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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A force F is applied on a square plate of side L. If the percentage error in the determination of L is 2% and that in F is 4%. What is the permissible error in pressure?A. 0.08B. 0.06C. 0.04D. 0.02 |
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Answer» Correct Answer - A `P = F/A = F/L^2 = FL^-2` Permissible error (%) in P = (% error in F)+2(% error in L). |
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| 2. |
The mass and density of a solid sphere are measured to be `(12.4 +-0.1)kg and (4.6 +- 0.2) kg//m^3`. Calculate the volume of the sphere with error limits . |
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Answer» Correct Answer - A::B::C::D Here, `m+- Delta m = (12.4 +-0.1)kg ` and ` rho +- Delta rho= ((4.6 +- 0.2)kg)/(m^(3))` Volume ` V = m/rho = 12.4/4.6` ` = 2.69m^3 = 2.7m^3` Now, ` DeltaV/V = +- ((Deltam)/(m) + (Deltarho)/(rho))` or `DeltaV = +- ((Deltam)/(m)+(Deltarho)/(rho)) xx V` `= +- ((0.1)/(12.4) + (0.2)/(4.6)) xx 2.7 = +-0.14` `:. V +- DeltaV = (2.7 +- 0.14)m^3` |
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| 3. |
A thin wire has length of 21.7 cm and radius 0.46 mm. Calculate the volume of the wire to correct significant figures? |
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Answer» Correct Answer - A::C::D `V = pi r^2 l ` `=(pi) (0.046 cm)^2 (21.7 cm)` `=0.1443112 cm^3 = 0.14 cm^3 ` (rounding off to two significant digits.) |
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| 4. |
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? |
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Answer» Correct Answer - B `g = 4 pie^2L//T^2` Here, `T= (t)/(n)` and `(DeltaT) = (Deltat)/(n)`. Therefore, `(DeltaT)/(t) = (Deltat)/(t)`. The errors in both L and t are the least count errors. Therefore, ` (Delta g)/g = (DeltaL)/L+2(DeltaT)/T ` `= (Delta L)/L + 2(Delta t)/t ` ` = (0.1)/(20.0) + 2((1)/(90)) = 0.027 ` Thus, the percentage error in g is ` `( Delta g)/gxx100 = 2.7%` |
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| 5. |
The radius of a sphere is measured to be `(1.2 +- 0.2)cm`. Calculate its volume with error limits. |
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Answer» Correct Answer - B::C Volume, `V = 4/3 pi r^3 = 4/3 (22/7)(1//2)^3` `= 7.24 cm^3 = 7.2 cm^3` Further, `(DeltaV)/V = 3 ((Deltar)/r)` ` DeltaV= 3((Deltar)/r)V = (3xx0.2xx7.2)/(1.2)` = 3.6`cm^3` V = `(7.2 +- 3.6)cm^3` . |
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| 6. |
Let us now use a millimeter scale (on which millimeter marks are there). This is also our normal meter scale which we use in our routine life. From the figure, we can see that length AB is more than 3.3 cm and less than 3.4 cm. If we note down the length, ` `l= AB = 3.4cm. ` Then, this measurement has two significant figures 3 and 4 in which 3 is absolutely correct and 4 is reasonably correct (doubtful). Least count of this scale is 0.1 cm because this scale can measure accurately only upto 0.1 cm. Futher, maximum uncertainly or maximum possible error in l can also be 0.1 cm. |
| Answer» Correct Answer - A | |
| 7. |
Radius of a wire is 2.50 mm. The length of the wire is 50.0 cm. If mass of wire was measured as 25g, then find the density of wire in correct significant figures. `[Given, pi = 3.14, exact]` |
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Answer» Correct Answer - B::C Given , r = 2.50 mm = 0.250cm Further given that, ` l = 50.0 cm ` m = 25gm `pi = 3.14 exact ` `rho = m/V = m/(pi r^2 l)` `= (25)/((3.14)(0.250)(0.250)(50.0))` `= 2.5477 g//cm^3` But in the measured values, least number of significant figures are two. Hence, we will round off the result to two significant figures. `:. rho = 2.5 g//cm^(3)` |
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| 8. |
Find, volume of a cube of side a = `1.4 xx (10^-2)` m. |
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Answer» Correct Answer - A::B::C Volume `V=a^3` `=(1.4 xx 10^(-2)) xx (1.4 xx 10^(-2)) xx (1.4 xx 10^(-2)) = 2.744 xx 10^(-6) m^(3)` Since, each value of `a` has two significant figures. Hence, we will round off the result to two significant figures. `V = 2.7 xx 10^(-6) m^(3)`. |
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| 9. |
Let us use a centimeter scale (on which only centimeter scales are there) to measure a length AB. From the figure, we can see that length `AB` is more than `7 cm` and less than `8cm` . In this case, Least Count `(LC)` of this centimeter scale is 1 cm, as it can measure accurately, upto centimeters only. If we note down the length (l) of line AB as l = 7cm then maximum uncertainly or maximum possible error in l can be 1 cm(=LC), because this scale can measure accurately only upto 1cm. |
| Answer» Correct Answer - A | |
| 10. |
If we measure a length l=3.267 cm with the help of a screw gauge, then (a) What is maximum uncertainty or maximum possible error in l? (b)How many significant figures are there in the measured length ? (c) Which digits are absolutely correct and which is/are doubtful? |
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Answer» Correct Answer - A::B::C::D (a) `16.235 xx 0.217 xx 5` `=17.614975` `=20` (rounding off to minimum one number of significant figure) (b) `0.00435 xx 4.6` `= 0.02001` `=0.020` (rounding off to minimum two number of significant figures) |
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| 11. |
A physical quantity `rho` is related to four variables `alpha, beta, gamma` and `eta` as `rho=((alpha^3)(beta^2))/(etasqrt( gamma))` The percentage errors of measurements in `alpha, beta, gamma` and `eta` are `1%, 3%, 4%` and `2%` respectively. Find the percentage error in `rho`. |
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Answer» Correct Answer - A::C `%` error in `rho = 3` (`%`error in `alpha`) `+ 2` (`%` error in `beta`) `+1/2` (`%` error in gamma) `+` (`%` error in `eta`) |
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| 12. |
Find the percentage error in specific resistance given by `rho = (pir^2R)/(l)` where r is the radius having value `(0.2+-0.02)`cm, R is the resistance of `(60+-2) ohm` and l is the length of `(150+-0.1)`cm. |
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Answer» Correct Answer - B::C::D `(Delta rho)/(p) xx 100=[2(Deltar)/(r)+(DeltaR)/(R)+(DeltaI)/(I)]xx100 ` ` =[(2xx0.002)/(0.2)+(2)/(60)+(0.1/150)]xx100 ` `23.4%` |
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| 13. |
Solve with due regards to significant figures `4.0 xx (10^-4) - 2.5 xx 10^(-6)`. |
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Answer» Correct Answer - A::D `(4.0 xx 10^(-4)) - (0.025 xx 10^-4)` `= 4.025 xx 10^(-4)` `=4.0 xx 10^(-4)` upto one decimal place. |
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| 14. |
Add `6.75 xx (10^3) cm` to `4.52 xx (10^2) cm` with regard to significant figures. |
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Answer» Correct Answer - A::B::C `a= 6.75 xx 10^3 cm` `b= 4.52 xx 10^2 cm` `=0.452 xx 10^(3) cm ` ` = 0.452 xx 10^(3) cm `(upto 2 places of decimal) `:. a+b = = (6.75 xx 10^(3)+ 0.45 xx 10^(3)) cm ` `=7.20 xx 10^3 cm ` |
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| 15. |
Evaluate`(25.2 xx 1374)/(33.3)`. All the digits in this expression are significant. |
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Answer» Correct Answer - A::D We have `(25.2 xx 1374)/(33.3) = 1039.7838` Out of the three numbers given in expression 25.2 and 33.3 have 3 significant digits and 1374 has four. The answer should have three significant digits. Rounding 1039.7838……. To three significant digits, it becomes 1040. Thus we write `(25.2 xx 1374)/(33.3) = 1040`. |
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| 16. |
Find density when a mass of 9.23 kg occupies a volume of `1.1m^3`. Take care of significant figures. |
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Answer» Correct Answer - C::D `rho=m/V = 9.23 /1.1` `= 8.3909090 kg//m^3 = 8.4 kg//m^3` (rounding off to two significant digits.) |
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| 17. |
Write down the number of significant figures in the following (a) 6428 (b)62.00 (c)0.00628 cm (d) 1200N |
| Answer» Correct Answer - A::B::C::D | |
| 18. |
The number of significant figures in 3400 isA. 3B. 1C. 4D. 2 |
| Answer» Correct Answer - D | |
| 19. |
Simplify and round to the appropriate number of significant digits (a) `16.235 xx 0.217 xx 5 ` (b) `0.00435 xx 4.6` |
| Answer» Correct Answer - A::B | |
| 20. |
Round to the appropriate number of significant digits (a) 13.214 + 234.6 +7.0350 +6.38 (b) 1247 + 134.5 + 450 + 78 |
| Answer» Correct Answer - A::B | |
| 21. |
Round 742396 to four, three and two significant digits. |
| Answer» Correct Answer - B::D | |
| 22. |
Round off 231.45 to four, three and two significant digits. |
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Answer» Correct Answer - B::C `231.5` , (four significant digits) `231` , (three significant digits) `230` , (two significant digits) |
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| 23. |
Length, breadth and thickness of a rectangular slab are 4.234 m, 1.005 m and 2.01 m respectively. Find volume of the slab to correct significant figures. |
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Answer» Correct Answer - C `V = (4.234 xx 1.005 xx 2.01)m^3 ` `=8.5528917 m^3 = 8.55 m^3` (rounding off to three significant digits.) |
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| 24. |
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. `1 xx 10^(2) cm^(3)`B. `2 xx (10^2)cm^3`C. `1.763 xx (10^2)cm^3`D. None of these |
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Answer» Correct Answer - B `V = lbt = 12xx 6 xx 2.45 ` ` = 176.4 cm^3 = 2 xx 10^(2) cm^3` . (rounding off to one significant digit of breadth) |
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| 25. |
Round the following numbers to 2 significant digits (a)3472 (b)84.16 (c)2.55 (d)28.5 |
| Answer» Correct Answer - A::B::C::D | |
| 26. |
The length and breadth of a metal sheet are 3.124m and 3.002m respectively. The area of this sheet upto correct significant figure isA. `9.378 m^(2)`B. `9.37 m^2`C. `9.4 m^2`D. None of these |
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Answer» Correct Answer - A `A = l xx b = 3.124 xx 3.002` `=9.378248 m^2` ` = 9.378 m^2`. (rounding off to four significant digits) |
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