InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1. |
A vernier calipers has `1mm `marks on the main scale. It has `20` equal divisions on the Verier scale which match with `16` main scale divisions. For this Vernier calipers, the least count isA. 0.02 mmB. 0.05 mmC. 0.1 mmD. 0.2 mm |
|
Answer» Correct Answer - d `16 MSD=20 VSD` `1 VSD=4//5 MSD` `LC=1 MSD - 1VSD` `=(1-4/5) MSD=(1-4/5) (1 mm)=0.2 mm` |
|
| 2. |
A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading =58.5 degree Vernier scale reading =0.9 divisions Given that 1 divisions on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. the angle of the prism from the above data is :A. `58.59^(@)`B. `58.77^(@)`C. `58.65^(@)`D. `59^(@)` |
|
Answer» Correct Answer - c `L.C. =0.5/30=(1/60)^(@)` `:.` Reading `=58.5^(@)+(1/60)^(@)xx9=58.65^(@)` |
|
| 3. |
In `u-v` method for finding the focal length of a concave mirror. The mirror is fixed at position `A` marked `20.0cm` on an optical bench and an object needle is placed at position `B` marked `45.0cm` on an optical banch. For no parallax between object needle and image needle the image needle at position `C 57.5cm` on optical bench. Then percentage error in the measurement of focal length of the mirror is |
|
Answer» Distance of the object, `u=45.0-20.0` `=25.0 cm` Distance of the image, `v=57.5-20.0=37.5 cm` Using mirror formula, `f=(uv)/(u+v)=(25xx37.5)/(25+37.5)` `=(25xx37.5)/(62.5)=15.0 cm` The maximum permissible error in focal length is given by `(Delta f)/f=(Deltau)/u+(Deltav)/v+(Delta(u+v))/((u+v))` `(Delta f)/f=(Delta u)/u+(Delta v)/v+(Deltau)/((u+v))+(Deltav)/((u+v))` `=0.2/25.0+0.2/37.5+0.2/62.5+0.2/62.5` `=0.008+0.005+0.003+0.003=0.019` `:. Deltaf=0.019xxf=0.019xx15.0=0.285 cm cong 0.3 cm` `:.` The focal length of concave mirror is `(15.0 pm 0.3) cm`. |
|
| 4. |
A student performs an experiment for determination of ` g [ = ( 4 pi^(2) L)/(T^(2)) ] , L ~~ 1m` , and he commits an error of `Delta L`. For `T` he takes the time of `n` oscillations with the stop watch of least count ` Delta T` . For which of the following data , the measurement of `g` will be most accurate ?A. `DeltaL=0.5, DeltaT=0.1, n=20`B. `DeltaL=0.5, DeltaT=0.1, n=50`C. `DeltaL=0.5, DeltaT=0.01, n=20`D. `DeltaL=0.5, DeltaT=0.05, n=50` |
|
Answer» Correct Answer - d `(Deltag)/g=(Deltal)/(l)+2(DeltaT)/(n T)` (`Deltal` and `DeltaT` are least, and the number of resdings are maximum.) |
|
| 5. |
In determination of value of acceleration due to gravity (g) by simple pendulum, the time period is measured by a topwatch (L.C. = 0.5 second) and length of the thread is measured with metre scale (L.C. 0.1 cm) and diametre of bob is measured with vernier callipers (L.C. = 0.1 cm). The following observations are recorded : Length of the thread `=105.3 cm` Diameter of the bob `=2.45 cm` Time period `=2.07 s` Number of oscillations `=10` Calculate the value of g, estimate error and write the result in proper significant figures. |
|
Answer» The time period of simple pendulum is given by `T=2pi sqrt(L/g)` Here, `L=l+r=105.3+1.225=106.525 cong 106.5 cm` or `g=(4pi^(2)L)/T^(2)` `=(4xx9.87xx106.5)/((2.07)^(2))=982 cm//s^(2)` The error in g is given by `(Deltag)/(g)=(DeltaL)/L+2 (Delta T)/(T)=(Delta(l+r))/((l+r))+2(Delta T)/(T)` `=((0.1+0.005))/((105.3+1.225))+2xx0.5/(2.07xx10)` `=0.00098+0.048 cong 0.048 cong 5%` `:. Deltag=0.048xxg=0.048xx982=47 cm//s^(2)` Thus, the value of g is `(982 pm 47) cm//s^(2)`. Note : Let us see the result if we increase the number of oscillations from 10 to 20. `(Delta g)/g=0.00098xx(2xx0.5)/(2.07xx20) cong 0.024 cong 2%` Thus, the relative error in g falls down from `5 %` to `2 %` by increasing number of oscillations from 10 to 20. |
|
| 6. |
A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?A. Error `DeltaT` in measuring T, the time period, is 0.05 sB. Error `DeltaT` in measuring T, the time period is, 1 sC. Percentage error in the determination of g is 5%D. Percentage error in the determination of g is 2.5 % |
|
Answer» Correct Answer - a, c `T=40/20=2s` `(DeltaT)/T=(Deltat)/t=1/40` `rArr DeltaT=T/40=2/40=0.05 s` `g=(4pi^(2)Ln^(2))/t^(2)" "`where `t=nT` `(Deltag)/g=(2 Delta t)/t` `rArr %` error `=(2 Delta t)/txx100=5%` |
|
| 7. |
Student `I ,II , and III` perform an experiment for measuring the acceleration due to gravity `(g) ` usinf a simple pendulum. They use lengths of the pendulum and // or record time for different number of oscillations . The observations are shown in the following table . Least count for length ` = 0.1 cm` `{:(underset(III)underset(II)underset(I)underset()underset()("Student"),underset(20.0)underset(64.0)underset(64.0)underset((cm))underset("Pendulam")("Length of "),underset(4)underset(4)underset(8)underset((n))underset("n Oscillation")("Number of"),underset(9.0)underset(16.0)underset(16.0)underset((s))underset("Period")("Time")):}` Least count for time `= 0.1 s`. If `E_(I), E_(II)` , and `E_(III)` are the percentage errors in `g` , i.,e., `((Delta g)/( g) xx 100)` for students I,II , and III, respectively , thenA. `E_(I)=0`B. `E_(I)` is minimumC. `E_(I)=E_(II)`D. `E_(II)` is maximum |
|
Answer» Correct Answer - b `g=4pi^(2)(l/T^(2))rArr (Deltag)/g=(Deltal)/l+2 (Delta T)/T` `rArr E=((Deltal)/l+2(DeltaT)/T)xx100` The greater the valu of T, the lesser the error. Hence fractional error in the first observation is minimum. |
|
| 8. |
A quantity X is given by `in_(0) L(DeltaV)/(Delta t)`, where `in_(0)` is the permittivity of free space, L is a length, `DeltaV` is a potential difference and `Delta t` is a time interval. The dimensional formula for X is the same as that of - |
|
Answer» As capacity of parallel plate capacitor if given by `C=(epsi_(0)A)/(d)` which has the same dimension as of `epsi_(0)L`, `:. X=epsi_(0)L. (DeltaV)/(Delta t)=(C Delta V)/(Delta t)=("charge")/("time")=` current Thus quantity X has the dimension of current. |
|
| 9. |
The dimensions of `1/2 in_(0) E^(2)` (`in_(0)`: permittivity of free space, E: electric field) is-A. `[MLT^(-1)]`B. `[ML^(2)T^(-2)]`C. `[ML^(-1)T^(-2)]`D. `[ML^(2)T^(-1)]` |
|
Answer» Correct Answer - c Here, `(1//2)epsi_(0)E_(2)` represents energy per unit volume. `|epsi_(0)|[E^(2)]=("Energy")/("Volume")=([ML^(2)T^(-2)])/([L^(3)])=[ML^(-1)T^(-2)]` |
|
| 10. |
The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2. A. 1.2 mmB. 1.25 mmC. 2.20 mmD. 2.25 mm |
|
Answer» Correct Answer - a Least count `=("Pitch")/("Number of division on circuit scale")=0.5/50` `=0.01 mm` Now, diameter of ball `=(2xx0.5)+(2.5-5)(0.01)=1.2 mm` |
|
| 11. |
The time dependence of a physical quantity P is given by `P=P_(0) exp (-alpha t^(2))`, where `alpha` is a constant and t is time. The constant `alpha`A. is dimensionlessB. has dimensions `[T^(-2)]`C. has dimensions `[T^(2)]`D. has dimensions of P |
|
Answer» Correct Answer - b `[alphat^(2)]=` dimensionless `[alpha]=[T^(-2)]` |
|
| 12. |
Turpentine oil is flowing through a tube of length `L` and radius `r`. The pressure difference between the two ends of the tube is `p` , the viscosity of the coil is given by `eta = (p (r^(2) - x^(2)))/(4 vL)`, where `v` is the velocity of oil at a distance `x` from the axis of the tube. From this relation, the dimensions of viscosity `eta` areA. `[M^(0)L^(0)T^(0)]`B. `[MLT^(-1)]`C. `[ML^(2)T^(-2)]`D. `[ML^(-1)T^(-1)]` |
|
Answer» Correct Answer - b `eta=(p(r^(2)-x^(2)))/(4vl)=([ML^(-1)T^(-2)][L^(2)])/([LT^(-1)][L])=[ML^(-1)T^(-1)]` |
|
| 13. |
The ratio of magnitudes of unit for viscosity in SI to that in CGS is x. Then the value of `(x+2)//2` is : |
|
Answer» Correct Answer - 6 `eta =[ML^(-1)T^(-1)]` `:. x= ((1 kg)/(1 g))^(1)((1 m)/(1 cm))^(-1)((1 s)/(1 s))^(-1) =10` `rArr " "(x+2)//2=6` |
|
| 14. |
The units and dimensions of impendance are :A. mho, `[ML^(2)T^(-1)Q^(-2)]`B. ohm, `[ML^(2)T^(-3)Q^(-2)]`C. ohm, `[ML^(2)T^(-2)Q^(-1)]`D. ohm, `[MLT^(-1) Q^(-1)]` |
|
Answer» Correct Answer - b Unit of imedance is ohm `(Omega) [ML^(2)T^(-3) A^(-2)]`. |
|
| 15. |
The dimensional formula for force per unit linear mass density of wire is the same as that for :A. velocityB. accelerationC. latent heatD. specific heat |
|
Answer» Correct Answer - c `F/(m//l)=(Fl)/m=([MLT^(-2)][L])/([M])=[L^(2)T^(-2)]` Lentent heat, `L=Q/m=([ML^(2)T^(-2)])/([M])=[L^(2)T^(-2)]` |
|
| 16. |
Match the physical quantities in Column -I with their dimensional formula in Column -II |
|
Answer» Correct Answer - `a rarr r; b rarr s; c rarr p; d rarr q` (a) Angular momentum `=I omega=[ML^(2)T^(-1)]` `:. (a) rarr (r)` (b) Coefficient of viscosity `=[ML^(1)T^(-1)]` `:. (b) rarr (s)` (c) Torque `= FR_(bot)=[ML^(2)T^(-2)]` `:. (c) rarr (p)` (d) Angular acceleration `=(d omega)/(dt)=([T^(-1)])/([T])=[M^(0)L^(0)T^(-2)]` `:. (d) rarr (q)` |
|
| 17. |
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurment of the current and the voltage difference are `3%` each, then error in the value of resistance of the wire iS :A. `6 %`B. zeroC. `1 %`D. `3 %` |
|
Answer» Correct Answer - a `R=V/I` `% (Delta R)/(R)= % (DeltaV)/(V) + % (DeltaI)/(I)` `=3%+3%=6%` |
|
| 18. |
The SI unit of electron mobility is :A. `m^(2) s^(-2) V^(-1)`B. `ms V^(-1)`C. `ms^(-1) V`D. `m^(2)s^(-2)V^(-2)` |
|
Answer» Correct Answer - a Electron mobility `=("Drift velocity")/("Electric field")` `mu_(e)=v/E=(ms^(-1))/(Vm^(-1))=m^(2)s^(-2)V^(-1)` |
|
| 19. |
The unit of charge is :A. coulombB. franklineC. faradayD. amphere `xx` sec |
|
Answer» Correct Answer - a, b, c, d `q=ixxt` Hence unit of change is coulomb and amp `xx` sec Another unit is frankline and faraday. |
|
| 20. |
Which of the following is not unit of length ?A. micronB. light yearC. angstromD. radian |
| Answer» Correct Answer - d | |
| 21. |
An atmosphere :A. is a unit of pressureB. is a unit of forceC. gives an idea of the composition of airD. is the height above which there is no atmosphere |
| Answer» Correct Answer - a | |
| 22. |
What is the SI unit of force?A. wattB. dyneC. newtonD. poundal |
| Answer» Correct Answer - c | |
| 23. |
Which of the following is not unit of time ?A. Solar yearB. Leap yearC. Light yearD. Tropical year |
| Answer» Correct Answer - c | |
| 24. |
`N kg^(-1)` is the unit of :A. velocityB. forceC. accelerationD. none of these |
| Answer» Correct Answer - c | |
| 25. |
Pressure is defined as :A. momentum per unit areaB. momentum per unit area per unit timeC. momentum per unit volumeD. energy per unit volume |
|
Answer» Correct Answer - b, d Reynold number `=(rho vD)/eta=([ML^(-3)][LT^(-1)][L])/([ML^(-1)T^(-1)])` `=[M^(0)L^(0)T^(0)]` Coefficient of friction `=f/R=([ML^(2)T^(-2)])/([ML^(2)T^(-2)])=[M^(0)L^(0)T^(0)]` (b) Latent heat, `L=Q/m=([ML^(2)T^(-2)])/([M])=[L^(2)T^(-2)]` Gravitational potential `=U/m=([ML^(2)T^(-2)])/([M])=[L^(2)T^(-2)]` (c) Curie `=3.7xx10^(7)` disintegration/sec `=[T^(-1)]` Frequency `=(1)/("Time period")=[T^(-1)]` |
|
| 26. |
SI unit of pressure isA. atmosphereB. barC. pascalD. mm of Hg |
| Answer» Correct Answer - c | |
| 27. |
One sec is defined to be equal to :A. 1650763/73 periods of krypton clockB. 652189.63 periods of krypton clockC. 1650763.73 periods of cesium clockD. 9192631770 periods of cesium clock |
| Answer» Correct Answer - d | |
| 28. |
The unit of impulse is the same as that ofA. energyB. forceC. angular momentumD. linear momentum |
| Answer» Correct Answer - d | |
| 29. |
Dyne-sec stands for the unit of :A. forceB. workC. momentumD. angular momentum |
| Answer» Correct Answer - c | |
| 30. |
The expression `[ML^(2)T^(-2)]` representsA. PowerB. kinetic energyC. momentumD. pressure |
| Answer» Correct Answer - b | |
| 31. |
Joule-second is the unit ofA. energyB. momentumC. angular momentumD. power |
| Answer» Correct Answer - c | |
| 32. |
If `P` represents radiation pressure , `C` represents the speed of light , and `Q` represents radiation energy striking a unit area per second , then non - zero integers `x, y, z` such that `P^(x) Q^(y) C^(z)` is dimensionless , find the values of `x, y , and z`.A. x = 1, y = 1, z = 1B. x = -1, y = 1, z = 1C. x = 1, y = -1, z = 1D. x = 1, y = 1, z = -1 |
|
Answer» Correct Answer - c `M^(0)L^(0)T^(0)=[P^(x)S^(y)C^(2)]` `[M^(x+y)L^(-x+2)T^(-2x-3 y-z)]=M^(0)L^(0)T^(0)` `x+y=0" "-x+2=0` `-2x-3y-2=0` `x=1" "," "y=-1` and `2 = 1` |
|
| 33. |
A system has basic dimensions as density [D], velocity [V] and area [A]. The dimensional representation of force in this system is :A. `[AV^(2)D]`B. `[A^(2)VD]`C. `[AVD^(2)]`D. `[A^(0)VD]` |
|
Answer» Correct Answer - a `F=D^(x)V^(y)A^(z)` Equate power of dimensions. |
|
| 34. |
Number of particles is given by `n=-D(n_(2)-n_(1))/(x_(2)-x_(1))` crossing a unit area perpendicular to `X`-axis in unit time, where `n_(1)` and `n_(2)` are number of particles per unit volume for the value of `x` meant to `x_(2)` and `x_(1)`. Find dimensions of `D` called as diffusion constantA. `[L^(2)T^(-1)]`B. `[LT^(-2)]`C. `[L^(2)T^(4)]`D. `[LT^(-3)]` |
|
Answer» Correct Answer - a `N=-D ((n_(2)-n_(1)))/((x_(2)-x_(1)))` `1/([L^(2)T])=D(1//[L^(3)])/([L])` `D=[L^(2)T^(-1)]` |
|
| 35. |
The dimensional formula of magnetic flux density is :A. `[M^(1)L^(0)T^(-2)A^(-1)]`B. `[M^(1)L^(2)T^(-1)A^(-1)]`C. `[M^(1)L^(2)T^(-1)A^(-2)]`D. `[M^(1)L^(2)T^(-2)A^(-1)]` |
|
Answer» Correct Answer - d Dimensional formula of magnetic flux is `[M^(1)L^(2)T^(-2) A^(-1)]` as `phi= BA` |
|
| 36. |
There is a concept that if unit of a physical quantity is large then the multiplier of the unit becomes less while expressing any fixed magnitude. If in a new system of unit, unit of time is 0.5 second and the unit of length is 4 metre. Answer the following questions. The value of speed of light becomes (c is intial value) :A. cB. `c//2`C. `c//4`D. `c//8` |
|
Answer» Correct Answer - b Since, `c=3xx10^(8) m//s` Hence, `3xx10^(8) m//s=n_(2) (8 m s^(-1))` Hence, `n_(2)=((3xx10^(8))/8)` |
|
| 37. |
There is a concept that if unit of a physical quantity is large then the multiplier of the unit becomes less while expressing any fixed magnitude. If in a new system of unit, unit of time is 0.5 second and the unit of length is 4 metre. Answer the following questions. If unit of time is doubled, the unit of length is halved and unit of mass is increased by a factor n such that a particular force retains its value in both th system of units then the value of n is :A. 8B. 6C. 4D. 2 |
|
Answer» Correct Answer - a Unit of force = kg `m s^(-2)` Hence, according to given condition `((nxxkgxxm)/(2xx4xxs^(2)))` Hence, n should be 8. So, that magnitude remains constant. |
|
| 38. |
Least count of a vernier callipers is `0.01 cm` When the two jaws of the instrument touch each other the `5th` division of the vernier scale coincide with a main scale division and the zero of the vernier scale lies to the left of the zero of the main scale. Furthermore while measuring the diameter of a sphere, the zero markof the vernier scale lies between `2.4 cm` and `2.5 cm` and the `6th` vernier division coincides with a main scale division. Calculate the diameter of the sphere. |
|
Answer» The instrument has a negative error, `e=(-5xx0.01)cm` or `e=-0.05 cm` Measured reading `=(2.4+6xx0.01)=2.46 cm` True reading = Measured reading `- e` `=2.46-(-0.05)` `=2.51 cm` Therefore, diameter of the sphere is `2.51` cm. |
|
| 39. |
Light year is :A. light emitted by the sum in one yearB. time taken by light to travel from the sum to the earthC. the distance travelled by light in one year, in free spaceD. the time taken by earth to go round the sun once |
| Answer» Correct Answer - c | |
| 40. |
The unit of atomic mass is :A. `O = 16.0000`B. `O^(16) = 16.0000`C. `C = 12.0000`D. `C^(12) = 12.0000` |
| Answer» Correct Answer - d | |
| 41. |
One nanometer is equal to :A. `10^(9) mm`B. `10^(-6) cm`C. `10^(-7) cm`D. `10^(-9) cm` |
| Answer» Correct Answer - c | |
| 42. |
Light year is a unit ofA. energyB. intensity of lightC. timeD. distance |
| Answer» Correct Answer - d | |
| 43. |
Express 1 micro in metre.A. `10^(-9) m`B. `10^(-12) m`C. `10^(-6) m`D. `10^(-5) m` |
| Answer» Correct Answer - c | |
| 44. |
How many wavelength of `Kr^(86)` are there is one metreA. `1553164.13`B. `1650763.73`C. `2348123.73`D. `652189.63` |
| Answer» Correct Answer - b | |
| 45. |
The dimensional formula for magnetic permeability `mu` is :A. `[MLT^(-2)A^(-2)]`B. `[M^(0)L^(-1)T]`C. `[M^(0)L^(2)T^(-1) A^(2)]`D. `[ML^(2)T^(-2)A^(-2)]` |
|
Answer» Correct Answer - a `(dF)/(dL)=mu_(0)/(4pi) (2i_(1)i_(2))/(d)` |
|
| 46. |
Which of the following is a unit of permeability ?A. H/mB. Wb/AmC. ohm `xx` s/mD. `V xx s//m^(2)` |
|
Answer» Correct Answer - a, b, c Permeability, `mu_(0)=((2pixxBxxr)/i)` Hence, unit of permeability is H/m and `Wb//m^(2)` |
|
| 47. |
Four lengths are measured as 18.425 cm, 7.21 cm, 5.04 cm and 10.3571 cm. taking significant figures into account, the sum of lengths should be :A. 41.0321 cmB. 41.03 cmC. 41.032 cmD. 41.0 cm |
|
Answer» Correct Answer - b The sum is expressed upto minimum number of digits after decimal point i.e., two digits in this case. |
|
| 48. |
Which of the following is the dimensional formula for capacitance `xx ("potential")^(2)`?A. `[ML^(2)T^(-1)]`B. `[ML^(2)T^(-2)]`C. `[ML^(-2)T^(-3)]`D. `[ML^(-1)T^(-2)]` |
|
Answer» Correct Answer - a Energy stored in a capacitor is `U=1/2 CV^(2)` `:.` Dimensional formula for `CV^(2)=[ML^(2)T^(-2)]` |
|
| 49. |
It is estimated that per minute each `cm^(2)` of earth receives about 2 calorie of heat energy from the sun. This constant is called solar constant S. Express solar constant in SI units. |
|
Answer» Given that `S = 2" cal/cm"^(2)` min But as 1 cal `=4.18 J, 1 cm=10^(-2) m` and 1 min = 60 s `S=(2xx4.18 J)/((10^(-2)m)^(2)(60s))` `~~1.4xx10^(3) J/(m^(2)s)` or `S=1.4xxkW//m^(2)` |
|
| 50. |
Given that mass of sun is `2 xx10^(30)` kg and radius of sun is `7 xx10^(8) m`. In what range do you expect the mass density of the sum to be ? Is it in the range of density of solid or liquid or gas ? Explain. |
|
Answer» Mass density of sun `=("Mass of sun")/("Volume of sun")` `=M/(4/3 pi R^(3))=(2xx10^(30)xx3xx7)/(4xx22xx(7xx10^(8))^(3))kg m^(-3)` `=1.39xx10^(3) kg m^(-3)` Thus, mass density of sun is in the range of density of solids (and not gases). At extremely high temperature of the sun (temp. of core `~10^(7) K` and surface temp.`cong 6000 K`), the gases matter exists in ionised from i.e. plasma. The high density of plasma is due to the gravitational attraction on outer layers due to inner layers of the plasma constituting the sun. |
|