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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.
| 1601. |
A ball is dropped from a height of 20 m. If coefficient of restitution is 0-5. The ball rebounds to a height of: |
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Answer» Solution :`h_(n)=E^(2N).h_(0)impliesh_(1)=e^(2).h_(0)` `implies h_(1)=(0.5)^(2)xx20=5 m` |
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| 1602. |
what we call the spectrum of absorbed light? |
| Answer» SOLUTION :ABSORPTION SPECTRUM | |
| 1603. |
The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is 60^(@), then the coefficient of friction is |
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Answer» `(1)/(3)`  Now , force of friction, f = `mu`N = `mu` mg cos `theta` and net retarding force, `(F_(1)) ` = mg sin `theta` + f `therefore` Net accelerating force down the inclined plane is F = mg sin `theta - mu mg cos theta"" ` .... (i) `therefore` EXTERNAL force NEEDED (up the inclined plane ) to maintain sliding motion is (net retarding force ) `F_(1) = mg sin theta + mu mg cos theta "" `....(ii) it is given as, `F_(1) = 2F""`.... (iii) From Eqs. (i) , (ii) and (iiI) we get mg sin `theta` + `mu mg cos theta` = 2 (mg sin `theta - mu mg cos theta` ) or `3mu mg cos theta = mg sin theta ` or `""(sin theta)/(cos theta) = 3mu` or `"" tan theta = 3 mu "" (because theta = 60^(@))` or `"" tna 60^(@) = 3mu` or `"" sqrt(3) = 3mu` or `"" mu = (1)/(sqrt(3))` When inclination of plane is `60^(@)` then the coefficient of friction , `mu = (1)/(sqrt(3))` |
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| 1604. |
A 85.0 mH inductor is connected as in Fig.to an ac generator with E_m= 30.0 V. What is the amplitude of the resulting alternating current if the frequency of the emf is (a) 1.00 kHz and (b) 5.00 kHz? |
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| 1605. |
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 xx 10^(-2)T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coll is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^(-1). What is the moment of inertia of the coil about its axis of rotation? |
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Answer» Solution :Magnetic moment of a current CARRYING coil is, `m= NIA` `= NI (pi R^(2) )` `= (16) (0.75) (3.14) (0.1)^(2)` `therefore m= 0.3768 "Am"^(2)` ( or`JT^(-1)` ) Periodic time of oscillation of above coil, `T= 2pi sqrt((I)/(mB) )` (Where, `I=` moment of inertia of given coil about AXIS is ROTATION) Frequency of oscillation is, `f- (1)/(T) = (1)/(2pi )sqrt((mB)/( I) )` `therefore f^(2) = (1)/( 4pi^(2) ) ((mB)/( (I)))` `therefore I= (mB)/( 4 pi^(2) f^(2) )` `therefore I= ((0.3768) (5 xx 10^(-2) ) )/( (4) (3.14)^(2) (2)^(2) )` `therefore I= 1.194 xx 10^(-4) (kg) (m^(2) )` |
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| 1606. |
A state when there is no more absorption of heat by a body to rise the temperature of any part of it, is called |
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Answer» Critical STATE |
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| 1607. |
A : On viewing the clear portion of the sky through a calcite crystal, the intensity of the transmitted light varies as the crystal is rotated R : The light coming from the sky is polarized due to scattering of sunlight by particles in the atmosphere. The scattering is largest for blue light. |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION of A |
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| 1608. |
In which of the following mirrors can a person not see a reflection larger than his height ? |
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Answer» Concave mirror Magnification for plane mirror is m = 1 Magnification for plane mirror is m `gt` -1 and m `gt` +1 For a convex mirror a person will not be ABLE to see a reflection larger than his height. |
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| 1609. |
The variation of current (I) and voltage (V) is as shown in figure. The variation of power P with currnet I is best shown by which of the following graph |
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Answer»
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| 1610. |
Letthe x-zplanebe theboundarybetweentwotransparentmedia. Medium1 inz le0has arefrativeindex of sqrt(2) and medium2 withz lt 0hasa refracticveindex of sqrt( 3) A rayof lightin medium 1 givenby thevectorvec(A)=6 sqrt(3) hat(i)+8 sqrt(8) hat(j)- 10 hat(k) isincident on theplaneof separation. Theangleofrefractionin medium2 is |
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Answer» `40^(@)` |
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| 1611. |
(A) : The trajectory of neutron when it is projected perpendicular to a magnetic field is a parabola . (R) : A moving charge entering parallel to the magnetic field lines moves in a circular path. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 1612. |
A proton, a deuteron ion and an alpha-particle of equal kinetic energy perform circular motion normal to a uniform magnetic field B. If the radii of their paths are r_(p),r_(d)andr_(alpha) respectively then _____ [Here, q_(d)=q_(p),m_(d)=2m_(p)] |
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Answer» `r_(alpha)=r_(p)ltr_(d)` `thereforev=(QBR)/m""...(1)` Kinetic ENERGY of electric charge = K = `1/2mv^(2)` = `1/2m((qBr)/m)^(2)` `thereforeK=(Q^(2)B^(2)r^(2))/(2m)` `thereforer^(2)=(2Km)/(q^(2)B^(2))rArrrpropsqrtm/q` `thereforer_(alpha):r_(p):r_(d)=sqrt4/2:sqrt1/(1)1:sqrt2/1` = `1:1:sqrt2` `thereforer_(alpha)=r_(p)ltr_(d)` |
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| 1613. |
A regularhexagon of sidea stand in a vertical planewith one sideon the horizontalground . A particle is projected suchthat it touches the four uppervertices of the hexagon beforereturningto the ground . Find theratio of the velocityof the particleon reachingthe groundto its minimum velocity. |
| Answer» SOLUTION :`SQRT((31)/(3))` | |
| 1614. |
The maximum wavelength of a beam of light can be used to produce photo electric effect on a metal is 250 nm. The energy of the electrons in Joule emitted from the surface of the metal when a beam of light of wavelength 200 nm is used[h=6.62xx10^(-34)Js,C=3xx10^(8)ms^(-1)] |
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Answer» `89.61xx10^(-22)` |
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| 1615. |
A ball is dropped from a height h on a inclined plane, ball collides elastically with the first plane. Find the value of 'h' (in meter) so that it strikes perpendicular on the second inclined plane. |
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| 1616. |
A step up transformer converts a low input voltage into a high output voltage. Does it violet law of conservation of energy Explain? |
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Answer» Solution :No. A step up transformer steps of the VOLTAGE while its steps down the current. So the INPUT and output power remain same (PROVIDED there is no LOSS). Hence there is no violation of the principle of ENERGY conservation. |
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| 1617. |
Five cells have been connected in parallel to form a battery. The emf and internal resistances of the cells have been shown in figure. A load resistance R is connected to the battery. (a) Which of the 5 cells will have maximum current flowing through it? (b) find the current through load resistance R. |
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Answer» (B) `I=(80 epsilon)/(31R+16r)` |
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| 1618. |
Define half-life period of a radioactive substance. Establish its relationship with the decay constant. |
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Answer» Solution :Half-life period of a radioactive substance is the time at which both number of nuclides of radioactive substance as WELL as its ACTIVITY (7.E., both N and R) have been reduced to one-half of their initial values. THUS, at `t =T_(1/2), N=N_(0)/2` Hence, from relation `N=N_(0)e^(-lambdat)`, we have `N_(0)/2= N_(0)e^(-lambdaT_(1/2))implies log_(e) 2 = lambda T_(1/2)` or `T_(1/2)=(log_(e)2)/(lambda)=0.693/(lambda)`. |
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| 1619. |
The lines of force are close together, when electric field E is ...................... . |
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| 1620. |
A sample of radioactive material decays simultaneously by two processes A and B with halfs -lives 1/2 and 1/4h , respectively. For first half hoạr it decays with the process A, next one hour with the process B and for further half an hour with both A and B. If originally there were N_0nuclei, find the number of nuclei after 2h of such decay |
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Answer» `(N_0)/((2)^8)` |
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| 1621. |
The penetrating power of X-ray increases with the increase in its : |
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Answer» velocity So PRETENDING power `prop v`. |
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| 1622. |
Explain Nuclear fission reaction with an example. |
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Answer» Example. When slow neutrons are bombarded on a heavy nucleus of `""_(92)U^(235)` it produces two lighter nuclei of Ba and Kr and tremendous amount of energy. The nuclear reaction can be represented as follows : `""_(92)U^(235)+n^(1) rarr ""_(56)Ba^(141)+""_(36)Kr^(92)+3 ""_(0)n^(1)+DELTA Q` Importance of nuclear fission 1. CONTROLLED nuclear fission is used in number of yields e.g. (i) to prepare radio isotopes. (ii) to generate power for propulsion of ships, submarine and aircrafts. (III) to produce plutonium for explosive purpose. 2. Uncontrolled nuclear fission is used in atom bomb. |
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| 1623. |
A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in figure. The separation of the plate is d. If b=d//2, then the ratio of capacities of the capacitor after and before inserting that slab will be |
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Answer» `SQRT(2):1` |
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| 1624. |
Two polarising sheets have their polarising directions parallel to each other so that the intensity of transmitted light is maximum. The angle through wihich the either sheet must be turned so that intensity becomes one half the intial valueis |
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Answer» `60^(@)" or "120^(@)` |
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| 1625. |
The electrical conductivity of a semiconductor of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm, is incident on it. The band gap in (eV) for the semi conductor is …… |
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Answer» 0.5 EV BAND gap energy `E_(g)=h_(f)` `=(hc)/(lambda)` `=(hc)/(lambda e)"" `(In eV) `=(6.63xx10^(-34)xx3xx10^(8))/(2480xx10^(-9)xx1.6xx10^(-19))` `=0.0050126xx10^(2)` `~~0.5eV` |
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| 1626. |
A magnifying glass is used, as the object to be viwed can be brought closer to the eye than the normal near point. This results in |
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Answer» a larger ANGLE to be subtended by the OBJECT at the eye and hence viewed in greater detail. |
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| 1627. |
Which of the folowing staements is not correct? |
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Answer» The FIRST ionization energies (in KJ `mol^(-1)`) of carbon,silicon,germanium,TIN, and lead are 1086, 786, 761, 708 and 715 respectively |
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| 1628. |
The phase difference between the alternating current and emf is pi/2 rad. Which of the following cannot be the constituent of the circuit ? |
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Answer» L,C |
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| 1629. |
STATEMENT-1: The capacitance of the capacitor remains constant irrespective of the charge present on it. because STATEMENT-2: Capacitance depends on the size and the shape of the capacitor and also on the surrounding medium. |
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Answer» Statement-1 is TRUE , Statement-2 is True , Statement-2 is a CORRECT EXPLANATION for Statement-1. |
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| 1630. |
When AC is switched on the thin metallic disc is found to thrown up in air. How will you explain the mechanism behind the movement of disc (##EXP_SPS_PHY_XII_C06_E03_002_Q01.png" width="80%"> |
| Answer» SOLUTION :Whenever the magnetic flux LINKED with a METAL block changes, induced currents are PRODUCED due to this current, disc bicomes a magnet. Hence | |
| 1631. |
A ball whose kinetic energy is E is projected at an angle of 45^(@) to the horizontal. The kinetic energy of the ball at the highest point of its flight willbe: |
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Answer» E K.E. at the throw `=(1)/(2)m upsilon^(2)cos^(2) 45^(@)` K.E. at the highest point = `(1)/(2)mupsilon^(2)cos^(2)45^(@)` `EXX((1)/(sqrt2))^(2)=(E)/(2)` Hence choice is (c) |
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| 1632. |
Find the virtual value of current through a capacitor of capacitance 10 mu F, when connected to a source of 110 volt at 50 cycles supply. What is its reactance ? |
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| 1633. |
किसी पूर्णांक m के लिए , प्रत्येक सम पूर्णांक निम्नलिखित रूप का होता है : |
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Answer» m |
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| 1634. |
In double slit experiment, the angular width of the fringes is 0.20^(@) for the sodium light (lambda = 5890 Å). In order to increase the angular width of the fringe by 10%, the necessary change in the wavelength is : |
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Answer» increase of `589 Å` `Deltaw_(0) = (Delta lambda)/(d)` `(Deltalambda)/(lambda) = (Deltaw_(0))/(w_(0)) = (10)/(100) = 0.1` `RIGHTARROW Delta lambda = 0.1 lambda = 0.1 xx 5890 Å` = 589 Å(increase). |
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| 1635. |
Special structures in the thallus of lichen for nitrogen fixation and retaining moisture |
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Answer» Cyphellae |
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| 1636. |
A charged 8mF capacitor having charge 5mC is connected to a 5mH inductor. What is : (i) the frequency of current oscillations? |
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Answer» SOLUTION :(i) FREQUENCY of current oscillations `V= (1)/(2 PI sqrt(LC) ) ` |
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| 1637. |
A solid which is transparent to visible light and whose conductivity increases with temperature is formed by |
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Answer» METALLIC BONDING |
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| 1638. |
Discuss about astronomical telescope. |
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Answer» Solution :(i) An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon ETC. the image formed by astronomical telescope will be inverted. (ii) It has an objective of long focal length and a much larger aperture than the eyepiece. (iii) LIGHT from a distant object enters the objective and a REAL image is formed in the tube at its second focal point. (iv) The · eyepiece magnifies this image producing a final inverted image. Magnification of astronomical telescope : The magnification m is the ratio of the angle `beta`SUBTENDED at the EYE by the final image to the angle `alpha`which the object subtends at the lens or the eye. ` m= beta/ alpha` From the diagram, m ` = (h//f_e)/(h//f_0)` `m = (f_0)/(f_e)` The length of the telescope is approximately, `L = f_0 + f_e` |
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| 1639. |
A dog weighing 5 kg is standing on a flat boat so that he is metre from the shore. He walks 4 metre on the boat toward shore and then halts. The boat weight 20 kg and one can assume that there is no friction between it and the water. How far is the dog from the shore at the end of this time ? |
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Answer» Solution :GIVEN that initially the system is at rest so initial momentum of the system (Dog + boat) is zero. Now as in motion of dog no external FORCE is applied to the system final momentum of the system zero. So, `m vec(v)_(1)+M vec(v)_(2)=0` [as (m+M)= Finite] or `m(Delta vec(r )_(1))/(dt)+M(Delta vec(r )_(2))/(dt)=0 "" [as vec(v)=(d vec(r ))/(dt)]` or `m Delta vec(r )_(1)+M Delta vec(r )_(2)=0` [as `Delta vec(r )=vec(d)=` displacement] `md_(1)-Md_(2)=0` [as `vec(d)_(2)` is opposite to `vec(d)_(1)`] i.e., `md_(1)=Md_(2) ""`......(1)  Now when dog moves 4 m towards shore relative to boat, the boat will SHIFT a distance `d_(2)` relative to shore opposite to the displacement of dog so, the displacement of dog relative to shore (towards shore) will be `d_(2)=4-d_(1)(because d_(1)+d_(2)=d_(rel)=4)` .....(2) SUBSTITUTING the value of `d_(2)` from Eqn. (2) in (1) `md_(1)=M(4-d_(1))` or `d_(1)=(M xx 4)/((m+M))=(20xx4)/(5+20)=3.2 m` As initially the dog was 10 m from the shore, so now he will be 10 - 3.2 = 6.8 m away from the shore. |
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| 1640. |
The radii of curvature of two spherical surfaces of a concave convex lens (mu =1.5) are 20 cm and 40 cm. (i) What is its focal length when it is in air? Also find its focal length when it is immersed in a liquid of refractive index (ii) mu=1.2 (iii) mu =2 |
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Answer» (80 CM, 160 cm, -160 cm) |
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| 1641. |
How is NOT gate different from AND or OR gate ? |
| Answer» SOLUTION :AND or OR gates can have TWO or more inputs while NOT GATE cannot be MADE from junctiondiodes. It is realised by using a transistor | |
| 1642. |
A laser beam has a wavelength of 6xx10^(-7)m and aperture of 6xx10^(-2)m. The beam is sent towards the moon which is at a distance of 4xx10^(8)m from the earth. Calculate (i) angular spread of the beam and (ii) the arel spread when it reaches the moon. |
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| 1643. |
Find the mutual inductance between the circular coil and the loop shownin figure. |
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Answer» SOLUTION :emf induced =(mu_(0)(a_1)^2 (a_2)^2 Erv)/(2L(a^(2)+x^(2))^(3//2)((Rx)+R^(2))` [From question 20] `(d phi)/(dt)=(ERV)/(L((Rx)/(L)+r)^2)` M=(E )/((DI)/(dt))=(N(mu_0)(pi)a^(2)a^(2))/(2 (a^(2)+x^(2))^(3//2))`. |
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| 1644. |
Assertion : When two lenses are used to make achromatic lens combination then the materials of the two lenses are always different. Reason : Dispersive power of the materials of the two lenses must be of opposite sign to become achromatic combination. |
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Answer» If both assertion and REASON are correct and RASON is a correct EXPLANATION of the assertion. |
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| 1645. |
Thetop of the atmosphere is at about 400kV with respect to the surface of the earth , corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100Vm^(-1). Why then do we not our house into the open? (Assume the house to be a steel cage so there is no field inside) |
| Answer» Solution :Since our body and the surface of EARTH both are conducting therefore and equipotential surface. As we STEP out into the open from our house the original equipotential SURFACES of open air CHANGE, keeping out body and the ground at the same potential that is why we donot GET an electric shock. | |
| 1646. |
The path of a positively charged particle 1 through a rectangular region of uniform electric field is as shown in the figure. What is the direction of electric field and the direction of deflection of particles 2,3 and 4? |
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Answer» TOP, down, top, down |
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| 1647. |
A car travels at 80 km//h on a level road in the positive direction of an x axis. Each tyre has a diameter of 66 cm. Relative to a woman riding in the car and in unit-vec notation, what are the velocity vecv at the (a) center, (b) top, and (c) bottom of the tyre and the magnitude a of the acceleration at the (d) center, (e) top, and (f) bottom of each tyre? Relative to a hitchhiker sitting next to the road and in unit-vec notation, what are the velocity v at the (g) center, (h) top, and (i) bottom of the tyre and the magnitude a of the acceleration at the (j) center, (k) top, and (I) bottom of each tyre? |
| Answer» Solution :(a) `v_("center")= 0`, (b) `vecv_("top")= (+22m//s)HATI`, (c) `vecv_("Bottom") (-22m//s)hati`,(d) `a_("center")=0`, (e) `a_("top")= 1.5 xx 10^(3)m//s^(2)`, (f) `a_("bottom")= 1.5 xx 10^(3)m//s^(2)`, (G)`vecv= (+22m//s)hati`, (h) `2v= +44m//s`, (i)zero (j) it does not accelerate, (k)the answer is as it was in part (e)that is , `a= 1.5 xx 10^(3)m//s^(2)`, (l) as EXPLAINED in part (k) that is , `a = 1.5 xx 10^(3)m//s^(2)` | |
| 1648. |
The strength of magnetic field around an infinitely long current carrying conductor is |
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Answer» same EVERY where |
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| 1649. |
Light from a sodium lamp (S) passes through two polaroid sheets P_(1) and P_(2) as shown in figure. What will be the effect on the intensity of the light transmitted (i) by P_(1) and (ii) by P_(2) on rotating polaroid P_(1) about the direction of propagation of light ? Justify your answer in both cases. |
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Answer» Solution :(i) When monochromatic light of Intensity lo passes through the polaroid Pi, the transmitted light becomes plane POLARISED whose intensity is `I_(1)= (I_(0))/(2)` . This intensity `I_(1)` does not change even on becomes plane polarised whose internsity is `I_(1)=(I_(0))/(2)` . This internsity `I_(1)` does not change even on rotating `P_(1)` about the DIRECTION of propagation of light. (ii) If pass axis of polaroid sheet `P_(2)` subtends an angle `theta` with that of polaroid `P_(1)`, then in accordance with MALUS law the intensity of transmitted light is `I_(2)=I_(1)cos^(2) theta=(1)/(2) I_(0)cos^(2) theta.` If polaroid sheet `P_(1)` is rotatecd about the direction of propagation of light, then VALUE of 0 and consequently intensity `I_(2)` varies between the limiting values `(I_(0))/(2)` and zero. |
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| 1650. |
Three molecules have velocities 0.5 km/s, 1 km/s and 2 km/s. The ratio of their r.m.s. velocities and average velocity is |
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Answer» 1 |
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