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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. |
The probability of finding a particle in space is represented by . |
| Answer» SOLUTION :MATTER WAVES. | |
| 2. |
(A) The force of repulsion between atomic nucleus and alpha-particle varies with distance accorcling to inverse square law. (R). Rutherford did alpha -particle scattering experiment. |
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Answer» Both 'A' and 'R' are true and 'R' is the correct explanation of 'A'. |
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| 3. |
A bar magnet is fixed in sucha waythat isits north poleis towardsgeomagneticnorth pole (NP) of the earth. The neutral point will be . |
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Answer» on the AXIAL LINE |
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| 4. |
As part of his discovery of the neutron in 1932, James Chadwick determined the mass of the neutron ( newly identified particle ) by firing a beam of fast neutrons , all having the same speed, as two different targets and measuring the maximum recoil speeds of the target nuclei. The maximum speed arise when an elastic head-on colliion occurs between a neutron and a stationary target nucleus. Represent the masses and final speeds of the two target nuclei as m_(1), v_(1) m_(2)and v_(2) and assume Newtonian mechanics applies. The neutron mass can be calculated from the equation : m_(n) = ( m_(1) v_(1)- m_(2) v_(2))/(v_(2) - v_(1)) Chadwick directed a beam of neutrons on paraffin, which contains hydrogen. The maximum speed of the protons ejected was found be 3.3 xx 10^(7) m//s. A second experiment was performed using neutrons from the same source and nitrogen nuclei as the target. The maximum recoil speed of the nitrogen nuclei was found to be 4.7 xx 10^(6) m//s. The masses of a proton and a nitrogen nucleus were taken as 1u and 14 u, respectively. What was Chadwick's value for the neutron mass ( in u ) ? |
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Answer» |
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| 5. |
A uniform wire is cut into 10 segments of increasing length. Each segment is having a resistace of 8 Omegamore than that of previous segment. If the resistance of shortest segment is R and largest segment is 2 R. The original resistance of wrie in ohm. |
| Answer» ANSWER :D | |
| 6. |
(I) Metal selected for electron emission should have low work function. (II) Potential barrier presents leaving free electron from the metallic surface (III) Potential barrier created by the positive nuclei of the metal. (IV) Electron in the outermost shells are tightly pound to the nucleus. |
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Answer» I and II only |
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| 7. |
Fuel is consumed at the rate of 50 kg s^(-1)in a rocket. Find the thrust on the rocket if the velocity of the exhaust gases is 2 km s^(-1). Also calculate the velocity of the rocket at the instant, when its mass is reduced to l/10th of its initial mass if its initial velocity is zero, (neglect gravity) |
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Answer» Solution :`(dm)/(dt) = 50 kgs^(-1) , U=2 KM s^(-1) = 2 xx 10^(3) ms^(-1)` Initial velocity, `u_(0)=0` (i) The thrust on the ROCKET. `F =u(dm)/(dt) = 2 xx 10^(3) xx 50 = 1 xx 10^(5)` N (ii) The velocity of the rocket. `v=v_(0) + ulog_(e) m_(0)/m` `=0+2 xx 10^(3) log_(e) [m_(0)/m_(0) xx 10] [THEREFORE m=m_(0)/10]` `=4.606 xx 10^(3) m//s` |
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| 8. |
What is a potentiometer ? |
| Answer» Solution :A POTENTIOMETER is an instrument for measuring, comparing or DIVIDING small potential DIFFERENCES. It is typically used to MEASURE the emf of a cell by the comparison method. It consists of a long uniform resistance wire along with a cell of extremely stable emf connected ACROSS its ends. | |
| 9. |
What is meant by the statement "charge is quantized". |
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Answer» Solution :The minimum charge that MAY be transferred from ONE body to the other body is equal TOT he charge of an electron `(1.6xx10^(-19)C)`. The charge is available in the multiples of charge on electron. HENCE the charge is said to be quantised. |
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| 10. |
One mole of an ideal gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K the heat required is (Given R=(8.3J)/("mole"xxK)) |
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Answer» `198.7` J |
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| 11. |
The frequency of a sinusoidal wave y=0.40cos{2000t+0.080} would be |
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Answer» `100piHz` |
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| 12. |
The wavelength of de-Broglie wave associated with a matter particle is 2Å. The momentum of the particle is |
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Answer» <P>`3.3xx10^(-24)kg" "ms^(-1)` |
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| 13. |
When a ray of light enters into the medium of refractive index of mu, angle of refraction is half of angle of incidence so angle of incidence = ...... |
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Answer» `2sin^(-1)(MU/2)` `=(2sin""1/2cos""1/2)/(sin""i/2)` `THEREFORE mu =2cos""i/2` `therefore mu/2=cos""i/2` `i/2=cos^(-1)(mu/2)` `therefore i= 2cos^(-1)(mu/2)` |
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| 14. |
A flywheel rotating about an axis experinces an angular retardation proportional to the angular through which it rotates its rotational kinetic energy gets reduced by DeltaE while it rotates through an angular theta then |
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Answer» `DeltaEproptheta^(2)` |
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| 15. |
Plane wavefronts are incident on a spherical mirror as shown, the reflected wavefronts will be |
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Answer»
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| 16. |
In the previous problem, when both the keys are closed, charge that will flow through key K_(1) will be : |
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Answer» `(t)/((d+t))Q` |
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| 17. |
Electromagnetic waves are produced by______ |
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Answer» CHARGES at rest |
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| 18. |
Magnetic susceptibility and chi_m are related as: |
| Answer» SOLUTION :`chi_m` = I/H | |
| 19. |
A constant retarding force of 20 N acts on a body of mass 5 kg moving initially with a speed of 10 ms^(-1). How long does the body take to stop? |
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Answer» 1.5 s |
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| 20. |
From Brewster.s law for polarisation , it follows that the angle polarisation depends upon |
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Answer» the WAVELENGTH of LIGHT |
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| 21. |
Threshold wavelength for metal is 10,000Å. If light of wavelength 5461Å is incident on it, then stopping potential is 1.02 V, then value of Planck's constant is : |
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Answer» `6.45xx10^(-34)J-sec` |
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| 22. |
In an atom when electrons transition from excited state to ground state, which of the statement is true ? |
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Answer» K increase and U and E decreases. In `Kprop(1)/(n^(2))` as n decreases K increases. In `Uprop-(1)/(n^(2))` as n decreases the NEGATIVE valueof U increases hence POTENTIAL ENERGY decreases. In `Eprop-(1)/(2n^(2))` as n decreases, the negative value of E increases hence total energy decreases. |
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| 23. |
What is the distance of closest approach of alpha-particle and Nucleus ? |
| Answer» SOLUTION :`(Ze^2)/(piepsilon_0mv^2)` | |
| 24. |
What are the characteristics of a hole ? |
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Answer» SOLUTION :i. Holes carry a UNIT positive charge . ii. Holes have the same charge as that of an ELECTRON . III. Energy of hole is high as compared to that of electron . iv. The MOBILITY of hole is smaller than that of electron . |
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| 25. |
Threshold frequency for metal surface is 4xx10^(14) Hz.When radiation of 5xx10^(14)Hz is incident on this surface ,photo-electric current obtained is 1.8 mA.If frequency of incident radiation is made half and intensity is made three times,value of photo-electric current will be..... |
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Answer» 0.9 mA `V_(0)=4xx10^(14)HZ` INITIALLY V=`5xx10^(14)`Hz Now frequency of incident radaition Is MADE half, `v.=(5xx10^(14))/(2)=2.5xx10^(14)`Hz For photo-electric effect v.`gtV_(0)` but here v.`ltV_(0)` hence photo-electric effect will not be obtained .Hence ,CURRENT will not be obtained . |
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| 26. |
Find the recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state(R=1.097xx10^(7) m^(-1)) : |
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Answer» `2.4 MS^(-1)` |
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| 27. |
Figure shows object O. Final image I is formed after two refractions and one reflection is also shown in figure. Find the focal length of mirror. (in cm) : |
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Answer» 10 |
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| 28. |
An electron moving around the nucleus with an angular momentum l has an orbital magnetic moment |
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Answer» `E/m L` |
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| 29. |
The velocity of electron v in the second orbit of hydrogen, then its velocity in the fifth orbit will be ...... |
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Answer» `(5)/(7)v` `vecv_(n)=(Ze^(2))/(2in_(0)nh)` But for H,Z=1 and `(e^(2))/(2in_(0)h)` constant `:.v_(n)prop(1)/(n)` `:.(v_(5))/(v_(2))=(2)/(5)` `:.v_(5)=(2)/(5)v_(2)` `:.v_(5)=(2)/(5)v""[:.v_(2)=v]` |
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| 30. |
Light passes through a glass slab of refractive index .n. and thickness .t.. If .C. is the speed of light in free space, then time taken to emerge | light out of glass slab will be ...... |
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Answer» `(t)/(NC)` `therefore` Time `=(nt)/(c)` |
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| 31. |
A spherical conductor of radius 12 cm has charge of 1.6 xx 10^(-7)C distributed uniformly on its surface. What is the electric field (a) insidethe sphere(b)just outside the sphere(c ) at a point 18 cm from the centre of the spere ? |
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Answer» Solution :Here `Q = 1.6 xx 10^(-7) C, R = 12 cm = 0.12 m` (a) ELECTRIC FIELD inside a SPHERICAL conductor having charge distributed uniformly on its surface is zero at all POINTS. (b) Just OUTSIDE the surface `E = 1/(4pi epsi_0). Q/r^2 = (9 xx 10^9 xx 1.6 xx 10^(-7))/((0.18)^2) = 4.44 xx 10^4 N C^(-1) = 4.4 xx 10^4 NC^(-1)`, (c) At a point 18 cm from the centre of sphere (i.e., r = 18 cm = 0.18 m), the electric field `E = 1/(4pi epsi_0) . Q/r^2 = (9xx10^9 xx 1.6 xx 10^(-7))/((0.18)^2)= 4.44 xx 10^4 N C^(-1) =4.4 xx 10^(4) N C^(-1)` |
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| 32. |
Electric field is given by vec(E) = (8hat(i)+4hat(j)+3hat(k))NC^(-1). Electric flux through Y-Z plane X-Z plane are in ratio |
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Answer» `4:3:8` |
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| 33. |
The coefficient of linear expansion of steel is 1 xx 10^(-5)/^@C . How much expansion should engineers anticipate in a 2000 m steel bridge, if it undergoes a change in temperature from 0^@C to30^@C ? |
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Answer» Solution :Here it is GIVEN that `alpha = 1 xx 10^(-5) PER^@C, Delta theta = 30^@ C-0^@C = 30^@C`. `l_1 = 2000 m` . HENCE the EXPANSION to be anticipated is given by `Delta l = 2000 m xx 30^@Cxx 1 xx10^(-5) per^@C=0.6m`. |
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| 34. |
y,(x,t) = 0.8//[(4x + 5t)^(2) + 5] represents a movine pulse, where x,y are in metre and t in second. Then |
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Answer» PULSE is MOVING in `+x` -direction |
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| 35. |
A body of mass m at rest gets exploded into 3 parts, having masses in the ratio 1: 1:3. Its two parts having equal masses move at right angles to each other with 15 m/s each. The velocity of third is: |
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Answer» `5sqrt2 ms^(-1)`  Now `p_(12)=sqrt(p_(1)^(2)+p_(2)^(2))=p_(1)sqrt2=m/5xx15sqrt2` ALSO `p_(3)=(3m)/(5)xxv` `:. (3m)/(5)v=(15m)/(5)xxsqrt2` `implies v+(15)/(3)xxsqrt2` or `v=5sqrt2 ms^(-1)` |
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| 36. |
Suppose that the upper half of the concave mirror.s reflecting surface as shown in figure is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror ? |
| Answer» SOLUTION :INTENSITY will be HALVED | |
| 37. |
In the circuit shown, a voltmeter of internal resistance 'R' when connected across B and C, reads (100)/(3) volts. Neglecting the internal resistance of the cell, the value of 'R' is |
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Answer» `100 k Omega ` |
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| 38. |
Which of the following system of unit is also called the metric system? |
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Answer» CGS |
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| 39. |
A small signal voltage V(t) = V_0 sinomegat is appliedacross an ideal capacitor C. |
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Answer» Over a FULL cycle the capacitor C does not consume any energy from the VOLTAGE source Power `P=V_(rms)I_(rms) cos rho` `=V_(rms) I_(rms) cos 90^@` `=V_(rms) I_(rms)xx0` `therefore` P=0 |
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| 41. |
A p-n junction is fabricated from a semiconductor with band gap energy of2.8eV. Can it detect a radiation of wavelength 5,000nm?Why? |
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Answer» SOLUTION :`E=hv=(HC)/LAMDA=(6.6xx10^(-34)xx3xx10^8)/(5000xx10^(-9))=3.96xx10^(-9)J` `E=(3.96xx10^(-19)/(1.6xx10^(-19))=0.248eV` Here `EltEg`(E=0.248eV and Eg=2.8eV),so diode cannot DETECT that photon. |
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| 42. |
What is the nature of nuclear force? |
| Answer» Solution :The NUCLEAR force is highly complex in nature and are very different from ORDINARY FORCES .The strong nuclear force between nucleons in a nucleus is APPROXIMATELY about 100 time as coulomb.s repulsive force between PROTONS. | |
| 43. |
A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction |
| Answer» Answer :B | |
| 44. |
A body is projected upwards with a velocity of 19.6 m/s. Find Greatest height it rises. |
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Answer» SOLUTION :u-19.6 m/s, G = `-9.8m/s^2` and v=0 (a) If S is the maximum height REACHED, then `2gs = v^2-u^2` `0^2-(19.6)^2 = -2x9.8 s or S = 19.6 m. |
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| 45. |
The acceleration of a particle is increasing linearly with time t as bt.The particle starts from origin with an initial velocity V_(0).The distance travelled by the particle in time t will be |
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Answer» `V_(0)t+(1)/(3)BT^(2)` |
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| 46. |
A moving coil galvanometer has a resistance of 9.8 Omega and gives a full scale deflection when a current of 10 mA is passed through it. The value of the shunt required to convert it into a milliameter to measure currents upto 500 mA is |
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Answer» `0.02 OMEGA` |
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| 47. |
Three containers C_(1), C_(2) and C_(3) have water at different temperatures. The table below shows the final temperature T when different amounts water (given in liters) are taken from each container and mixed (assume no loss of heat during the process) The value of theta (in .^(@)C to the nearest integer) is ------------ |
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Answer» |
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| 48. |
A point object is placed on the axis in front of a convex lens of focal length 30 cm. The distance of the object from the lens is 45 cm. If the lens is moved away from object by 1 mm and also the object is moved apart from the line joining the object and optical centre by 1 mm the displacement of the image has magnitude- |
| Answer» Answer :A | |
| 49. |
The collector resistance and the input resistance of a CE amplifier are respectively 10 kOmega and 2 k Omega . If betaof the transistor is 49 , the voltage gain of the amplifier is |
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Answer» A) 125 |
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| 50. |
Name the phenomenon which proves transverse wave nature of light. Give two uses of the devices whose functioning is based on this phenomenon. |
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Answer» Solution :POLARIZATION. Uses : Polaroids can be used in sunglasses, window panes, PHOTOGRAPHIC CAMERAS, 3D movie cameras. |
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