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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.
| 41401. |
A g of hydrogen is converted into 0.993g of helium in a thermonuclear reaction. The energy released is |
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Answer» `63 XX 10^(7)J` |
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| 41402. |
The bob of simple pendulum having length 1, is displaced from mean position to an angular position o with respect to vertical. If it is released, then velocity of bob at equilibrium position is |
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Answer» `sqrt(2g(1- cos theta))` `therefore AB = 1(1- cos theta) =h` At point C, the velocity of bob =0. The vertical ACCELERATION =G `therefore v^(2) = 2gh` or, `v=sqrt(2gl(1- cos theta))` 
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| 41403. |
As per Coulomb's law, the force of attraction or repulsion between two point charges is directly proportional to the |
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Answer» SUM of the MAGNITUDE of charges |
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| 41404. |
Howdoes theelectric flux due toa pointchargeenclosedby asphericalGaussiansurface get aaffected whenitsradiusis increased ? |
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Answer» Solution :As per Gauss’s theorem, Electric`PHI=q/epsilon_0`where q is the CHARGE enclosed by a closed SURFACE through which flux is `phi` Electric flux does not depend on the radius, hence it REMAINS unaffected. |
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| 41405. |
The graphs of the equations 5x - 15y = 8 and 3x- 9y = 24/5 are two line which are |
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Answer» Coincident |
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| 41406. |
Tuning fork A of frequency 258 cycle/sec gives 8 beats with a tuning fork B. When prongs of B are cut and again A and B are sounded the no. of beats heard remains same. The frequency of B in cycles/sec is |
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Answer» 250 |
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| 41407. |
Two equal bar magnets are kept as shown in the figure. The direction of resultant magnetic field, indicated by arrow head at the point P is (approximately) |
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Answer»
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| 41408. |
An explosion of atomic bomb release an energy of 7.6 xx 10^(13) J. If 200 Mev energy is released on fission of one ""_235Uatom calculate (i) the number of uranium atoms undergoing fission, (ii) the mass of uranium used in the bomb. |
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Answer» `4.375 xx10^(24), 926 .66G` |
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| 41409. |
Figure shows a parallel plate capacitor and the current in the connecting wires that is dischargingthe capacitor. |
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Answer» The displacement current is LEFTWARD 
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| 41410. |
An electric dipole with dipole moment 4xx10^(9) Cm is aligned at 30^(@) with the direction of a uniform electric field of magnetude 5xx10^(4)NC^(-1). Calculate the magnitude of the torque acting on the dipole. |
| Answer» SOLUTION :`10^(-4)` NM | |
| 41411. |
In the previous problem find the loss of kinetic energy of the system during collision if the particles are moving unidirectionally. |
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Answer» Solution :`DELTA KE=(1)/(2)(m_(1)m_(2))/(m_(1)+m_(2))v_(1)^(2)(1-e^(2))` If they are UNIDIRECTIONALLY, the velocity of approach has the magnitude of `|u_(1)-u_(2)|=|5-10|=5` m/sec. `RARR Delta K.E. =((1)/(2)xx3xx6)/(3+6)=(5)^(2){1-(1//2)^(2)}=75//4 J`. |
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| 41412. |
What is a wave front? |
| Answer» Solution :A wavefront is the locus of points which are in the same state or PHASE of VIBRATION. | |
| 41413. |
A coil of wire of radius r has 600 turns and a self inductance of 108 mH. The self inductance of similar of 500 turns is |
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Answer» 108 mH |
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| 41414. |
Two cells of emfs E_1 and E_2 and of negligibleinternal resistances are connected with two variable resistors as shown in Fig. A2.4. When the galvanometer shows no deflection, the values of the resistances are P and Q. What is the value of the ratio E_2/E_1? |
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Answer» `P/Q` by `V_P = (E_1/(P+Q))P.` Potential `V_P` across P as determined from `E_2` is same as `E_2` because no current is drawn, i.e., `V_P = E_2` Therefore, `E_2 = E_1(P/(P+Q)) or E_2//E_1 = (P/(P+Q))` . |
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| 41415. |
An acid catalysed hydrolysis of ester follows rate law,R=k[CH_3COOC_2H_5][H^+].Half life in presence of 0.1M HA (ka=10^(-5)) is 100 min. Then half life (in min) in presence of 0.01M HCl. |
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Answer» `H^+=sqrt(0.1xx10^(-5))=10^(-3)` `100="In2"/(kxx10^(-3))` `t_(1//2)="In2"/(kxx10^(-2))`=10 MIN |
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| 41416. |
A defiection magnetometer works on |
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Answer» COULOMB's law |
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| 41417. |
How will the applied voltage be divided among the capacitors? |
| Answer» SOLUTION :DIRECTLY PROPORTIONAL to CAPACITY. | |
| 41418. |
State Lenz's law. Explain, by giving examples that Lenz's law is a consequence of conservation of energy. |
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Answer» Solution :Lenz's Law : Statement : The polarity of induced emf is such that it opposes the cause which produced it . i.e. `in=-N(dQ_(B))/(dt)-ve ` sign indicates induced emf opposes the change in MAGNETIC field. A closed wire is connected to G as shown and bar magnet is kept over the coil when the magnet MOVES towards it the FLUX linked with it goes on INCREASING and induced emf is produced which causes deflection in the coil. (Applying Len's Law) to oppose the cause the upper face acquires N-Polarity which repel N pole of the coil. To move it towards the coil work is required to be done to counteract repulsive force. This mechnical work done is converted into elecrical energy which is converted intoheat energy due to Joul's effect when magnet moves away from the coil it gains s polarity in the upper face and induced e.m.fopposes upper movement of the target. If magnet is not moved no mechnical work is done and no emf i.e. no electric energy is produced. 
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| 41419. |
Two coils have a mutual inductance of 5 xx 10^(-3) H. The current in the first coil changes according to the relation I = 10 sin (100 pit). What is the value of maximum e.m.f. induced in the second coil ? |
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Answer» `12 pi V` |
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| 41420. |
Detemine how many times the intensity of a narrow beam of thermal neutrons will decrease after passing through the heavy water layer of thickness d=5.0 cm. The effective cross-section are of neutrons quitting the beam due to scattering if the thickness of the plate is d= 0.50 cm. The effective cross-section of interaction of deuterium and oxygen nuclei with thermal neutrons are equal to sigma_(1_= 7.0 band sigma_(s)= 11 b respectivley. |
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Answer» Solution :Here `(1)/(eta)=E^(-n_(2)sigma_(2)+n_(1)sigma_(1)d)` where `1` refers to `O^(1)` and 2o r `D` NUCLEI Using `n_(2)=2N,n_(1)=N=` concentration of `O` nuclei in heavy water we get `(1)/(eta)= e^(-2sigma_(2)+sigma_(1)nd)` Now using the data for heavy water `n=(1.1xx6.023xx10^(23))/(20)= 3.313xx10^(22)PER c c` Thus substituting the values `eta= 20.4=(I_(0))/(I)` |
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| 41421. |
a. Which is more stable , ""_(3)^(7)Li "or" ""_(3)^(4)Li ? b. Give reason for your answer. |
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Answer» SOLUTION :a. `""_(3)^(7)LI`, is stable. It contains large number of NEUTRONS than the other. |
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| 41422. |
If the temperature of a perfectly black-body increases two times then the rate of radiation of the body also increases by |
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Answer» EIGHT times |
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| 41423. |
Two point charges q AND 2q are placed at (a,0) and (0,a) A point charge q^(1) is placed at a P on the quarter circle of radius a as shown in the diagram so that the electric field at the origin becomes zero |
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Answer» the POINT P is `((a)/(sqrt(3)),(sqrt(2)a)/(sqrt(3)))` |
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| 41424. |
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionableuranium would our country need per year by 2020? Take the heatenergy per fission of 235U to be about 200MeV. |
| Answer» SOLUTION :`3.076 XX 10^(4) KG`. | |
| 41425. |
A projectile is projected horizontally with 8 ms^(-1) , its velocity after 1/4 second is equal to : |
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Answer» 8.37 m`s^(-1)` |
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| 41426. |
The equation of a plane Progressive wave is Y = 1.5 sin (314t - 12.56 x) metres. Then amaximum particle velocity is : |
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Answer» 25 m/s here r = 1.5 m,w= 314and K = 12.56 . `m^(-1)` max PARTICLE VELOCITY `V_(max) = rw = 1.5 xx 314 = 471 ` m/s hence the CORRECT CHOICE is (b) . |
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| 41427. |
Find the mean electrostatic potential produced by an electron in the centre of a hydrogen atom if the electron is in the ground state for which the wave function is Psi(r )=Ae^(-r//r_(1)), where A is a certain constant, r_(1) is the first Bohr radius. |
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Answer» Solution :We find `A` by normalization as above. We get `A=(1)/(sqrt(pir_(1)^(3)))` Then the electronic charge density is `rho=e|Psi|^(2)=-ee^(-2r//r_(1))/(pir_(1)^(3))~=rho (VEC(r ))` The potential `Psi(vec(r ))` due to this charge density is `varphi(vec(r ))=(1)/(4piepsilon_(0)) INT(rho(vecr ))/(|vec(r )-vec(r )|)d^(3)r'` so at the origin `varphi(0)=(1)/(4pi epsilon_(0))int_(0)^(oo)(rho(r'))/(r') 4pi r'^(2) dr'=(-e)/(4pi epsilon_(0))int_(0)^(oo)(4r')/(r_(1)^(3))e^(-2r//r_(1))dr` `=-(e)/(4pi epsilon_(0)r_(1))int_(0)^(oo)xe^(-X)DX= -(e )/((4pi epsilon_(0)r_(1)))` |
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| 41428. |
What is the name given to that part of electromagnetic spectrum which is used in Radar ? |
| Answer» Solution :MICROWAVES (or RADIO WAVES of extremely short WAVELENGTHS). | |
| 41429. |
The energy of the highest energy photon of Balmer series of hydrogen atom is close to. |
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Answer» 13.6 eV |
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| 41430. |
In MKS system of units epsilon_(0)equals - |
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Answer» `9xx10^(9) N-m^(2)//C^(2)` `:.epsilon_(0) =(1)/(4pi xx 9xx10^(9)) ("coul")^(2)//N-m^(2)`. |
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| 41431. |
A capacitor is charged from a battery through a resistance of 1MOmeg. If it takes 1 sec for the charge to reach one-half of its final value then capacity of capacitor is |
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Answer» `0.3muF` |
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| 41432. |
A ray of light strikes a glass plate at an angle of 60^@. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is : |
| Answer» Answer :A | |
| 41433. |
What is a rectifier ? With suitable circuit describe the action of a full wave rectifier by drawing input and output waveforms. |
Answer» SOLUTION :For diagram See S.A.T.Q.-I. 8.For positive half cycle of input a.c., one of the two diodes gets forward biased and conducts and OUTPUT current is obtained across the load `R_L`, For negative half cycle of input a.c., the other DIODE gets forward biased and thus output current is obtained due to it. Therefore, output is obtained for both the CYCLES of input a.c. 
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| 41434. |
Derive an expression for potential energy of a system of two charges in the absence of the external electric field. |
Answer» Solution : Imagine the charges `q_1` and `q_2`are initially at infinity. First, the charge `q_1`is brought from infinity to the point A and no work is done for this. This is because there is no electric FIELD to oppose this charge. Consider another point B at a distance from A. Electric potential at B due to r is given by, `V_1 = (1)/(4pi epsi_0 )(q_1)/(r )` Next the charge `q_2` is brought from infinity to the point B. When the charge 92 is MOVED, the electric field due to `q_1` opposes it. Hence work has to be done. Work done in bringing the charge `q-2`from infinity to the point B. `W = V_1 xx q_2` [since `V = (W)/(q_0) rArr W = qV`] `W = (1)/(4pi epsi_0) (q_1)/(r ) xx q_2` This work done is stored in the charges as electric potential energy and given by `PE=U = (1)/(4pi epsi_0) (q_1 q_2)/(r )` |
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| 41435. |
An electron of charge 'e' moves in a circular orbit of radius 'r' around the nucleus at a frequency 'n'. The magnetic moment associated with the orbital motion of the electron is |
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Answer» `pi n er^2 ` |
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| 41436. |
Which of the following molecule do not give Br_(2)/Water test |
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Answer» Phenol `rarr` In Benzene `PI` bonds are stable due to aromaticity. `rarr` Aniline & phenol REACT with `Br_(2)`/water due to high REACTIVITY. |
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| 41437. |
In the given reaction C + O_2 → CO_2 |
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Answer» 2 mole carbon REACTS with 1 mole `O_2` |
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| 41438. |
In Fig 16-46, a sinusoidal wave moving along a string is shown twice as crest A travels in the positive direction of any x axis by distance d=6.0 cm in 3.0 ms. The tick marks along the axis are separated by 10cm, height H=6.00MM. The equation for the wave is in the form y(x,t) =y_(m) sin (kx pm omegat). so what are (a) y_(m), (b) k, (c) omega and (d) the correct choice of sign in front of omega? |
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Answer» |
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| 41439. |
Which of the following expressions represents a simple harmonic progressive wave |
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Answer» `y = A SIN OMEGAT` |
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| 41440. |
The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents. |
| Answer» Solution :SINCE relative permeability of a given MATERIAL `(mu_r = 0.9983)` is SLIGHTLY less than 1, the material is a DIAMAGNETIC material. | |
| 41441. |
The ratio of the forces between two charges placed at a certain distance apart in air at half of the distance apart in medium of dielectric .k. is |
| Answer» Answer :B | |
| 41442. |
The Milky Way galaxy is a part of a group of galaxies called the Local Group. The proper distance from the Milky Way, on one side of the Local Group, to the M31 galaxy on the other side is approximately 2.4xx10^(6) light-years. How long (in xx10^(5) years) would it take a spaceship traveling at 0.999c to travel this distance according to travelers onboard? |
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| 41443. |
Assertion:Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars. Reason:In fusion heavier nuclei split to form lighter nuclei. |
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Answer» If both assertion and REASON are true and reason is the CORRECT explanation of assertion . |
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| 41444. |
Find the maximum energy thatbeta- particle may have in the followingdecay: ""_8O^(19)to ""_9F^(19)+""_(-1)e^0+vecv Givenm(""_8O_(19))=19.003576 a.m.u. m(""_9F^(19))=18.998403 a.m.u. m(""_(-)e^0)=0.000549 a.m.u. |
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| 41445. |
A uniform chain of length L and mass m is lying on a smooth table. One third of its length is hanging verti cally down over the edge of the table. How much work need to be done to pull the hanging part back to the table ? |
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Answer» `(MGL)/(3)` |
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| 41446. |
Two beams, A and B of plane polarised light with mutually perpendicular plane of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity) a rotation of polaroid through 30^(@) makes the two beams appear equally bright. If the initial intensities of the two beams are I_(A) and I_(B) respectively, then (I_(A))/(I_(B))=..... |
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Answer» `(3)/(2)` `:.I_(A)cos^(2)30^(@)=I_(B)cos^(2)60^(@)` `:.(I_(A))/(I_(B))=(cos^(2)60^(@))/(cos^(2)30^(@))=(((1)/(2))^(2))/(((SQRT(3))/(2))^(2))` `:. (I_(A))/(I_(B))=(1)/(3)` |
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| 41447. |
A balloon from rest accelerates uniformly upward with 'a' ms^(-2), for t seconds of time. A stone is released from the balloon. Now, read the following statements to pick the right ones. (a) The stone's initial velocity is zero, relative to balloon (b) The stone's initial velocity is nonzero, relative to earth ( c) The time taken to reach the ground from the balloon's frame of reference is inversely proportional to sqrt((a+g)) (d) The time take to reach the ground from earth's frame of reference is directly proportional to sqrt((a+g)) |
| Answer» Answer :A | |
| 41448. |
Spheres of iron and lead having same mass are completely immersed in water. Density of lead is more than that of iron. Apparent loss of weight is W_(1) for iron sphere and W_(2) for lead sphere. Then (W_(1))/(W_(2)) is : |
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Answer» 1 `m_(1)=m_(2)` `V_(1)p_(1)=V_(2)p_(2)` `(V_(1))/(V_(2))=(p_(2))/(p_(1))rArrprop1/p` Since density of iron is less than density of lead. `therefore` Iron sphere has more volume and will displace more water. Now apparent loss of weight is equal to weight of water displaced. So apparent loss of weight `W_(1)` for iron is more than loss of wt. `W_(2)` for lead. `therefore(W_(1))/(W_(2))gt1` So correct CHOICE is (d). |
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| 41449. |
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r lt lt h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V_0 and the top plate at -V_0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) |
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Answer» (a) zero Here `QpropV_0` …(ii) ALSO `S=ut+1/2at^2` `h=1/2(QE)/(m)t^2=1/2((Qxx2V_0)/(mh))xxt^2` `:.` `tprop(1)/(V_0)` …(iii) `[:' Q prop V_0]` From (i), (ii) and (iii) `I_(av)prop (V_0)/(1/V_0)=I_(av)propV_0^2` |
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| 41450. |
A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density sigma at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period T of the oscillations of the cylinder will be : |
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Answer» Smaller than `2pi[(M)/((K+A sigma g))]^((1)/(2))` |
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