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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.
| 351. |
Round off `20.96` to three significant figuresA. `20.9`B. `20`C. `21.0`D. `21` |
| Answer» Correct Answer - C | |
| 352. |
Round off to four significant figures (A) 49.687 ,(B) 2.0095 |
| Answer» Correct Answer - `(A) to 49.69, (B) to 2.010` | |
| 353. |
STATEMENT-1 Number of significant figures in 0.007 s one and that in 0.700 is true STATEMENT 2 : All zeros in a number are not significant STATEMENT 3: Length of a rod is measured 100 m .Number significant figures in this measurement 1A. TFTB. FFTC. TFFD. FTT |
| Answer» Correct Answer - 3 | |
| 354. |
The number of significant figures in the measured value 26000 isA. FiveB. TwoC. ThreeD. Infinite |
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Answer» Correct Answer - B |
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| 355. |
The number of significant figures in the measured value 0.0204 isA. FiveB. ThreeC. FourD. Two |
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Answer» Correct Answer - B |
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| 356. |
A measured value 0.00274 expressed in two Significant digits isA. `0.0`B. `0.0028`C. `0.0027`D. `2.8xx10^(-3)` |
| Answer» Correct Answer - 3 | |
| 357. |
A : If displacement y of a particle executing simple harmonic motion depends upon amplitude a angular frequency `omega` and time t then the relation `y=asinomegat` cannot be dimensionally achieved. R : An equation cannot be achieved by dimensional analysis, if it contains dimensionless expressions.A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).C. If Assertion is true statement but Reason is false, then mark (3).D. If both Assertion and Reason are false statements, then mark (4). |
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Answer» Correct Answer - A |
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| 358. |
Round off the value 2.845 to three significant figuresA. 2.85B. 2.84C. 2.8D. 2.83 |
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Answer» Correct Answer - B |
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| 359. |
The number of significant figures in the measured value 4.700 m is the same as that in the valueA. 4700 mB. 0.047 mC. 4070 mD. 470.0 m |
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Answer» Correct Answer - D |
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| 360. |
A : An exact number has infinite number of significant digits. R : A number, which is not a measured value has infinite number of significant digits.A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).C. If Assertion is true statement but Reason is false, then mark (3).D. If both Assertion and Reason are false statements, then mark (4). |
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Answer» Correct Answer - B |
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| 361. |
If a calculated value 2.7465 g contains only three significant figures, the two insignificant digits in it areA. 2 and 7B. 7 and 4C. 6 and 5D. 4 and 6 |
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Answer» Correct Answer - C |
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| 362. |
If a measured quanityh has n significant figures, the relibale digits in it areA. `n`B. `n-1`C. `n+1`D. `n//2` |
| Answer» Correct Answer - B | |
| 363. |
A force of 8.26 N is applied normally over an area of `4.2 m^2`. Calculate the pressure exerted over the area to appropriate significant figures. |
| Answer» `8.26 -: 4.2 =1.96666 N//m^2`. Round off the value to two significant figure. Or `2.0 N//m^2` | |
| 364. |
The voltage across a lamp is 6.82 V when the current passing through it is 4.1 A. Find the power consumed to appropriate significant figures. |
| Answer» P = VI = (6.82) (4.1) W = 27.962 W Round off to two significant figures. Or 28 W | |
| 365. |
The voltage across a lamp is `(6.0 +- 0.1) V` and the current passing through it is `(4.0 +- 0.2)` ampare. Find the power consumed by the lamp. |
| Answer» (Power = V x l). Take the upper and the lower limits and find the maximum of the two or `(24.0 pm 1.6 ) ` W | |
| 366. |
One astronomical unit (AU) is equal toA. `9.46xx10^(15)m`B. `1.496xx10^(11)m`C. `3xx10^(8)m`D. `3.08xx10^(16)m` |
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Answer» Correct Answer - B |
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| 367. |
Check the dimensional correctness of the following relation. Volume of a sphere =`4/3pi"(radius)"^2` |
| Answer» Correct Answer - Not consistent | |
| 368. |
The relability of a measurement depends onA. PrecisionB. accuracyC. systematic errorD. random error |
| Answer» Correct Answer - B | |
| 369. |
The number of significant figures in 0.06900 isA. `1.5`B. `4`C. `2`D. `3` |
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Answer» Correct Answer - B In `ul(0). ul(0)6900`, the underlined zeroes are not significant figures are four (6900). |
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| 370. |
The temperature of two bodies measured by a thermometer are `t_1=20^@C pm 0.5^@C ` and `t_2=50^@C pm 0.5^@C` . Calculate the temperature difference and error there in .A. `30^(@)C+-1^(@)C`B. `70^(@)C+-0.5^(@)C`C. `30^(@)C+-0.5^@C`D. `70^(@)C+-1^(@)C` |
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Answer» Correct Answer - A Here, `t_(1)=20^(@)C+-0.5^(@)C,t_(2)=50^(@)C+-0.5^(@)C` The temperature difference of two bodies is `t=t_(2)-t_(1)=50^(@)C-20^(@)C=30^(@)C` the error in temperature difference of two bodies is `Deltat=(Deltat_(1)+Deltat_(2))=(0.5^(@)C+0.5^(@)C)=1^(@)C` `therefore` The temperature difference is `30^(@)C+-1^(@)C`. |
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| 371. |
Find the relative error in Z, if `Z=A^(4)B^(1//3)//CD^(3//2)`.A. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)+(DeltaC)/(C)+(3)/(2)(DeltaD)/(D)`B. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)-(DeltaC)/(C)-(3)/(2)(DeltaD)/(D)`C. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)+(DeltaC)/(C)-(3)/(2)(DeltaD)/(D)`D. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)-(DeltaC)/(C)+(3)/(2)(DeltaD)/(D)` |
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Answer» Correct Answer - A `Z=(A^(4)B^(1//3))/(CD^(3//2))`. The relative error in Z is given by `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)+(DeltaC)/(C)+(3)/(2)(DeltaD)/(D)` |
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| 372. |
A new unit of length is chosen such that the speed of light in vecuum is unity. What is the distance between the sun and the earth in terms of the new unit, if light takes 8 min and 20 sec. to cover the distance ?A. 300B. 400C. 500D. 600 |
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Answer» Correct Answer - C Here, Speed of light in vaccum `c=1` nw unit of length `s^(-1)` time taken by light of sun to reach the earth `t=8` min and 20 seconds `=(8xx60+20)s=500s` `therefore` Distance between the sun and the earth `S=ct=1` new unit of length `s^(-1)xx500s` `=500` new unit of length. |
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| 373. |
Light from the sun reaches the earth approximately inA. 5sB. 50sC. 500sD. 0.5s |
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Answer» Correct Answer - C Time taken by light to reach the earth from the sun is `t=8` min `20 s `=8xx60s+20s=480s+20s=500`s |
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| 374. |
The dimensional formula for energy isA. `[MLT^(-2)]`B. `[ML^2T^(-2)]`C. `[M^(-1) L^2T]`D. `[ M L^2 T]` |
| Answer» Correct Answer - 2 | |
| 375. |
The dimensional formula for refractive index isA. `[M^1L^1T^1]`B. `[M^0L^0T^0]`C. `[M^1L^0T^0]`D. `[MLT^(-1)]` |
| Answer» Correct Answer - 2 | |
| 376. |
What are the dimensions of the change in velocity ?A. `[M^(0)L^(0)T^(0)]`B. `[LT^(-1)]`C. `[MLT^(-1)]`D. `[LT^(-2)]` |
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Answer» Correct Answer - B |
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| 377. |
The dimensional formula for energy isA. `[MLT^(-2)]`B. `[ML^(2)T^(-2)]`C. `[M^(-1)L^(2)T]`D. `[ML^(2)T]` |
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Answer» Correct Answer - B |
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| 378. |
The pair of the quantities having same dimensions isA. Displacement, velocityB. Time, frequencyC. Wavelength focal lengthD. Force, acceleration |
| Answer» Correct Answer - 3 | |
| 379. |
The pair of the quantities having same dimensions isA. Displacement, velocityB. Time, frequencyC. Wavelength, focal lengthD. Force, acceleration |
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Answer» Correct Answer - C |
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| 380. |
STATEMENT 1: Two quantities with different dimensions may have same unit. STATEMENT-2 Two quantities with different units may have same dimensions STATEMENT-3 Unitless quantities must be dimensionless tooA. TFTB. TFFC. TTTD. FTT |
| Answer» Correct Answer - 4 | |
| 381. |
Which of the following units can possibly be a unit of time?A. Leap yearB. Tropical yearC. Sidereal yearD. Light year |
| Answer» Correct Answer - 1,2,3 | |
| 382. |
Which of the following functions of A and B may be performed if A and B possess different dimensionsA. A+BB. A-BC. `A/B`D. None of these |
| Answer» Correct Answer - 1,2 | |
| 383. |
The radius of a spherical ball is (10.4 `pm` 0.4) cm Select the correct alternativeA. The percentage error in radius is 3.9%B. The percentage error in radius is 0.4%C. The percentage error in volume is 11. 5%D. The absolute error in volume is 1.2 cm |
| Answer» Correct Answer - 1,3 | |
| 384. |
A packet contains silver powder of mass 20.23 g `pm` 0.01 g Some of the powder of mass 5.75 g `pm` 0.01 g is taken out from it The mass of the powder lell back isA. `14.48 g pm 0.00 g`B. `14.48 pm 0.02 g`C. `14.5 g pm 0.1 g `D. `14.5 g pm 0.2 g` |
| Answer» Correct Answer - 2 | |
| 385. |
Zero error of an instrument introducesA. Systematic errorB. Random errorC. Least count errorD. Personal error |
| Answer» Correct Answer - 1 | |
| 386. |
The currect voltage relation of diode is given by `1=(e^(1000 V//T)-1) mA`, where the applied voltage V is in volt and the temperature T is in degree Kelvin. If a student makes an error measuring `+-0.01 V` while measuring the current of `5 mA` at `300 K`, what will be error in the value of current in mA? |
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Answer» `I = (e^(1000V//T) - 1) mA` `dV = +- 0.01V, T = 300K,I = 5mA` `I + 1 = e^(1000V//T)` `log(I + 1) = (1000V)/(T)` `(dI)/(I + 1) = (1000)/(T) dV implies dI = 0.2mA` |
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| 387. |
For two resistors `R_1` and `R_2`, connected in parallel, find the relative error in their equivalent resistance. if `R_1=(50 pm2)Omega ` and `R_2=(100 pm 3) Omega`. |
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Answer» `R_"eq"=(R_1R_2)/(R_1+R_2)=33.33 Omega` and `1/R_"eq"=1/R_1+1/R_2` we have `(DeltaR_"eq")/(R_"eq"^2)=(DeltaR_1)/(R_1^2)+(DeltaR_2)/(R_2^2)` `rArr (DeltaR_"eq")/R_(eq)=((DeltaR_1)/(R_1^2)+(DeltaR_2)/(R_2^2))R_"eq"` or 0.036 |
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| 388. |
Assertion: The number `1.202 `has four significant figure and the number `0.0024` has two significant figure. Reason: All the zero digits are significant.A. if both assertion and reason are true reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation fo assertion.C. If assertion is true but reaso is false.D. IF both assertion and reason are false. |
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Answer» Correct Answer - B b |
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| 389. |
Two resistors of resistances `R_(1)=(300+-3) Omega` and `R_(2)=(500+-4) Omega ` are connected in series. The equivalent resistance of the series combination isA. `(800+-1)Omega`B. `(800+-7)Omega`C. `(200+-7)Omega`D. `(200+-1)Omega` |
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Answer» Correct Answer - B The equivalent resistance of series combination is `R_(s)=R_(1)+R_(2)=300Omega+500Omega=800Omega` the error in equivalent resistace is given by `DeltaR(DeltaR_(1)+DeltaR_(2))=(3+4)Omega=7Omega` hence, the equivalent resistance along with error is `(800+-7)Omega` |
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| 390. |
Two resistances `R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega` are connected in series . Find the equivalent resistance of the series combination.A. `(66.7+-1.8)Omega`B. `(66.7+-4.0)Omega`C. `(66.7+-3.0)Omega`D. `(66.7+-7.0)Omega` |
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Answer» Correct Answer - A Here, `R_(1)=(100+-3)Omega,R_(2)=(200+-4)Omega` The equivalent resistance is given by `(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2)),(1)/(R_(p))=(1)/(100)+(1)/(200)=(3)/(200),R_(p)=(200)/(3)=66.7Omega` The error in equivalent resistance given by `(DeltaR_(p))/(R_(p)^(2))=(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2)),DeltaR_(p)=DeltaR_(1)((R_(p))/(R_(1)))^(2)+DeltaR_(2)((R_(p))/(R_(2)))^(2)` `=3((66.7)/(100))^(2)+4((66.7)/(200))^(2)=1.8Omega` |
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| 391. |
The velocity fo a body is expressed as `V = G^(a) M^(b) R^(c)` where `G` is gravitational constant. `M` is mass, `R` is radius. The values of exponents `a, b` and `c` are:A. `(1)/(2),(1)/(2), -(1)/(2)`B. `1,1,1`C. `(1)/(2),(1)/(2), (1)/(2)`D. `1,1, (1)/(2)` |
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Answer» Correct Answer - A `[LT^(-1)]alpha [M^(-1) L^(3) T^(-2)]^(a) M^(b) L^(c)` |
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| 392. |
The velocity fo a spherical ball through a visocous liquid is given by `v = v_(0) (1- e^(kt))`, where `v_(0)` is the initial velocity and `t` represents time. If `K` depends on radius of ball `(r)`, coefficient of viscosity `(eta)` and mass fo the ball `(m)`, tehnA. `k = mr//eta`B. `k = eta m//r`C. `k = r eta//m`D. `k = mr eta` |
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Answer» Correct Answer - C `k alpha r^(a) eta^(v) m^(c), T^(-1) alpha L^(a) [M^(-1) T^(-1)]^(b) M^(c)` |
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| 393. |
If force `(F)`, work `(W)` and velocity `(V)` are taken as fundamental quantites then the dimensional formula of Time `(T)` isA. `[W^(1) F^(1) V^(1)]`B. `[W^(1) F^(1) V^(-1)]`C. `[W^(-1) F^(-1) V^(-1)]`D. `[W^(-1) F^(-1) V^(-1)]` |
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Answer» Correct Answer - D `T alpha F^(x) W^(y) V^(z), M^(0) L^(0) T^(1) alpha[MLT^(-2)]^(x) [ML^(2) T^(-2)]^(y) [LT^(-1)]^(z)` |
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| 394. |
If force `F`, Length `L` and time `T` are chosen as fundamental quantites, the dimensional formula for Mass isA. `[FLT]`B. `[F^(-1) L^(-1) T^(-2)]`C. `[F^(-2) L^(-2) T^(-2)]`D. `[F^(1) L^(-1) T^(2)]` |
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Answer» Correct Answer - D `M prop F^(a) L^(b) T^(c)` |
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| 395. |
The error in the measurement of the length of the sample pendulum is `0.2%` and the error in time period `4%`. The maximum possible error in measurement of `(L)/(T^(2))` isA. `4.2%`B. `3.8%`C. `7.8%`D. `8.2`%` |
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Answer» Correct Answer - D Let `x = (L)/(T^(2)) , (Delta x)/(x)= (Delta L)/(L) + 2 (Delta T)/(T)` |
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| 396. |
If the length of a simple pendulum is recorded as `(90 +- 0.2)cm` and period as `(1.9 +- 0.02)s`, the percentage fo error in the measurement of acceleration due to gravity isA. `4.2`B. `2.1`C. `1.5`D. `2.8` |
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Answer» Correct Answer - B `g = 4pi^(2) (l)/(T^(2)), (Delta g)/(g) xx 100 = (Delta l)/(l) xx 100 + (2Delta T)/(T) xx 100` |
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| 397. |
Which of the following statements is incorrect regarding mass?A. it is a basic property of matter.B. The SI unit of mass is kg.C. The mass of an atom is expressed in u.D. It depends upon the temperature, pressure or location of the object in space. |
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Answer» Correct Answer - D Mass does not depend on the temperature, pressure or location of the object in space. |
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| 398. |
If minutes is the unit of time. `10ms^(-2)` is the unit of acceleration and `100 kg` is the unit of mass, the new unit of work in joule isA. `10^(5)`B. `10^(6)`C. `6xx10^(6)`D. `36xx10^(6)` |
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Answer» Correct Answer - D `w alpha Ma^(2) T^(2) , (w_(2))/(w_(1)) = (M_(2) a_(2)^(2) T_(2)^(2))/(M_(1) a_(1)^(2) T_(1)^(2))` |
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| 399. |
If the units of mass, tieme and length are `100g, 20cm` and 1 minute respectivelty then equivalent energy for `1000 erg` in the new system will beA. `90`B. `900`C. `2xx10^(6)`D. `300` |
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Answer» Correct Answer - A `n_(1)[M_(1) L_(1)^(2) T_(1)^(-2)] = n_(2) [M_(2) L_(2)^(2) T_(2)^(-2)]` |
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