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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Energy levels showing discrete energies possessed by the electrons in an isolated atom are ? |
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Answer» |
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| 2. |
In the given figure, the wedge is acted upon by a constant horizontal force 'F'. The wedge is moving on a smooth horizontal surface. A ball of mass 'm' is at rest relative to the wedge. The ratio of forces exerted on 'm' by the wedge when 'F' is acting and 'F' is withdrawn assuming no friction between the edge and the ball, is equal to : |
Answer» Solution : `N_(1)=mg COS THETA+F SIN theta` `N_(2)=mg cos theta (N_(1))/(N_(2))=1+(F sin theta)/(mg cos theta)` `(N_(1))/(N_(2))=1+(cancel(m)cancel(G) tan theta sin theta)/(cancel(m)cancel(g)cos theta)` `(N_(1))/(N_(2))=1+Tan^(2)theta=sec^(2)theta` |
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| 3. |
The dimensional formula of the emf of a source of current are same as that of |
| Answer» Answer :D | |
| 4. |
One metallic loop is placed inside non-uniform magnetic field. Will the emf be induced in the loop? |
| Answer» SOLUTION :Non-uniform magnetic field MEANS, there is non-uniform DISTRIBUTION of magnetic flux over the loop. But emf can be generated only when flux is CHANGED with time. So no emf will be induced in this case. | |
| 5. |
A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it as shown in the figure. The net field vec(E) at the center O is |
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Answer» `(q)/(2pi^(2) epsilon_(0) R^(2)) hat(J)` |
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| 6. |
A thin conducting rod AB is introduced in between the two point charges +q_(1) and -q_(2) as shown in figure. For this situation mark the correct statement(s). |
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Answer» The total FORCE experienced by `q_(1)` is vector SUM of electric force experienced by `q_(1)` due to `q_(2)` and due to induced CHARGES on rod. |
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| 7. |
A smooth massless string passes over a smooth fixed pulley. Two masses m1 and m_(2) (m_(1) gt m_(2)) are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is |
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Answer» `(m_(1)+m_(2))g`  Let a be the common acceleration of the system and T be the tension in the string. The equations of motion of two masses are `m_(1)g-T=m_(1)a` …(i) `T-m_(2)g=m_(2)a` …(II) Adding eqn. (i) and eqn. (ii), we get `a=((m_(1)-m_(2)))/(m_(1)+m_(2))g` ...(iii) Acceleration of centre of mass of the system is `a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=(m_(1)a-m_(2)a)/(m_(1)+m_(2))` (`becausea_(1)` and `a_(2)` are equal in magnitude but opposite in direction) `a_(CM)=((m_(1)-m_(2))/(m_(1)-m_(2)))a=((m_(1)-m_(2))/(m_(1)+m_(2)))((m_(1)-m_(2))/(m_(1)+m_(2)))g` (Using (iii)) `=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g` The TOTAL external force ACTING on the two masses is `F_("ext")=(m_(1)+m_(2))a_(CM)=(m_(1)+m_(2))((m_(1)+m_(2))/(m_(1)+m_(2)))^(2)g` `=((m_(1)-m_(2))^(2))/(m_(1)+m_(2))g` |
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| 8. |
Which of the following in non linear |
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Answer» `I_(3)` |
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| 9. |
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm. a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass ? |
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Answer» Solution :Here, f = 5 CM, U = ? For the closest distnace, `V=-25cm` As `(ul(1))/(upsilon)-(ul(-1))/(upsilon)=(-1)/(f)` `(1)/(u)=(1)/(v)-(1)/(f)=(1)/(-25)-(1)/(5)=(-1+5)/(25)` `u'=(25)/(-6)-4.2cm` This is the closest DISTANCE at which he can read the book. Forthe farthest distance, `v'=oo, u'=?` As `(1)/(f)=(1)/(v')-(1)/(u')` `(1)/(upsilon)=(1)/(v')-(1)/(f)=(1)/(oo)-(1)/(5)=(-1)/(5)` `u'=-5cm` This is the farhest distance at which he can real the book. |
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| 10. |
The binding for deuterium (""_(1)^(2)H) and helium (""_(4)^(2)He) are 1. IMeV and 7.0 MeV respectively. The energy released when two deutrons fuse t form a helium nucleus (""_(4)^(2)He) is K(10^(7)) eVm where K= _____ (Round of to nearest integer) |
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Answer» 2 |
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| 11. |
Incorrect order of boiling point is - |
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Answer» `HF gt HI gt HBR gt HCl` |
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| 12. |
Assertion A charged particle is rotating in a circular path in uniform magnetic field. Then, plane of circle is perpendicular to the magnetic field. Reason Circular motion is a two-dimensional motion. |
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Answer» If both Assertion and Reason are true and Reason is the CORRECT EXPLANATIONOF Assertion. |
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| 13. |
If 3 kg of mass is converted into energy. Energy released is |
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Answer» `9XX10^(8)J` |
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| 14. |
A tungsten filament bulb is connected to a variable voltage supply. The potential difference, V is varied and the current, I and steady temperature T of the filament is recorded. A graph is plotted for ln (VI) V_(s) ln (T). Find the slope of the graph. Assume that temperature of filament T gt gt T_(0) = atmospheric temperature |
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| 15. |
Let vecF be the force acting on a particle having position vector vecr and vec(tau) be the torque of this force about the origin. Then |
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Answer» `VECR.vec(TAU)ne0 and vecF.vec(tau)ne0` `vec(tau)" is PERPENDICULAR to "vecr. THEREFORE vec(tau) .vecr=0 or vecr.vec(tau)=0` `vec(tau)" is perpendicular to "vecF. therefore vec(tau) .vecF=0 or vecF.vec(tau)=0` |
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| 16. |
Can two separate candle or electric bulbs produce interference pattern? Why? |
| Answer» SOLUTION :No, DUE to the RANDOM CHANGES in PHASE. | |
| 17. |
The three stable isotopes of neon: " "_(10)^(20)Ne, " "_(10)^(21)Ne and " "_(10)^(22)Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. |
| Answer» Solution :Average atomic mass of NEON is the WEIGHTED average of masses of THREE isotopes `(90.51 xx 19.99 + 0.27 xx 20.99 + 9.22 xx 21.99)/100 u=20.1771u=20.18u`. | |
| 18. |
A body of mass 5 kg is placed on a rough horizontal surface of coefficient of static friction 1/3 . The least pulling force to the applied on the body at an angle 45^(@) with the horizontal to slide it, it (g=10ms^(-2)) |
| Answer» Answer :B | |
| 19. |
Show that a projectile fired at an angle 0 with the horizontal crosses a certain height at two timings t_(1) and t_(2) and the sum of these two is equal to : |
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Answer» TOTAL TIME of flight |
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| 20. |
For the same, kinetic energy the momentum will be maximum for : |
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Answer» a proton SINCE MASS in this case is maximum for an `alpha`-particle `:.` It will possess maximum momentum. |
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| 21. |
The electric flux through a Gaussian surface that encloses three charges given byq_1 = -14 nC,q_2 = 78.85 nC, q_3 = - 56 nC |
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Answer» `10^(3) NM^2 C^(-1)` |
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| 22. |
White light is used to illuminate the two slits in Young's double slit experiment, separation between the slits is b and the screen is at a distance d ( gt gt b) from the slits.At a point on the screen, directly in front of the slits, certain wavelengths are missing.Some of these missing wave lengths are : |
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Answer» `((b^(2)))/(d), ((b^(2)))/(3D)` For the missing wavelengths `(2n - 1)(lambda)/(2) = (b^(2))/(2d)` when `n = 1, (lambda)/(2) = (b^(2))/(2d) , lambda = (b^(2))/(d)` `n = 2, 3 xx lambda/2 = (b^(2))/(2d), lambda = (b^(2))/(3d)`. |
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| 23. |
An inductor used to reduce the high frequency components of an alternating signal by presenting a higher impedance, is called: |
| Answer» Answer :D | |
| 24. |
A block of mass M slides along the sides of bowl as shown in the figure. The walls of the bowl are frictionless and the base has coefficient of friction 0.1, and length 0.5 m. The block is released from the point A which is 0.2 m high as shown in figure. Then the block comes to rest. |
Answer» Solution : From law of conservation of energy mgh `=(1)/(2)mv^(2)` `therefore v^(2)=2(0.2)g=0.4g` `therefore (1)/(2)m(0.4)g=0.1(m)(g)(0.5)+mgh_(1)` `therefore h_(1)=0.15 m` Similarly when it reaches to P again it RISES to 0.1 m. And then COMES to Q and rises to 0.5 m and then FINALLY comes to REST at P. |
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| 25. |
Plot a graph to show variation of the angle of deviation as a function of angel of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism. |
| Answer» Solution :A TRANSFORMER converts low voltage ac into HIGH voltage ac and vice VERSA | |
| 26. |
A stone is thrown with a velocity v at an angle theta with the horizontal. Its speed when it makes an angle beta with the horizontal is : |
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Answer» V COS `THETA` |
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| 27. |
Resistivity of semiconductor of approximately ……….. . |
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Answer» `(10^(-2)" to "10^(-8))OMEGA m` |
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| 28. |
Statement - I : Quantization of charge is invalid after the presence of quark particles. Statement - II :Quark particle has charge greater than electron. |
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Answer» If both STATEMENT-I and Statement-II are TRUE, and Statement-II is the correct EXPLANATION of Statement -I. |
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| 29. |
Why are the equipotential surfaces about a single charge are not equidistant ? |
| Answer» Solution :ELECTRIC FIELD at a point distant r from a point charge Q is given as `E = (Q)/(4 pi in_(0) r^(2))` . As `E prop (1)/(r^(2))` , the field is non-uniform . Therefore , equipontential SURFACES are not equidistant . | |
| 30. |
shows the following three vectors: veca = (4.2 m) hati - (1.5 m) hatj, vecb = (-1.6m) hati + (2.9 m) hatj, andvecc = (-3.7m) hatj. What is their vector sum vecr which is also shown ? |
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Answer» Solution :For x axis, we add the x components of `veca, VECB and vecc` to get the x component of the vector sum `vecr:` `r _(x) = a _(x) + b _(x) + c_(x)` `= 4.2 m - 1. 6m + 0= 2.6 m.` Similarly, for the y axis, `r _(y) = a _(y) + b _(y) + c _(y)` `=-1.5 m + 2.9 m - 3.7 m =-2.3m.` We then combine these components of `vecr` to write the vector in unit-vector notation: ` vecr = (2.6m ) hati - (2.3 m) hatj,` where `(2.6m)hati` is the vector component of `vecr` along the x axis and `- (2.6 m) hatj` is that along the y axis. shows one way to arrang ethese vector components to from `vecr.` (Can you sketch the other way?) We can also answer the equation by GIVING the magnitude and an angle for `vecr.` the magnitude is `r = sqrt ((2.6m) ^(2) + (-2.3 m) ^(2))~~ 3.5 m` and the angle (MEASURED from the + x direction) is `theta = tan ^(-1) ((-2.3 m)/( 2.6m)) =- 41 ^(@),` where the minus SIGN means clockwise. 
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| 31. |
Electrons from n = 2 to n= 1 in hydrogen atom is made to fall on a metal surface with work function 1.2ev. The maximum velocity of photo electrons emitted is nearly equal to |
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Answer» `6XX10^(5)` m/s |
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| 32. |
The structure of crystals can be studied using |
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Answer» DIFFRACTION of VISIBLE light |
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| 33. |
भारत किस महासागर के मध्य मे स्थित है |
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Answer» प्रशांत महासागर |
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| 34. |
When two capacitors C_1 and C_3 are connected in series, the ratio of charges and potential differences across two capacitors are _________ and __________ respectively. |
| Answer» SOLUTION :`1 : 1, C_2 : C_1.` In series grouping charge on each CAPACITOR is same and potential DIFFERENCES are inversely proportional to the capacitance of individual CAPACITORS. | |
| 35. |
A body is thrown with the velocity 20ms^(-1)at an angle of 60^@with the horizontal. Find the time gap between the two positions of the body where the velocity of the body makes an angle of 30^@with horizontal. |
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Answer» 1.15 s |
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| 36. |
Which of the following is a unit of mobility ? |
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Answer» `m^(2) V^(-1) S^(-1)` From ` MU = (v_(d))/(E) , " unit " (ms^(-1))/(Vm^(-1)) = m^(2) V^(-1) s^(-1)` From `mu = (sigma)/("ne"), "unit " = (ʊ)/(m m^(-3) C) = m^(2)ʊ C^(-1)` `"" m^(2) ʊ C^(-1)` `mu = (sigma)/("ne") = ("ne"^(2) TAU)/("mne") = ( e tau)/(m) = (Cs)/(kg) [ because (m)/("ne"^(2) tau)THEREFORE sigma = ("ne"^(2) tau)/(m) ]` `therefore m^(2) Omega C^(-1) ` is not a unit of mobility. |
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| 37. |
If there are n free electrons per unit volume in a conductor, if mu represents the mobility of conductor the resistivity of the conductor is |
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Answer» `mune` |
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| 38. |
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. It its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively. |
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Answer» 2qE and MINIMUM For q = `0^(@)`, U = -pE, which is minimum |
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| 40. |
Draw the graph showing the variation of binding energy per nucleon with mass number. Give the reason for the decrease of binding energy per nucleon for nuclei with high mass numbers. |
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Answer» Solution :For GRAPH, see Fig. 13.04. For nuclei of high mass number, the number of PROTONS inside the nucleus is more. These protons repel each other. Due to larger FORCE of repulsion on ACCOUNT of greater number of protons PRESENT in the nucleus, the value of binding energy per nucleon decreases. |
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| 42. |
A ray of light passes from vacuum into a medium of refractive index mu, the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is: |
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Answer» `COS^(-1)((MU)/2)` |
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| 43. |
A wave propagates on a string in positive x-direction with a speed of 40 cm/s. The shape of string at t = 2 s is y=10 cos x/5where x and y are in centimeter. The wave equation is |
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Answer» `y = 10 COS (x/5 - 8t) ` |
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| 44. |
A magnifying glass is used , as the object to be viewed 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 details |
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| 46. |
Plane polarised light is inciden on a piece of quartz cut parallel to the axis Find the least thickness for which the O and E-rays suffer a phase change of pi//2. Given mu=1.5442 and mu_(E)=1.5533 and lambda=5xx10^(-7)m. |
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| 47. |
Germanium is doped one part million with indium at room temperature. Calculate the conductivity of doped germanium. Given: concentration of Ge atoms =4.4xx10^(28)m^(-3), intrinsic carrier concentration (n_(i))=2.4xx10^(19)m^(-3), mu_(e)=0.39m^(2)V^(-1)s^(-1) and mu_(k)=0.19m^(2)V^(-1)s^(-1) |
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| 48. |
Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t = 0. Find the number of active nuclei at time t. |
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| 49. |
Fundamental frequency of pipe is 100 Hz and other two frequencies are 300 Hz and 500 Hz then |
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Answer» PIPE is open at both the ends |
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