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. |
If error in measuring diameter of a circle is 4 %, the error in the radius of the circle would beA. 0.02B. 0.08C. 0.04D. 0.01 |
|
Answer» Correct Answer - C `because D=2R:.(DeltaD)/(D)=(DeltaR)/(R)` |
|
| 2. |
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure isA. 0.01B. 0.02C. 0.06D. 0.08 |
|
Answer» Correct Answer - D `F=(P)/(A)=(P)/(l^(2))rArr(DeltaF)/(F)=(DeltaP)/(P)+2(Deltal)/(l)` |
|
| 3. |
If error in measuring diameter of a circle is 4 %, the error in circumference of the circle would be :-A. 0.02B. 0.08C. 0.04D. 0.01 |
|
Answer» Correct Answer - C Circumference `l=2piR=piDrArr l propD` |
|
| 4. |
A wire has a mass `0.3+-0.003g`, radius `0.5+-0.005mm` and length `6+-0.06cm`. The maximum percentage error in the measurement of its density isA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - D `rho=(m)/(pir^(2)l)rArr(Deltarho)/(rho)=(Deltam)/(m)+2(Deltar)/(r)+(Deltal)/(l)rArr(Delta rho)/(rho)xx100` `=(0.003)/(0.3)xx100+2x(0.005)/(0.5)+(0.06)/(6)xx100=4%` |
|
| 5. |
Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in d remains constantB. the absolute error in d increasesC. the fractional error in d remains constantD. the fractional error in d decreases |
|
Answer» Correct Answer - D `d = (lamda)/(2 sin theta)` `delta(d)=((lamda)/(2sin^(2)theta))cos theta delta theta (delta theta ="constant")` as `theta` increasesm `(cos theta)/(sin^(2)theta)` decrease so Absolute error `|delta(d)|` decreases Also fractional error `=|(delta(d))/(d)|=(((lamdacos theta)/(2 sin^(2)theta))deltatheta)/((lamda)/(2 sin theta))=(cos theta)delta theta` as `theta` increases, `cot theta` decreases, so fractional error decreases. |
|
| 6. |
Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in d remains constantB. the absolute error in d increasesC. the fractional error in d remains constantD. the fractional error in d decreases |
|
Answer» Correct Answer - D `d = (lamda)/(2 sin theta)` `delta(d)=((lamda)/(2sin^(2)theta))cos theta delta theta (delta theta ="constant")` as `theta` increasesm `(cos theta)/(sin^(2)theta)` decrease so Absolute error `|delta(d)|` decreases Also fractional error `=|(delta(d))/(d)|=(((lamdacos theta)/(2 sin^(2)theta))deltatheta)/((lamda)/(2 sin theta))=(cos theta)delta theta` as `theta` increases, `cot theta` decreases, so fractional error decreases. |
|
| 7. |
A student measures the time period of `100` ocillations of a simple pendulum four times. The data set is `90 s`, 91 s, 95 s, and 92 s`. If the minimum division in the measuring clock is `1 s`, then the reported men time should be:A. `92 pm 3s`B. `92 pm 2s`C. `92 pm 5.0 s`D. `92 pm 1.8 s` |
|
Answer» Correct Answer - B `T_(AV)=92s` `(|DeltaT|)_("mean")=1.5s` since uncertainity is 1.5 s so digit 2 in 92 uncertain so reported mean time should be `92 pm 2` |
|
| 8. |
in an experiment the angles are required to be using an instrument, `29` divisions of the main scale exactly coincide with the `30` divisions of the vernier scale. If the sallest division of the main scale is half- a degree `(= 0.5^(@)`, then the least count of the instrument is :A. One degreeB. Half degreeC. One minuteD. Half-minute |
|
Answer» Correct Answer - C Least count of vernier callipers L.C = 1 MSD - 1 VSD But here `29 MSD = 30 VSD rArr 1 VSD = (29)/(30)MSD` `rArr L.C=1MSD-(29)/(30)MSD=` `(1)/(30)MSD=(1)/(30)xx0.5^(@)=((1)/(60))^(@)=1` minute |
|
| 9. |
A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : `58.5 degree` Vernier scale reading : `09` divisions Given that `1` division on main scale correspods 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:A. 59 degreeB. 58.59 degreeC. 58.77 degreeD. 58.65 degree |
|
Answer» Correct Answer - D `1 VSD = (29)/(30)xx0.5^(@) = ((29)/(60))^(@)` and `1 MSD = ((1)/(2))^(@)=((30)/(60))^(@)` Least count `= 1 MSD - 1 VSD = ((1)/(60))^(@)` Reading `= 58.5^(@)+9x((1)/(60))^(@)=58.65^(@)`. |
|
| 10. |
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. `Delta L = 0.5, Delta T - 0.1, n = 20`B. `Delta L = 0.5, Delta T = 0.1,n = 50`C. `DeltaL = 0.5, Delta T = 0.01, n = 20`D. `Delta L= 0.1, Delta T = 0.05, n = 50` |
|
Answer» Correct Answer - D `g=(4pi^(2)l)/(T^(2))rArr(Deltag)/(g)=(Deltal)/(l)+(2 DeltaT)/(T)`Here `T = 2sec` For (A) `(Deltag)/(g)=(0.1)/(1)+(0.2(0.1))/(2) = 0.6` For (B) `(Deltag)/(g)=0.5+(0.2)/(2)=0.6` For (C) `(Deltag)/(g)=0.5+(0.02)/(2)=0.51` For (D) `(Deltag")/(g)=0.1+(0.1)/(2)=0.15` `rArr(Deltag)/(g)` minimum for (D). Also number of observations are maximum in (D) |
|
| 11. |
The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?A. 0.01B. 0.05C. 0.02D. 0.03 |
|
Answer» Correct Answer - D `d_(1)=0.5+8xx0.01+0.03` `d_(2)=0.5+4xx0.01+0.03` `d_(3)=0.5+6xx0.01+0.03` `d_(n)=(d_(1)+d_(2)+d_(3))/(3)=0.5+0.03+(18xx0.01)/(3)` `-0.59` |
|
| 12. |
Internal micrometer is a measuring intrument used to measure internal diameter (ID) of a large cylinder bore with high accuracy. Construction is shown in figure. There is one fixed rod B (to the right in figure) and one moved rod A (to the left in figure). It is based on the particle of advancement of a screw when it is rotated in a nut with internal threads. Main scale reading can be directly seen on the hub which is fixed with respect to rod B. When the cap is rotated rod A moves in or cut depending on direction of rotation. The circular scale reading is seen by checking which division of circular scale coincide with the references line This is to be multiplied by LC to get circular scale reading. Least count = value of 1 circular scale division `= ("pitch")/("number of division on circular scale")` Length of rod A is chosen to match the `ID(PQ)` to be measured. Zero error is checked by taking reading between standard blocks fixed at normal value of `ID` to be measured. Zero error is positive if cap end is one the right of the main scale and negative it is on the left side. In the above instrument, while measuring an internal diameter. `ID` is set of 321 mm with no zero error. It cap end is after `n^(th)` division adn `17^(th)` division of main scale coincides with the reference line, the `ID` is -A. 321.717 mmB. 321.87 mmC. 328.17 mmD. 324.67 mm |
|
Answer» Correct Answer - D `IR=321+7(0.5)+17(0.01)=324.67mm` |
|
| 13. |
Internal micrometer is a measuring intrument used to measure internal diameter (ID) of a large cylinder bore with high accuracy. Construction is shown in figure. There is one fixed rod B (to the right in figure) and one moved rod A (to the left in figure). It is based on the particle of advancement of a screw when it is rotated in a nut with internal threads. Main scale reading can be directly seen on the hub which is fixed with respect to rod B. When the cap is rotated rod A moves in or cut depending on direction of rotation. The circular scale reading is seen by checking which division of circular scale coincide with the references line This is to be multiplied by LC to get circular scale reading. Least count = value of 1 circular scale division `= ("pitch")/("number of division on circular scale")` Length of rod A is chosen to match the `ID(PQ)` to be measured. Zero error is checked by taking reading between standard blocks fixed at normal value of `ID` to be measured. Zero error is positive if cap end is one the right of the main scale and negative it is on the left side. During zero setting of the above instrument, the end of the cap is on left side of the zero of main scale (i.e. zero of main scale is not visible) and `41^(th)` division of circular scale coincides with the reference line, the zero error is -A. `-0.09 mm`B. `+0.41 mm`C. `-0.41 mm`D. `+0.09 mm` |
|
Answer» Correct Answer - A Zero error `=(50 - 41)(0.01)=0.09 mm` |
|
| 14. |
Consider the MB shown in the diagram, let the resistance X have temperature coefficient `alpha_(1)` and the resistance from the RB have the temperature coefficient `alpha_(2)`. Let the reading of the meter scale be 10 cm from the LHS. if the temperature of the two resistance increase by small temperature `DeltaT` then what is the shift in the position of the null point ? Neglect all the other changes in the bridge due to temperature rise.A. `9(alpha_(1)- alpha_(2))Delta T`B. `9(alpha_(1)+alpha_(2))Delta T`C. `1//9(alpha_(1) + alpha_(2))Delta T`D. `1//9(alpha_(1)-alpha_(2))Delta T` |
|
Answer» Correct Answer - A `(X)/(10)=(90)/(90)rArr X=10Omega` `rArr (10(1+alpha_(1)DeltaT))/(10+Deltal)=(90(1+alpha_(2)DeltaT))/(90-Deltal)` `rArr(1+alpha_(1)DeltaT)(1+(Deltal)/(10))^(-1)=(1+alpha_(2)DeltaT)(1-(Deltal)/(90))` `rArr (1+alpha_(1)DeltaT)(1-(DeltaT)/(10))=(1+alpha_(2)DeltaT)(1+(Deltal)/(90))` `rArr1+alpha_(1)DeltaT-(Deltal)/(10)=1+alpha_(2)DeltaT+(Deltal)/(90)` `rArr(alpha_(1)-alpha_(2))DeltaT=(Deltal)/(9)rArrDeltal=9(alpha_(1)-alpha_(2))DeltaT` |
|
| 15. |
In a vernier callipers, `N` divisions of the main scale coincide with `N + m` divisions of the vernier scale. what is the value of `m` for which the instrument has minimum least count.A. 1B. NC. infinityD. `N//2` |
|
Answer» Correct Answer - A `L.C = 1 MSD - 1 VSD` but `(N) MSD = (N + M)(VSD)` `rArrL.C=1MSD-((N)/(N+m))MSD` `= ((m)/(N+m))=((1)/((N)/(m)+1))MSD` For minimum least count, m should be minimum so `m =1` |
|
| 16. |
In a vernier calliper, N divisions of vernier scale coincide with (N-1) divisions of main scale (in which division represent 1mm). The least count of the instrument in cm. should beA. NB. `N-1`C. `(1)/(10 N)`D. `(1)/(N-1)` |
|
Answer» Correct Answer - C Least count = 1 MSD - 1 VSD `(N -1)MSD=N(VSD) rArr VSD = (1-(1)/(N))MSD` Least count `= 1MSD - (1-(1)/(N))MSD` `= ((1)/(N))(1mm)=(1)/(10 N)cm` |
|
| 17. |
If in an experiment for determination of velocity of sound by resonance tube method using a tuning fork of `512 Hz`, first resonance was observed at` 30.7 cm` and second was obtained at `63.2 cm`, then maximum possible error in velocity of sound is (consider actual speed of sound in air is `332m//sA. 256 cm/secB. 92 cm/secC. 102.4 cm/secD. 204.8 cm/sec |
|
Answer» Correct Answer - D `v=flamda=2f(l_(2)-l_(1))` `=512xx2(63.2-30.7)xx10^(-2)=332.8ms^(-1)` `(Deltav)/(v)=((Deltalamda)/(lamda))=((Deltal_(2)+Deltal_(1))/(l_(2)-l_(1)))` `rArr Deltav=v((Deltal_(2)+Deltal_(1))/(l_(2)-l_(1)))=(332.8)((0.1+0.1)/(32.5))=2.048 m//s`. |
|
| 18. |
A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,A. The intensity of the sound heard at the first resonance was more than that at the second resonanceB. the prongs of the tuning fork were kept in a horizontal plane above the resonance tubeC. the amplitude of vibration of the ends of the prongs is typically around 1 cmD. the length of the air-column at the first resonance was some what shorter than 1/4th of the wavelength of the sound in air. |
|
Answer» Correct Answer - A::D For a longer air column, absorption of energy is more. Due to end correction `l + e = (lamda)/(4)` |
|
| 19. |
In a vernier callipers, n divisions of its main scale match with (n+1) divisions on its vernier scale. Each division of the main scale is a units. Using the vernier principle, calculate its least count. |
|
Answer» Correct Answer - `(a)/(n+1)` Least count = 1 MSD - 1 VSD Here `n(MSD) = (n+1)(VSD) rArr 1 VSD = (na)/(n+1)` `rArr` Least count `= a- (na)/(n+1)=(a)/(n+1)` |
|
| 20. |
In the experiment to determine the speed of sound using a resonance column -A. prongs of the tuning fork are kept in a vertical planeB. prongs of the tuning fork are kept in a horizontal planeC. In one of the two resonance observed, the length of the resonating air column is close to the wavelength of sound in airD. In one of the two resonance obsvered, the length of the resonating air column is close to half of the wavelength of sound in air. |
| Answer» Correct Answer - A | |
| 21. |
In a meter bridge set up, which of the following should be the properties of the one meter long wire?A. High resistivity and low temperature coefficientB. Low resistance and low temperature coefficientC. Low resistivity and high temperature coefficientD. High resistivity and high temperature coefficient |
| Answer» Correct Answer - A | |
| 22. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. B and CB. C and DC. A and DD. `B_(1)` and `C_(1)` |
| Answer» Correct Answer - C | |
| 23. |
The relative error in resistivity of a material where resistance = 1.05 + 0.01 W diameter = 0.60 + 0.01 mm length = 75.3 + 0.1 cm isA. `0.04`B. `0.40`C. `0.08`D. `0.01` |
| Answer» Correct Answer - A | |
| 24. |
The diameter of a cylinder is measured using a vernier callipers with no zero error . It is found that the zero of the vernier scale lies between `5.10 and 5.15 cm` of the main scale . The `24 th` division of the vernier scale exactly coincides with one of the main scale divisions . The diameter of the cylinder isA. 5.112 cmB. 5.124 cmC. 5.136 cmD. 5.148 cm |
|
Answer» Correct Answer - B 50 divisions = 2.45 cm `rArr 1` division `= (2.45)/(50)=0.049 cm` `rArr` least count `= 1 MSD - 1 VSD = 0.05 - 0.049 = 0.001 cm` So vernier reading `= 0.001 xx 24 = 0.024 cm` Therefore diameter of cylinder `= 5.10 + 0.024 = 5.124 cm` |
|
| 25. |
The following observations were taken for dtermining the surface tension of water by capillary tube method: diameter of capillary , `D = 1.25 xx 10^(-2) m` and rise of water in capillary , `h = 1.45 xx 10^(-2) m`. Taking ` g = 9.80 m s^(-2)` and using the relation `T = ( r gh//2) xx 10^(3) N m^(-1)`, what is the possible error in measurement of surface tension `T`? (a) `2.4%` (b) `15%` (c) `1.6%` (d) 0.15%`A. 0.0016B. 0.016C. 0.16D. 0.024 |
|
Answer» Correct Answer - B `T=(rhg)/(2)rArr(DeltaT)/(T)=(Deltar)/(T)=(Deltah)/(h)+(Deltag)/(g)` `(DeltaT)/(T)xx100=((0.01xx10^(-2))/(1.25xx10^(-2))+(0.01xx10^(-2))/(1.45xx10^(-2))+(0.01)/(9.80))xx100` `=0.80+0.69+0.10=1.59%=1.6%` |
|
| 26. |
The radius of a disc is 1.2 cm. Its area according to idea of significant figures, will be given by :-A. `4.5216 cm^(2)`B. `4.521 cm^(2)`C. `4.52 cm^(2)`D. `4.5 cm^(2)` |
|
Answer» Correct Answer - A Area of disk `=piR^(2)=(3.14)(1.2)^(2)=4.5216=4.5cm^(2)` |
|
| 27. |
The volume of a sphere is `.176 m^3` What will be the volume of 25 such spheres taking into account the significant figures.A. `0.44 xx 10^(2) cm^(3)`B. `44.0 cm^(3)`C. `44 cm^(3)`D. `44.00 cm^(2)` |
|
Answer» Correct Answer - D Volume of 25 spheres `=25 xx 1.76 = 44.00` |
|
| 28. |
What is the fractional error in g calculated from `T = 2 pi sqrt((l)/(g))` ? Given that fractional error in `T` and `l` are `pm x` and `pm y` respectively.A. x + yB. x - yC. 2x +yD. 2x - y |
|
Answer» Correct Answer - C `T=2pisqrt((l)/(g))rArrg=(4pi^(2)l)/(T^(2))` `rArr(Deltag)/(g)=(Deltal)/(l)+2(DeltaT)/(T)=2x+y` |
|
| 29. |
Graph of position of image vs position of point object from a convex lens is shown. Then, focal length of the lens is A. `0.50 pm 0.05 cm`B. `0.50 pm 0.10 cm`C. `5.00 pm 0.05 cm`D. `5.00 pm 0.10 cm` |
|
Answer» Correct Answer - C From `(1)/(f)=(1)/(v)-(1)/(u)=(1)/(10)-(1)/(-10)=(1)/(5)rArr f=5cm` From graph `Delta u = 0.1 cm, Delta v = 0.1 cm` But `(Deltaf)/(f^(2))=(Deltav)/(v^(2))+(Deltau)/(u^(2))` so `Deltaf=((0.1)/(100)+(0.1)/(100))(25)=0.05` `rArrf = (5 pm 0.05)cm` |
|
| 30. |
A cube has a side of length `1.2 xx 10^(-2)m`. Calculate its volume.A. `1.7 xx 10^(-6) m^(3)`B. `1.70 xx 10^(-6) m^(3)`C. `1.70 xx 10^(-7) m^(3)`D. `1.78 xx 10^(-6) m^(3)` |
|
Answer» Correct Answer - A `V=a^(3)=(1.2xx10^(-2))^(3)=1.728xx10^(-6)m^(3)` `=1.7 xx10^(-6)m^(3)` |
|
| 31. |
A physical quantity `X` is represented by `X = (M^(x) L^(-y) T^(-z)`. The maximum percantage errors in the measurement of `M , L , and T`, respectively , are `a% , b% and c%`. The maximum percentage error in the measurement of `X` will beA. `(alpha a + beta b + gamma c) %`B. `(alpha a- beta b + gamma c)%`C. `(alpha a - beta b- gamma c)%`D. None of these |
|
Answer» Correct Answer - A `X=M^(a)L^(b)T^(c)` `rArr(DeltaX)/(X)=a((DeltaM)/(M))+b((DeltaL)/(L))+c((DeltaT)/(T))=a alpha+b beta+c gamma` |
|
| 32. |
The area of a rectangle of size `1.23 xx 2.345 cm` isA. `2.88 cm^(2)`B. `2.884 cm^(2)`C. `2.9 cm^(2)`D. `2.88435 cm^(2)` |
|
Answer» Correct Answer - A Area `= 1.2xx2.345=2.884=2.9cm^(2)` |
|
| 33. |
What is vernier constant?A. It is the value of the one main scale division by the total number of divisions on the main scaleB. It is the value of one vernier scale division divided by the total number of division on the vernier scaleC. It is the difference between value of one main scale division and one vernier scale divisionD. It is not the least count of vernier scale |
|
Answer» Correct Answer - C 1 MSD - 1 VSD = Vernier Constant |
|
| 34. |
The period of oscillation of a simple pendulum in the experiment is recorded as `2.63 s , 2.56 s , 2.42 s , 2.71 s , and 2.80 s`. Find the average absolute error.A. 0.1 sB. 0.11 sC. 0.01 sD. 1.0 s |
|
Answer» Correct Answer - B Average time period `=(2.63+2.56+2.42+2.71+2.80)/(5)` `(13.12)/(5)=1.624=2.62s` Average absoulte error `=(0.01+0.06+0.20+0.09+0.18)/(5)=(0.54)/(5)` `=0.108=0.11s`. |
|