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.
| 201. |
Which is the greatest? 0.007 g, 70 mg, 0.07 cg ……….(i) 0.07 cg(ii) 0.007 g(iii) 70 mg(iv) all are equal |
|
Answer» The correct option is (iii) 70 mg. |
|
| 202. |
1006 g is equal to ………(i) 1 kg 6 g(ii) 10 kg 6 g(iii) 100 kg 6 g(iv) 1 kg 600 g |
|
Answer» (i) 1 kg 6 g 1006 g is equal to 1 kg 6 g. |
|
| 203. |
Find the railway time for 12 : 05 a.m, 4 : 20 pm |
|
Answer» 12 : 05 am is 00 : 05 hrs 4 : 20 p.m. is (12 + 4) : 20 hrs = 16 : 20 hrs |
|
| 204. |
True or False(i) Pugazhenthi ate 100 g of nuts which is equal to 0.1 kg.(ii) Meena bow it 250 ml of butter milk which is equal to 2.5 l.(iii) Karkuzhali’s bag 1 kg 250 g and poong- kodi’s bag 2 kg 750 g. The total weight of their bags 4 kg.(iv) Vanmathi bought 4 books each weighing 500 g. Total weight of 4 books is 2 kg.(v) Gayathiri bought 1 kg of birthday cake. She shared 450 g with her friends.The weight of cake remaining is 650 g |
|
Answer» (i) True (ii) False (iii) True (iv) True (v) False |
|
| 205. |
Geetha brought 2 l and 250 ml of water in a bottle. Her friend drank 300 ml from it. How much of water is remaining in the bottle? |
|
Answer» Total Capacity of water = 2 l 250 ml = (2 x 1000 + 250) l = 2000 + 250 ml = 2250 ml Water Consumed = 300 ml Water remaining in the bottle = (2250 – 300) ml = 1 litre 950 ml Total Capacity of water = 2 l 250 ml= (2 x 1000 + 250) l = 2000 + 250 ml = 2250 ml Water Consumed = 300 ml Water remaining in the bottle = (2250 – 300) ml = 1950 ml = 1 litre 950 ml |
|
| 206. |
If `E` = energy , `G`= gravitational constant, `I`=impulse and `M`=mass, then dimensions of `(GIM^(2))/(E^(2)` are same as that ofA. timeB. massC. lengthD. force |
|
Answer» Correct Answer - A `(GIM^(2))/(E^(2))=([M^(-1)L^(3)T^(-2)][MLT^(-1)][M^(2)])/([ML^(2)T^(-2)]^(2))` `[M^(0)L^(0)T]` |
|
| 207. |
The dimensions of impulse are equal to that ofA. forceB. linear mometumC. pressureD. angular momentum |
|
Answer» Correct Answer - B Impluse is equal to change un linear momentum. |
|
| 208. |
Which one of the following system of units is not based on the units of length, mass and time alone?A. C.G.S.B. F.D.S.C. S.I.D. M.K.S. |
|
Answer» Correct Answer - c S.I. system. It is based on 7 fundamental units. |
|
| 209. |
Convert 7 kg 250 g in to mg, g and cg? |
|
Answer» 7 kg 250 g. (i) 7 kg 250 g = 7 × 1000 + 250 g. = 7000 + 250 = 7250 g. (ii) 7 kg 250 g = 7250 g. = 7250 × 1000 mg = 72,50,000 mg. (iii) 7 kg 250 g = 7250 g. = 7250 × 100 cg = 7,25,000 cg. |
|
| 210. |
If momentum (P), area (A) and time (T) are assumed to be fundamental quantities, then energy has dimensional formula (A) [P1 A-1/2T-1] (B) [P1 A1/2T-1] (C) [P2 A-1 T-1] (D) [P1 A-1 T1] |
|
Answer» Answer is (B) [P1 A1/2T-1] [Energy] = [M1 L2 T-2 ] = [M1 L1 T-1] [L1 ] [T-1] = [P1 A1/2T-1] |
|
| 211. |
Which of the following set have different dimensions?(A) Pressure, Young’s modulus, stress (B) e.m.f, potential difference, electric potential (C) Heat, work done, energy (D) dipole moment, electric flux, electric field |
|
Answer» Answer is (D) dipole moment, electric flux, electric field |
|
| 212. |
Dimension of angular momentum is(A) [M1 L2 T–2] (B) [M1 L–2T–1] (C) [M1 L2 T–1] (D) [M1 L0 T–1] |
|
Answer» Answer is (C) [M1 L2 T–1] |
|
| 213. |
If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities, then the dimension of gravitational constant is (A) [c2 g0 p-2] (B) [c0 g2 p-1] (C) [cg3 p-2] (D) [c-1g0 p-1] |
|
Answer» Answer is (B) [c0 g2 p-1] Let G ∝ cx gy pz Substituting dimensions, [M-1 L3 T-2]= [M0L1T-1]x [M0L1T-2] y [M1 L-1 T-2]z Comparing powers of M, L, T - 1 = z, x + y - z = 3 and - x - 2y - 2z = - 2 Solving, x = 0, y = 2 |
|
| 214. |
The wavelength of sodium light is 0.0000005893 m. This can be expressed in powers of ten asA. `5.893xx10^(-6)m`B. `5.893xx10^(-7)m`C. `5.893xx10^(5)m`D. `5.893xx10^(-8)m` |
|
Answer» Correct Answer - b As the number is less than one, start shifting the decimal point from left to right. Each shift is equivalent to multiplying by ten. To keep the value of the number constant, we have to divide by ten or multiplying by `10^(-1)`. In this case when we shift the decimal point to the right seven times, we get the number `0000005.893xx10^(-7)`. As the 6 zeros before 5 are of no use, we write the number as `5.893xx10^(-7)m` |
|
| 215. |
(a) What are the universally accepted basic metric units?(b) A cow gives 10 litres of milk in the morning and 8 litres in the evening. Find the total milk it gives for a week in ml? |
|
Answer» (a) Universally accepted basic metric units are: (b) Litres of milk the cow gives in the morning = 10 l. Litres of milk the cow gives in the evening = 8 l. Total milk per day = 10 + 8 = 18 l Total milk for a week = 18 × 7 l = 126 l = 126 × 1000 ml = 126000 ml. Total milk the cow gives for a week = 1,26,000 ml. |
|
| 216. |
An equation of a gas is given as `(P+(a)/(V^(2))=(bT)/(V)`, where P = pressure, V = volume and T= absolute temperature of the gas and a and b are constants. What is the dimensional formula for a?A. `[M^(-1)L^(5)T^(2)]`B. `[M L^(5)T^(-2)]`C. `[M L^(-5)T^(-1)]`D. `[M L^(5) T]` |
|
Answer» Correct Answer - b From the principle of homogeneity of dimensions, all terms in a physical equation must have the same dimensions. `therefore` Dimensions of P = Dimensions of `((a)/(V^(2)))` `therefore [a]=[PV^(2)]` `=[(F)/(A)xxV^(2)]=[(M^(1)L^(1)T^(-2))/(L^(2))xx(L^(3))^(2)]` `therefore [a]=[M^(1)L^(5)T^(-2)]` |
|
| 217. |
In a system of units, the units of force and energy are 10 N and 50 J respectively. What is the unit of length in this system?A. 3 mB. 5 mC. 7 mD. 10 m |
|
Answer» Correct Answer - b `[F]=[M^(1)L^(1)T^(-2)]=10N" ... (1)"` `[E]=[M^(1)L^(2)T^(-2)]=50J" ... (2)"` Dividing (2) by (1), we get `([M^(1)L^(2)T^(-2)])/([M^(1)L^(1)T^(-2)])=[L]=(50J)/(10N)=(50Nm)/(10N)=5m` |
|
| 218. |
The velocity (V) of a particle (in cm/s) is given in terms of time (t) in sec by the equation `V=at+(b)/(c+t)`. The dimensions of a, b and c areA. `{:(a,b,c),(L^(2),T,L^(1)T^(-2)):}`B. `{:(a,b,c),(L^(1)T^(2),L^(1)T,L^(1)):}`C. `{:(a,b,c),(L^(1)T^(-2),L^(1),T^(1)):}`D. `{:(a,b,c),(L^(1),L^(1)T,T^(2)):}` |
|
Answer» Correct Answer - c L.H.S. is `V=[L^(1)M^(0)T^(-1)]` Every term has the dimensions of velocity. `therefore at=V therefore a=(V)/(t)` `therefore [a]=(M^(0)L^(1)T^(-1))/(T^(1))=[L^(1)T^(-2)]and(b)/(c+t)=V` But as we can add only the terms with the same dimensions `therefore` C has the dimensions of t. Thus, `C=t therefore b=V" (time)"therefore [b]=[L^(1)T^(-1)xxT^(1)]=L^(1)` Thus, `[a]=L^(1)T^(-2),[b]=L^(1),[c]=T^(1)` |
|
| 219. |
A box contains 20 kg 280 g of sugar. After transferring into a smaller box of 2 kg 720 g 7 times find the remaining sugar left in the big box in cg. |
|
Answer» Total sugar in the box = 20 kg 280 g. Measurement of smaller box = 2 kg 720 g. 7 times the smaller box = 2 kg 720 g × 7 = 19 kg 040 g. Remaining sugar = 20 kg 280 g – 19 kg 40 g = 1 kg 240 g = 1240 g.= 1240 × 100 cg = 124000 eg. Sugar left in the box = 1,24,000 eg. |
|
| 220. |
Can you add 2 m 20 cm and 4 kg 80 g? Why? |
|
Answer» No, we can’t. To add or subtract, the measures should be in the same unit. |
|
| 221. |
A force is given by `F= at + bt^(2) ` where , t is the time .The dimensione of a and b areA. `[M^(1)L^(1)T^(3)],[M^(1)L^(1)T^(-4)]`B. `[M^(2)L^(1)T^(2)],[M^(1)L^(1)T^(-2)]`C. `[M^(1)L^(-1)T^(-2)],[M^(1)L^(2)T^(-3)]`D. `[M^(-1)L^(1)T^(-4)],[M^(1)L^(1)T^(-3)]` |
|
Answer» Correct Answer - A In a dimensional equation, all terms must have the same dimensions. In this case, they should have the dimensions of force. `[F]=[M^(1)L^(1)T^(-2)]` (i) `[a]=[(F)/(t)]=[(M^(1)L^(1)T^(-2))/(T^(1))]=[M^(1)L^(1)T^(-3)]` (ii) `[b]=[(F)/(t^(2))]=[(M^(1)L^(1)T^(-2))/(T^(2))]=[M^(1)L^(1)T^(-4)]` |
|
| 222. |
The whole length of a metre scale is divided into 500 equal parts then the smallest measurement that can be measured by using the scale is `"_______"`.A. 0.5 mB. 0.005 mC. 50 mmD. 2 mm |
| Answer» Correct Answer - D | |
| 223. |
Which of the following statements is /are correct?A. The weight of a body can be zero.B. The weight of a body can be greater than zero.C. The mass of a body can be zero.D. both a and b |
| Answer» Correct Answer - C | |
| 224. |
Convert 20 m 20 cm into cm, m and km. |
|
Answer» 20 m 20 cm. (i) 20 m 20 cm = 20 × 100 + 20 cm = 2000 + 20 cm = 2020 cm (ii) 20 m 20 cm = 2020 cm = 2020 × 1100 m = 20.20 m (iii) 20 m 20 cm = 20.20 m = 20.20 × 11000 km = 0.02020 km |
|
| 225. |
7 km – 4200 m is equal to(a) 3 km 800 m(b) 2 km 800 m(c) 3 km 200 m(d) 2 km 200 m |
|
Answer» (b) 2 km 800 m 7 km – 4200 m is equal to 2 km 800 m. |
|
| 226. |
Say the time in two ways: |
|
Answer» (i) 10 : 15 hours; quarter past 10; 45 minutes to 11 (ii) 6 : 45 hours; quarter to 7; 45 minutes past 6 (iii) 4 : 10 hours; 10 minutes past 4; 50 minutes to 5 (iv) 3 : 30 hours; half-past 3; 30 minutes to 4 (v) 9 : 40 hours; 20 minutes to 10; 40 minutes past 9. |
|
| 227. |
Match the following:(i) 9.55 – (a) 20 minutes past 2(ii) 11.50 – (b) quarter past 4(iii) 4.15 – (c) quarter to 8(iv) 7.45 – (d) 5 minutes to 10(v) 2.20 – (e) 10 minutes to 1? |
|
Answer» (i) 9.55 – (d) 5 minutes to 10 (ii) 11.50 – (e) 10 minutes to 12 (iii) 4.15 – (b) quarter past 4 (iv) 7.45 – (c) quarter to 8 (v) 2.20 – (a) 20 minutes past 2 |
|
| 228. |
Find whether 2000,2018 and 2012 are leap years or not. |
|
Answer» 2000 is a leap year (divisible by 400) 2018 is a not leap year (not divisible by 4) 2012 is a leap year (divisible by 4) |
|
| 229. |
David left home at 4 : 30 p.m to meet his friend. He came back after 3 hr 25 min At what time did he came back? |
|
Answer» Time at which David left home = 4 : 30 p.m. Time he spent outside = 3 hr : 25 min Time at which he came back = 7 : 55 p.m |
|
| 230. |
Which of the following scale gives more accurate reading?A. One metre in a scale is divided into 1000 equal parts.B. One cm in a scale is divided into 2 equal parts.C. One mm in a scale is divided into 2 equal parts.D. One cm in a scale is divided into 5 equal parts. |
|
Answer» Correct Answer - C The scale which is having th esmallest least count is more accurate. When 1 mm is divided into two equal parts `L.C=1/2mm =0.5" mm "`. |
|
| 231. |
What is the order of magnitude of the distance of the sun from the earth in SI unit?A. 12B. 11C. 13D. None of these |
|
Answer» Correct Answer - B Distance of sun from the earth` =AU=1.496xx10^(11)`m order of magnitude=11 |
|
| 232. |
If x = an then relative error is (where n is power of a)(A) \(\frac{Δa}{a}+n\) (B) n\(\frac{Δa}{a}\)(C) \(\frac{Δa}{a}-n\) (D) \(\frac{Δa}{na}\) |
|
Answer» Answer is (B) n\(\frac{Δa}{a}\) |
|
| 233. |
The formula for percentage error is(A) Percentage error = \(\frac{|Δa_m|}{a_m}\) x 100%(A) Percentage error = \(\frac{1}{n}\)\(\displaystyle\sum_{i=1}^{n} |Δa_i|\) x 100%(C) Percentage error = \(\frac{a_m}{|Δa_m|}\) x 100%(D) Percentage error = \(\frac{1}{n}\)\(\displaystyle\sum_{i=1}^{n} a_i\) x 100% |
|
Answer» Answer is (A) Percentage error = \(\frac{|Δa_m|}{a_m}\) x 100% |
|
| 234. |
Given: l1 = 44.2 ± 0.1 and l2 = 23.1 ± 0.1, the uncertainty in l1 + l2 is (A) 0 (B) 0.1 (C) 0.2 (D) 0.4 |
|
Answer» Answer is (C) 0.2 |
|
| 235. |
If x = a/b , hen maximum relative error in the measurement is(A) \(\frac{Δa/a}{Δb/b}\)(B) \(\frac{Δa}{a}+\frac{Δb}{b}\)(C) \(\frac{Δa}{a}-\frac{Δb}{b}\)(D) \(\frac{Δb/b}{Δa/a}\) |
|
Answer» Answer is (B) \(\frac{Δa}{a}+\frac{Δb}{b}\) |
|
| 236. |
The random error in the arithmetic mean of 100 observations is x, then random error in the arithmetic mean of 400 observations would beA. 4xB. `(1)/(4)x`C. 2xD. `(1)/(2)x` |
|
Answer» Correct Answer - B Since error is measured for 400 observations instead of 100 observation. So, error will reduce by `1//4` factor. |
|
| 237. |
The radius of a ball is `(5.2pm0.2)` cm. The percentage error in the volume of the ball is (approximately).A. 0.11B. 0.04C. 0.07D. 0.09 |
|
Answer» Correct Answer - A Radius of ball=5.2cm `V=(4)/(3)piR^(3)implies(DeltaV)/(V)=3((DeltaV)/(R))` `((DeltaV)/(V))xx100=3((0.2)/(5.2))xx100=11%` |
|
| 238. |
If the error in the measurement of momentum of a particle is (+ 100%), then the error in the measurement of kinetic energy isA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - C Since error in measurement of momentum is +100%.`p_(1)=p,p_(2)=2p` `K_(1)=(p^(2))/(2m),K_(2)=((2p)^(2))/(2m)` `%"in" K=((K_(2)-K_(1))/(K_(1)))xx100` `=((4-1)/(1))xx100=300%` |
|
| 239. |
1 kg `m^(-3)`=_____.A. 1000 g `m^(-3)`B. `(1)/(1000)g" "cm^(-3)`C. 10000 kg `cm^(-3)`D. 1 g `cm^(-3)` |
| Answer» Correct Answer - B | |
| 240. |
1 m`^(3)=`_____litre.A. 1B. 10C. 100D. 1000 |
| Answer» Correct Answer - D | |
| 241. |
The realtive density of mercury is 13.6. Find the ratio of the mass of mercury of a certain volume to that of water having ten times the volume of mercury. |
|
Answer» Let `V cm^(3) "be the volume of mercury".` `therefore "volume of water" = 10 V cm^(3)` `"Density of mercury" = 13.6 g cm^(-3)` `therefore "Mass of mercury", m_(1)+(13.6) V g` `"Density of water" = 1 g cm^(-3)` ` therefore "Mass of water", m^(2) = (10 V) (1) g = 10 V g.` `"Required ratio is" m_(1), m_(2) = 13.6 V : ` `10 V = 34 : 25 or 1.36 : 1 ` |
|
| 242. |
A vernier scale has 10 divisions. It slides over the main scale, whose 1 M.S.D. is 1.0 mm. If the number of division to the left side of the zero of the vernier scale on the main scale is 49 and the 8th vernier scale division coincides with the main scale, calculate teh length in centimetres. |
|
Answer» N = 10 1 M.S.D = 1 mm M.S.R = 49 mm V.C = 8 `L = M.S.R + V.C xx L.C` L.C= `(1MSD)/(10) = (1) /(10)` = 0.1 mm l= `49 + `xx` 0.1` =49.8 mm = 4.98 cm |
|
| 243. |
Which of the following is true with respect to a standard vernier callipers?A. N main scale divisions = (N-1) vernier scale divisions.B. N vernier scale divisions = (N-1) main scale divisionsC. N vernier scale divisions = (N+1) main scale divisions.D. None of these. |
|
Answer» Correct Answer - B For a standard vernier callipers N vernier Scale divisions = (N-I) main scale divisions |
|
| 244. |
The zeroeth division of a vernier scale lies between the 40th and 41th main scale divisions. Then, the M.S.R. = 4 cm if M.S.D = 1 mm. |
| Answer» Correct Answer - 1 | |
| 245. |
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 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(A) 58.59 degree (B) 58.77 degree (C) 58.65 degree (D) 59 degree |
|
Answer» Answer is (C) 58.65 degree 30 VSD = 29 MSD 1 VSD = \(\frac{29}{30}\) MSD Least count of vernier = 1 M.S.D. – 1 V.S.D. = 0.5° - \(\Big(\frac{29}{30}\times0.5°\Big)\) = \(\frac{0.5°}{30}\) Reading of vernier = M.S. reading + V.S. reading x L.C. = 58.5° + 9 x \(\frac{0.5°}{30}\) = 58.65° |
|
| 246. |
The zeroeth divisions of the vernier and main scales of a standard vernier callipers coincide, thenA. The external jaws are in contact with each other.B. there is some distance between the internal jaws.C. Both (a) nor (b)D. Neither (a) nor (b). |
|
Answer» Correct Answer - A If the zeroeth devision of the vernier callipers main scales of a standard veraier callipers coincide, then the external jaws are in contact with each other. |
|
| 247. |
A screw guage gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions The diameter of wire from the above data is(A) 0.52 cm (B) 0.052 cm (C) 0.026 cm (D) 0.005 cm |
|
Answer» Answer is (B) 0.052 cm Least count of screw gauge = \(\frac{1}{100}\) mm = 0.01 mm Diameter = Main scale reading + (Divisions on circular scale x least count) = 0 + \(\Big(52\times\frac{1}{100}\Big)\) = 0.52 mm Diameter = 0.052 cm |
|
| 248. |
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 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is(A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm |
|
Answer» Answer is (B) 5.124 cm For a given vernier callipers, 1 MSD = 5.15 - 5.10 = 0.05 cm 1 VSD = \(\frac{2.45}{50}\) = 0.049 cm ∴ L.C = 1 MSD - 1VSD = 0.001 cm Thus, the reading = 5.10 + (0.001 x 24) = 5.124 cm ⇒ diameter of cylinder = 5.124 cm |
|
| 249. |
In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count (in cm) of the callipers is(A) \(\Big(\frac{n-1}{n}\Big)x\) (B) \(\frac{nx}{(n - 1)}\) (C) \(\frac{x}{n}\) (D) \(\frac{x}{(n-1)}\) |
|
Answer» Answer is (C) \(\frac{x}{n}\) One main scale division, 1 M.S.D. = x cm One vernier scale division, 1 V.S.D. = \(\frac{(n-1)x}{n}\) Least count = 1 M.S.D. – 1 V.S.D. = \(\frac{nx-nx+x}{n}\) = \(\frac{x}{n}\) cmm. |
|
| 250. |
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?(A) A metre scale (B) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm (C) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm (D) A screw guage having 50 divisions in the circular scale and pitch as 1 mm |
|
Answer» Answer is (B) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm As per the question, the measured value is 3.50 cm. Hence the least count must be 0.01 cm = 0.1 mm For vernier scale, where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm 1 MSD = 1 mm and 9 MSD = 10 VSD, Least count = 1 MSD – 1 VSD = 0.1 mm Hence, correct option is (B). |
|