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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. |
The angle of banking `theta` for a cyclist taking a turn on a curved road of radius r, with a velocity v is given by tan `theta=(v^(n))/(rg)`. What is the value of n?A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - c `tantheta=(v^(n))/(rg) therefore [tantheta]=[(v^(n))/(rg)]` The trignometrical ratio tan `theta` has no dimensions and is represented as `[L^(0)M^(0)T^(0)]` `[L^(0)M^(0)T^(0)]=([L^(1)T^(-1)]^(n))/([L^(1)L^(1)T^(-2)])=([L^(1)T^(-1)]^(n))/([L^(1)T^(-1)]^(2))` `therefore [L^(1)T^(-1)]^(n)=[L^(1)T^(-1)]^(2) therefore n=2` `therefore` The correct relation is tan `theta=(v^(2))/(rg)`. |
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| 2. |
To find the volume of an irregularly shaped wooden piece, a sinker made of steel (relative density = 7.85) having a mass of 157 g was used. If the level of water before and the after immersing the sinker-wood combination are 102 ml and 170 ml, respectively, find the volume of the wooden piece. |
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Answer» ` Volume= ("mass")/("density")` (ii) Will wooden piece float on water? The relative density of the sinker = 7.85 So, the density of the sinker = `7.85 g cm^(-3)` Given the mass of the sinker = 157 g. Then, the volume of the sinker =` (d)/(m)` Given the volume of (sinker + wooden piece), find the volume of the sinker. what is the change in the volume of water due to the immersion of sinker and the wooden piece? Then, find the volume of the wooden piece |
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| 3. |
A combined double screw gauge as shown in the figure is used as If the readings on the two scales with the tips of the screws touching each other are 2.97 mm and 6.04 mm when a rod is held between the screws readings are 12.55 mm and 5.96 mm what is the diameter of the ord ?A. 9.5 mmB. 9.6 mmC. 9.58 mmD. Not possible to determine |
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Answer» Correct Answer - D `{:(,,"Scale 1",,"Scale 2 ",,"Total"),("With rod ",,12.55,,5.96,,18.51),("Without rod ",,2.97,,6.04,,9.01),("Movement " ,,9.58 ,,-0.08,,=9.5"mm"):}` Diameter of the rod =9.5 mm |
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| 4. |
The least counts of a vernier calipers and a screw gauge are in the ratio of `5: 1` . The main scales of both the instruments are marked in mm. The zero error on the vernier calipers is `+0.15 `mm where as that on screw gauge is `-0.06` mm. If the diameter of a rod lies between 0.9 cm and 1.0 cm and x and y are the VCD and CSR on the two instruments relation between x and y given that the number of C.S.D = 100 isA. `x-5y =9`B. `5x -y =9`C. `x-5y=21`D. `5x-y =21` |
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Answer» Correct Answer - A Least count of screw gauge `=(1)/(100) =0.01 mm` Least count of vernier calipers `=5xx 0.01 =0.05 mm` Let d be the diameter of the rod `9 mm lt d lt 10 mm` `d=9+ y xx 0.01 +0.06 ` by screw gauge ` 5x -15 =y+6` `5x -y =21` |
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| 5. |
What is the ratio of 1 KWh (kilowatt hour) to 1 MeV (million electron volt)?A. `2.25xx10^(19)`B. `2.25xx10^(17)`C. `2.25xx10^(23)`D. `2.25xx10^(12)` |
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Answer» Correct Answer - a Both of them are the units of electric energy. `1kWh=1000Wxx1h=1000(J)/(s)xx3600s` `=36xx10^(5)=3.6xx10^(6)J` and 1 eV = `1.6xx10^(-19)J` `therefore 1MeV=1.6xx10^(-19)xx10^(6)J=1.6xx10^(-13)J` `therefore (1kWh)/(1MeV)=(3.6xx10^(6)J)/(1.6xx10^(-13)J)=2.25xx10^(19)` |
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| 6. |
If the units of length, mass and time are doubled, then the new unit of work expressed in terms of L, M and TA. becomes four timesB. becomes two timesC. becomes eight timesD. is not changed |
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Answer» Correct Answer - b `[W]=[M^(1)L^(2)T^(-2)]=[M.(L^(2))/(T^(2))]` If the units of L, M and T are doubled, then the new unit of work = `(2M).((2L)^(2))/((2T)^(2))=2M[(L^(2))/(T^(2))]` `therefore` The unit becomes two times. |
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| 7. |
A vernier calipers and a screw gauge have the same least count and zero error with the difference that the vernier calipers has positive zero error and the screw gauge has negative zero error. The number of circular scale divisions on the screw gauge is twice the number of vernier scale divisions on the vernier calipers and the pitch of the screw gauge is 1mm. If the vernier coinciding division and the circular are 3 and 97, respectively then find the number of vernier scale divisions on the vernier calipers.A. 10B. 200C. 50D. 100 |
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Answer» Correct Answer - B The smallest weight that can be measured accurately using a physical balance is 1 mg i.e., 0.001 g. |
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| 8. |
If the zero error correction of a screw gauge with least of count 0.01 mm is +0.05 mmA. The number of C.S.D is 100 and the zero of the circular scale is 5 divisions above the index line.B. The number of C.S.D is 100 and the zero of the circular scale is 5 divisions below the index line.C. the number of C.S.D. Is 50 and the zero of the circular scale is 5 divisions above the index line.D. Both (1) and (3) |
| Answer» Correct Answer - D | |
| 9. |
A student measured the diameter of a wire using a micrometer screw gauge of least count 0.001 cm. He recorded the following measurements. The correct measurement isA. 5.3 cmB. 5.320 cmC. 5.32 cmD. 5.3200 cm |
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Answer» Correct Answer - b As the least count is 0.001 cm, the correct measurement is (b) i.e. 5.320 cm. |
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| 10. |
Constant error can be caused due to (A) faulty construction of instrument. (B) wrong setting of instrument. (C) lack of concentration of observer. (D) wrong procedure of handling the instrument. |
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Answer» Answer is (A) faulty construction of instrument. |
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| 11. |
Instrumental error can be minimised by (A) taking large number of readings. (B) using different accurate instrument for the same reading. (C) adjusting zero of the instrument. (D) maintaining the temperature of the surrounding. |
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Answer» Answer is (B) using different accurate instrument for the same reading. taking large number of mesurment minimised the instrumental error
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| 12. |
Error due to non-removal of parallax between pointer and its image in case of magnetic compass needle causes ....... (A) instrumental error (B) persistant error (C) personal error (D) random error |
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Answer» Answer is (C) personal error |
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| 13. |
What are the clocks used to measure a smaller. duration of time? |
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Answer» Electronic clock, Stop watch. |
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| 14. |
Give some examples of larger length measures. |
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Answer» 1. Height of the building, 2. Length of a banner, 3. Height of lamp post |
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| 15. |
In earlier days, which instruments are used to measure time? |
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Answer» Sand clock and Sun dial. |
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| 16. |
What are the unit of measurements of very small length? |
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Answer» 1. Millimetre 2. Centimetre. |
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| 17. |
What are the clock used by ancient people to measure day time? |
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Answer» 1. sand clock 2. sun clock. |
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| 18. |
What is meant by parallax? |
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Answer» Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight. |
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| 19. |
What formula is used to measure area of your class room? |
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Answer» My class room area = Length × Breadth |
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| 20. |
Where are the electronic balances used? |
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Answer» Electronic balances are used in grocery shops and jeweleries. |
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| 21. |
What are the materials needed to find the length of a banana? |
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Answer» A meter scale, a string or thread, sketch pen.. |
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| 22. |
Analogy .1. Potatoes : Kilogram; Water?2. Stop clock : accurate time; _______ : accurate weight3. Mass : Balance; Length?4. Amount of matter: mass; Gravitational Pull? |
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Answer» 1.Litre. 2. Electronic balance. 3. measuring tape. 4. Weight. |
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| 23. |
Fill in the blanks. 1. SI Unit of length is symbolically represented as _____ 2. 500 gm = _____ kilogram3. Distance between Delhi and Chennai can be measured in _____ 4. 1 m = _____ cm. 5. km = _____ m. |
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Answer» 1. m 2. 0.5 3. Kilometre 4. 100 5. 5000 |
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| 24. |
While measuring the length of a sharpened pencil, reading of the scale at one end is 2.0 cm and at the other end is 12.1 cm. What is the length of the pencil? |
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Answer» The difference between the two readings is the length of the pencil. = 12.1 cm – 2.0 cm = 10.1 cm or 10 cm and 1 mm. |
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| 25. |
Analogy1. Sugar: Beam balance; Lime juice?2. Height of a person : cm; length of your sharpened pencil lead?3. Milk: volume; vegetables? |
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Answer» 1. Measuring Jar. 2. mm (mull metre) 3. mass |
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| 26. |
V is the volume of a liquid flowing per second through a capillary tube of length l and radius r, under a pressure difference (p). If the velocity (v), mass (M) and time (T) are taken as the fundamental quantities, then the dimensional formula for `eta` in the relation `V=(pipr^(4))/(8etal)`A. `[M V^(-1)]`B. `[M^(1)V^(-1)T^(-2)]`C. `[M^(1)V^(1)T^(-2)]`D. `[M^(1)V^(-1)T^(-1)]` |
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Answer» Correct Answer - c `V=(pipr^(4))/(8etal) therefore eta=(pipr^(4))/(8Vl)` `pi` and 8 have no dimensions. Writing the dimensional formulae for all quantities, we get `[eta]=([M^(1)L^(-1)T^(-2)][L^(4)])/([L^(3)T^(-1)][L^(1)])[because V="Volume/sec"]` `[eta]=[M^(1)L^(-1)T^(-1)]` But velocity `v=(L)/(T) therefore L=vT` `therefore [eta]=[M^(1)V^(-1)T^(-1)T^(-1)]=[M^(1)V^(-1)T^(-2)]` |
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| 27. |
The potential difference across the ends of a wire is found to be `(100pm2)` volt and the current flowing in the wire is found to be `(10pm0.1)`A. What is the maximum percentage error in the measurement of resistance?A. `2%`B. `3%`C. `4%`D. `5%` |
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Answer» Correct Answer - B `R=(V)/(I),V=100V,deltaV=2V,I=10A,delta1=0.1A` `therefore` Max. percentage error in R is `((deltaR)/(R))_(max)xx100=((deltaV)/(V)xx100)+((deltaI)/(I)xx100)` `=(2)/(100)xx100+(0.1)/(10)xx100=2%+1%=3%` Note : Errors are to be added. |
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| 28. |
V is the volume of a liquid flowing per second through a capillary tube of length l and radius r, under a pressure difference (p). If the velocity (v), mass (M) and time (T) are taken as the fundamental quantities, then the dimensional formula for `eta` in the relation `V=(pipr^(4))/(8etal)`A. `[Mv^(-1)]`B. `[M^(1)v^(-1)T^(-2)]`C. `[M^(1)v^(1)T^(-2)]`D. `[M^(1)v^(-1)T^(-1)]` |
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Answer» Correct Answer - C `V=(piPr^(4))/(8etal) therefore eta=(piPr^(4))/(8Vl)` `pi` and 8 have no dimensions. Writing the dimensional formulae for all quantities, we get `[eta]=([M^(1)L^(-1)T^(-2)][L^(4)])/([L^(3)T^(-1)][L^(1)])[because V="Volume/sec"]` `[eta]=[M^(1)L^(-1)T^(-1)]` But velocity `v=(L)/(T) therefore L=vT` `therefore [eta]=[M^(1)v^(-1)T^(-1)T^(-1)]=[M^(1)v^(-1)T^(-2)]` |
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| 29. |
In the reading 2.614 cm of measurement with a vernier calliper, only uncertain figure is (A) 1 (B) 2 (C) 4(D) 6 |
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Answer» Answer is (C) 4 A vernier calliper has a least count 0.01 cm. Hence measurement is accurate only upto three significant figures. |
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| 30. |
If the number of divisons on the vernier scale ofa standard vernier calliper is N, then N is always equal to (N- 1) main scale divisions. Would the vernier calliper be rendered useless if N is equal to fewer number of main scale divisions, for example N = ( N-3) M.S.D.? Why? How would the least count calculations be modified? |
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Answer» (i) What is least count? Should the value of least count be large or small/ Will the accuracy of the venier calipers be increased by decreasing the value of the least count? The least count, L.C.=`(1 M.S.D.)/(N)=1 M.S.D-1` V.S.D. Given the principle pf vernier os N V.S.D.=(N-1) M.S.D. This implies that, 1 V.S.D. =`((N-1)/(N))` M.S.D. On substituting(2) in(1) , find the value of L.C. Consider, if N V.S.D. = (N-3) M.S.D. `rArr 1 V. S.D. =`((N-1)/(N))` M.S.D Substituting (3) in (1), find the value of L.C. In which case is the L.C. is teh least? |
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| 31. |
Two metal blocks A and B were placed on the two pans ofa common balance and the beam of the balancec was horizontal. However, when the blocks were interchanged and block B was placed in the left pan and block A in the right pan, the beam was not balanced and tilted down toward the left. Discuss teh various factors that could be the cause of this. Are teh masses of two blocks equl? |
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Answer» (i) Consider the weights of the empty pans. How would the beam be balanced if the empty pans have unequal masses? (ii)`m_(B)gtm_(A)` |
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| 32. |
Two identical containers A and B are filled with two different liquids of equal masses. The level of the liquid in container A is found to be one-fourth of the level of the liquid in container B. What is the ratio of thedensity of two liquids? If the density of he liquid in the container A is 2 " g "` cm^(-3)`, then find the density of the mixture of the two liquids. |
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Answer» Mass of liquid A, `(m_(A))` = mass of liquid B, `(m_(B))` Volume = `area xx height` Area of cross section of A = Area cross section of B Height of liquid column A = `1/4 xx` " height " of liquid column A = `1/4 " (Volume of B) " Density of `"mass/volume"` `(density " of " A)/(density " of " B) = m_(A)/V_(A)xx V_(B)/m_(B)=V_(B)/V_(B)=4/1=4 :1` If density of A is 2 " g " `cm^(-3)` ,then density of B = " g " `cm^(-3)` = 0.5 " g " `cm^(-3)` Density of mixture = `"mass of mixture"/"volume of mixture"` `(m_(A)+m_(B))/(v_(A)+v_(B))=(m_(A)=m_(A))/(v_(A)=4v_(A)) ( :. m_(B) = m_(A) and v_(B)= 4v_(A))` `(2m_(A))/(5v_(A)) = (2xx d_(A)xx v_(A))/(5V_(A)` `(2xx2)/5=4/5=0.8 " g "cm^(-3)` |
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| 33. |
If the each of the units of force velocity and frequency are doubled how many times has the unit of mass become ?A. doubledB. four timesC. halvedD. one-fourth |
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Answer» Correct Answer - C (i) M , L and T in the dimensional formula are replaced by the units of mass length and time respectively in the corresponding system (ii) 1 kg `m^(2) s^(-3) =10^(7) g cm^(2) s^(-3)` |
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| 34. |
1 kilogram is equal to _____.A. 1000 gramB. 100 gramC. 1000 milligramD. 100 milligram |
| Answer» Correct Answer - A | |
| 35. |
The smallest length that can be accurately measured by using a metre scale is_____ cm. |
| Answer» Correct Answer - 0.1 | |
| 36. |
5 litre of a liquid weights 5kg. The density of liquid is_____.A. 1 kg `m^(-3)`B. 1 g `cm^(-3)`C. `100kg" "m^(-3)`D. `100g" "m^(-3)` |
| Answer» Correct Answer - B | |
| 37. |
If the density of a substance is `2xx10^(3)kg" "m^(-3)`, then the mass of 5 `m^(3)` of this substances isA. 1000 kgB. 10000 gC. 10000 kgD. both a and b |
| Answer» Correct Answer - C | |
| 38. |
Unit of weight is (are)_____.A. `N`B. g fC. kg fD. All the above |
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Answer» Correct Answer - D N, kgf and gf are units of weight. |
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| 39. |
The mass of a certain amount of salt determined by two different persons was found to differ by 200 mg. One using the left pan for the weight s and the other using the right pan. Which of the following statement is false?A. There is a zero error in the balanceB. The true mass of is the mean of the two weighingsC. The true mass is 100 mg less than the higher value.D. The true mass is 200 mg more than the lower value. |
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Answer» Correct Answer - D `l_(1):l_(2) =2:3` `g_(1):g_(2)=3:2` `(T_(1))/(T_(2)) =(sqrt(l_(1)//l_(1)))/(sqrt(l_(2)//l_(2))) = sqrt((4)/(9))` `(T_(1))/(T_(2)) =sqrt((4)/(9))=(2)/(3) : (f_(1))/(f_(2)) =3:2` |
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| 40. |
The amount of heat absorbed by a body while raising its temperature is given by the equation. Q = ms 0, here Q is the amount of heat, m is the mass of the body, s is the specific heat capacity and 0 is the rise tmemperature. Dertermine the S.I. unit and C.G.S. unit of heat is erg and S.I. unit is jouble. |
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Answer» (i) S.I. unit ofheat is `1 J (= 1"kg" m^(2) s^(-2))` "and its CGS unit is" `1 "erg" (= 1 g cm^(2) s^(-2))` (ii) erg `g^(-1)" "^(0)C^(-1),J kg^(-1) k^(-1)` |
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| 41. |
A ruler marked is inches is used to measure the diameter of a rod as shown in the figure below. Find the diameter if the rod in inches. If one inch is approximately equal to 25. 4m, express tehis diameter in cm ( correct to 2 decimal places). |
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Answer» (i)0 inch to 2 inch = 32 division 2 inch to 3 inch = 8 division (ii) `1(13)/(16)"inch",4.60 cm` |
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| 42. |
If x = (a – b), the maximum percentage error in the measurement of x will be(A) \(\Big[\frac{Δa}{a-b}-\frac{Δb}{a-b}\Big]\)x 100(B) \(\Big[\frac{Δa}{a-b}+\frac{Δb}{a-b}\Big]\)x 100(C) \(\Big[\frac{Δa}{a}+\frac{Δb}{b}\Big]\)x 100(D) \(\Big[\frac{Δa}{a}-\frac{Δb}{b}\Big]\)x 100 |
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Answer» Answer is (B) \(\Big[\frac{Δa}{a-b}+\frac{Δb}{a-b}\Big]\)x 100 |
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| 43. |
Assertion: The error in the measurement of radius of the sphere is 0.3%. The permissible error in its surface area is 0.6%. Reason: The permissible error is calculated by the formula \(\frac{ΔA}{A}\) = \(\frac{4Δr}{r}\)(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion (B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion (C) Assertion is True, Reason is False (D) Assertion is False, Reason is False. |
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Answer» Answer is (C) Assertion is True, Reason is False A = 4πr2 ∴ Fractional error \(\frac{ΔA}{A}\) = \(\frac{2Δr}{r}\) \(\frac{ΔA}{A}\) x 100 = 2 x 0.3% = 0.6% |
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| 44. |
Planck’s constant has same dimensions as(A) energy. (B) angular momentum. (C) mass. (D) force |
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Answer» Answer is (B) angular momentum. |
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| 45. |
Which among the following physical quantities does not posses a unit?A. AreaB. VolumeC. DensityD. Specific gravity |
| Answer» Correct Answer - D | |
| 46. |
The unit of solid angle is steradian. What is the dimensional formula for steradian?A. `[L^(1)M^(1)T^(-1)]`B. `[L^(0)M^(0)T^(0)]`C. `[L^(2)M^(-1)T^(1)]`D. `[L^(-2)M^(1)T^(0)]` |
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Answer» Correct Answer - b Angle has no dimensions. Hence the dimensional formula is `[L^(0)M^(0)T^(0)]`. |
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| 47. |
R, L and C represent the physical quantities resistance, inductance and capacitance respectively. Which one of the following combination has dimensions of frequency?(A) \(\frac{R}{\sqrt{RC}}\) (B) \(\frac{R}{L}\)(C) \(\frac{1}{LC}\) (D) \(\frac{C}{L}\) |
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Answer» Answer is (B) \(\frac{R}{L}\) |
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| 48. |
From the following physical quantities, which is the only quantity which has negative dimensions of mass?A. Linear momentumB. Gravitationsl constantC. ForceD. Work |
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Answer» Correct Answer - b `[p]=[mv]=[L^(1)M^(1)T^(-1)],[F]=[L^(1)M^(1)T^(-2)]` (c ) Gravitational constant [G] `because F=G(m_(1)m_(2))/(r^(2))` `therefore [G]=[(Fr^(2))/(m_(1)m_(2))]=[(L^(1)M^(1)T^(-2)xxL^(2))/(M^(2))]` `therefore [G]=[L^(3)M^(-1)T^(-2)]` Thus G has the -ve dimension of mass. (d) `[W]=[L^(2)M^(1)T^(-2)]` |
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| 49. |
Refractive index mu is given as `mu=A+B/lambda^2,` where A and B are constants and lambda is wavelength, then dimensions of B are same as that ofA. WavelengthB. AreaC. VolumeD. Velocity |
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Answer» Correct Answer - b The refractive index has no dimensions. `therefore (B)/(lambda^(2))` has no dimensions. i.e. B has the same dimensions as that of `lambda^(2)` or `("Length")^(2)` or (Area). |
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| 50. |
Which of the following physical quantities have the same dimensions? (1) Energy density (2) Refractive index (3) Magnetic induction (4) PressureA. 1 and 2B. 2 and 3C. 1 and 4D. 1 and 3 |
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Answer» Correct Answer - c (1) Energy density = `[("Energy")/("Volume")]` `[(L^(2)M^(1)T^(-2))/(L^(3))]` `=[L^(-1)M^(1)T^(-2)]` (2) Refractive index has no dimensions. (3) Magnetic induction (B) `because F=BIl` `therefore [B]=[(F)/(Il)]=[(M^(1)L^(1)T^(-2))/(A^(1)L^(1))]=[M^(1)A^(-1)T^(-2)]` (4) `therefore` [Pressure] `=[(F)/(A)]=[(L^(1)M^(1)T^(-2))/(L^(2))]` `=[L^(-1)M^(1)T^(-2)]` Thus (1) and (4) have the same dimensions. |
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