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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.
| 5201. |
A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40 cm apart. A point object is placed between the mirrors, on their common axis and 15cm from the concave mirror, Find the position, nature of the image, and over all magnification produced by the successive reflections, first at concave mirror and then at convex mirror. |
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Answer» Solution :According to given problem, for concave mirror, u=-15 cm and f=-10 cm So `(1)/(v ) +(1)/(-15 )=(1)/(-10) , i.e., V=-30cm` i.e., concave mirror will form real, inverted and enlarged IMAGE I, of object O at a DISTANCE 30 cm from it, i.e., at a distance 40 - 30 = 10 cm from convex mirror.  For convex mirror the image I, will act as an object and so for it u=-10 cm and f=+15cm. `(1)/(v) +(1)/(-10) =1/(15) ,i.e., v=+6cm` So final image `I_2` is formed at a distance 6 cm behind the convex mirror and is virtual as shown in figure Over all magnification `= m_1 xx m_2=-2 xx 6//10 = -6//5` Negative INDICATES final image is inverted w.r.t given object. |
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| 5202. |
A child standing on the edge of a freely spinning merry go-round moves quickly to the center. Which one of the following statements is necessarily true concerning this event and why? |
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Answer» The angular speed of the SYSTEM decreases because the moment of inertia of the system has increased. |
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| 5203. |
Three concentric metallic shells A , B and C of radii a, b and c (a lt b lt c) have surface densities + sigma ,- sigma and +sigma respectively as shown in Fig. If shells A and C are at the same potential , obtain the relation between a , b and c. |
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Answer» SOLUTION :If `V_(A) = V_(C)` , then we have `[(sigma)/(in_(0)) [a - B + c] = (sigma)/(in_(0)) [a^(2) - b^(2))/(c) + c] IMPLIES a - b + c = (a^(2) - b^(2))/(c) + c implies (a- b) = (a^(2) - b^(2))/(c) implies c = a + b` |
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| 5204. |
Three concentric metallic shells A , B and C of radii a, b and c (a lt b lt c) have surface densities + sigma ,- sigma and +sigma respectively as shown in Fig. Obtain the expressions for the potential of three shells A , B and C . |
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Answer» Solution :We know that potential at a POINT either on the surface or inside a charged SPHERICAL shell having surface density `sigma` is `V = (sigma R)/(in_(0))` , where R is radius of shell . The potential at a point OUTSIDE the charged shell at a distance (where `r GT R`) is V = `(sigma R^(2))/(in_(0) r)` In present question `sigma_(A) = + sigma , sigma_(B) = - sigma` and `sigma_(c) = + sigma` , moreover `a lt b lt c` `therefore` Electric potential at shell A , `V_(A) = (sigma A * a)/(in_(0)) + (sigma_(B) * b)/(in_(0)) + (sigma_(C) * c)/(in_(0)) = (sigma a)/(in_(0)) - (sigma b)/(in_(0)) + (sigma_(c) )/(in_(0)) = (sigma)/(in_0) [a -b + c]"" ..... (i)` Electric potential at shell B , `V_(B) = (sigma_(A) * a^(2))/(in_(0) b) + (sigma _(B) * b)/(in_(0)) + (sigma_(c) * c)/(in_(0)) = (sigma * a^(2))/(in_(0) b) - (sigma *b)/(in_(0)) + (sigma * c)/(in_(0)) = (sigma)/(in_(0)) [ (a^(2))/(b) - b + c] ""..... (ii) ` and electric potential at shell C . `V_(C)= (sigma_(A) * a^(2))/(in_(0) c) + (sigma_(B) * b^(2))/(in_(0) c) + (sigma_(c) * c)/(in_0)) = (sigma a^(2))/(in_(0) c) - (sigma b^(2))/(in_(0) c) + (sigma_(c))/(in_(0)) = (sigma)/(in_(0)) [ (a^(2) - b^(2))/(c) + c] "" ... (iii)` |
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| 5205. |
Two blocks of same mass 'm' are connected by the cable passing over a pulley of mass 'm' and radius 'R'. There is no slipping between the pulley and the cable. The coefficient of friction between the block 'A' and the horizontal surface is 1//4 . At the moment shown in the figure,the tension in the cable AB is |
| Answer» Answer :D | |
| 5206. |
Which of these provisions were passed by the Assembly on the night of 4 August, 1789? |
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Answer» ABOLITION of FEUDAL SYSTEM of obligations |
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| 5207. |
A vibration magnetometer consists of two identical bar magnets placed one over the other such thatthey are perpendicular and bised each other. The time period of oscillation in a horizontal magneticfield is 2^(5//4) sec. one of the magnets is removed and if the other magnet oscillates in the same field, then time period in seconds is :- |
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Answer» `2^(1//4)` `M_"eff"=sqrt(M^2+M^2)=Msqrt2` and the moment of inertia is 2I. Thus, `T=2pisqrt(I/(Msqrt(2H)))` When one of the magnets is withdrawn, the time `T'=2pisqrt(I(MH))` `therefore (T')/T=sqrt(1/2^(1//2))` or `T'=T/2^(1//4) = 2^(5//4-1//4)` = 2 sec |
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| 5208. |
A parallel plate capacitor having plate area 50cm^(2) and separated by a distance 2mm with dielectric material has a capacity of 100pF. The plates are charged to p.d. of240V. Calculate the charge stored in it and the dielectic constant. |
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Answer» |
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| 5209. |
The refractive index of the core of an optical fibre is mu_(2) and that ofth cladding is mu_(1). The angle of incidence on the face of the core of that the light ray just under goes total internal reflection at the cladding is |
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Answer» `SIN^(-1)((mu_(1))/(mu_(2)))` |
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| 5210. |
Assertion : It is advantageous to transmit electric power at high voltage. Reason: High voltage inmplies high current. |
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Answer» |
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| 5211. |
The specimen of an intrinsic semiconductor contains 1.2xx10^15If it is doped by phosphorous atoms in a small proportion, then the number of holes/m3 in the doped semiconductor will |
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Answer» slightly INCREASE |
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| 5212. |
A capacitor of capacitance C is connected to a voltage source of p.d. V. The capacitor gets charged. a. What happens to the voltage across the capacitor as charge increases? b. What is the net work done when the capacitor is fully charged? |
| Answer» SOLUTION :a. INCREASES B. `W=1/2 CV^(2)` | |
| 5213. |
In optical communication system operating at 1200nm only 2% of the source frequency is available for TV transmission having a bandwidth of 5MHz. The number of TV channels that can be transmitted is |
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Answer» 2 millions |
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| 5214. |
Principle of perpendicular axes there states I_ z= _____ + _____. |
| Answer» SOLUTION :[`I_x + I_y`] | |
| 5215. |
Which of the following colors is scattered minimum |
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Answer» Violet |
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| 5216. |
Two capacitors C_1and C_2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero Then : |
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Answer» `5C_1=3C_2` |
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| 5217. |
The equivalent capacity between the points A and B in the adjoining circuit will be |
| Answer» ANSWER :B | |
| 5218. |
In Young's double slit experiment if d, D and lamda represent the distance between the slits, the distance of the screen from the slits and wavelenfth of light used respectively, then the band width is width is inversely proportional to |
| Answer» ANSWER :B | |
| 5219. |
In a uranium ore the ratio of U^(238) nuclei to Pb^(206) nuclei is eta= 2.8. Evaluate the age of the ore, assuming all the lead Pb^(206) to be final decay product of the uranium series. The half-life of U^(238) nuclei is 4.5.10^(9) years |
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Answer» Solution :What this implies is that in the time since the ORE was formed, `(ETA)/(1+ eta)U^(238)` nuclei have remained undecayed. Thus `(eta)/(1_eta)= e^(-txx(IN2)/(T_(1//2))` or `t= T_(1/2)(In (1+eta))/((eta)/(In2))` SUBSTITUTING `T_(1/2)= 4.5xx10^(9)years, eta= 2.8` we GET `t=1.98xx10^(9)`years. |
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| 5220. |
A fission reaction is given by _(92)^(236) U rarr_(54)^(140) Xe + _(38)^(94)St + x + y , where x and y are two particle Consider_(92)^(236) U to be at rest , the kinetic energies of the products are deneted by k_(xe) K _(st) K _(x) (2MeV ) and Ky(2MeV) repectively . Let the binding energy per nucleus of _(92)^(236) U, _(54)^(140) Xeand _(38)^(94)St be 7.5 MeV , 8.4 MeV and 8.5 MeV,respectively Considering different conservation laws, the correct option (s) is (are) |
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Answer» x=n, y=n, `K_"Sr"`=129 MEV, `K_"Xe"`=86 MeV `K_x`=2 MeV , `K_y`=2 MeV , `K_"Xe"`=? , `K_"Sr"`=? By conservation of CHARGE number and mass number , `x-=y-=n` B.E. per nucleon of `._92^236U`=7.5 MeV B.E. per nucleon of `._54^140Xe`or `._38^94Sr`=8.5 MeV Q value of reaction, Q=Net Kinetic energy gained in the PROCESS `=K_(Xe)+K_(Sr)+2+2-0=K_(Xe)+K_(Sr)+4`...(i) As number of NUCLEONS is conserved in a reaction , so Q = DIFFERENCE of binding energies of the nuclei `=140xx8.5+9.4xx8.5-236xx7.5`=219 MeV ...(ii) From Eqns.(i) and (ii) `K_"Xe"+K_"Sr"`=219-4=215 MeV Xe and Sr have momentum of same magnitude but in opposite directions. Hence, lighter body has larger kinetic energy So, from options, `K_"Sr"`=129 MeV , and `K_"Xe"` =86 MeV Hence, option (a) is correct . |
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| 5221. |
Supposethetaisthe polarising anglefor a transparentmediumand thespeed of light in that medium is w. (Then according to Browster law ) |
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Answer» `theta = COT^(-1) (v//c)` |
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| 5222. |
In a transistor if (I_C )/(I_E) = alpha and(I_C )/( I_B ) = beta If alphavaries between (20)/( 21) and(100)/(101)then the value of betalies between |
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Answer» `1-10` |
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| 5223. |
Linear polarized light is incident at Brewster angle on the surface of a medium . If the incident beam is polarized parallel to the plane of incidence then the parallel component of light is |
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Answer» COMPLETELY refracted |
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| 5224. |
The equal resistances, 400 Omega each are connected in series with a 8V battery. IF the resistance of first one increases by 0.5%, the charge required in the resistance of the second one in order to keep the potential difference across it unaltered is to |
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Answer» increase it by `1 OMEGA` Initially, the emf 8V will be divided equally between the two resistances.SO the voltage across each resistance will be 4V. When the first resistance is INCREASED, the second resistance should also be inceased by `2 Omega` to KEEP the voltage across it unchanged. |
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| 5225. |
State the principle of a transformer. Explain its construction and working. Derive an expression for the ratio of e.m.f's in terms of number of turns in primary and secondary coil. Two diametrically opposite pointsof a metal ring are connected to two terminals of the left gap of meter bridge. The resistance of 11Omega is connected in right gap. If null point is obtained at a distance of 45 cm from the left end, find the resistance of metal ring. |
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Answer» Solution :Numerical : Given : `R_(2)=11Omega,l_(1)=45" cm",l_(2)=100-45=55" cm"` RESISTANCE of METAL ring `(R_(1))=?` `(R_(1))/(R_(2))=(l_(1))/(l_(2))` Resistance of each half segment of the metal ring `=(R_(1))/(2)`. These half segments are connected in parallel in the left gap. `R_(eff.)=((R_(1))/(2)xx(R_(1))/(2))/((R_(1))/(2)+(R_(1))/(2))=((R_(1))/(2)xx(R_(1))/(2))/(R_(1))=(R_(1)^(2))/(4R_(1))` `R_(eff.)=(R_(1))/(4)OMEGA` From the formula, `(R_(eff.))/(R_(2))=(l_(1))/(l_(2))` `((R_(1))/(4))/(11)=(45)/(55)` `(R_(1))/(44)=(45)/(55)` `R_(1)=(45)/(55)xx44` `R_(1)=36Omega` `:.` The resistance of metal ring is `36Omega`. |
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| 5226. |
Two charges - q and + q are located at points A(0, 0, -a) and B(0, 0, +a) respectively. How much work is done in moving a test charge from point P(7,0,0) to Q(-3,0,0) ? |
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Answer» Solution :The charges –Q and + located at points A(0,0,-a) and B(0, 0, +a) constitute an electric dipole. Point P(7,0,0) and Q(-3,0,0) both lie at the equatorial line of this dipole. Therefore `V_P = V_Q` =0. As a result WORK DONE in moving a test charge go from point P to Q `W = q_0[V_Q-V_p] = q_0[0-0]=0` |
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| 5227. |
If the positive centre of charge coincides with negative centre of charge, what is its dipole moment? |
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Answer» |
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| 5228. |
वायुमंडलीय गैस का मिश्रण है |
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Answer» ठोस द्रव का |
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| 5229. |
ठोस विलयन का उदाहरण है |
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Answer» मिश्र धातु |
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| 5230. |
A piston of mass m divides a cylinder containing gas into two equal parts. Suppose the piston is displaced to the left to a distance x and let go Fig. Assuming the process to take place at a constant temperature, find the frequency of the piston's oscillations. |
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Answer» `p_(1)(d-x)S=p_(2)(d+x)S=pdS` The force acting on the PISTION is `F=(p_(1)-p_(2))S=(2pxSd)/(d^(2)-x^(2))=(2pVx)/(d^(2)-x^(2))` where V is the volume of one half of the vessel. As can be seen, the force does not conform to Hooke.s law, and the OSCILLATIONS are not harmonic. But for small deflections of the piston (when `xltltd` ), the force will be quasi-elastic: `F=2pVx//d^(2)` and the oscillations of the piston will be harmonic. The RIGIDITY of the system is`k=F//x=2pV//d^(2)`. |
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| 5231. |
The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half time. |
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Answer» SOLUTION :If the half-life is `T` days `(2)^(-7//T)=(1)/(2.5)` Hence `(7)/(T)= (In 2.5)/(In 2)` or `T=(7In 2)/(In 2.5)= 5.30 days` |
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| 5232. |
When 500μA curre nt is passed through a galvanometer of 20Omega , it gives a full scale deflection. The external resistance need to be connected to measure 5V is |
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Answer» `98Omega` PARALLEL |
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| 5233. |
Calculate the electric dipole moment for the following charge configurations . |
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Answer» Solution :(i) The electric field LINES start at `q_(2)` and end at `q_(1)` . In figure (a) `q_(2)` is positive and `q_(1)` is negative . The number of lines starting from `q_(2)` is 18 and number of the lines ending at `q_(1)` is 6. So `q_(2)` has greater magnitud. The ratio of `|(q_(1))/(q_(2))|=(N_(1))/(N_(2))=(6)/(8)=(1)/(3)` . It implies that `|q_(2)|=3|q_(1)|`. (ii) In figure (b) , the number of field lines emanating from both positive charges are equal (N=18). So the charges are equal . At point A the electric field line are denser compared to the lines at point B. So the electric at point A is greater in magnitude compared to the field at point B . Further no electric field line passes through C which implies that the resultant electric field at C due to these two charges is zero . (iii) In the figure (c) the electric field lines start at `q_(1)` and `q_(3)` and end at `q_(2)` and end at `q_(2)` , This implies that `q_(1)` and `q_(3)` are positive charges. The ratio of the number of field lines is `|(q_(1))/(q_(2))|=(8)/(16)=|(q_(3))/(q_(2))|=(1)/(2)` implying that `q_(1)` and `q_(3)` are half of the magnitude of `q_(2)` . So `q_(1)=q_(3)=+10n` C. The water molecule `(H_(2)O)` has this charge configuration . The water molecule has three ATOMS (two H atom and one O atom ). The centres of positive (H) and negative (O) charges of a water molecule lie at different point hence it possess permanent dipole moment . The O-H bond length is `0.958xx10^(-10)` m due to which the electric dipole moment of water molecule has the magnitude p =`6.1xx10^(-30)` Cm. The electric dipole moment `vecP` is DIRECTED from center of negative charge to the center to the center of positive charge asd shown in the figure. 
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| 5234. |
A composite block is made of slabs A,B,C,D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat 'Q' flows only from left to right through the blocks. Then in steady state |
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Answer» heat flow through A and E SLABS are same.  Let width of each rod is d `R_(1)=(1)/(8kd),R_(2)=(4)/(3kd)` `R_(3)=(1)/(2kd),R_(4)=(4)/(5kd)` `R_(5)=(1)/(24kd)`, Correct CHOICE is (a,c,d) |
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| 5235. |
A progressive wave of frequency 500 HZ is travelling with a velocity of 360ms^(-1). The distance between the two points, having a phase difference of 60^(@) is ……… |
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Answer» Solution :We know that for a wave `V=flambda` so `lambda=(v)/(F)=(360)/(500)=0.72 CM` |
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| 5236. |
A deflection of 24 divisions ofa ballistic galvanometer is obtained either by charging a capacitor of 3 mu F capacitance to a potential difference of 2V and discharging through the galvanometer or by connecting the ballistic galvanometer in series with a flat circular coil of 80 turns, each of diameter 1cm, the combined resistance of coil and galvanometer being 4000 ohm and quickly thrusting the coil into a strong magnetic field so that the plane of the coil is perpendicular to the direction of the field. Calculate the sensitivity of the galvanometer and calculate the strength of the magnetic field. The strength of the earth's magnetic field may be neglected. |
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Answer» |
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| 5237. |
द्विनिषेचन क्रिया होती है: |
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Answer» शैवालों में, |
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| 5238. |
A thin copper ring of radius 'a' is charged with q units of electricity. An electron is placed at the centre of the copper ring. If the electron is displaced a little, it will have frequency |
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Answer» `(1)/( 2pi ) sqrt(( eq)/(4 PI in_0ma^3))` |
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| 5239. |
For an LCR circuit driven at frequency omega, the di equation reads L (di)/(dt)+Ri+ q/C=V_i=V_m sin omegat (a) Multiply the equation by i and simplify where possible. (b) Interpret each term physically. (c) Cast the equation in the form of a conservation of energy statement. (d) Integrate the equation over one cycle to find that the phase difference between v and i must be acute. |
Answer» Solution :Let circuit L-C-R as shown in figure.  From Kirchhoff.s law of closed circuit, `V_L+V_C+V_R=V_m sin omegat` `L(di)/(dt)+q/C+iR=V_m sin omegat`…(1) Multiplying I on both side, `Li(di)/(dt) + q/C i+i^2 R=V_m I sin omegat`...(2) where `Li(di)/(dt)=d/(dt)(1/2Li^2)` indicates the rate of change of ENERGY stored in an inductor. `i^2R=P` JOULE heat loss `q/C i=d/(dt)(q^2/(2C))` indicates the rate of change of energy stored in the CAPACITOR and Vi = rate at which driving force pours in energy. Hence, equation (2) is in the form of conservation of energy statement. Integrating both sides of equation (2) with respect to time over one full cycle (0 - T) we may write `int_0^T d/(dt)(1/2Li^2+q^2/(2C))dt+int_0^TRi^2dt=1/2int_0^T Vidt` where, `V_m sin omegat=V` `THEREFORE [0+1/2i^2Rt]_0^T =1/2 int_0^T Vi dt` `therefore [0+1/2i^2RT]=1/2int_0^T Vi dt` `therefore` 0+(Positive)=`int_0^T Vi dt [because i^2RT gt 0]` `therefore int_0^T Vi dt gt 0` if phase difference between V andi is a constant and acute angle. |
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| 5240. |
The magnetic induction at O due to a current in conductor shaped as shown in figure |
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Answer» `(mu_(0)i)/(4pi)[(3PI)/(2a)+(sqrt(2))/(B)]` `:. B=B_(1)+B_(2)` |
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| 5241. |
Explain classification of material on basis of resistivity. |
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Answer» Solution :`rArr`Material is classified in three categories : conductor, semiconductor and insulator. `rArr` For perfect conductor value of resistivity is zero and conductivity is infinite. For METAL resistivity is of ORDER of `10^(-6) Omega `m to `10^(-8) Omega` m. `rArr`For perfect insulator resistivity is infinite and conductivity is zero. `rArr` For materials like CERAMIC, rubber and plastic value of resistivity is about `10^(18)` times value of resistivity of metals. `rArr`Resistivity of semiconductor is more than resistivity of metaJs and less than resistivity of insulators. `rArr` Resistivity of semiconductor decrease with INCREASE in temperature. `rArr` Resistivity of non-conductor depends on impurities added in them. |
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| 5242. |
What is the number of atoms in an elementary cell ef a simple cubic lattice? |
Answer» 
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| 5243. |
In series L-C-R circuit, L=2.0H, C=32muF and R=10Omega find Q - factor of resonance circuit. |
| Answer» Solution :A 100 mH INDUCTOR, a `20muF` CAPACITOR and a `10Omega` resistor are connected in series to a 100V ac source. Calculate (i) impedance of circuit at resonance (II) CURRENT at resonance (iii) resonant frequency. | |
| 5244. |
Theexcess pressure in a soapbubble of diameter 8 cm and surface tension 0.02 N/m, is |
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Answer» `2N//m^2` |
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| 5245. |
A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the centre of mass of the system (trolley + child) will move with a speed : |
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Answer» V `therefore` speed of CENTRE of mass will not change |
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| 5246. |
Convex lens of focal length A and concave lens of focal length B are kept in contact, then the effective focal length of system is ....... |
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Answer» A+B `1/f=(1)/(f_1)+(1)/(f_2)` `=1/A+(1)/(-B)` `(-B+A)/(-AB)` `therefore-((A-B)/(AB))` `thereforef=(AB)/(B-A)` |
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| 5247. |
Show that the decay rate R of a sample of radionuclide is related to the number of radioactive nuclei N at the same instant by the expression R=lambdaN. |
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Answer» SOLUTION :`N=N_0e^(-lambdat)` differentiating both sides we get `(DN)/(DT) =-lambdaN_0e^(-lambdat)=-lambdaN` i.e, decay rate `R=- (dN)/(dt)=lambdaN` |
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| 5248. |
Can two equipotential surfaces intersect each other? Give reasons. |
| Answer» Solution : Two equipotential surfaces cannot intersect each other. Because if they do then at the POINT of intersection there are two POSSIBLE values of ELECTRIC POTENTIAL, which is not possible. | |
| 5249. |
1 gram of radioactive elements reduces to 1//3 gram in 2 days. The mass of the element remaining at the end of days is : |
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Answer» 1/6 g |
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| 5250. |
If the time of flight of a body is T and horizontal range is R, then the angle of inclination of direction of projection with the horizontal is : |
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Answer» `TAN^(-1)((GT^(2))/R)` |
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