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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.
| 5851. |
The logic circuit and its truth table are given, what is the gate X in the diagram |
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Answer» SOLUTION :From the truth table we not that `Y=A+B` i.e., it is for OR gate (or) `A+X=A+B=A+B.(A+BARA)(because A+barA=1)` then, `A+X=A+B.A+B.barA=A.(1+B)+barA.B=A+barA.B` So `X=barA.B`, which is AND gate with INPUTS as `barA` and B |
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| 5852. |
Plane polarised light is incident on an analyser. The intensity then becomes three fourth. The angle of the axis of the analyser with the beam is |
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Answer» `30^(@)` |
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| 5853. |
All along we have been considering the apparent depth of an object viewed normally. Suppose that the object is being viewed at an angle theta to the surface. If the actual depth is d and angle of ray inside water is phi, what is the apparent depth ? |
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Answer» |
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| 5854. |
A capacitor of capacitance 8 mu F is connected across the terminals of a battery of emf 24V. Find the work done by the battery during the process. |
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Answer» `2.3mJ` |
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| 5855. |
A sphere of radius R is exposed to a parallel beam of radiation of intensity I as shown in figure. Choose the correct option (s) of the following. |
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Answer» If the surface of the sphere is completely reflecting, radiation force in the sphere is `(2IpiR^(2))/(c )`  CONSIDER a circular strip of radius `R sin theta` and of width `Rd theta`. AMOUNT of energy falling on the strip per sec `=IdAcos theta` Change in momentum due to reflection `dp=(2I)/(c )(DACOS theta)cos theta` Force on the strip `dF=(2I)/(c )dAcos^(2)theta=(2I)/(c )xx2piRsin thetaxxRd thetaxxcos^(2)theta` Net force on the sphere `F=int_(0)^((pi)/(2))dF costheta=int_(0)^((pi)/(2))(4piIR^(2))/(c )cos^(3)thetasinthetad theta` `=-(4piIR^(2))/(c )int_(0)^((pi)/(2))cos^(3)theta(-sinthetad theta)` `=-(4piIR^(2))/(c )[(cos^(4)theta)/(4)]_(0)^((pi)/(2))=(piIR^(2))/(c )[cos^(4)theta]_((pi)/(2))^(0)=(IpiR^(2))/(c )` When surface is completly absorbing `dF=(I(dAcostheta))/(c )` Net force `F=intdF=int_(0)^((pi)/(2))(I2piRsintheta(Rd theta)costheta)/(c )` `=(IpiR^(2))/(c )int_(0)^((pi)/(2))sin2thetad theta=(IpiR^(2))/(c )[(-cos2theta)/(2)]_(0)^((pi)/(2))` `=(IpiR^(2))/(2c )[cos2theta]_((pi)/(2))^(0)=(IpiR^(2))/(2c )xx2=(IpiR^(2))/(c )` when surface is partially reflecting with reflection coefficent `0.3` and absorbtion coefficent `0.7` net force on the sphere is `F=int_(0)^((pi)/(2))(2(0.3l)(dAcostheta)costheta)/(c )xxcostheta+int_(0)^((pi)/(2))((0.7l)(dAcostheta))/(c )` `= 0.3(IpiR^(2))/(c )+0.7(IpiR^(2))/(c )=(IpiR^(2))/(c )` Note: In all the above cases radiation force=radiation pressure (due to absorbtion)xxeffective area perpendicular to the flow of energy`=(I)/(c )xxpiR^(2)` |
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| 5856. |
A motorboat covers a given distance in 6 hours moving downstream on a river. It covers the same distance in 10 hours moving upstream. The time it takes to cover the same distance in still water is |
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Answer» 6.5 hours `t=(2t_1t_2)/(t_1+t_2)` `t=(2xx6xx10)/(6+10)=(120)/(16)=7.5 hrs` |
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| 5857. |
What is frequency ? How we express it ? |
| Answer» Solution :The number of REVOLUTIONS per SECOND by ELECTRONS around the nucleus and it is GIVEN by `f_n = (Ze^2)/4piepsilon_0nhr_n`. | |
| 5858. |
Find the total charge stored in the network of capacitors connected between A and B as shown in figure. |
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Answer» SOLUTION :The GIVEN network can be redrawn as shown here . The equivalent capacitance of the network is given as : `1/C=(1)/((C_1+C_2))+(1)/((C_3+C_4))` `=1/((4+6)MUF)+1/((2+3)muF)=(1/10+1/5)=3/10 muF` `IMPLIES C = 10/3muF` `:.` Total CHARGE on the network `Q=CV=(10/3muF)xx3V=10muC` 
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| 5859. |
Give an example for beta^+- decay process. |
| Answer» SOLUTION :`""_11Na^22 to ""_10Ne^22 + ""_(+1)e^0 + V` | |
| 5860. |
Four identical point charges are fixed at the four corners of a square of a side length l. Another charged particle of mass m and charge +q is projected towards centre of square from a large, distance along the line perpendicular to plane of square. The minimum value of initial velocity V_(0) in m/s required to cross the square is? (m = 1gm.l=4sqrt(2)m,Q=1muc,q=0.5muc) |
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Answer» |
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| 5861. |
A parallel plate capacitor with circular plates of radius 0.8 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Omega across a 4V battery. Calculate the magnetic field at a point P, halfway between the centre and the perpendicular of the plates after t=10^(-3)s. (The charge on the capacitor at time t is q (t) = CV [1 - exp (-t//tau)], where the time constant tau is equal to CR.) |
| Answer» SOLUTION :`i_(d)=0.3679xx10^(-6)A, B=1.84xx10^(-13)T`. | |
| 5862. |
Two long coaxical insulated solenoid's S_1 and S_2 of equal lengths are would one over the other as shown in figure. A steady current 'T' flows through the inner solenoid S_1 to the other end B, which is connected to the outer solenoid S_2 through which the same current flows in opposite direction so as to come out at end A. If n_1 and n_2 are the number of turns per unit length, find the magnitude and direction of net magnetic field at a point (i) inside on the axis and (ii) outside the combined system. |
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Answer» Solution : Magnetic field, `B=mu_0nI` Magnitude of NET magnetic field inside the combined system on the AXIS, `B = B_1 – B_2` `=mu_0n_1I-mu_0n_2I` `=mu_0(n_1-n_2)I`Outside the combined system, net magnetic field is ZERO. |
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| 5863. |
Light is incident from a medium A to medium B. The graph of sine of angle of incidence i versus sine of angle of refraction r is shown in fig. Which of the following is/are correct? |
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Answer» Total INTERNAL REFLECTION occurs above a CERTAIN value of i. |
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| 5864. |
Discuss some points about Gauss's law. |
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Answer» Solution :(i) The total charge contained in the closed surface is zero, then the net electric flux through a closed surface is zero. (ii) Gauss.s law is true for any closed surface, no matter what its shape or size. (iii) The charges may be located ANYWHERE inside the surface. (iv) In the situation when the surface is so chosen that there are some charges inside and some outside the electric field, whose flux appears on the left side of Equ : `PHI = (sumq)/epsilon_(0)` is due to all the charges both insideand outside S. The term Q on the right side of Gauss.s law, however, represents only the total charge inside S. (v) The surface that we choose for the application of Gauss.s law is called the Gaussian surface. (vi) Gauss.s law is OFTEN USEFUL towards a much easier calculation of the electrostatic field when the system has some symmetry, (vii) Gauss.s law is based on the inverse square dependence on distance. |
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| 5865. |
The weight of the body at earth's surface is W. At a depth half way to the centre of earth its weight will be (Density of earth is uniform) : |
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Answer» W `therefore W.=W(1-((R )/(2))/(R ))=(W)/(2)` Hence correct CHOICE is (B). |
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| 5866. |
What are Isomers?Give examples. |
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Answer» Solution :Isomers are nuclids with same NUMBER of PROTONS and NEUTRONS, but having DIFFERENT energy states and spins. Theyhave radioactive property. e.g `""^(99)TC & ""^(99m)Tc,` m - STANDS for metastable state & Tc -Teohnetium. |
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| 5867. |
In the diagram above, assume that the tabletop is frictionless. Determine the acceleration of the blocks once they're released from rest. |
Answer» Solution :There are two blocks, so we need to DRAW two free-body diagrams.  To get the acceleration of each one, we use Newton's Second Law, `F_("net")=ma`.  Notice that there are two unknowns, `F_(y)` and a , but we can eliminate `F_(T)` by adding the two equations, and then we can solve for a.  A quicker way of solving for the acceleration is to treat the entire SYSTEM (blocks plus string) as one object. Since we are only concerned with forces acting on the object, we can ignore tension. The string is part of the object. Then we only need to CONSIDER forces acting in the direction of motion (Mg) and forces opposite the direction of motion (none). Our mass in Newton's Second Law becomes M+ m, so `F_("net")=ma` becomes Mg=(M+m)a , giving us the same ANSWER for acceleration. |
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| 5868. |
Calculate the radius of the third Bohr orbit of hydrogen atom and the energy of electron in that orbit. |
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Answer» |
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| 5869. |
By what fraction does the mass of an H atom decreases when it makes an n=3 to n=1 transition ? |
| Answer» SOLUTION :`-1.29xx10^(-8)` | |
| 5870. |
In the previous problem, calculate the effective conductance if all the three resistors are connected in series. |
| Answer» SOLUTION :`((G_1G_2G_3)/(G_2G_3+G_1G_3+G_1G_2))` | |
| 5871. |
Young.s double slit experiment is made in liquid .the tenth bright fringe in liquid lies in screen where 6^(th)dark fringe lies in vacuum. The refractive index of the liquid lies in screen where 6^(th) dark fringe lies in vacumm . The refractive index of the liquid is approximately . |
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Answer» `1.8` |
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| 5872. |
The diameter of an object of a telescope, which can just resolve two stars situated an angular displacement of 10^(-4) degree, should be |
| Answer» Answer :B | |
| 5873. |
Most materials have the refractive index, n lt 1. So, when a light ray from air enters a naturally occuring material, then by Snell's law, (sin theta_(1))/(sin theta_(2)) = (n_(2))/(n_(1)), it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n=((c)/(V))=pmsqrt(epsilon_(r)u_(r)), where c is the speed of electromagnetic waves is vacuum, v its speed in the medium , epsilon_(r) and mu_(r) are relative permittivity and permeability of the medium respectively. In normal materials, both epsilon_(r) and mu_(r) are positive, implying positive n for the medium. When both epsilon_(r) and mu_(r) are negative, one must choose the negative root of n. Such negative refractive index material can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of proporgation of the refracted light. However, similar to normal materials, the frequency of light remais unchanged upon refraction evenin metal - materials. |
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Answer» The speed of LIGHT in the meta - material is V = c|N| |
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| 5874. |
A point object is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of convex mirror ? |
Answer» Solution : GIVEN : `-u=0.12m` `x=0.10m, f=0.10m` using the FORMULA `(1)/(f)=(1)/(-u)+(1)/(v)` we write `(1)/(0.10)=(1)/(0.12)+(1)/(v)` `therefore (1)/(v)=(10)/(1)-(100)/(12)=(120-100)/(12)=(20)/(12)` `thereforev=(12)/(20)=(3)/(5)=0.60m` From the fig., radius of curvature of the mirror is `0.60-0.10=0.50`m so that the focal length of the convex mirror `f=(r)/(2)=(0.50)/(2)=0.25m`. |
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| 5875. |
Most materials have the refractive index, n lt 1. So, when a light ray from air enters a naturally occuring material, then by Snell's law, (sin theta_(1))/(sin theta_(2)) = (n_(2))/(n_(1)), it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n=((c)/(V))=pmsqrt(epsilon_(r)u_(r)), where c is the speed of electromagnetic waves is vacuum, v its speed in the medium , epsilon_(r) and mu_(r) are relative permittivity and permeability of the medium respectively. |
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Answer»
`sin theta_(2) = (n_(1))/(n_(2)) sin theta_(1)` `rArr n_(2) "is - ve". THEREFORE theta_(2) is - ve.` |
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| 5877. |
In carbon-nitrogen nuclear fusion cycle, protons are fused to form a helium nucleus, positrons and release some energy. The number of protons fused and the number of positrons released in this process respectively are |
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Answer» 4,4 |
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| 5878. |
A long straight horizontal cable carries a current of 2.5 A in the direction 10^@ south of west to 10^@ north of east. The magnetic meridian of the place happens to be 10^@ west of the geographic meridian. The earth's magnetic field at the location is 0.33 G and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). |
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Answer» Solution :Here I = 2.5 A , `B_E = 0.33 G = 3.3 xx 10^(-5) T`, angle of dip `delta=0^@` . As shown in Fig. 5.02 , the CABLE is lying along magnetic west-east direction. Horizontal component of earth.s magnetic field `B_H= B_E * cos delta= 3.3 xx 10^(-5) cos 0^@ = 3.3 xx 10^(-5) T`. let neutral points lie at a distance r from the cable , they by definition of neutral point, field due to current carrying WIRE must be equal and opposite to `B_H` . Thus, `B= (mu_0 I)/(2pir) = B_H`  `impliesr=(mu_0 I)/(2pi B_H) =( 4pi xx 10^(-7) xx 2.5)/(2pi xx 3.3 xx 10^(-5)) = 1.5 xx 10^(-2) m ` or 1.5 cm As per right hand rule, the neutral points lie on a straight line PARALLEL to the cable at a perpendicular distance of 1.5 cm above the plane of cable (i.e , plane of paper). |
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| 5879. |
Why did other countries broke off diplomatic relations with South Africa? |
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Answer» WHITE rulers |
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| 5880. |
Who thought of improving (मे सुधार ) the sound of the Pungi? |
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Answer» a musician |
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| 5881. |
Beta particles before emission from a radioactive element |
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Answer» EXIST OUTSIDE the atom |
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| 5882. |
Four resistors of resistance 15 Omega, 12 Omega, 4 Omega and 10 Omega arc connected in cyclic order to form a Wheatstone bridge. The resistance (in Q) that should be connected in parallel across the 10 Omega resistor to balance the Wheatstone bridge is |
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Answer» `10 Omega`  suppose P = 15 `Omega, , Q = 12 Omega` `R = 4 Omega , S = (10r)/(10 + r) ` Suppose, bridge achieves balanced condition by connecting RESISTOR r in parallel to `10 Omega` For balaned condition of Wheatstone bridge, `therefore (P)/(Q)= (S)/(R)` ` (15)/(12) = (10r)/((10 + r) xx 4)` `therefore (5)/(4)= (10r)/(40 + 4R)` `therefore 5 = (10r)/(10 + r)` `therefore 50 + 5r + 10 r` `therefore 50 = 5rtherefore r = 10 Omega` |
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| 5883. |
Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50 V. A particle of mass 10^(-15) and charge 10^(-11) C enters in it with a velocity 10^(7)m//s. The acceleration of the particle will be |
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Answer» `10^(8)m//s^(2)` |
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| 5884. |
The refractive index of diamond is much greater than that of glass. How does a diamond cutter makes use of this fact ? |
| Answer» SOLUTION :As refractive index of diamond is much greater hence critical angle for diamond-air interface is quite SMALL (In fact it is 24.5). Thus, a diamond cutter has large OPTIONS to cut the diamond faces so that light suffers TOTAL internal reflection from these faces. | |
| 5885. |
Relation between vertical component v and horizontal component H at aplace where angle of dip is 60^(@) |
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Answer» V=H |
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| 5886. |
If C, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure in M.K.S units are the fundamental units then the dimension of length will be : |
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Answer» <P>`C/G` `:.` CORRECT choice is `(d)`. |
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| 5887. |
What is susceptibility of a substance ? |
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Answer» SOLUTION :Magnetic susceptibility of a substance is defined on the ratio between INTENSITY of magnetisation to magnetic FIELD intensity , GIVEN by `chi = I/H`, where I = intensity magnetisation and H = magnetic field intensity. |
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| 5888. |
Assume that a molecule is moving with Root mean square speed at temperature 300K. The de Broglie wavelength of nitrogen molecule is |
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Answer» `2.75xx10^-11m` |
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| 5889. |
In a photo cell, the photo-electrons emission takes place |
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Answer» After 10 SEC on INCIDENT of LIGHT RAYS |
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| 5890. |
The simple harmonic motion of a particle is given by y = 3 sin omegat + 4 cos omegat. Which one of the following is the amplitude of this motion? |
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Answer» 1 |
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| 5891. |
Draw a schematic diagram of a step-up transformer . Explain itsworking principle . Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils . In an ideal transformer , how is this ratio related to the currents in the two coils ? How is the transformer used in large scale transmission and distribution of electrical energy over long distances ? |
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Answer» Solution :Step-up transformer : Principle : It is a DEVICE which converts low voltage A.C. into high voltage A.C. It is based UPON the principle of mutual induction . When alternating current passed through a coil , an induced e.m.f. is set up in the neighbouring coil . Construction :A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core . One of the coils known as primary (P) is connected to A.C. supply . The other coil known as secondary (S) is connected to the load. Working : When an alternating current is passed through the primary , the magnetic flux through the iron core changes which does two things.It produces e.m.f in the primary and an induced. e.m.f is also set up in the secondary . If the primary is negligible , the back e.m.f will be equal to the voltage applied to the primary. `therefore "" V_p = - N_p (d phi)/(dt ) and V_s = -N_s (d phi)/(dt)` where `N_p and N_5` are number turns in the primary and secondary respectively .`V_p and V_5`are their respective voltages. `therefore "" (V_s)/(V_p)=(N_s)/(N_p)` This ratio`(N_s)/(N_p)` is CALLED the turns ratio In a step- up transformer : `N_s gt N_p` So,`"" V_s gt V_p` Large scale transmission : The large scale transmission and distribution of electrical energy over long DISTANCES is done with the USE of transformer . The voltage output of the generator is stepped-up-so that current is reduced and power loss `I^(2) R` is cut down .It is then transmitted over long distances to an area sub- stations and utility poles before a power supply of 240 V reaches our homes . 
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| 5892. |
A train is moving forward at a velocity of 2.0m/s. At the instant the train begins to accelerate at 0.80 m//s^(2) a passenger drops a coin which takes 0.50s to fall to the floor. Relative to a spot on the floor directly under the coin at release, it lands. |
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Answer» 1.1 m TOWARDS the REAR of the TRAIN |
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| 5893. |
The distance between any two successive dark bands is given by : |
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Answer» `BETA=(lambdaD)/d` |
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| 5894. |
Discuss atom as a magnetic dipole. Hence define Bohr magneton. |
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Answer» Solution :Atom as a Magnetic Dipole Every atom consists of a central part, called nucleus, where the ENTIRE mass and positive charge of the atom are concentrated. An equal number of negative charges (i.e. electrons) revolve around the nucleus in circular orbits. The revolution of electron i.e., negative charge in anticlockwise direction is equivalent to conventional current I in the clockwise direction. The upper FACE with clockwise current acts as south pole and the lower face acts as north pole. Thus an electron revolving in an orbit is equivalent to a magnetic dipole.  SUPPOSE e is the charge on an electron revolving in a circular orbit of radius r with a uniform angular velocity, `omega`. If T is the time period of revolution of the electron, then Current = `"charge"/"time"` or `I=e/T` = `e/(2pi//omega)""(becauseT=(2pi)/omega)` or `I=(eomega)/(2pi)""...(1)` If M is the magnetic dipole moment of the magnetic dipole M = IA Using (1), and putting `A=pir^(2)`, we get `M=(eomega)/(2pi)pir^(2)` `M=1/2eomegar^(2)""...(2)` Relation between orbital angular momentum and magnetic dipole momentum. Because of orbital MOTION, the revolving electron has an orbital angular momentum L given by `L=mvr=mromegar=momegar^(2)` `therefore` From (2), `M=1/2e/mmomegar^(2)` `M=1/2e/mL` In vector FORM, `vecM=-(e/(2m))vecL""...(3)` This shows that like `vecL`, magnetic dipole moment `vecM` of the atom is perpendicular to the plane of the orbit. Negative sign shows that direction of `vecM` is opposite to the direction of `vecL` , as is clear from the figure. According to Bohr.s theory, an electron cannot revolve in any orbit it likes. It can revolve only in those orbits for which the total angular momentum is an integral multiple of `h//2pi`, where h is the Planck.s constant i.e. Angular momentum of the electron, `L=(nh)/(2pi)` or `mr^(2)omega=(nh)/(2pi)` (angular momentum, `L=mvxxr=mromegaxxr=mr^(2)omega`) where n = 1, 2, 3, ........ `therefore""r^(2)omega=(nh)/(2pim)""...(4)` From (2) and (4), `M=e/2*(nh)/(2pim)=n*(eh)/(4pim)=nmu_(B)""...(5)` It is obvious that `(eh)/(4pim)` is the least value of the magnetic moment. `(mu_(B)=9.27xx10^(-24)JT^(-1)orAm^(2))`. Bohr Magneton It is defined as the magnetic dipole moment associated with an electron due to its orbital motion in the first orbit of hydrogen atom. |
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| 5895. |
A working transistor with its three legs marked P,Q, and Ris tested using a multimeter to Rand the other (positive) terminal to P or Q . Some resistance is seen on the muiltimeter. Which of the following is true for the transistor? |
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Answer» It is PNP TRANSISTOR with R as emitter |
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| 5896. |
Consider a coil of wire carrying current I, forming a magnetic dipole placed in an infinite plane. If phi_(1)is the magnitude of magnatic flux through the inner region and phi_(0)is magnitude of magnetic flux through outer region then which of the following is correct ? |
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Answer» `phi_(1) lt phi_(0)` `THEREFORE phi_(1) =-phi_(0)` |
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| 5897. |
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square . They are placed in a uniform magnetic field and same electric current is passed through them. Whichof the following loop configuration will experience greater torque ? |
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Answer» CIRCLE |
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| 5898. |
What type of wave front will emerge from a distant source of light ? |
| Answer» SOLUTION :PLANE WAVEFRONT | |
| 5899. |
Hertz is the unit of |
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Answer» ELECTRIC current. |
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| 5900. |
A motor car weighing 500 kg is moving with a speed of 15 m/s and the speed reduces to 5 m/s after brakes are applied. Calculate the value of retarding force after brake is applied ? |
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Answer» 500 N |
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