)H.m^(-1) . Thus the relative magnetic permeability of the mediummu_(r)=(mu)/(mu_(0)) . To define the magnetic field at a point in a medium another vector needs mention , which is magnetisationvec(M). The magnetic field is known as the magnetisation of the point . In most cases vec(M)propvec(H) "or,"vec(M)=kvec(H),this k is called the magnetic susceptibility of the medium . The relation among these vectors is expressed as vec(B)=mu_(0)(vec(H)+vec(M)).Magnetic permeability at a point in the medium is0.002H.m^(-1). The magnetic susceptibility of the medium is |
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Answer» `-0.03` |
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| 4306. |
Figure 22-26a shows a plastic rod with a uniform charge -Q. It is bent in a 120^(@) circulararc of radius r and symmetrically paced across an x axis with the origin at the center of curvature P of the rod. In terms of Q and r, what is the electric field vecE due to the rod at point P? |
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Answer» Solution :KEY IDEAS Because the ROD has a continuous charge distribution, we must find an expression for the electric fields due to differential elements of the rod and then sun those fields via calculus.  An element :Consider a differential element having arc length ds and located at an angle `theta` above the x axis (Figs. 22-26b and c). IF we let `lambda` represent the linear charge density of the rod, our element ds has a differential charge of magnitude `dq=lambda ds. "" (22-35)` The element.s field: Our element produces a differential electric field`d vecE` at point P, which is a distance r from the element. Treating the element as a point charge, we can rewrite Eq. 22-3 to express the magnitude of `d vecE` as `dE=(1)/(4pi in_(0)) (dq)/(r^(2))=(1)/(4pi in_(0)) (lambda ds)/(r^(2)). ""(22-36)` The direction of `d vecE` is toward ds because charge dq is negative. Symmetric PARTNER: Our element has a symmetrically located (mirror image) element ds. in the bottom half of the rod. The electric field `d vecE.` set up at P by ds. also has the magnitude given by Eq. 22-36, but the field vector points toward ds. as shown in Fig. 22-26d. If we resolve the electric field vectors of ds and ds. into x and y components as shown in Figs. 22-26e and f, we see that their y components cancel (because they have equal magnitudesand are in opposite directions). We also see that their x components have equal magnitudes and are in the same direction. Summing: Thus, to find the electric field set up by the rod, we nened sum (via integration) only the x components of the differential electric fields set up by all the differential elements of the rod. From Fig. 22-26f and Eq. 22-36, we can write the component `dE_(x)` set up by ds as `dE_(x)=dE cos theta=(1)/(4 pi epsilon_(0)) (lambda)/(r^(2)) cos theta ds."" (22-37)` Equation 22-37 has TWO variables, `theta` and s. Before we can integrate it, we must eliminate one variable. We do so by REPLACING ds, using the relation `ds=r d theta`, in which `d theta` is the angle at P that includes are length ds (Fig. 22-26g). With this REPLACEMENT, we can integrate Eq. 22-37 over the angle made by the rod at P, from `theta= -60^(@)` to `theta=60^(@)`, that will give us the field magnitude at P: `E= int dE_(x)= int_(-60^(@))^(60^(@)) (1)/(4pi epsilon_(0)) (lambda)/(r^(2)) cos theta r d theta` `=(lambda)/(4 pi epsilon_(0)r) int_(-60^(@))^(60^(@)) cos theta d theta = (lambda)/(4 pi epsilon_(0)r) [sin theta]_(-60^(@))^(60^(@))` `=(lambda)/(4pi epsilon_(0)r) [sin 60^(@)-sin(-60^(@))]=(1.73lambda)/(4pi epsilon_(0)r).""(22-38)` (If we had reversed the limits on the integration, we would have gotten the same result but with a minus sign. Since theintegration gives only the magnitude of `vecE`, we would then have discarded the minus sign.) Charge density: To evaluate `lambda`, we note that the full rod subtends an angle of `120^(@)` and so is one-third of a full circle. It arc length is then `2pi r//3`, and its linear charge density must be `lambda=("charge")/("length")=(Q)/(2pi r//3)=(0.477Q)/(r )` Substituting this into Eq. 22-23 and simplifying give us `E=((1.73)(0.477Q))/(4pi epsilon_(0)r^(2))=(0.83Q)/(4pi epsilon_(0)r^(2)). "" ` (Answer) The direction of `vecE` is toward the rod, along the axis of symmetry of the charge distribution. We can write `vecE` in unit-vector notation as `vecE=(0.83Q)/(4pi epsilon_(0)r^(2))hati`. |
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| 4307. |
Ohmic resistance of conductor ..... |
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Answer» DEPENDS on V only. |
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| 4308. |
Figure 10-55 shows an early method of measuring the speed of light that makes use of a rotating slotted wheel. A beam of light passes through one of the slots at the outside edge of the wheel, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.0 cm and 500 slots around its edge. Measurements taken when the mirror is L = 500 m from the wheel indicate a speed of light of 3.0xx10^(5)km//s. (a) What is the (constant) angular speed of the wheel? (b) What is the linear speed of a point on the edge of the wheel? |
| Answer» SOLUTION :(a) `3.8xx10^(3)rad//s`, (B) `1.9xx10^(2)m//s` | |
| 4309. |
How the king saved his kingdom? |
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Answer» By GIFTING rings worth 3 LAKH RUPEES to Duraisaini |
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| 4310. |
A Cassegrain telescope uses two mirrors of radii of curvature 220 mm and 140 mm. The distance b/w the two mirrors is 20 mm. Wherewill the final image of an object at infinity be? |
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Answer» Solution :By USING the formula for mirrors, `(1)/(f)=(1)/(u)+(1)/(v)` we get Since, `f=r//2=(220)/(2)=110` mm hence, `(1)/(110)=(1)/(oo)+(1)/(v)` `THEREFORE v=110`mm virtual object distance for the second MIRROR `=(110-20)=90` mm For the second mirror, Hence `(1)/(-70)=(1)/(-90)+(1)/(v)` `therefore(1)/(v)=(1)/(-70)+(1)/(90)=(-90+70)/(6300)` or`v= -(6300)/(20)= -315` mm The image is formed at 315 mm from the SMALLER mirror in the direction of light. |
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| 4311. |
There is a actual transfer of energy in |
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Answer» LONGITUDINAL WAVES |
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| 4312. |
Define self-inductance and give its SI unit. Derive an expression for self-inductance of a long, air cored solenoid of length l, radius r and having N number of turns. |
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Answer» SOLUTION :Self INDUCTANCE of a coil is NUMERICALLY equal to the flux linked with the coil when the current through the coil is 1 A. S.I. UNIT of Self inductance is henry |
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| 4313. |
1MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if (a) power is transmitted at 220V. Comment on the feasibility of doing this. (b) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V. (rho_(cu)=1.7xx10^(-8) SI unit ) |
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Answer» Solution :(a) Length of copper wire = 2 x 10 km =20 km `therefore` L=20000 m RESISTANCE of copper wire, `R=rho l/A=rho l/(pir^2)` `=(1.7xx10^(-8)xx20000)/(3.14xx(0.5xx10^(-2))^2)` `=(3.4xx10^(-4))/(3.14xx0.25xx10^(-4))` `=4.3312 Omega approx 4 Omega` P=VI `rArr I= P/V=10^6/220 = 0.45 xx 10^4` A POWER loss = `I^2R=(0.45xx10^4)^2xx4` `=0.8263xx10^8` `approx 82.63xx10^8` W Power loss is greater than 1 MW, so this method cannot be used for transmission. (b) When power 1 MW is transmitted at 11000 V, V.I.=`10^6` W =11000 I. but CURRENT, `I.=P/(V.)=10^6/11000=1000000/11000=1/11xx10^3` A power loss `=(I.)^2R= 1/121 xx 10^6 xx 4` `=0.03305xx10^6` `approx` 33050 W FRACTION of power loss = `33050/10^6xx100%` =3.3 % |
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| 4314. |
A bullet loses 1/10 th or its velocity when it passes through a plank. Then the maximum number of planks through which it can pass |
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Answer» 9 |
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| 4315. |
The loudest painless sound produces a pressure amplitude of 28 Nm^2. Calculate the intensity of this sound wave at S.T.P. Density of air at ST.P.=1.3 kg m^(-3) speed of sound at S.T.P.= 332 m/s. |
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Answer» Solution :`y=a sin (omegat -kx)` is the equation of PROGRESSIVE wave of 1/2 `rhoa^2omega^2v` is its intensity `P=(Edy)/(dx), v=sqrt(E/RHO)` `=v^2 rho (-ak) (cos omegat-kx)=v^2 rho ak sin (omegat-kx-pi//2)` `therefore P_(max)=v^2 rho ak` `therefore l=1/2rho(P_"max"^2)/(v^4 rho^2 k^2) v^2k^2 v "" {omega/k=v}` `=1/2 rho_"max"^2/(rhov) =1/2 xx28^2/(1.3xx332)=0.91 W-m^2` |
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| 4316. |
If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0 to t=tau s, then tau may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming dampling is small) in secods : |
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Answer» `(2)/(B)` Equation (1) has solution of TYPE `x=e^(alpha t)` from (1) `m alpha^(2)+b alpha+k=0` `alpha=(-b+-sqrt(b^(2)-4mk))/(2m)` on solving for x, we get `x=e^((-bt)/(2m))` `omega_(1)=sqrt(omega_(0)^(2)-alpha^(2))` where `omega_(0)=sqrt((k)/(m))` `alpha=(b)/(2)` `:.` average life `=(2)/(b)`. So, correct choice is (a). |
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| 4317. |
Figure shows crosssection view of a infinite cylindrical wire with a cavity, current density is uniform vec(j)=-j_(0)hat(k) as shown in figure |
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Answer» magnetic field inside cavity is uniform |
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| 4318. |
What happens if the magnitude of capacitance of capacitor are large ? |
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Answer» Solution :If the MAGNITUDE of capacitance of capacitor" is large potential difference V is small for a given charge. `[" Because" C= (Q)/(V)]` This means a capacitor with large capacitance can hold large amount of charge Q at a relatively small potential difference. High potential difference implies strong electric field around the conductors. A strong electric field can ionise the surroundingair and accelerate the charges so produced to the oppositely charged plates, thereby neutralising conductors. the charge on the capacitor plates at least partly. The charge of the capacitor leaks away due to I capacitor. the reduction in insulating power of the inversing medium and capacitor becomes useless. The phenomenon of leakage of charge from capacitor is called Breakdown. The sharp ends of the conductor (here plates) have a large electric charge density. The electric field near such region is very strong . This strong electric field can strip the electrons from the surface of metal this EVENT is called dielectric breakdown and is also called corona discharge . Themaximum electric field UPTO which an insulating medium can maintain its insulating property is called the dielectric strength . (March - 2020) For air the value of dielectricstrength is about `3 xx 10^(6)` V /m and the electric field corresponds to a potential difference of `3 xx 10^(4)` V between the conductors for 1 cm distance between them. Thus, for a capacitors to STORE a large amount of charge without leaking its capacitance should be high. |
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| 4319. |
A galvanometer of resistance 10Omegais to be used to measure current upto 1A .What must be the value of shunt resistance if galvanometer gives full scale deflection for current of 10mA ? |
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Answer» `100/99 OMEGA` |
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| 4320. |
A galvanometer has a resistance of 100Omega. A difference of potential of 1.0 V between its terminals gives a full scale deflection. Calculate the shunt resistance which will enable the instrument to read upto 2A. |
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Answer» |
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| 4321. |
If normal reaction is halved, then the value of mu_s |
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Answer» DECREASES |
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| 4322. |
What is the principle of potentiometer ? Why it is so named ? |
| Answer» Solution :On the PRINCIPLE that when a steady current flows through a wire of uniform CROSS SECTION and COMPOSITION, the potential drop ACROSS any length because, it is used to measure P.D. | |
| 4323. |
A particle is projected from the surface of earth with velocity equal to its escape velocity , at 45^(@) with horizontal . What is the angle of its velocity with horizontal at height h= R . (Here horizontal at some point means a line parallel to tangent on earth just below that point . ) |
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Answer» `30^(@)` |
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| 4324. |
The surface of a metal is illuminated with the light of 440 nm. The kineticenergy of the ejected photo electrons was found to be 1.68 e V. The work function of the metal is (hc=1240 e Vnm) |
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Answer» `3.09 E V` |
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| 4325. |
Assertion : Nuclei having number about 60 are most stable. Reason : When two or more light nuclei are combined into a heavier nucleus, then the binding energy per nucleon will increase. |
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Answer» If both ASSERTION and reason are true and reason is the correct explanation of assertion |
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| 4326. |
'The person standing behind the open door inside the room can hear the voice of this person standing on the other side of the door but they cannot see each other". Giveexperiment based on the diffraction for this statement. |
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Answer» SOLUTION : Take the average wavelength of VISUAL light `lamda=6xx10^(-7)` m and he WIDTH of our window or door is d=1 m, then `(lamda)/(d)=6xx10^(-7)`. Because of its small value, the waves of light does not bend. That is diffraction of light is negligible. While in normal conversation the sound frequency is between 100 Hz to 400 Hz. For the sake of simplicity, suppose the frequency of sound is 330 Hz and velocity of sound in air is 330 m/s. So wavelength `lamda=(v)/(v)=(330)/(330)=1m` and width of window or door d = 1 m then `(lamda)/(d)=1` which is the maximum value of diffraction. Hence, sound waves have a diffraction range of `0^(@)` to `90^(@)`. So sound can be heard in normal conversation but diffraction of light is negligible so we cannot see the person. |
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| 4327. |
A: Stars twinkle in sky at night, planets don.t.R: Volume of stars is much greater than planets. |
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Answer» Both ASSERTION and REASON are TRUE and the reason is CORRECT EXPLANATION of the assertion. |
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| 4328. |
How a swimmer jumping from a height is due to increase the number of loops made in the air ? |
| Answer» Solution :The SWIMMER can increase the number of LOOPS by PULLING his LEGS and hands inwards i.e. by decreasing the moment of inertia. | |
| 4329. |
The both of a pendulum has mass m = 1 kg and charge q = 40mu C. Length of pendulum is = 0.9 m. The point of suspensison also has the same charge 40 muC. What is the minimum speed u (in m/s) should be imparted to the bob so that it can complete vertical circle ? |
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| 4330. |
Explain the construction of refracting telescope by figure and obtain the equation of its magnification. |
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Answer» Solution :To observe very huge celestial bodies an Astronomical Telescope is used. Its ray diagram is shown in the figure. In this telescope two convex lenses are kept in such a way that their principal axis coincide. The lens facing the object is called objective and the lens near the eye is known as eye-piece. Here the diameter and the focal length of the objective are greater than that of the eyepiece.  The cyc-piccc can move to & fro in the telescope tube when the telescope is focussed on a distant object, parallel rays coming from this object form a real, inverted and small image AB on the second principal FOCUS of the objective. This image is th object for the eye-piece. Eye-piece is moved to & fro to get the final an MAGNIFIED inverted image A.B. of the origin object at a certain distance. In such telescope, rays from the object ai refracted by the objective and the image FORMED. Such telescope is called refractin telescope. Figure of Astronomical Telescope is show above. Magnification of the telescope, `m=("angle subtended by the final image with eye")/("angle subtended by the object with the objective or eye")` `therefore m= (beta)/(alpha)` But TAN`beta=(h)/(f_e)` and for small angle `tan beta~~beat` `therefore beta=(h)/(f_e)` thus tan`alpha=(h)/(f_0)` and for small angle tan`alpha~~alpha` `therefore alpha=(h)/(f_0)` `therefore` Telescope magnification `m=(beta)/(alpha)=(h//f_e)/(h//f_0)` `therefore m=(f_0)/(f_e)` This equation shows that to increase the magnification of the telescope, focal length of the objective should be increased and focal length of the eye-piece should be reduced. `f_0+f_e` is the optical length of the telescope. So, length of the tube `L ge f_0+f_e` For a telescope, light gathering power and resolving power are very important. |
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| 4331. |
The element of a heater is rated (P, V). If it is connected across a source of voltage V/2, then thepower consumed by it will be |
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Answer» P |
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| 4332. |
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she // he is carrying anything made of metal, the metal detector emits a sound. On what principal does this detector work? |
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Answer» Solution :At airport entrance , there is kept a metal detector of DOOR way type. Frame of this door way, consists of a coil having large no. of tightly wounded turns, there is connected with a capacitor. This LC circuit is in resonance with ac SOURCE connected with it. Suppose the coil has inductance, `L_(0) = ( mu_(0) N^(2) A)/( l)` Now, when any person carrying some metallic OBJECT, walks through this door way, inductance of coil, walks through this door way, inductance of coil changes from `L_(0)` to L where `L = ( mu N^(2) A)/( l ) `. Here L` gt gt gt gtL_(0)`. Because of this, there is a sudden change in the impedance of a circuit and hence sudden change in the current passing through the LC circuit. This sudden change in current is immediatly NOTED and electronic circuit kept in the metal detector SOUNDS "Beep... Beep..". This alarm gives indirect proof that the person walking through the door way is carrying some metallic object. |
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| 4333. |
The driver of a three - wheeler moving with a speed of 36 km/hr sees a child standing in the middle of the road and brings his vehicle at rest in 0.4s just in time to save the child. The average retarding force on the vehicle is (The mass of the three - wheeler is 400kg and the mass of the driver is 65kg) |
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Answer» `1.2 xx 10^(3) N` |
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| 4334. |
Two bodies begin to move from rest in the same straight line at the same instant of time from the same point. The first moves with constant velocity of 40 ms and the second starts with uniform acceleration of 4 ms. The time which the distance between them is maximum is : |
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Answer» 20 s |
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| 4335. |
An L-shaped body having two parts of mass 1/3 kg and 2/3 kg and length 6m each rotates about an axis passing through the centre of mass and perpendicular to its plane. Find the moment of inertia of the body (in kg//m^(2)) about this axis. |
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Answer» `y_(C)=(1/3xx3+0xx2)/1=1` `I_(C)+1xx(X_(C)^(2)+y_(C)^(2))=(1xx6^(2))/(3xx3) +(2XX2)/(3xx3)` `I_(C)+1(1^(2)+2^(2))=4+8` `I_(C)+5=12` `I_(C)=7 KG m^(2)` |
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| 4336. |
Relation between the magnetic field vectorvec(B) and magnetic intensityvec(H) at a point in a magnetic field vec(B)=muvec(H) wheremuis the mangnetic permeability of the medium in which the point is situated . Magnetic permeability of vacuum mu_(0)=4pixx10^(-7)H.m^(-1) . Thus the relative magnetic permeability of the mediummu_(r)-(mu)/(mu_(0)) . To define the magnetic field at a point in a medium another vector needs mention , which is magnetisationvec(M). The magnetic field is known as the magnetisation of the point . In most cases vec(M)propvec(H) "or,"vec(M)=kvec(H),this k is called the magnetic susceptibility of the medium . The relation among these vectors is expressed as vec(B)=mu_(0)(vec(H)+vec(M)). Magnetic field at the same point(inWb.m^(-2) )is |
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Answer» `4.2xx10^(-7)` |
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| 4337. |
A tuning fork produces 4 beats/sec. with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. the unknown frequency is : |
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Answer» 286 cps since on loading no. Of beats decreases. `THEREFORE v_(2) = 292 ` Hz. CORRECT choice is (c). |
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| 4338. |
In a circuit consisting of two resistors ( 2 ohm and 7 ohm) and a cell of current of 0.9 A flows through 2 ohm resistor and 0.3 A through 7 ohm resistor. The internal resistance of the cell is: |
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Answer» `0.5Omega` |
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| 4339. |
Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is |
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Answer» `Q/(epsilon_0)` |
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| 4340. |
What is meant by hysteresis? |
| Answer» SOLUTION :The phenomenon of LAGGING of magnetic induction BEHIND the magnetising field is CALLED HYSTERESIS. Hysteresis means .lagging behind.. | |
| 4341. |
A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string |
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Answer» 25.03 N |
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| 4342. |
A beam of converting rays of light meets at a point on a screen. A parallel plane glass slab is kept in the path of the converging rays. How for will the intersecting point of the rays be shifted? Draw a diagram to show it. |
Answer» Solution :If a PARALLEL plane glass slab is placed on the pair of the beam of converging rays, the intersecting point of the rays will be shifted away from the PREVIOUS point. We know that, a ray of light incident obliquely on a parallel glass slab emerges with a lateral DISPLACEMENT. In Fig. 271, the PATH of the rays has been drawn and the displacement of the intersecting point has been shown. In the absence of glass slab the rays PQ, XY and RS would meet at O. But due to the presence of the glass slab, the rays meet at O.. Hence, the displacement of the intersecting point = o o.. 
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| 4343. |
The value of wavelength in the lyman series is given as lambda=(913.4 n_i^2)/(n_i^2-1)Å Calculate the wavelength corresponding to transition from energy level 2,3 and 4. Does wavelength decreases or increase. |
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Answer» SOLUTION :`lambda_(21) =(913.4 XX 2^2)/(2^2-1)=1218 Å` `lambda_(31)=(913.4 xx 3^2)/(3^2-1)=1028Å` `lambda_(41)=(913.4xx4^2)/(4^2-1)=974.3 Å` `lambda_(41) LT lambda_(31) lt lambda_(21)` |
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| 4344. |
When impurity (atom) is added to an intrinsic semiconductor , then it becomes extrinsic semiconductors . What is the above process called ? b. What is the advantage of doing so ? c. Distinguish between intrinsic semiconductor and extrinsic semiconductor. |
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Answer» SOLUTION :a . Doping b. It increases THECONDUCTIVITY of SEMICONDUCTOR . C. Semiconductor (original) WITHOUT impurity is called intrinsic and the one doped is called extrinsic semiconductors. |
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| 4345. |
A point source of light is placed at a depthof h below the surface of water of refractive indexmu. Afloatingopaque disc is placed on the surface of water so thatlight from the source is not visible from the surface . The minimum diameter of the disc is |
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Answer» `(2H)/((mu^(2)-1)^(1/2))` Now, ` SIN theta_(c) = 1/mu ` So that ` tan theta_(c) = 1/((mu^(2) -1)^(1//2))` Fromfigure,` tan theta_(c) = r/h` Where r is the radiusof the disc.Therefore of the disc is ` 2 r = 2 h tan theta_(c) =(2h)/((mu^(2) -1)^(1//2)` ` (##MTG_WB_JEE_CHE_MTP_01_E01_023_S01.png" width="80%"> |
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| 4346. |
Let omegabea complex number which satisfies iomega=(u^(2)-1)/(u^(2)+1) for some suitable complex number u, then value of tan ^(-1)(omega) is -. |
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Answer» LNU |
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| 4347. |
A man wishes to row across a riverflowingto therightwith speed of 2m/s. If they velocity that the boat canhave V_(B) = 4 m//s , how should the man rowso as to reach across in . (a) shortest path (b)shortest time . |
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Answer» Solution :(a) Shortest Path . If the mainhasto move alongthe shortest path , then the path of thisboat should beAB ( perpepndicularto river). If THEMAN rows vertically , the riverwill carry him to right, hencethe MAN shouldinclinehis boat to theleft. Whenhis VELOCITYIS inclined ,it has two components , `V_(x)` and `V_(y)` [comparewith inclinedvelocity of cat].  Thehorizontal component `V_(x)` should cancel the SPEED the river . Thus`V_(x) = V_(B) cos theta = V_(R)` `therefore4 costheta=2 , theta = 60^(@)` Alternatively , we can use resultant vertical as in cat mouseproblem (Method 3). (b) Shortest time : Notethat whenthe man rows alongshortest distance , he DOES NOT cross the river in the shortest time , becausetime takento cross the river ` = ("widthe of river")/("vertical velocity")` As longas the boat is inclined at ANGLE `theta` , vertical velocityis `V_(B)`. To row in shortesttime ,theenitrevelocityof boatshould be used invertical direction i.e., Man shouldrowperpendicular to stream . Note: Whenthe man rows in shortest time , he does notreach pointB but instead point B. Shortest distanceis analogus to reachingapointon opposite bank(say house ) where shortest time is linked with onlyreachingopposite bank. 
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| 4348. |
What happens if the electrons in an atom were stationary ? |
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Answer» |
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| 4349. |
Klystrons, ___________and_________ are employed to produced microwaves. |
| Answer» SOLUTION :MAGNETRONS, GUNN DIODES | |
| 4350. |
The 'Facsimile Transmission (FAX)' involves |
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Answer» SPEECH communication |
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