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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.
| 41751. |
Two point charges +Q and -Q are separated by a certain distance. The resultant electric field is parallel to the line joining the charges at the points |
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Answer» on the line joining the charges |
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| 41752. |
A thread carrying a unifrom charge lambda per unit length has the configuraations shown in a) and (b). Assuming a curvature radius R to be considerably less than length of the thread find the magnitude of the electric field strenght at the point O. |
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Answer» `(lambdasqrt(2))/(4piepsilon_(0)R),0` |
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| 41753. |
Find the probability of an electron tunneling through a 5Å wide and 0.4 eV high potential barrier, if it is accelerated by a field of 0.3 V |
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Answer» `w=D//D_(0)=E^(-a),"where "a=(2L)/hsqrt(2M(U_(0)-ePhi))` |
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| 41754. |
A block placed on a smooth horizontal floor is connected to a springs as shown. Initially an external force of 100N keeps it stretched by 1cm beyond natural lenjgth i equilibrium. This force is now removed and another force F is applied on this block, which slowly moves it from this position to a position where the spring is finally compressed by 3 cm. Find the work done by this force F (in joules). |
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Answer» `W=1/2 k(x_("FINAL")^(2)-x_("initial")^(2))=4J` |
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| 41755. |
findoutmotionoftree, boyandoldmanas seenbybird |
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Answer» SOLUTION :withrespectto bird `:V_("TREE ") = 12 m//s ( DARR )and 16 m//s( darr)` ` V_(" old MAN") = 18 m//s( darr ) and 12m//s( darr )` ` V_("boy") = 12m//s (darr)` |
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| 41756. |
A conducting wire of given length is used to prepare the 'coil' of a moving coil galvanometer. To have maximum sensitivity the shape of the coil should be : |
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Answer» circular |
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| 41757. |
The amplitude of a wave disturbance propagating in the positive x-direction is given by at time in and by v=(1)/(1+x^(2)) at time t=0 and by y=(1)/([1+(x-1)^(2)]) at t=2 seconds, where x and y are in metres. The shape of the wave disturbance does not change during the propagation. What is the velocity of the wave? |
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Answer» |
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| 41758. |
Obtain the equation of electric potential energy of a dipole from equation of potential energy of a system of two electric charges. |
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Answer» Solution :The equation of potential energy of a system a two CHARGES is, `U (theta) = q_(1)V(r_(1))+q_(2)V(r_(2))+(kq_(1)q_(2))/(r_(12))` For DIPOLE `q_(1)=+q,q_(2)=-q` and their positior vectors are `vecr_(1)` and `vecr_(2)` . `:. U.(theta)=q[V(r_(1))-V(r_(2))]-(kq_(1)q_(2))/(r_(12))` The potential difference between positions `r_(1)`and `r_(2)` equals the work done in bringing a unit positive charge against field from `r_(2)` to `r_(1)`. The DISPLACEMENT parallel to the force is 2acose `theta` . `:. V(r_(1))-V(r_(2))=-Exx2a cos theta[ because `W=Fd] `=- p E COSTHETA[because` p =q(2a)=2a] `:. U.(theta)=-pEcostheta-(kq^(2))/(2a)` `:.U.theta=-(vecp.vecE)-(kq^(2))/(2a)` For a given dipole U(6) differs from U(`theta`) only by a constant. If `theta_(0)=(pi)/(2) ` be taken then `q[V(r_(1))-V(r_(2))]=0` If we drop the second term in equation (2) then, `U.(theta)=(vecP.vecE) ` is like `U =- (vecP.vecE).` |
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| 41759. |
What is the mood of the poem? |
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Answer» Joyous |
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| 41760. |
A wire of 4 Omega resistance is bent at 180° at its mid point and twisted together. Then the resistance between the ends is |
| Answer» Answer :B | |
| 41761. |
Calculate the electric field intensity which would be just sufficient to balance the weight of an electron. If this electric field is produced by a second electron located below the first one what would be the distance between them? [Given: 1.6 xx 10^(-19) C, m = 9.1 xx 10^(-31) kg and g = 9.8 m//s^2]. |
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Answer» Solution :As force on a charge e in an electric field E `F_e = eE` So according to GIVEN PROBLEM, `F_e = W` i.e., `eE = MG` `E = (mg)/(e ) = (9.1 xx 10^(-31) xx 9.8)/(1.6 xx 10^(-19)) = 5.57 xx 10^(11) V/m` As this intensity E is produced by another ELECTRON B, located at a distance r below A.  `E = 1/(4 pi epsilon_0) e/(r^2) "i.e., " r = sqrt((e)/(4 pi epsilon_0 E)` So, `r = [(9 xx 10^(9) xx 1.6 xx 10^(-19))/(5.57 xx 10^(-11))]^(1//2) ~= 5m` . |
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| 41762. |
In germanium crystal, the forbidden energy gap in joule is |
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Answer» `1.6xx10^(-19)` `E_g=0.7eV=0.7xx1.6xx10^(-19)J=1.12xx10^(-19)J` |
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| 41763. |
You are given a cube, a ring, a cylinder and a sphere, all having equal masses and cross sections with equal heights and maximum widths. Which one has the largest and which one has smallest moment of inertia about an axis perpendicular to the cross section and passing through the centre of mass. |
| Answer» Solution :We KNOW that `I sum(mr^2)`. Here ring has the largest moment of inertia because it.s mass is situated farthest from the AXIS of rotation while the sphere has least moment of inertia as it.s mass is NEAREST to the axis of rotation. | |
| 41764. |
मान लीजिए f(x)=3x-1 द्वारा परिभाषित फलन f:RrarrR है ,तब f- |
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Answer» एकैकी |
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| 41765. |
A bar magnet of magnetic moment 1.5 JT^(-1) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? |
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Answer» Solution :(a) (i) Required amount of work done, `W= mB ( cos theta_(1)- cos theta_(2) )` `= (1.5 ) (0.22) ( cos 0^(@) -cos 90^(@) )` `=(1.5)(0.22) (1-0)` `=0.33` J (ii) In this case, `W= mB ( cos theta_(1)- cos theta_(2))` `THEREFORE W= (1.5) ( 0.22)( cos0^(@) - cos180^(@) )` `therefore= (1.5) (0.22) (1- ( -1) )` `therefore W= (1.5) (0.22) (2)` `therefore W= 0.66` J (B) (i) Torque exerted, `tau = m B sin theta` `therefore tau = (1.5) (0.22) sin 90^@` `therefore tau= (1.5) (0.22) (1)` `therefore tau = 0.33` Nm (ii) In this case, `tau= m B sin theta` `therefore tau= (1.5) (0.22) (0)` `therefore tau=0` |
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| 41766. |
What happened after the first indication of the disaster? |
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Answer» a wave APPEARED VERTICAL and ALMOST TWICE the height of other waves |
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| 41767. |
When the angle of inclination of on inclinedpalneis theta, an objectdownwith uniformvelocity , If the same objectis pushed up withan intialvelocityu onthe sameinclinedplane ,is goesup theplaneand stopsat a certain distance on the plane. There after the body. |
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Answer» Slides downthe inclinedplaneand REACHESTHE ground . Withvelocity"u" |
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| 41768. |
Answer the following questions: (a) Long distance radio broadcasts use short-wave bands. Why? (b) It is necessary to use satellites for long distance TV transmission. Why? (c) Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? (f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction? |
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Answer» SOLUTION :(a) IONOSPHERE reflects waves in these bands. (b) Television signals are not properly reflected by the ionosphere (see text). Therefore, reflection is effected by satellites. (c) ATMOSPHERE absorbs X-rays, while visible and radiowaves can penetrate it. (d) It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth’s surface and causing damage to life. (e) The temperature of the earth WOULD be lower because the Greenhouse effect of the atmosphere would be absent. (f) The clouds produced by global nuclear WAR would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’. |
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| 41769. |
1 ohm resistance is in series with an Ammeter which is balanced by 75cm of potentiometer wire. A standard cell of 1.02V is balanced by 50cm. The Ammeter shows a reading of 1.5A. The error in the Ammeter reading is |
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Answer» 0.002 A |
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| 41770. |
किसी परमाणु का amu मे निकाला गया भार कहलाता है |
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Answer» परमाणु भार |
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| 41771. |
A string with tension T and mass per unit length mu is clamped down at x=0 and at x=L. at t=0, the string is at rest and displaced in the y-direction y(x,0)=2"sin"(2pix)/(L)+2"sin"(pix)/(L) Q. At what time t will the string for the first time have exactly the same shape as it did at time t=0? |
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Answer» `(L)/(2v)` |
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| 41772. |
(a) In the example 2 the current flowing through the wire is l_((l))=l_(0) sin omegat. What is the flux through the loop at t=(pi)/(2omega)? (b) What is the maximum flux when current l_(t)=l_(0)sin omegat |
| Answer» SOLUTION :(a) `phi_(B)=-(mu_(0)l_(0)l)/(2PI)log_(e).(a+l)/(a), (b) phi_("max")=(mu_(0)l_(0)l)/(2pi)log_(e).(a+l)/(a)` | |
| 41773. |
What physical quantity is the same for X-rays of wavelength 10^(–10)m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? |
| Answer» SOLUTION :The speed in vacuum is the same for all: `c=3xx10^(8) MS^(-1)`. | |
| 41774. |
Binding energy of a body . |
| Answer» SOLUTION :The BINDING energy of a body is defined as the MINIMUM energy that should be PROVIDED to a body , so that it can escape from the Earth's gravitational field . | |
| 41775. |
What is velocity head.What it implicates? |
| Answer» SOLUTION :The VELOCITY HEAD of a LIQUID is `1/2 v^2/g` and it is VARIABLE. | |
| 41776. |
An electron is accelerated through a potential difference of 1000 volts .Its velocity is nearly |
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Answer» `3.8 xx 10^(7) m//s` |
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| 41777. |
If force (F),length (L) and time (T) be considered fundamental units, then units of mass will be : |
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Answer» `[FL^(-1)T^(2)]` Hence correct CHOICE is `(a)`. |
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| 41778. |
Artificial magnet in the form of a rectangular or cylinderical bar is called .............. . |
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Answer» |
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| 41779. |
Two light sources with amplitudes 5 units and 3 units respectively interfere with each other. Calculate the ratio of maximum and minimum intensities. |
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Answer» Solution :Amplitudes, `a_(1) = 5, a_(2) 3` RESULTANT amplitude, `A=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cosphi)` Resultant amplitude is MAXIMUM when, `phi=0,cos0=1,A_(MAX)=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2))` `A_(min)=sqrt((a_(1)-a_(2))^(2))=sqrt((5-3)^(2))=sqrt((2)^(2))=2units` `1 prop A^(2)` `(I_(max))/(I_(min))=((A_(max))^(2))/((A_(min))^(2))` Substituting `(I_(max))/(I_(min))=((8)^(2))/((2)^(2))=(64)/(4)=16(or)I_(max):I_(min)=16:1` |
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| 41780. |
The binding energy per nucleon is almost constant for the nuclei having atomic mass number………. |
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Answer» `30lt A lt lt 170` |
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| 41781. |
In a meter bridge, the left and right gaps are closed by resistances 2 ohm and 3 ohm respectively. The value of shunt to be connected to 3 ohm resistor to shift the balancing point by 22.5 cm is |
| Answer» ANSWER :D | |
| 41782. |
The minimum potential difference between the base and emitter required to switch a silicon transistor 'ON' is approximately: |
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Answer» `1V` |
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| 41783. |
What will be ratio if the particles in above examples had same kinetic energies |
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Answer» |
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| 41784. |
Figure 10-61 is an overhead view of a rod of length 1.0 m and mass 1.0 kg that is lying stationary on a frictionless surface when three bullets hit it simultaneously. The bullets move along paths that are in the plane of the rod and per-pendicular to the rod. Bullet 1 has mass 10 g and speed 2.0 m/s. Bullet 2 has mass 20 g and speed 3.0 m/s. Bullet 3 has mass 30 g and speed 5.0 m/s. The labelled distance are alpha=10cm, b = 60 cm, and c = 80 cm. As a result of the impacts, the rod-bullets system rotates around its center of mass while the center of mass moves in a straight line over the frictionless surface. (a) What is the linear speed of the system's center of mass? (b) What is the distance between the rod's center and the system's center of mass ? ( c) What is the rotational inertia of the system about the system's center of mass? |
| Answer» Solution :(a) `6.6xx10^(-2)m//s` (B) `~~1.9xx10^(-3)m`, ( C) `8.8xx10^(-2)KG*m^(2)` | |
| 41785. |
Two closed pipes when sounded together five 4 beats/sec. If longer pipe has length 1m, the length of other pipe is ( velocity of sound 300m/s) |
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Answer» 80.5cm |
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| 41786. |
A Carnot enginetakes 300 calories of heat from a source at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is |
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Answer» 125 K `W=Q_(1)-Q_(2)=150cal` `eta=W/Q_(1)=150/300=1/2` `eta=1-T_(2)/T_(1)` `rArr1/2=1-T_(2)/500rArrT_(2)/500=1/2` `T_(2)=250K` |
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| 41787. |
A block of mass m=1kg is attached to a free end of a spring whose one end is fixed with as shown in the figure. The block is performing simple harmonic motion. The position of the block from O is given by x=2+(1)/(sqrt(2))sin2t, where x is in meter and t is in second. A particle of same mass is released from the circular path at a height h=80cm. then particle collidic and stricks to the block. The collision takes place when velocity of the block is zero and spring is elongated. Find (i) Time period of new SHM. (ii) Maximum velocity of the system |
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Answer» <P> Solution :`x=2+(1)/(sqrt(2))sin2t``rArr 2=sqrt((k)/(m)) rArr k=4N//m` `omegaL=(1)/(omegaC)` Just before collision, velocity of the SHELL, `2omegaL=(1)/(2omegaC')` Now , `P_(i)=P_(r)` `0+mx4=(m+m)xv_(0)rArrv_(0)=2M//s` Using conservation of energy, `:. C'=(1)/(4omega^(2)L)=(LC)/(4L)=(C )/(4)kx^(2)+ :. (DeltaC)/(C )=(3)/(4)(2m)(2)^(2)` `=(1)/(f_(1))=((1.5-1))/(R )+((4)/(3)-1)(-(1)/(R )-(1)/(R ))+((1.5-1)/(R ))=(1)/(R )-(2)/(3R)=(1)/(3R)(2m)v'^(2)` `=(1)/(f_(2))=((1.5-1))/(R )+((4)/(3)-1)(-(1)/(R ))+((1.5-1)/(R ))=(1)/(R )-(1)/(3R)=(2)/(3R)` 
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| 41788. |
How sounds of different frequencies are produced by opening or closing the different holes of a flute ? |
| Answer» Solution :A fine acts as pipe open at the ends. Resonance OCCURS when ever there is a pressure NODE at both ends. Therefore, the STANDING waves with all the holes closed corresponds to a half-wavelength or the lowest mode. Opening a finger hole acoustically SHORTENS the TUBE by releasing the pressure (pressure node) before the open end, thereby changing the length of the standing waves and its pitch. | |
| 41789. |
Obtain a relation for the magnetic induction at a point along the axis of a circular coil carrying current. |
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Answer» Solution :Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by TAKING two diametrically OPPOSITE LINE elements of the coil each of length `vec(dl)` at C and D. Let `vecr` be the vector joining the current element `(Ivec(dl))` at C to the point P. `PC=PD=r=sqrt(R^(2)+Z^(2))`and angle `/_CPO=DPO=theta` According to Biot-Savart.s law, the magnetic field at P due to the current element `Ivec(dl)` is `dvecB=(mu_(0))/(4pi)(Ivec(dl)xxhatr)/(r^(2))` The magnitude of magnetic field due to current element `Ivec(dl)` at C and D are equal because of equal distance from the coil. The magnetic field `dvecB` due to each current element I`vec(dl)` is resolved into two components, DB sin `theta` along y-direction and dB cos theta along z-direction Horizontal components of each current element cancels out while the vertical components `(dB cos thetahatk)` alone contribute to total magnetic field at the point P.  If we integrate `vecdl` AROUND the loop, `dvecB` sweeps out a cone, then the net magnetic field `vecB` at point P is `vecB=intdvecB=intdbcosthetahatk` `vecB=(mu_(0)I)/(4pi)int(dl)/(r^(2))costhetahatk` But cos `theta=(R)/((R^(2)+Z^(2))^(1/2))` Using Pythagorous theorem `r^(2)=R^(2)+Z^(2)` and integrating line element from 0 to `2piR`, we get `vecB=(mu_(0)I)/(4pi)=(R^(2))/((R^(2)+Z^(2))^(3/2))hatk`...(4) Note that the magnetic field `vecB` points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O. |
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| 41790. |
The period of a satellite moving very close to the surface of the earth of radius R is 84 minute. What will be the period of the same satellite, if is is taken at a distance of 3R from the surface of the earth ? |
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Answer» 84 MIN |
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| 41791. |
In the figure shown for an angle of incidence i at the top of the surface, what is the minimum refractive index for total internal reflection at the vertical surface. |
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Answer» SOLUTION :The RAY will total INTERNALLY reflect at the vertical surface if `theta gt theta_(c)` Now, `r=(90^(@)-theta)` and Snell.s law is `sin i=mu sin r` `(sini)/(mu)=sin(90^(@)-theta) implies cos theta=(sin i)/(mu)` or `sin theta=sqrt(1-cos^(2)theta)=sqrt(1-(sin^(2)i)/(mu^(2)))`  If `theta gt theta_(C)` , then `sin theta gt sin theta_(C)` ( As `sin theta` is an increasing function for `0 lt theta lt 90^(@))` `sqrt(1-(sin^(2)i)/(mu^(2)))gt(1)/(mu)` `1-(sin^(2)i)/(mu^(2))gt(1)/(mu^(2))` `mu^(2)-sin^(2)i gt 1 "or" (mu^(2)-1)gtsin^(2)i` If total internal reflection has to be larger for all value, the above inequality must be SATISFIED for all `(sin^(2)i)_("MAX")=1` `impliesmu^(2)-1gt1 "or" mu gtsqrt(2)` This total internal reflection phenomenon is used in fibre optics to bend light in a curved path. 
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| 41792. |
A car is rounding a flat curve of radius R=220 m at the curve's maximum design speed v=94.0km/h. What is the magnitude of the net force on the seat cushion from a passenger with mass m=85.0 kg? |
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Answer» |
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| 41793. |
A body of mass m = 4 kg starts moving with velocity v_(0) in a straight line is such a way that on the body is being doen at the rate which is proportional to the square of velocity as given by P = beta v^(2) where beta = ( 0.693)/( 2). Find the time elapsed in seconds before velocity of body is doubled. |
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| 41794. |
Condensation is reverse of …………… |
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Answer» Fusion |
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| 41795. |
The value of 1.999... In the form of p/q , where p and q are integers and q!=0, is |
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Answer» 19/10 |
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| 41796. |
De-Broglie wavelengths of the particle increases by 75% then kinetic energy of particle becomes. |
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Answer» `(16)/(49)` times |
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| 41797. |
Abody of mass in thrown up vertically with velocity v_(1) reaches a maximum height h_(1)in t_(1) seconds. Another body of mass 2 m is projected with a velocity v_(2) at an angle theta. The second body reaches a maximum height h_(2) in time t_(2) seconds. If t_(1) = 2t_(2), then ratio ((h_(1))/(h_(2)) is |
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Answer» 0.043055555555556 `h_(1) = (v_(1)^(2))/(2g)` Another body of mass 2m is projected with a velocity `v_(2)` at an angle `theta`. `therefore "Height attained" (h_(2)) = (v_(2)^(2)sin^(2)theta)/(2g)` `therefore "" (h_(1))/(h_(2)) = (v_(1)^(2))sin^(2) theta` `"But" "" t_(1) - 2t_(2)` `therefore "" (v_(2))/(g) = 2 [(v_(2)sintheta)/(g)]` `rArr "" v_(1) = 2v_(2)sintheta` `therefore "" (h_(1))/(h_(2)) = ((2)/(1))^(2) = (4)/(1)` |
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| 41798. |
A body of mass 100 gm is suspended from a light spring and stretches the spring by 10 cm. What is the force constant ? |
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Answer» Solution :We have F=kx `k=F/X=(MG)/x=(100xx980)/10` `=9800"dyne"/CN` |
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| 41799. |
2. Give te General formulas of C, E and V of capacitor? |
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Answer» (II) (II) (L) |
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| 41800. |
Two identical coaxial circular coils are carrying current .i. each in the same direction. If the coils approach each other, then |
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Answer» The CURRENT in each increases |
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