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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.
| 40051. |
An alpha particle of energy 5 MeV is scattered through 180^(@) by a found uramiam nucleus . The distance of closest approach is of the order of |
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Answer» `5.3 xx 10^(-12) m` |
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| 40052. |
A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, find the force exerted by the beam on the sphere. |
Answer» Solution : LET O be the centre of the sphere and OZ be the line opposite to the incident beam. Consider a radius OP of the sphere making an angle theta with OZ. Rotate this radius about OZ to get a circle on the sphere. Change theta to theta + d theta and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area (`2 pi (r^2) sin theta d theta)`. Consider a small part `Delta A` of this ring at P. Energy of the light falling on this part in time `(Delta t )` is ` Delta U= I (Delta t) (Delta A cos theta).` ` The momentum of this light falling on Delta A is (Delta U/c) ALONG QP. The light is reflected by the sphere along PR. The change in momentumis ` (Delta p = 2 (Delta U/ c ) cos theta = (2/c) I Delta t (Delta A (cos^2) theta)).` along the inward normal. Theforce on (Delta A ) due to the light falling on it, is ` (Delta p/ Delta t ) = 2/c I (Delta A(cos^2)theta)` This force is along PO, the RESULTANT force on the ring as WELL as on the sphere is along ZO by symmetry. The component of the force on ` Delta A` , along ZO is. ` (Delta p/ Delta t) cos theta = 2/c (I (Delta A)cos ^3) theta. ` ` The force ACTING on the ring is , ` dF= (2/c) I (2 pi (r^2)(sin theta) d theta) cos^3 theta.` The force on the entire sphere is ` F= (int_(0)^(pi/2) (4 pi (r^2) I/c) (cos^3)theta sin theta dtheta)` ` = - (int_(theta =0)^(pi/2)(4 pi (r^2)I/c) cos^3 theta d (cos theta). ` ` = - (4 pi (r^2)I/c) ([ cos^4 theta/ 4]_(0)^( pi/ 2) = (pi(r^2)I/c).` Note that integration is done only for the hemisphere that faces the incident beam. |
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| 40053. |
Direction of magnetic field of Earth is ...... |
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Answer» from NORTH to SOUTH |
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| 40054. |
What physical quantity is the same for X - rays of wavelength 10^(-10)m, red light of wavelength 6800Å and radiowaves of wavelength 500m? |
| Answer» SOLUTION :As all the WAVES/rays MENTIONED here are electromagnetic waves, their speed in vacuum is same having a VALUE `c=3xx10^(8)ms^(-1)`. | |
| 40055. |
Two coherent point sources S_1 and S_2 are separated by a small distance 'd' as shown. The fringes obtained on the screen will be : |
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Answer» straigth LINES  It will be circle because path DIFFERENCE will be same on circle. Fringes will make concentric circles. |
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| 40056. |
For linearly polarized light the plane of polarization is |
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Answer» Perpendicular to both the DIRECTION of POLARIZATION and the direction of propagation |
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| 40057. |
Name the types of electromagnetic radiations which (i) are used in destroying cancer cells, (ii) cause tanning of the skin, and (iii) maintain the earth.s warmth. Write briefly a method of producing any one of these waves. |
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Answer» Solution :(i) `gamma-`rays are used in destroying cancer cells. These are produced in nuclear reactions and are emitted by CERTAIN radioactive materials. (ii) UV radiations cause tanning of the SKIN. These are produced by SPECIAL UV lamps. (iii) IR rays maintain the earth.s WARMTH. These are produced by all HOT bodies. |
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| 40058. |
A convex mirror offocal length fforms an image which is 1/n times the object. The distance of the object from the mirror is: |
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Answer» (n-1)FF Here `m=+1/n, f to +f` So, `+ 1/n= (+f)/(+f-u) IMPLIES u=-(n-1)f` |
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| 40059. |
A ray of light incident on an equilateral glass prism shows minimum deviation of 30^(@). Calculate the speed of light through the prism. |
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Answer» SOLUTION :Here, A ` = 60^(@) , D = 30^(@)` REFRACTIVE INDEX, `n = sin((A+D)/(2))/sin((A)/(2))=sin((60+30)/(2))/(sin((60)/(2)))` `n = (sin45^(@))/(sin30^(@))=((1)/sqrt(2))/((1)/(2))=sqrt(2)` n = 1.414 Velocity of light in GLASS, `v = (c )/(n) = (3xx10^(8))/(1.414)` `v = 2.12 xx 10^(8) ms^(-1)` |
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| 40060. |
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_(g))_(C),(E_(g))_(Si) and (E_(g))_(Ge). Which of the following statements is true? |
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Answer» `(E_(g))_(Si) LT (E_(g))_(Ge) lt (E_(g))_(C)` |
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| 40061. |
In young double slit experiment, if the width of 4^(th) bright fringe is 2 times 10^-2 cm, then the width of 6^(th) bright fringe will be….cm. |
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Answer» `10^-2` |
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| 40062. |
A hydrogen - like neutral species in some excited state A, on absorbing a photon of energy 3.066 eV get excited to a new state B. When the electron from state B returns back, photons of a maximum ten different wavelengths can be observed in which some photons are of energy smaller than 3.066 eV, some are of equal energy and only four photons are having energy greater than 3.066 eV. The ionization energy of this atom is |
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Answer» 14.6 eV |
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| 40063. |
An object of mass 2 kg has linear momentum magnitude 6 kg*m//s. What is this object's kinetic energy? |
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Answer» 3J |
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| 40064. |
What is meant by critical size? |
| Answer» Solution :An uncontrolled nuclear chain reaction can occur only when the lump of URANIUM (TARGET) is GREATER than a certain MINIMUM size. This is CALLED critical size. | |
| 40065. |
What is earthing or grounding in household electrical circuit ? |
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Answer» Solution :When we bring a charged body in contact with the earth, all the excess charge on the bod disappears by causing a momentary current fr pass to the ground through the connectin conductor (such as our body). This process o sharing the charges with the earth is called grounding or earthing. A thick metal plate is buried DEEP into the earth and thick WIRES are drawn from this plate, these are USED in buildings for the purpose of earthing NEAR the mains supply. The ELECTRIC wiring in our houses has three wires : live, neutral and earth. The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate. Metallic bodies of the electric appliances such as electric iron, refrigerator, TV are connected to the earth wire. When any fault occurs or live wire touches the metallic body, the charge flows to the earth without damaging the appliance and without causing any injury to the humans. Earthing provides a safety measure for human body, electrical circuits and appliances. |
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| 40066. |
What is conservative field ? Give an example. |
| Answer» Solution :If the line integral of a field around a CLOSED PATH is zero, then such a field is CALLED CONSERVATIVE field. EXAMPLE : Electric field. | |
| 40067. |
An engine is moving on a circular track with a constant speed It .is blowing a whistle of frequency500 Hz.The frequency recived by an overser standing stationary at the centre of the track is (##TRG_PHY_MCQ_XII_C07_E02_048_Q01.png" width="80%"> |
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Answer» 500 Hz |
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| 40068. |
A wheel rolls smoothly along a floor while being accelerating horizontally by a force vecF_(n) applied at its top. The wheel has radius R = 0.500 m, mass m = 1.60 kg, and acceleration magnitude a = 2.00 m//s^(2). The frictional force on the wheel is in the forward direction and has magnitude f = 1.80 N. What is the rotational inertia of the wheel? |
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Answer» Solution :The (linear) acceleration a of a rolling object is the acceleration of the CENTER, which is the center of mass com. (2) Newton.s second law written for x components relates the forces on the wheel to the resulting acceler ation. (3) Newton.s second law written in angular form relates the torques on the wheel to the resulting angu lar acceleration. (4) The linear acceleration and the angular acceleration are related by the radius according to Eq. 11-6 (`a_("com") = alpha R`). Calculations: Newton.s second law for motion (`F_("net, y")= ma_(x)`) along the x axis gives us `F_(n)+ f= ma` `F_(a)= ma-f= (1.60kg)(2.00m//s^(2))-1.80N= 1.40N` Next, USING Newton.s second law for angular motion (`tau_("net") = Ialpha`), with torques and the angular acceleration = measured around the center of the wheel, we have `-RF+Rf= Ialpha` On the left SIDE of this equation we have the torque due to the applied force, at distance R from the wheel.s center and perpendicular to that distance. We insert a minus sign because were that torque to act alone, it would cause a clockwise rotation. We also have the torque due to the frictional force. However, this torque acting alone would cause a COUNTERCLOCKWISE rotation, and so we use a plus sign. We can relate the angular acceleration to the lin ear acceleration of the center with Eq. 11-6 (`a_("com") = alphaR`). Substituting for `alpha` and solving for I, we find `I= (R^(2)(-F+f))/(a)= ((0.5)^(2)(-1.40N+ 1.80N))/(2.00m//s^2)= 0.050kg.m^(2)` |
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| 40069. |
Describe briefly with the help of a necessary circuit diagram, the working principle of a solar cell. |
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Answer» Solution :Asolar cell is basically a p-n JUNCTION which generates emf when solar radiation falls on the p - nJunction. The generation of emf by a solar cell, when light falls on it, is due to following three basic processes: (i) Generation of e-h pairs due to incident light (photon energy E = hv `gt E_(g)` ) close to the junction. (II)Separation of electrons and holes due to electric field of the depletion region. As a result electrons are swept to n-side and holes to p-side of junction. (iii) COLLECTION of electrons reaching the n-side and holes reaching p-side by metal contacts provided at the two sides of solar cel. As a result of these three processes, an emf is developed such that p-side becomes positive andn side negative. The voltage obtained is known as "photovoltage" When an external load is connected, as shown in , a PHOTOCURRENT flows through the load. 
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| 40070. |
For the circuit shown in the figure, calculate the equivalent resistance for the two cases given as : (1) V_(A) gt V_(B) and (2) V_(B) gt V_(A). Here consider D_(1) and D_(2) to be ideal diodes. |
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Answer» SOLUTION :(1) When `V_(A)gt V_(B)` then `D_(1)` is in forward bias and `D_(2)` is in reverse bias CONNECTION. But second diode is iedeal therefore RESISTANCE `R_(2)` of `D_(2)` is infinite. Therefore CURRENT is flowing throgh only `D_(1)`, so effective resistance is `R_(1)`. In this situation total resistance of the circuit, `R_(AB)=50+0` `therefore R_(AB)=50OMEGA` (2) When `V_(B) gt V_(B)` then, `D_(1)` is in reverse bias and `D_(2)` is in forward bias. Therefore resistance `R_(1)` of `D_(1)` is infinite, so current flowing through `D_(1)` is zero. But effective resistance is only `R_(2)`. `R._(AB)=50+0` `therefore R._(AB)=50Omega` |
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| 40071. |
Show that m=(f)/(u+f) for a lens. |
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Answer» SOLUTION :By definition, `m=(V)/(u)` so that `v=m u`. `therefore (1)/(f)=(1)/(v)-(1)/(u)` `(1)/(f) = (1)/(m u)-(1)/(u)` `(1)/(f)+(1)/(u)= (1)/(m u)` `(u+f)/(uf)=(1)/(m u)` `therefore m=(f)/(u+f)` |
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| 40072. |
A person holds a block weighing 2kg between his hands & keeps it from falling down by pressing it with his hands. If the force exerted by each hand horizontally is 50N, the co-efficient of friction between the hand & the block is (g=10ms^(-2)) |
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Answer» 0.2 |
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| 40073. |
A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogengas in ground state (a)Assume that when a photon suppose some of its energy to a hydrogen atom the rest of the energy appears as another photon moving in the same direction as the incident photon Neglecting the light emitted by the excited hydrogen ato in the direction of the incident beam, ? (b) A radiation detector is placed near the gas to detect radiation coming perpenducular to the incident beam find the wevelength of radiation that may be detected by the detector |
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Answer» `E=(hc)/lambda = 1212/100` `=12.42 EV` (a) The Possible transitions may be `E_1` to `E_2`. Energy absorbed in `E_1` to `E_2` `=10.2 eV`. Energy LEFT `=12.42 - 10.2 = 2.22 eV` `2.22 eV = (hc)/lambda = 1242/lambda` or `lambda = 559.45 =560 nm` Energy absorbed in `E_1` to `E_3` `=12.1 e V` Energy left `=12.42 -12.1 = 0.31 eV` `0.32 = (hc)/lambda=1242/lambda` `lambda = 1242/0.32` `=3881.2 = 3881 nm` Energy absorbed in `E_3` to `E_4=0.65` Energy left `=12.42 -0.65 = 11.77` `11.77 = (hc)/lambda` or, `lambda = 1242/11.77 = 105.52` (b) The energy absorbed by the 'H' atom is now radiated perpendicular to the incident beam. `rarr10.2 =(hc)/lambda` or `lambda = 1242/10.2 = 121.76` `rarr 12.1 =(hc)/lambda` or `lambda = 1242/12.1 = 102.64 nm` `rarr 0.65 = (hc)/lambda` or `lambda = 1242/0.65 = 1910.76 nm` |
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| 40074. |
Charge passing through a conductor carrying current is given by Q = 5t^(2) + 3t +1, How much amount of current will pass in t = 5 s ? |
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Answer» 9A I = `(DQ)/(dt ) = 10 `By TAKING t = 5 in t + 3 I = 53 A |
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| 40075. |
A capacitor is charged with 9muC and has a 100V p.d. between the plates. Calculate its capacity and energy stored in it, if airis trapped between the plates. |
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Answer» SOLUTION :Data supplied, `Q=9muC =9 XX 10^(-6)C ""V=100V` Capacity, `C=Q/V=(9 xx 10^(-6))/(100)=9 xx 10^(-8)F` Energy, `U=1/2 CV^(2)=1/2 QV=1/2 Q^(2)/C` `therefore U=1/2 xx 9 xx 10^(-6) xx 100=4.5 xx 10^(-4)J` |
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| 40076. |
Space wave propagation is used in |
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Answer» MICROWAVE COMMUNICATION |
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| 40077. |
Resultant of which of the following may be equal to zero ? |
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Answer» 10N,10N, 30N |
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| 40078. |
The energy levels of an atom of element X are shown in the diagram. Which one of the level transitions will result in the emission of photons of wavelength 620 nm? Support your answer with mathematical calculations. |
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Answer» |
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| 40079. |
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light ? Refractive index of water is 1.33 . |
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Answer» Solution :Here wavelength of incident LIGHT in AIR `lamda=589nm=589xx10^(-9)m`, and speed of light in air `c=3xx10^(8)ms^(-1)` `therefore` Frequency of light `v=(c)/(lamda)=(3xx10^(8))/(589xx10^(-9))=5.09xx10^(14)HZ`. (a) Wavelength frequency and speed of reflected light are exactly same as that of incident light. (b) `because` REFRACTIVE index of water n=1.33 `therefore` Wavelength of refracted light `lamda.=(lamda)/(n)=(589nm)/(1.33)=444nm` Frequency of refracted light `v.=v=5.09xx10^(14)Hz`. and speed of refracted light `v=(c)/(n)=(3xx10^(8))/(1.33)=2.26xx10^(8)ms^(-1)` |
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| 40080. |
A : Standard optical diffraction gratings can not be used for discriminating between different X-ray wavelengths. R : The grating spacing is not of the order of X-ray wavelengths. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 40081. |
When the length and area of cross-section both are doubled, then it's resistance |
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Answer» will BECOME half |
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| 40082. |
In order to continue emission of electromagnetic waves from LC oscillations, we should |
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Answer» INCREASE the capacitance of a capacitor. |
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| 40083. |
Choose the correct statement out of the following. |
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Answer» Work done by the static friction on a BODY will ALWAYS be positive |
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| 40084. |
The relation between U,P and V foran ideal gas in an adiabatic process is given by U = 2 + 3 PV. The gas is |
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Answer» monoatomic Fromthe firstlawof THERMODYNAMICS, we get or ` 0- dU + PdV` From the given equation `dU = 3 ( PdV + VdP)+ PdV` ` or 4P (dV) +3 V (dP) = 0or 4 (dV)/V)=-3 ((dP)/P)` On integrating , we get ` "In"(V^(4) + "In"(P^(3))="constantor " PV^(4//3)` = constantCompareit with standardequation of an adiabatic process ` PV^(y) ` = constant,we get ` gamma = 4/3 `i.e,GASIS polyatomic . |
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| 40085. |
A coil of area 'A' , carrying a steady current I, has a magnetic moment vecm associated with it. Write the relation between vecm, I and A in vector form. |
| Answer» SOLUTION :`VECM = I VECA`. | |
| 40086. |
Two batteries of emfs 24V and 6V and internal resistance 2Omega and 1Omega respectively are joined in parallel, negative to negarive by a resistor of 2Omega and positive to positive by another resistor of 4Omega. Calculate the terminal voltages of the two batteries. |
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Answer» |
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| 40087. |
Consider a uniform eleceric field =3 xx 10^(3) hati N//C. a. What is the flux of this field through a square of 10cm on a slide whose plane is parallel to the yz plane? b. What is the flux through the same square if the moral to its plane makes a 60^@ and with the satis? |
| Answer» Solution :CHARGES 1 AMD 2 are NEGATIVE charge 3 is positiveparticle 3 has the HIGHEST charge to mass ratio | |
| 40088. |
A Rowland ring of mean radius 15 cm has 3500 tums of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A? |
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Answer» Solution :When toroidal space in a toroid is FILLED with some magnetic material, it becomes Rowland ring. Magnetic field inside this material is given by, `B = munI` `therefore B= mu_(R) mu_(0) ((N)/( 2PI r) ) I` (Where, `N=` total no. of turns, `mu_(r) =`Relative permeability, `r=` Average radius of a toroid) `therefore B= (800)(4pi xx10^(-7) ) ((3500)/( 2pi XX 0.15))(1.2)` `therefore B = 4.48` T |
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| 40089. |
If omega_c and omega_m are angular frequencies of carrier wave and modulating signal respectively, then Band width of amplitude modulated waves is equal to |
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Answer» `(omega_c+omega_m)/2` |
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| 40090. |
An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of 1 mA at both 200 Hz and 800 Hz frequency. What is the resonance frequency of the circuit ? |
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Answer» 600 Hz For maximum value of current `omega L-(1)/(omega C)=0`or`omega L=(1)/(omega C) ""`…..(i) Resonance frequency of series LCR ciruit is `omega = (1)/(sqrt(LC))` From (i),`200 L=(1)/(800 C) ""` [`because` current reaches a value of 1 mA] or `(1)/(LC)=200xx800` or `(1)/(sqrt(LC))=sqrt(200xx800)=400 Hz` |
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| 40091. |
As per Ampere's circular law oint vecB.vecd l=……………….. . |
| Answer» SOLUTION :`mu_0 SUM I` ENCLOSED | |
| 40092. |
A convex and a concave mirror of radius 10 cm each are placed 15 cm apart, facing each other. An object is placed mid-way between them. Find the position of final image if the reflection first takes place in the concave and then in the convex mirror: |
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Answer» Final image is formed on the pole of concave MIRROR. |
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| 40093. |
Who are crafty men? |
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Answer» Those who DISREGARD books |
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| 40094. |
How much will a steel wire 2m long and 1mm diameter stretch under a 5 kg wt. load ? (Y = 2 xx 10^11 N/m^2) |
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Answer» a)`62.42 XX 10^(-7) m` |
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| 40095. |
Rest mass of a photon is zero but dynamic mass of a photon is (E)/(c^(2)). |
| Answer» Solution :True: Rest mass of photon is zero but DYNAMIC mass is `m=(E)/(c^(2))` in accordance with mass ENERGY EQUIVALENCE. | |
| 40096. |
Which of the following statement is true for an-type semiconductor? |
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Answer» The donor LEVEL lies just below the bottom of the conduction bond. |
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| 40097. |
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields on an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves. |
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Answer» Solution :Electromagnetic waves are produced when CHARGED particles oscillte. The VIBRATIONS of charged particles result in ENERGY which is emitted as electromagnetic radiation. The mathematical expressions for electric and MAGNETIC fields ALONG the z-axis are `vecE=E_0sin(omegat -kz)hati` and `vecB=B_0sin(omegat-kz)hatj` Properties of electromagnetic waves: (i) They carry energy through space and this energy is distributed equally between the electric and magnetic fields at the time of propagation of electromagnetic waves. (ii) They do not require any material medium for their propagation |
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| 40098. |
ABC is a small, isosceles right angled triangle of hypotenuse 1cm. A charge of +40 pC (picocoulomb) is placed at the right angled corner A and -20 pC and -20 pC at B and C respectively. Show that this system of charges may be treated as a dipole for all external points at large distances. Calculate the potential due to this system of charges at a point on the extended side AC at a distance 40 cm from A. |
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Answer» |
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| 40099. |
If the velocity and wavelength of light in air is V_a and lamda_aand that in water is V_w and lamda_w then the refractive index of water is, |
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Answer» `(V_(W))/(v_(a))` |
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| 40100. |
A long, vertical magnet is placed at a perpendicular distance of 0.8m from the centre of a horizontal magnetic needle of length 0.12 m ad pole strength 0.6Am. If the moment of the couple acting upon the needle is 72 xx 10^(-7)Nm, Find the pole strength of the long magnet. [Hint: Since the magnet is long, the effect of the distance pole may be neglected.] |
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Answer» |
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