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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.
| 201. |
What is the value of `(5.0 xx 10^(-6))(5.0xx10^(-8))` with due regards to significant figures ?A. `25xx10^(-14)`B. `25.0xx10^(-14)`C. `2.50xx10^(-13)`D. `250xx10^(-15)` |
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Answer» Correct Answer - A The product must have two significant digit as both the figures being multiples have two significant digits. |
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| 202. |
The radius of a thin wire is `0.16mm`. The area of cross section taking significant figures into consideration in square millimeter isA. 0.08B. `0.080`C. `0.0804`D. `0.080384` |
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Answer» Correct Answer - B `R`=0.16mm Hence, `A=piR^(2)=(22)/(7)xx(0.16)^(2)=0.080384` Since radius has two significant figure so answer also will have two significant figures. |
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| 203. |
The diameter of a wire is measured to be 0.0250 `xx` `10^(-4)` m. The number of significant figures in the measurement isA. fiveB. fourC. threeD. nine |
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Answer» Correct Answer - C |
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| 204. |
The voltage across a lamp, is 6.32V when the current passing through it is 3.4 A. Find the power consumed to appropriate significant figures. |
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Answer» Voltage across a lamp=6.32 V (3 significant figure) Current flowing through lamp=3.4 A (2 significant figure) `therefore` Power consumed, `P=VI=(6.32)(3.4)=21.488 W` Answer should have minimum number of significant figure. Here, the minimum number of significant figure is 2 `therefore` Power consumed=2 W |
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| 205. |
(a) the displacement of a particle is given by `s - a sin (omega t = kx)`, where `t` is in second and `x` is in meter. Find the dimensions of `omega` and `k`. (b) The velocity of a praticle is given by `v = v_(0) e^(-lambda t)`, where `t` is time. Find the dimensions of `v_(0)` and `lambda`. |
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Answer» (a) `s = a sin (omega t - kx)` Angles have no dimensions `(omega t = kx)` is dimensionless ltbr `omega t`is dimensions, `[k] = L^(-1)` (b) Power of exponential is dimensionsless. `lambda t` is dimensions, `[lambda] = T^(-1)` Dimensions of `v_(0) =` Dimensions of `v`, i.e., `LT^(-1)` `[V_(0)] = LT^(-1)` |
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| 206. |
If the length of rod A is 3.25 `pm` 0.01 cm and that of B is 4.19 `pm` 0.01 cm then the rod B is longer than rod A byA. `0.94+-0.00cm`B. `0.94+-0.01cm`C. `0.94+-0.02cm`D. `0.94+-0.005cm` |
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Answer» Correct Answer - C |
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| 207. |
If `x=at+bt^(2)`, where `x` is the distance travelled by the body in kilometres while `t` is the time in seconds, then the units of `b` areA. `km//s`B. `km-s`C. `km//s^(2)`D. `km-s^(2)` |
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Answer» Correct Answer - C |
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| 208. |
What is the number of significant figures in `0.310xx10`A. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - B |
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| 209. |
The unit of percentage error isA. Same as that of physical quantityB. Different from that of physical quantityC. Percentage error is unit lessD. Errors have got their own units which are different from that of physical quantity measured |
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Answer» Correct Answer - C |
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| 210. |
The decimal equivalent of `(1)/(20)` up to three significant figures isA. 0.05B. 0.05C. 0.005D. `5.0xx10` |
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Answer» Correct Answer - A |
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| 211. |
`MLT^(-1)` represents the dimensional formula ofA. PowerB. MomentumC. ForceD. Couple |
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Answer» Correct Answer - B |
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| 212. |
Hertz is the unit forA. FrequencyB. ForceC. Electric chargeD. Magnetic flux |
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Answer» Correct Answer - A |
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| 213. |
Unit of moment of inertia in MKS systemA. `kgxxcm^(2)`B. `kg//cm^(2)`C. `kgxxm^(2)`D. `"Joule"xxm` |
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Answer» Correct Answer - C |
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| 214. |
Unit of magnetic moment isA. `"Ampere-metre"^(2)`B. `"Ampere-metre"`C. `"Weber-metre"^(2)`D. `"Weber"//"metre"` |
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Answer» Correct Answer - A |
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| 215. |
The units of modulus of rigidity areA. `N-m`B. `N//m`C. `N-m^(2)`D. `N//m^(2)` |
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Answer» Correct Answer - D |
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| 216. |
The dimensional formula for the modulus of rigidity isA. `ML^(2)T^(-2)`B. `ML^(-1)T^(-3)`C. `ML^(-2)T^(-2)`D. `ML^(-1)T^(-2)` |
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Answer» Correct Answer - D |
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| 217. |
Find the dimensions of stress, strain and modulus of elasticity.A. `MLT^(-1)`B. `ML^(2)T^(-1)`C. `MLT(-2)`D. `M^(0)L^(0)T^(0)` |
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Answer» Correct Answer - D |
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| 218. |
A highly rigid cubical block `A` of small mass `M` and side `L` is fixed rigidly on the other cubical block of same dimensions and of modulus of rigidity `eta` such that the lower face of `A` completely covers the upper face of `B`. The lower face of `B` is rigidly held on a horizontal surface . `A` small force `F` is applied perpendicular to one of the side faces of `A`. After the force is withdrawn , block `A` executes faces of `A`. After the force is withdrawn , block `A` exceutes small oscillations , the time period of which is given byA. `2pisqrt((M eta)/L)`B. `2pisqrt(L/(M eta))`C. `2pisqrt((ML)/(eta))`D. `2pisqrt(M/(egaL))` |
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Answer» Correct Answer - D |
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| 219. |
Whose dimensions is `ML^(2)T^(-1)`A. TorqueB. Angular momentumC. PowerD. Work |
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Answer» Correct Answer - B |
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| 220. |
If `E=mc^(2)` wher `m=` mass of the body `c=` speed of light Guess the name of physical quantity `E`.A. EnergyB. PowerC. MomentumD. None of these |
| Answer» Correct Answer - (a) | |
| 221. |
One bar is equivalent of `10^(5)N//m^(2)`. The atmosphere pressure is `1.0313xx10^(5)N//m^(2)`. The same in bar isA. 1.88 barB. 1.013 barC. 2.013 barD. none of these |
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Answer» (b) 1 bar `=10^(5)N//m^(2)` `=11.013xx10^(5)N//m^(2)=(1.013xx10^(5))/(10^(5))"bar" =1.013"bar"` |
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| 222. |
The time between human heart beat is `8xx10^(-1)s`. How many heart beats are measured in one minute.A. 75B. 60C. 82D. 64 |
| Answer» (a) `n=1/(8xx10^(-1)s)=(1"minute")/(8xx10^(-1)s)=(60s)/(8xx10^(-1)s)=600/8=75` | |
| 223. |
One nautical mile is `6080ft`. The same is kilimetre isA. `0.9km`B. `0.8km`C. `1.85km`D. None of these |
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Answer» (c) `:. n_(1)u_(1)=n_(2)u_(2)` or `6080ft=n_(2)km` `6080xx12xx2.54cm=n_(2)xx10^(3)xx100cm` `:.n_(2)=(6080xx12xx2.54)/(10^(5))=1.85km` |
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| 224. |
The acceleration of a car is 10 mile per hour per second. The same is `(ft)//(s^(2))` isA. `1.467(ft)/(s^(2))`B. `14.67(ft)/(s^(2))`C. `40ft//s^(2)`D. none of these |
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Answer» (a) `n_(1)u_(1)=n_(2)u_(2)` or `(10"mile")/("hour second")=n_(2)(ft)/(s^(2))` or `(10xx1760xx3ft)/(60xx60s^(2))=n_(2)(ft)/(s^(2))` `:. n_(2)=1.467ft//s^(2)` |
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| 225. |
The height of the building is `50ft`. The same in millimetre isA. `560mm`B. `285mm`C. `1786.8mm`D. `15240 mm` |
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Answer» (d) `n_(1)u_(1)=n_(2)u_(2)` `:. n_(2)=(n_(1)u_(1))/(u_(2))implies(50 ft)/(mm) =(50xx12xx2.54cm)/(0.1cm)=15240mm` `:. 50ft=15240mm` |
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| 226. |
Which of the following is the most recise device for measuring length?A. A vernier calliper with 20 divisioins of the sliding scaleB. An optical insturment that can measure length within wavelength of lightC. A screw gauge of pitch `1mm` and 100 division on the circular scaleD. None of the above |
| Answer» (b) An optical instrument gives most precise measurement. | |
| 227. |
Which one is not a unit of time?A. Leap yearB. YearC. ShakeD. Light yeat |
| Answer» (d) leap year, year and shake are units of time and light year is the unit of distance. | |
| 228. |
If the force is given by `F=at+bt^(2)` with `t` is time. The dimensions of `a` and `b` areA. `[MLT^(-4)]` and `[MLT^(-2)]`B. `[MLT^(-3)` and `[MLT^(-4)]`C. `[ML^(2)T^(-3)]` and `[ML^(2)T^(-2)]`D. `[ML^(2)T^(-3)]` and `[ML^(3)T^(-4)]` |
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Answer» (b) Dimension of at `=` Dimension of `F` `[at]=[F]implies[a]=[F/t]` `[b]=[(MLT^(-2))/T]implies[a]=[MLT^(-3)]` Dimension of `bt^(2)=` Dimension of `F` `[bt^(2)]=[F]=[b]=[F/(t^(2))]` `[b]=[(MLT^(-2))/(T^(2))]implies[b]=[MLT^(-4)]` |
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| 229. |
A physical quantity is given by `X=[M^(a)L^(b)T^(c)]`. The percentage error in measurement of `M,L` and `T` are `alpha, beta, gamma` respectively. Then the maximum % error in the quantity `X` isA. `a alpha+b beta+c gamma`B. `a alpha+b beta-c gamma`C. `a/(alpha)+b/(beta)+c/(gamma)`D. None of the above |
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Answer» (a) `X=[M^(a)L^(b)T^(c)]` Maximum percentage error `=a alpha +b beta + cgamma` |
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| 230. |
A cube has a side of length `1.2xx10^(-2)m`. Calculate its volumeA. `1.7xx10^(-6)m^(3)`B. `1.73xx10^(-6)m^(3)`C. `1.70xx10^(-6)m^(3)`D. `1.732xx10^(-6)m^(3)` |
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Answer» (a) Volume `V=l^(3)=(1.2xx10^(-2)m)^(3)=1.728xx10^(-6)m^(3)` since length `(l)` has two significant figure, the volume `(V)` will also have two significant figure. Therefore, the corect answer is `V=1.7xx10^(-6)m^(3)` |
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| 231. |
The dimensions of the quantity hc (where `h=h/(2pi)`) isA. `[ML^(2)T^(-1)]`B. `[MLT^(-1)]`C. `[ML^(3)T^(-2]`D. `[ML^(2)T^(-1)]` |
| Answer» (c) `hc=Elamda=[ML^(2)T^(-2)][L]=[ML^(3)T^(-2)]` | |
| 232. |
The dimensions of frequency isA. `[T^(-1)]`B. `[M^(0)L^(0)T]`C. `[M^(0)L^(0)T^(-2)]`D. none of these |
| Answer» Correct Answer - (a) | |
| 233. |
The dimensions of wavelength isA. `[M^(0)L^(0)T^(0)]`B. `[M^(0)LT^(0)]`C. `[M^(0)L^(-1)T^(0)]`D. none of these |
| Answer» Correct Answer - (a) | |
| 234. |
The optical path difference is defined as `Deltax=(2pi)/(lamda)`. What are dimensions of optical path difference?A. `[M^(0)L^(-1)T^(0)]`B. `[M^(1)L^(1)T^(0)]`C. `[ML^(0)T^(1)]`D. `[ML^(-2T]` |
| Answer» Correct Answer - (a) | |
| 235. |
It is claimed that the two cesium clocks, if allowed t run for 100 yr, free from any disturbance, may differ by only about `0.02s`. Which of the following is the corret fractional error?A. `10^(-9)`B. `10^(-5)`C. `10^(-13)`D. `10^(-11)` |
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Answer» (d) Time interval `=100` years `=100xx365xx24xx60xx60s` `=3.155xx10^(9)s` Differene in time `=0.2s` `:.` Fractional error `=("Difference in time" (s))/("Time interval"(s))` `=0.2/(3.155xx10^(9))=6.34xx10^(-12)=10xx10^(-12)~~10^(-11)` |
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| 236. |
The dimensional formula for magnetic flux isA. `ML^(2)T^(-2)A^(-1)`B. `ML^(0)T^(-2)A^(-2)`C. `M^(0)L^(-2)T^(-2)A^(-3)`D. `ML^(2)T^(-2)A^(3)` |
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Answer» Correct Answer - A |
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| 237. |
If the value of resistance is 10.845 ohm and the value of current is 3.23 amp, the value of potential with significant numbers would beA. 35.0 VB. 3.50 VC. 35.029 VD. 35.030 V |
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Answer» Correct Answer - A |
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| 238. |
The length, breadth and thickness of a block are given by l=12 cm , b=6cm and t=2.45 cm. The volume of the block according to the idea of significant figures should beA. `1xx10^(2)" cm"^(3)`B. `2xx10^(2)" cm"^(3)`C. `1.764xx10^(2)" cm"^(3)`D. None of these |
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Answer» Correct Answer - B Using the relation for volume, `V` = Length `xx` Breadth `xx` Thickness =12`xx`6`xx`2.45=176.4 `cm^(3)` `V` =1.764`xx` `10^(2)` `cm^(3)` The minimum number of significant figure is 1 in breadth. Hence, the volume will contain only one significant figure. Therefore, `V=2xx10^(2)cm^(3)`. |
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| 239. |
A gas bubble from an explosion under water oscillates with a period `T` proportional to `P^(a) d^(b) E^(c )`, where `P` is the pressure, `d` is density of water and `E` is the total energy of the explosion. Find the value of a,b and c`.A. `a=1, b=1, c=2`B. `a=1, b=2, c=1`C. `a=(5)/(6), b=(1)/(2), c=(1)/(3)`D. `a=-(5)/(6), b=(1)/(2), c=(1)/(3)` |
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Answer» Correct Answer - D Given, `Tpropp^(a)d^(b)E^(c)` We have `["M"^(0)"L"^(0)"T"]=k["ML"^(-1)"T"^(-2)]^(a)["ML"^(-3)]^(b)["ML"^(2)"T"^(-2)]^(c)` where, `k` is a constant. On comparing dimensions of similar terms, we have `["M"^(0)"L"^(0)"T"]=k["M"^(a+b+c)"L"^(-a-3b+2c)"T"^(-2a-2c)]` On comparing powers of `M` we have `" "0=a+b+c" "...(i)` On comparing powers of `L`, we have `" "0=-a-3b+2c" "...(ii)` On comparing powers of `T`, we have `" "1=-2a-2c" "...(iii)` On solving Eqs. (i), (ii) and (iii), we have `" "a=-(5)/(6),b=(1)/(2),c=(1)/(3)` |
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| 240. |
A gas bubble from an explosion under water oscillates with a period `T` proportional to `P^(a) d^(b) E^(c )`, where `P` is the pressure, `d` is density of water and `E` is the total energy of the explosion. Find the value of a,b and c`. |
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Answer» `T prop P^(a) d^(b) E^(c )` `[T] prop [ML^(-1) T^(-2)]^(a) [ML^(-3)]^(b) [ML^(2) T^(-2)]^(c )` `M^(0) L^(0)T^(1) prop M^(a + b + c)L^(-a-3b+2c) T^(-2a - 2c)` Comparing powers of `M, L` and `T` `a + b + c = 0` `- a - 3b + 2c = 0` `- 2a - 2c = 1` From (iii), `a + c = - (1)/(2)` in (i), we get `b = (1)/(2)`, putting the value of `b` in (ii) `-a-3 ((1)/(2)) + 2c = 0 implies - a + 2c = (3)/(2)` From (iii), `a + c = - (1)/(2)` `- a + 2c = (3)/(2)` Adding `(v)` and `(vi)`, `3c = 1 implies c = (1)/(3)` `a + c = - (1)/(2)` ` a + (1)/(3) = - (1)/(2) implies = - (5)/(6)` `a = - (5)/(6), b = (1)/(2), c = (1)/(3)` |
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| 241. |
Joule-second is the unit ofA. WorkB. MomentumC. PressureD. Angular momentum |
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Answer» Correct Answer - D |
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| 242. |
The unit of specific resistance isA. `Ohm//cm^(2)`B. `Ohm//cm`C. `Ohm-cm`D. `(Ohm-cm)^(-1)` |
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Answer» Correct Answer - C |
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| 243. |
The binding energy of a nucleon in a nucleus is of the order of a fewA. eVB. ErgsC. MeVD. Volts |
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Answer» Correct Answer - C |
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| 244. |
`1eV` isA. Same as jouleB. `1.6xx10^(-19)J`C. `1V`D. `1.6xx10^(-19)C` |
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Answer» Correct Answer - B |
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| 245. |
`{:(,"Column I",,"Column II"),(("A"),"Work",(p),["A"^(1//2)"T"^(-1)]),(("B"),"Moment of inertia",(q),["FA"^(1//2)]),(("C"),"Velocity",(r),["FA"^(1//2)"T"^(2)]):}` |
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Answer» Correct Answer - `("A"to"q","B"to"r","C"to"p")` [A]=`["L"^(2)]` `therefore" "["L"]=["A"^(1//2)]` `["T"]=["T"]rArr["F"]=["MLT"^(-2)]` `therefore" "["M"]=["FL"^(-1)"T"^(2)]=["FA"^(1//2)"T"^(2)]` Now, `" "["W"]=["FL"]=["FA"^(1//2)]` `["I"]=["ML"^(2)]["FA"^(-1//2)"T"^(2)"A"]=["FA"^(1//2)"T"^(2)]` `["V"]=["LT"^(-1)]=["A"^(1//2)"T"^(-1)]` |
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| 246. |
A physical quantity `X` is defined by the formula `X=(IFv^(2))/(WL^(3))` where `I` is moment of inertia, `F` is force, `v` is velocity, `W` is work and `L` is length, the dimensions of `X` areA. `["MLT"^(-2)]`B. `["MT"^(-2)]`C. `["ML"^(2)"T"^(-3)]`D. `["LT"^(-1)]` |
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Answer» Correct Answer - B We knows, Dimensions of moment of inertia = `["ML"^(2)]` Dimensions of moment of force `F=["MLT"^(-2)]` Dimensions of moment of velocity `(v)=["LT"^(-2)]` Dimensions of moment of work `(W)=["ML"^(2)"T"^(-2)]` Dimensions of moment of length `(l)=["L"]` Here, Dimensions of `[X]=("Dimensions of "[IFv^(2)])/("Dimensions of "[WL^(3)])` `" "=(["ML"^(2)]["MLT"^(-2)]["LT"^(-2)]^(2))/(["ML"^(2)"T"^(-2)]["L"^(3)])` `(["M"^(2)"L"^(5)"T"^(-4)])/(["ML"^(-5)"T"^(-2)])=["MT"^(-2)]` |
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| 247. |
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. `0.1s`B. `0.11s`C. `0.01`D. `1.0s` |
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Answer» Correct Answer - B |
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| 248. |
The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g isA. `0.1%`B. `1%`C. `0.2%`D. `0.8%` |
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Answer» Correct Answer - C |
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