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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.
| 44851. |
A person hears the sound of a jet aeroplane after it has passed over his head. The angle of the jet plane with the horizontal when the sound appears to be coming vertically downwards is 60^(@) . If the velocity of sound is v, then the velocity to the jet plane is : |
| Answer» ANSWER :B | |
| 44852. |
If the resistance R_(1) is increased (See figure), how will the readings of the ammeter and voltmeter change? |
| Answer» SOLUTION :Here when `R_(1)` is INCREASED in the input circuit, input current `I_(B)` decreases. Now, ouputcurrent in common emitter circuit is `I_(C )=beta I_(B)` and so `I_(C )` decreases because of decrease in `I_(B)`. Hence READING in ammeter in the OUTPUT circuit decreases. Hence, voltage `I_(C )R_(2)` across `R_(2)` in the output decreases (`because I_(C )` decreases). Hence reading in voltmeter, connected across `R_(2)`, will also decrease. | |
| 44853. |
The energy equivalent of one atomic mass unit (a.m.u.) is : |
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Answer» `1.6xx10^(-19)` J |
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| 44854. |
Whenresistanceof2Omegaisconnected across the terminals of a battery, the current is 0.5A. When the resistance across the terminal is5 Omega,thecurrent is0.25A. (i) Determine the emf of the battery (ii) What will be current drawn from the cell when it is short circuited. |
| Answer» SOLUTION :E = 1.5 V, I = 1.5 A | |
| 44855. |
The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40m/s can go without hitting the ceiling of the hall? |
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Answer» SOLUTION :`H= 25 m u= 40 m//s`, `H=u^2 sin^2 THETA/(2G)` i.e 25 = `(40^2) sin^2 theta / (2xx9.8)` on solving `theta = 33.6^@` and ` R=u^2 sin 2 theta/(g) = (40)^2 sin 2 (33.6)/(9.8)` m= 150.5m |
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| 44856. |
A technician has only two resistance coils. By using them singly in series or in parallel, he is able to obtain the resistance 1.5, 2, 6 and 8 ohms. The resistance of two coils are |
| Answer» Answer :C | |
| 44857. |
Derive an expression for electrostatic potential energy of the dipole in a uniform electric field . |
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Answer» Solution : Electrostatic potential energy of a dipole in a uniform electric Field: place in the uniform electric field `overlineE`.A dipole experiencesa torque when kept ina uniform electric field `overline E`.This torque rotates the dipole to align it with the direction of the electric field.To rotate the dipole (at constant angular velocity ) fromits initial angle `theta` another angle `theta` against the torque exerted by the electric field ,an equal and opposite EXTERNAL torque must be applies on the dipole. The work done by the ecternal torque to rotate the diplole from the angle `theta` to `theta` at constant angular velocity is `W=int_(theta)^(theta)tau_(EXT)d theta` Since `tau_(ext)` is equal and opposite to `tau_E=overlinePxxoverlineE`,we have `|overlinetau_(ext)|=|overlinetau_E|=|overlinePxxoverlineE|` Subsituting equation (2) in equation (1) we get, `W=int_(theta)^(theta)pEsin thetad theta=pE(costheta.-costheta)` This work done is equal to the potential energy difference between the angular potions`theta` and `theta U(theta)-(U theta)=triangleU=-pE cos theta +pE cos theta` If the initial aangle is `=theta.=90^(@)` and is taken as reference POINT ,then `U(theta.)+pE cos 90^(@)=0`.The potential energy stored in the system of dipole kept in the uniform electric field is GIVEN by `U=-pEcos theta=-overlinep .overlineE` In addition to P and E ,the potential energy ALSO depends on the orientation `theta` of the electric dipole with respect to the external electric field The potential energy is maxmimum when the dipole is aligned anti-parallel `(theta=pi)` to the external electric field and maximum when th dipole is aligned parallel `(theta=0)` to the external electric field. 
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| 44858. |
Figure shows a rectangular pulse and triangular pulse approaching each other. The pulse speed is 0.5 cm/s. Sketch the resultant pulse at t = 2s. |
Answer» SOLUTION :
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| 44859. |
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble. [Ignore air resistance](a) during its upward motion(b)during its downward motion(c ) at the highest point where it is momentarily at rest. Do your answer change if the pebble was thrown at an angle of 45^(@) with the horizontal direction ? |
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Answer» Solution :When a BODY is thrown vertically upwards (or) it moves vertically downwards, gravitational pull of earth gives it a uniform ACCELERATION `a=+g=9.8 ms^(-2)` in the downward direction. Therefore, the net FORCE on the pebble in all the three cases is vertically downwards. As m = 0.05 kg and `a=+9.8 m//s^(2)` `therefore` In all the three CASE, `F = ma = 0.05^(@)9.8 =0.49 N`. Vertically downwards. If the pebble were thrown at an angle of `45^(@)` with the horizontal direction, it will have horizontal and vertical components of velocity. These components do not affect the force on the pebble. Hence our answere do not alter in any case. However in each case (C ), the pebble will not be at rest. It will have horizontal COMPONENT of velocity at highest point. |
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| 44860. |
What was the name of the Boat |
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Answer» The TALK Walker |
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| 44861. |
Theperiod of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be : |
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Answer» `2T` Correct choice is (b). `:.` TIME period DECREASES. HENCE correct choice is (b). |
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| 44862. |
In the given network, calculate the potential difference between the points B and D |
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Answer» SOLUTION :a, B are in SERIES. `therefore R_(1)=2+3=5Omega` c, d are in series `therefore R_(2)=20+5=25Omega` `I_(1)=I[(R )/(R_(1)+R_(2))]=4[(25)/(30)]=(10)/(3)A` `I_(2)=I[(R_(1))/(R_(1)+R_(2))]=4[(5)/(30)]=(2)/(3)A` `V_(A)-V_(B)=I_(1)[R_(a)]=(10)/(3)[2]=(20)/(3)" volt"` `V_(A)-V_(D)=I_(2)[R_(d)]=(2)/(3)[5]=(10)/(3)" volt"` `therefore R_(B)-V_(D)=(10)/(3)-(20)/(3)=-(10)/(3)" volts"` |
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| 44863. |
(i) Why does the Sun appear reddish at sunset or sunrise ?(ii) For which colour the refractive index of prism material is maximum and minimum ? |
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Answer» Solution : (i) At SUNSET or sunrise, the Sun.s RAYS have to pass through a larger distance in the atmosphere. Most of the BLUE and other shorter wavelengths are removed by scattering. HOWEVER, red COLOURED rays, being of maximum wavelength, are scattered the least and travel straight and enter the observer.s eye. As a result, the Sun appears reddish. (ii) For violet colour the refractive index of prism material is maximum. For red colour the refractive index of prism material is minimum. |
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| 44864. |
What is "" i. neutrino "" ii. antineutrino? Explain with examples. |
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Answer» Solution :i. `-BETA` (i.e. electron) emission is along with the emission of an antineutrino. `""_(0)^(1)N to ""_(1)^(1)p + ""_(-1)^(0)e + barupsilon` ii. `+beta` (POSITRON) emission is along with neutrino. `""_(1)^(1)p to ""_(0)^(1)n + ""_(+1)^(0)e + upsilon` Neutrino and antineutrino have no charge and no mass. |
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| 44865. |
In the broadcast of communication modulation is necessary:-What do you mean by modulation? |
| Answer» Solution :Modulation is the PROCESS of superprosing a LOW ferquency signal on a high frequency CARRIER wave | |
| 44866. |
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 16000 Å in vacuum. |
| Answer» SOLUTION :SPEED of microwaves as WELL as UV radiations remain same in vacuum at `3xx10^(8)ms^(-1)`. | |
| 44867. |
A polythene piece rubbed with wool is found to have a negative charge of 3 xx 10^(-7)C. (a) Estimatethenumberofelectrons transferred (from which to which ?) (b) Is there a transfer of mass from wool to polythene ? |
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Answer» SOLUTION :When N no. of electron get TRANSFERRED from wool to polythene, it gets a negative charge, q = - Ne(Where e = magnitude of charge of one electron) `therefore N =-q/e` `=-(-3.2 xx 10^(-7))/(1.6 xx 10^(-19))` `N = 2 xx 10^(12)` Here, N no. of electrons are transferred from wool to polythene. Hence, wool loses mass of N no. of electrons and polythene gains mass of N no. of electrons. Thus, mass transferred in this process. `= Nm_(e)` `=2 xx 10^(12) xx 9.1 xx 10^(-31)` kg `=1.82 xx 10^(-18)` kg (Above mass is negligibly small) |
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| 44868. |
The graph between applied force and change in the length of wire within elastic limit is a |
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Answer» straight line with POSITIVE SLOPE |
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| 44869. |
The switch S has been closed for long time and the electric circuit shown caries a steady current. Let C_1 = 3.0muF, C_2 = 6.0 muF, R_1 = 4.0 kOmega, and R_1 = 7.0 k Omega The power dissipated in R_2 is 2.8 W. |
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Answer» The power dissipated to the resistor `R_1 is 1.6W.` `i = 2 xx 10^(-2) A ` `P_(R1) = i^2R_1 = (2xx10^(-2))^2 xx 4 xx 10^(3) = 1.6W` `Q_(C1) = V_(R1) xx C_1 = 80 xx 3 xx 10^(-6) = 240 muC` `Q_(C2)= V_(R2) xx C_2 = 140 xx 6 xx 10^(-6) = 840 muC` `Q_(C1) = E_(C1) = 220 xx 3 xx 10^(-6) = 660 muC` 
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| 44870. |
Show that an ideal inductor does not dissipate power in an ac circuit. |
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Answer» SOLUTION :We know that average power consumed in an a.c. circuit is given as : `P_(av)=V_(rms) cos PHI` where `phi=` phase difference between instantaneous VALUES of voltage and current. For an ideal inductor current lags behind the voltage by `(pi)/(2)` . As `phi = (pi)/(2)` , so cos `phi` and CONSEQUENTLY the average power `P_(av)=V_(rms)xxI_(rms)xxcos""(pi)/(2)` =0 . So an ideal inductor does not dissipate power in an a.c. circuit. |
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| 44871. |
Assertion : Five charges +q each are placed at five vertices of a regular pentagon. A sixth xharge -Q is placed at the centre of pentagon. Then, net electrostatic force on -Q is zero. Reason : Net electrostatic potential at the centre is zero. |
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Answer» If both ASSERTION and Reason are true and Reason is the correct EXPLANATION of Assertion. |
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| 44872. |
Using a diagram , explain the working of a cyclotron. Hence obtain the expression for cyclotron frequency. |
| Answer» SOLUTION :REFER TEXT, TOPIC CYCLOTRON | |
| 44873. |
In a resonance-column experiment, a long tube, open at the top, is clamped vertically. By a separate device water level inside the tube can be moved up or down. The section of the tube from the open end to the water level acts as closed organ pipe. A vibrating tuning fork is held above the open, and the water level is gradually pushed down. The first and the second resonances occur when the water levels 24.1 cm and 74.1cm respectively below the open end. The diameter of the tube is |
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Answer» 2cm |
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| 44874. |
A ball thrown by one player reaches the other in 2 seconds. The maximum height attained by the ball above the point of projection will be (g = 10m//s^2) |
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Answer» 10m max height `(u^2 sin^2 theta)/(2g)g//2` or `((u^2sin^2 theta)//g^2) 10/2xx1^2 = 5m` |
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| 44875. |
A carnot engine has an efficiency of 20%. The energy is supplied to the engine at the rate of 2 KW What is the output power of the engine ? |
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Answer» 300 W |
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| 44876. |
A body is thrown vertically up reaches a maximum height of 78.4 m. After what time it will reach the ground from the maximum height reached ? |
| Answer» ANSWER :A | |
| 44877. |
The higher frequency TV broad casting bands range is |
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Answer» `54-72 MHZ` and `76` to `88 MHz` |
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| 44878. |
The density of a cube is measured by measuring mass and the length of its sides. If the maximum error in the measurement of mass and length are 3% and 2% respectively, then the maximum error in the measurement of the density of cube is : |
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Answer» `7%` `:.(Deltarho)/(rho)xx100=(DELTAM)/(m)xx100+3(DELTAL)/(l)xx100` `=3%+3xx2%` `=9%` Hence `(d)` is correct. |
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| 44879. |
IF have energy of a 100 mu F capacitor charge to 6kV could all be used to lift a 50 kg mass, then the greatest vertical height through which mass could be raised in…… m |
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Answer» 3.6 `therefore H=(CV^2)/(2MG)=(100 times 10^-6 times (6 times 10^3)^2)/(2 times 50 times 10)` `=3.6m` |
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| 44880. |
Demagnetisation of a magnet can be done by |
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Answer» ROUGH HANDLING |
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| 44881. |
When air is replaced by dielectric medium of constant K, the maximum force of attraction between two charges separated by distance d, .......... |
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Answer» BECOMES `K_(2)`times `F =(k_(q_(1)q_(2))/R^(2)), F. = k/K.(q_(1)q_(2))/r^(2)` |
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| 44882. |
A bar magnet having a magnetic moment of 2.0 xx 10^(-4) J//T is free to rotate in a horizontal plane. A horizontal magnetic field B = 5 xx 10^(-5) T exists in the space. Then the work done in rotating the magnet slowls from a direction parallel to the field to a direction 60^(@) from the field is |
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Answer» 0.1 J `= (-MB costheta_(2)) - (-MBcostheta_(1))` `=-M(cos60^(@)-cos0^(@))` `=1/2MB = 1/2xx(2.0xx10^(4)J//T)(5xx10^(-5)T = 0.5J` |
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| 44883. |
A thin and light thread of sufficient length L is attached to a wire frame. The frame (along with the thread) is dipped into soap solution. When the frame is taken out of the solution, the thread takes the form of a semicircle with a soap film extending between the frame and the thread as shown in the figure (a). Now the thread is deformed into two semicircles by applying a force of F at the middle of the thread, as a result the film gets strectched as shown in the figure. (b). Calculate the surface tension of the soap solution. |
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Answer» `(PIF)/(2L)` `F+2T=2sigmax`, where `T=2(SIGMA(X)/(4))` `rArr F=sigmax` and `2((pix)/(4))=L` `rArrsigma=(piF)/(2L)` 
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| 44884. |
A particle performs SHM along a straight line and its position is vec(R ), acceleration is vec(a), velocity is vec(v) is and force on particle is vec(f). Then which of the following statement are true? (i) vec(v). vec(a) is always + ve (ii) vec(v).vec(R ) may be -ve (iii) vec(f).vec(R ) is always -ve (iv) vec(v) is parallel to vec(f) sometimes |
| Answer» Answer :A | |
| 44885. |
Write about electron microscope. |
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Answer» Solution :Electron Microscope : Principle : This is the direct application of wave nature of PARTICLES. The wave nature of the electron is used in the construction of microscope called electron microscope. The resolving power of a microscope is inversely proportional to the wavelenth of the radiation used for illuminating the OBJECT under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths. De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes. As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope. Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.  Working: The electron beam passing across a suitably ARRANGED either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done. The electrons emitted from the source are accelerated by high potentials. The beam is MADE parallel by magnetic CONDENSER lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample. With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science. |
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| 44886. |
A wooden block floats in a liquid with 40% of its volume inside the liquid. When the vessel containing the liquid starts rising upwards with acceleration a = g/2, the percentage of volume inside the liquid is : |
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Answer» 0.2 Thus correct CHOICE is (d). |
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| 44887. |
The emissive power of a body does not depend upon___________of the body |
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Answer» area |
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| 44888. |
Do the two electric bulbs connected to the same electrence? |
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Answer» |
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| 44889. |
What is the advantage of using thick metallic strips to join wires in a potentiometer ? |
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Answer» Solution :In potentiometer instead of using a very long wire, wire of different LENGTH are connected by using thin metal strip such that NECESSARY equivalent length is OBTAINED. RESISTANCE of these metal strip is very small and can be neglected. There is no need to calculate length of strips. In EXPERIMENT instead of using one long wire, different wire of l m length are joined together to obtain necessary length of potentiometer. In potentiometer centimeter scale or meter scale are used hence null point can be accurately obtained. |
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| 44890. |
Pick out the correct statements from the following: I. Electron emission during beta -decay is always accompanied by neutrino. II. Nuclear force is charge independent. III. Fusion is the chief source of stellar energy. |
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Answer» I, II are correct |
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| 44891. |
A wheeel having mass has charges +q and q on diametrically opposite points. It remains in equilibrium on a rough inclined places in the presence of uniform vertical electric field E = |
| Answer» Answer :B | |
| 44892. |
A coil having an area of 2m^2 is placed in a magnetic field which changes from 2Wb/m^2 to 5Wb/m^2 in 3 sec. What will be the e.m.f. induced in the coil? |
| Answer» SOLUTION :E = `(-dphi)/DT` = `[A[B_2-B_1]]/t` = `2xx3/3` = 2V | |
| 44893. |
Two identically charged particles are fastened to the two ends of a spring ofspring constant 100 N m ^(-8) C, find the extension In the length of the apring. Assume that the extension is small as scompared to the natural length . Justify this assumption after you solve the problem. |
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Answer» SOLUTION :`ELECTRIC force = spring force ` ` 1/(4piepsilon_0)(q^2 )/(x+0.1)^2=K(0.1+x) ` ` or (0.1+x)^3=(9xx10^9xx(2.0xx10^-8)^2)/10^-2 ` ` = (36xx10^9xx10^-16) /10^-2 ` `= 36xx10^-5 ` ` or x=3.6xx10^-6m ` |
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| 44894. |
One of two parallel metallic plates is uniformly charged with charge +q, and the other one is charged with charge -q. In this case, the electric field between them is E. When the negatively charged plate is discharged and the recharged with a +ve charge 4q, electrical field between the plates becomes: |
| Answer» ANSWER :B | |
| 44895. |
A certain capacitor is charged to a potential V. By what percentage the potential to be increased, so that energy increases by 10%? |
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Answer» Solution :Energy `alpha (p.d)^(2)` `U alpha V^(2) and U. alpha (V.)^(2)` `(U.)/(U)=(110)/(100) =1.1, U.=1.1U` `(U.)/(U)=((V.)^(2))/(V^(2)) therefore (V.)^(2)=(1.1)^(2)` `V.=(sqrt(1.1))V=1.049V` % increase in potential `=((V.-V)/(V)) XX 100 =((1.049V-V)/(V)) xx 100=4.9%` |
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| 44896. |
What is space wave propagation ? Give two examples of communication system which use space wave mode. A TV tower is 80 m tall. Calculate the maximum distance upto which the signal transmitted from the tower can be received. |
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Answer» Solution :When wave travel in space in a straight line from the transmitting antenna to the receiving antenna, this mode of proagation is called the space wave propagation. Examples : Television broadcast, microwave links, satellite communication. `d= sqrt(2HR) = sqrt( 2 xx 80 xx 6.4 xx 10^(6)) = 32 km` |
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| 44897. |
Two capacitors of 3muF and 6 muF are connected in series and a potential difference of 900 V is applied across the combination. They are then disconnected and reconnected inparallel. The potential difference across the combination is |
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Answer» Zero `thereforeCalpha1/VthereforeC_(1)/C_(2)=V_(2)/V_(1)rArr3/6=V_(2)/V_(1)rArrV_(1)=2V_(2)` ALSO `V_(1)+V_(2)=900V` `V_(2)=300VandV_(2)=600V` COMMON potential V = `(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` = `(3xx10^(-6)xx600+6xx10^(-6)xx300)/(3xx10^(-6)+6xx10^(-6))` = `(1800+1800xx10^(-6))/(9xx10^(-6))=3600/9` = 400 V 
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| 44898. |
There are two metal verticalrails connected to a capacitor of capacitance 10 mF as shown in figure. Magnetic field strength of 1 T is applied perpendicular to theplaneofrails. Separation between the rails is 10 cm. Ratio of current flowing through the loop and acceleration of the rod is found to be n x 10^(-3) in MKS system. Find the value of n. |
Answer»  Motional emf developed across the LENGTH of the rod is BVL, when rod moves with velocity v. Let q be the instantaneous charge on capacitor. q= CBvl `I= dq//dt` `I= "CBL"(dv//dt)` I= CBla `(I)/(a)= CBl` `(I)/(a)= 10 xx 10^(-3) xx 1 xx 0.1 = 1 xx 10^(-3)` Hence value of n = 1. |
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| 44899. |
Chromatic aberrations in the formation of image by a lens arise becauses |
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Answer» of non-paraxial rays |
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| 44900. |
An object 1 cm tallis placed 4 cm infront of a mirror. In order to produce an upright image of 3 cm height one needs a |
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Answer» Convex mirror of RADIUS of CURVATURE 12 cm |
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