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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.
| 51. |
The relation `tan theta = v^(2)// r g` gives the angle of banking of the cyclist going round the curve . Here `v` is the speed of the cyclist , `r` is the radius of the curve , and `g` is the acceleration due to gravity . Which of the following statements about the relation is true ?A. It is both dimensionally as well as numerically correct.B. It is neither dimensionally correct correct.C. It is dimensionally correct but not numerically.D. It is numerically correct but not dimensionally. |
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Answer» Correct Answer - A Here , `[ tan theta] = [(V^(2))/( rg)] = M^(0) L^(0)T^(0)` . Also , in actual expression for the angle of banking of a road , there is no numerical factor involved. Therefore , the relation is both numerically and dimensionally correct. |
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| 52. |
Which of the following length measurements is most precise and why? (a) `2.0 cm` , (b) `2.00 cm` , ( c ) `2.000 cm` |
| Answer» (c ) `2.000 cm` , because it contains maximum number of significant figures , `4`. | |
| 53. |
Which of the following product of `e , h , mu , G` ( where ` mu` is permeability ) be taken so that the dimensions of the product are same as that of the speed of light ?A. `he^(-2) mu^(-1) G^(0)`B. `h^(2) e G^(0) mu`C. `h^(0)e^(2) G^(-1) mu`D. `he^(-2) mu^(0)` |
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Answer» Correct Answer - A Here `v = e^(a) h^(b) mu^( c ) G^(d) `. Taking the dimensions, `M^(0) LT^(-1) A^(0)` ` = [AT^(1)] ^(a) [ML^(2) T^(-1) ]^(b) [ MLT^(-2) A^(-2)]^( c ) [ M^(-1) L^(3) T^(-2) ]^(d)` There will be four simultaneous equations by equating the dimensions of `M,L , T, and , A`. These are ` a-2c = 0 , a - b - 2c -2d = -1 , b+ c - d = 0 and 2b + c + 3d = 1`. Solving for `a , b , c and d`, we get ` a = -2 , b = 1 , c = -1 , d = 0` Thus , `v = e^(-2) h mu^(-1) G^(0)` |
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| 54. |
Given that `T` stands for time and `l` stands for the length of simple pendulum . If `g` is the acceleration due to gravity , then which of the following statements about the relation `T^(2) = ( l// g)` is correct?A. It is correct both dimensionally as well as numerically.B. It is neither dimensionally correct nor numerically.C. It is dimensionally correct but not numerically.D. It is numerically correct but not dimensionally. |
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Answer» Correct Answer - C The correct relation for time period of simple pendulum is `T = 2 pi ( l// g)^(1//2)`. So the given relation is numerically incorrect as the factor of ` 2 pi` is missing. But it is correct dimensionally. |
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| 55. |
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. VolumeC. PressureD. Area |
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Answer» Correct Answer - D As `mu = ("Velocity of light in vaccum")/("Velocity of light in medium")`, hence `mu` is dimensionless. Thus , each term on the RHS of given equation should be dimensionless , I,e, `B // lambda^(2)` is dimensionless, i.e., `B` should have dimension of `lambda^(2)`, i.e., `cm^(2)`, i.e., area. |
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| 56. |
The resistance `R= (V)/(I)`, where `V= (100+-5.0) V` and `I=(10+-0.2)A`. Find the percentage error in `R`. |
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Answer» Correct Answer - `7%` The percentage error in V is 5% and in I it is 2%. The total error in `R` would therefore br ` 5% + 2% = 7%` |
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| 57. |
In a number without decimal , what is the significant of zeros on the right of non - zero digits? |
| Answer» Not significant . In a number without decimal , the zeros on the right of non- zero digits are not significant. | |
| 58. |
Convert `1 MW` power on a new system having basic units of mass , length , and time as `10 kg , 1 dm , and 1 min` , respectively . |
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Answer» `[P] = [ML^(2)T^(-3)]` Using the relation , `n_(2) = n_(1) [(M_(1))/(M_(2))]^(x) [(L_(1))/(L_(2))]^(y) [(T_(1))/(T_(2))]^(z)` ` = 1xx 10^(6) [( 1kg)/( 10 kg) ]^(1) [ ( 1 m)/( 1 dm)]^(2) [ ( 1 s)/( 1 min) ]^(-3)` `[ As 1 MW = 10^(6) W ]` ` = 10^(6) [( 1 kg)/( 10 kg)] [ ( 10 dm)/( 1dm)]^(2) [ ( 1 s)/( 60 s)]^(-3) = 2.16 xx 10^(12) unit` |
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| 59. |
Convert `54 km h^(-1)` into `m s^(-1)`. |
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Answer» Let `v = 54 km h^(-1) = n_(2) m s^(-1)`. `[v] = LT^(-1) , a = 0 , b = 1, c = -1` `n_(2) = 54 [(kg)/(g) ]^(0) [ ( km)/(m)]^(1) [(h) /( s)]^(-1)` `= 54 xx 1 xx 1000 xx [ 3600 ]^(-1) = ( 54 xx 1000)/( 3600) = 15 ` Hence , `54 km h ^(-1) = 15 m s^(-1)`. |
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| 60. |
The displacement covered by a body in time `(5.0 +- 0.6) s is ( 40.4 +- 0.4 ) m`. Calculate the speed of the body . Also determine the percentage error in the speed. |
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Answer» Here ` s = 40.0 +- 0.4 m and t = 5.0 +- 0.6 s`. Therefore, speed , `v = ( s) /( t) = ( 40.0)/(5.0) = 8.0 m s^(-1)` As ` v = (s)/( t) , therefore , ( Delta v)/( v) = ( Delta s)/( s) + ( Delta t)/(t)` Here `Delta s = 0.4 m , s = 40.0 m , Delta t = 0.6 s, t = 5.0 s`. Therefore, or `( Delta v)/( v) = ( 0.4)/( 40.0) + ( 0.6) / (5.0) rArr Delta v = [ 0.01 + 0.12 ] xx 8.0 = 1.04` Hence , `v = ( 8.0 +- 1.04) ms^(-1)`, percentage error `(( Delta v)/( v) xx 100) = 0.13 xx 100 = 13%` |
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| 61. |
The length of a rectangular sheet is `1.5 cm` and breadth is ` 1.023 cm` . Find the area of the face of a rectangular sheet to the correct number of significant figures. |
| Answer» `Area = 1.5 xx 1.203 = 1.8045 cm^(2) = 1.8 cm^(2)` ( Upto correct number of significant figure). | |
| 62. |
The length and breadth of a field are measured as : `l = ( 120 +- 2) m and b = ( 100 +- 5) m` , respectively. What is the area of the field? |
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Answer» Now `(Delta A)/( A) = ( Delta l)/(l) + ( Delta b)/( b) = ((2)/(120) + (5)/(100)) = 0.0667,` `Delta A = 0.0667 xx A ` Now `A = l . B = 120 xx 100 = 12000 m^(2)` rArr `Delta A = 0.0667 xx 12000 = 800.4 m^(2)` Area of the field `= A +- Delta A` `= 12000 +- 800.4 = ( 1.2 +- 0.08) xx 10^(4) m^(2)` |
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| 63. |
The position of a particle at time `t` is given by the relation `x(t) = ( v_(0) /( alpha)) ( 1 - c^(-at))`, where `v_(0)` is a constant and `alpha gt 0`. Find the dimensions of `v_(0) and alpha`. |
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Answer» From the principle of dimensionsal homogenity , `[ alpha t] =` dimensionless :. `[ alpha] = [(1)/( t)] = [ T^(-1)]` Similarly , `[x] = ([v_(0)])/([ alpha]) = [L] [T^(-1)] = [LT^(-1)]` |
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| 64. |
The position of a particle at time `t` is given by the relation `x(t) = ( v_(0) /( alpha)) ( 1 - c^(-at))`, where `v_(0)` is a constant and `alpha gt 0`. Find the dimensions of `v_(0) and alpha`.A. `M^(0) LT^(-1) and T^(-1)`B. `M^(0) LT^(1) and T^(-1)`C. `M^(0) LT^(-1) and LT^(-2)`D. `M^(0) LT^(-1) and T` |
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Answer» Correct Answer - A Here `at` is dimensionless. So ` a = (1)/( t) = [(1)/(T)] = [T^(-1)] x = (V_(0))/(a) and V_(0) = xa = [ LT^(-1)]` ` = [M^(0)LT^(-1)]` |
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| 65. |
The time dependence of a physical quantity `P` is given by `P = P_(0)e^(-alpha t^(2))` , where `alpha` is a constant and `t` is time . Then constant `alpha` is//hasA. DimensionlessB. Dimensions of `T^(-2)`C. Dimensions of `P`D. Dimensions of `T^(2)` |
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Answer» Correct Answer - B Here `alpha t^(2)` is a dimensionless. Therefore , `alpha = (1)//(t^(2))` and has the dimension of `T^(-2)`. |
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| 66. |
The mass of a box is `2.3 kg`. Two marbles of masses `2.15 g and 12.39 g` are added to it . Find the total mass of the box to the correct number of significant figures. |
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Answer» Total mass `= 2.3 + 0.00215 + 0.01239 = 2.31 kg` The total mass in appropriate significant figures will be `2.3 kg` |
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| 67. |
Each side of a cube is measured to be `7.203 m` . Find the volume of the cube up to appropriate significant figures. |
| Answer» `Volume = a^(3) = ( 7.023)^(3) = 373.715 m^(3)` | |
| 68. |
Each side of a cube is measured to be `5.402 cm`. Find the total surface area and the volume of the cube in appropriate significant figures. |
| Answer» Total surface area ` = 6 xx ( 5.402)^(2) = 175.09 cm^(2) = 175.1 cm^(2)` (Upto correct number of significant figure). | |
| 69. |
Taking into account the figures , what is the value of `9.99 m + 0.0099 m`? |
| Answer» ` 9.99 m + 0.0099 m = 9.999 m = 10.00 m `( In proper significant figure). | |
| 70. |
A research worker takes `100` observations in an experiment . If he repeats the same experiment by taking `500` observation , how is the probable error affected? |
| Answer» Probable error reduces to `1//5` as the number of observations is made `5` times. | |
| 71. |
Which quantity in a given formula should be measured most accurately Why? |
| Answer» Quantity having higher powers, because errors get multiplied by powers. | |
| 72. |
A sperical body of mass `m` and radius `r` is allowed to fall in a medium of viscosity `eta`. The time in which the velocity of the body increases from zero to `0.63 ` times the terminal velocity `(v)` is called constant `(tau)`. Dimensionally , `tau` can be represented byA. `(mr^(2))/( 6 pi eta)`B. `sqrt((6 pi m r eta)/( g^(2)))`C. ` (m)/( 6 pi eta r v)`D. None of these |
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Answer» Correct Answer - D `[ ( m r^(2))/( 6 pi eta)] = [ (ML^(2))/( ML^(-1) T^(-1))] = [L^(3) T]` As we have `[ eta] = [ML^(-1) T^(-1)]` `[(( 6 pi m r eta)/( g^(2)))^(1//2) ] = [ (( MLML^(-1) T^(-1))/( L^(2) T^(-4)))^(1//2)]` ` [ (m) /( 6 pi eta r v)] = [ (M) /( ML^(-1) T^(-1) LLT^(-1))] = [L^(-1) T^(2)]` Thus , none of the given expressions have the dimensions of time . |
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| 73. |
A student writes four different expressions for the displacement `y` in a periodic motion . Which of the following can be correct?A. ` y = a T sin ( 2 pi t)/(T)`B. ` y = a sin V t`C. ` y = (a)/( T) sin ( t)/(a)`D. ` y = (a)/ (sqrt(2))[ sin (2pi t)/(T) + cos (2pi t)/(T)]` |
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Answer» Correct Answer - D Since `LHS` is displacement , so `RHS` should have dimensions of displacement . In `(a) , aT` does not have the dimensions of displacement . Also the argument of a trigonometric function should be dimensionless . In `(b)` , argument is not dimensionless and in `( c) , a//T` does not have the dimensions of displacement . Hence , the correct choice is `(d)`. |
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| 74. |
A student when discussing the properties of a medium ( except vaccum) writes Velocity of light in vaccum = Velocity of light in medium This formula isA. Dimensionally correctB. Dimensionally incorrectC. Numerically incorrectD. Both `a and c` |
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Answer» Correct Answer - D The formula can be written as `("Velocity of light in vaccum") /("Velocity of light in medium") = 1` This formula is dimensionally correct as both the sides are dimensionless. Numerically , this ratio is equal to refractive index which is greater than `1`. Hence , the equation is numerically incorrect. |
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| 75. |
A famous relation in phyics relates the moving mass ` m` to the rest mass `m_(0)` of a particle in terms of its speed `v` and the speed of light `c`.( This relation first arose as a consequence of the special theory of relativity due to Albert Einstein). A body recalls the relation almost correctly but forgets where to put the constant `c` . He writes `m = (m_(0))/((1- V^(2))^(1//2))`. Guess where to put the missing `c`. |
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Answer» According to the principle of homogenity of dimensions , powers of `M,L , T` on either side of the formula must be equal. For this , on `RHS` , the denominator `(1 - v^(2))^(1//2)` should be dimensionless. Therefore, instead of `(1 - v^(2))^(1//2)` , we should write `( 1 - v^(2)// c^(2))^(1//2)`. Hence, the correct formula would be `m = (m_(0))/(( 1 - v ^(2)// c^(2))^(1//2))` |
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| 76. |
The `SI` unit of inductance, the henry can be written asA. Weber/ampereB. Volt -second/ ampereC. `"Joule"// ("ampere")^(2)`D. Ohm - second |
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Answer» Correct Answer - A::B::C::D `L = ( phi)/(I) , L = - e//((dI)/( dt)) , L = ( 2U)/(I^(2)) , L = R xx t` |
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| 77. |
The length and breadth of a rectangle are `( 5.7 +- 0.1 ) cm and ( 3.4 +- 0.2 ) cm`, respectively calculate the area of rectangle with error limits. |
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Answer» Correct Answer - `( 19 +- 1.5) cm^(2)` `A = l b = 5.7 xx 3.4 = 19.38 cm^(2) ~ 19 cm^(2)` ( two significant figures) `(Delta A )/( A) = (Delta l )/( l) + ( Delta b)/(b) rArr Delta A = ((0.1)/(5.7) + (0.2)/(3.4)) ( 19.38)` `Delta A = 1.48 = 1.48 ~ 1.5` ( two significant figures) So area `= ( 19 +- 1.5) cm^(2)` |
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| 78. |
The sides of a rectangle are `(10.5 +- 0.2) cm and ( 5.2 +- 0.1 ) cm`. Calculate its perimeter with error limits . |
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Answer» Correct Answer - `0.6 cm ; 31.4 +- 0.6 cm` ` l = 10.5 +- 0.2 cm , b = 5.2 +- 0.1 cm` `P = 2 l + 2 b = 31.4 cm, DeltaP = 2( Delta l + Delta b) = 0.6 cm` So `P = 31.4 +- 0.6 cm` |
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| 79. |
A body travels uniformly a distance of `( 13.8 +- 0.2) m` in a time `(4.0 +- 0.3) s`. Find the velocity of the body within error limits and the percentage error. |
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Answer» Here `S = ( 1.38 +- 0.2 ) cm , t = ( 4.0 +- 0.3 ) s` ` V = (13.8)/(4.0) = 3.45 ms^(-1)` `(Delta V)/( V) = +- (( Delta S )/( S) + ( Delta t)/( t) ) = +- ((0.2)/(13.8) + (0.3)/(4.0)) = +- 0.0895` `Delta V = +- 0.0895 xx 3.45 = +- 0.3` ( rounded off to one place of decimal) `V = ( 3.45 +- 0.3) ms^(-1)` Percentage error `= ( Delta V)/( V) xx 100 = 0.0895 xx 100 = 8.95 %` |
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| 80. |
The radius of a sphere is measured to be `(2.1 +- 0.02) cm`. Calculate its surface area with error limits. |
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Answer» Correct Answer - `55.4 +- 26.4 cm^(2)` `r = 2.1 +- 0.5 cm` rArr `A = 4 pi r^(2) = 4 xx (22)/(7) (2.1)^(2)` `= 4 xx 13.86 = 55.4 cm^(2)` `( Delta A)/( A) = 2( Delta r )/( r ) rArr Delta A = ( 2 xx 0.5)/(2.1) xx 55.44 = 26.4` Area ` = 55.4 +- 26.4 cm^(2)` |
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| 81. |
10 rotations of the cap of a srew gauge is equivalent to 5 mm . The cap has 100 divisons. Find the least count . A reading taken for the diameter of wire with the srew gauge shows four complete rotations and 35 divisions on the circular scale . Find the diameter of the wire. |
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Answer» Least count ` = ( 0.5)/(100) = 0.005 mm` The diameter of the wire ` = ( 4 xx 0.5 + 35 xx 0.005 ) mm` `= 2.175 mm` |
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| 82. |
The length of one rod is `2.53 cm` and that of the other is `1.27 cm`. The least count of the measuring instrument is `0.01 cm`. If the two rods are put together end to end , find the combined length. |
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Answer» Length `l = 2.53 + 1.27 = 3.80 cm, Delta l = 0.01 + 0.01 = 0.02` (Most probable errors of both the rods are added) Hence true value `= ( 3.80 +- 0.02) cm` |
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| 83. |
The values of measurement of a physical quantity in five trails were found to be ` 1.51 , 1.53 , 1.53, 1.52 , and 1.54`.A. Average absolute error is `0.01`.B. Relative error is `0.01.`C. Percentage error is `0.01%`.D. Percentage error is `1%`. |
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Answer» Correct Answer - A::B::D Mean value ` = ( 1.51 + 1.53 + 1.53 + 1.52 + 1.54)/(5) = 1.53` Absolute errors are `( 1.53 - 1.51 = 0.02)`, `( 1.53 - 1.53 = 0.00) , ( 1.53 - 1.53) = 0.00)`, `( 1.53 - 1.52 = 0.01 ) and ( 1.54 - 1.53 = 0.01)` Mean absolute error is `( 0.02 + 0.00 + 0.00 + 0.01 + 0.01)/(5) = ( 0.04)/( 5) = 0.008 ~~ 0.01` So choice (a) is correct. Relative error ` = ( 0.01)/( 1.53) = 0.00653 ~~ 0.01 % error = (0.01)/( 1.53) xx 100 ` ` = 1%` |
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| 84. |
Find the dimensions of physical quantity `X` in the equation `"Force" = ( X) /("Density")`. |
| Answer» `[ X] = ["Force"] xx ["Density"] = [MLT^(-2)] xx [ ML^(-3)] = [ M^(2) L^(-2) T^(-2)]` | |
| 85. |
In resonance tube experiment , the velocity of sound is given by `v = 2 f_(0) ( l_(2) - l_(1))` . We found `l_(1)=25.0 cm` and `l_(2)=75.0 cm`. If there is no error in frequency, what will be the maximum permissible errror in the speed of sound ? (Take `f_(0) = 325 Hz`) |
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Answer» `V = 2f_(0)(l_(2) - l_(1))` `(dV) = 2f_(0) ( dl_(2) - dl_(1))` `(dV)_(max) = max of [ 2f_(0) (+- Delta l_(2) +- Deltal_(2)]` `l_(1) = 25.0 cm rArr l_(1) = 0.1 cm` ( place value of last number) `l_(1) = 75.0 cm rArr l_(2) = 0.1 cm` ( place value of last number) So the maximum permissible error in the speed of sound `(dV)_(max)` `= 2( 325 Hz) ( 0.1 cm + 0.1 cm ) = 1.3 m s^(-1)` Value of `V = 2f_(0) (l_(2) - l_(1) = 2(325 Hz) (75.0 cm - 25.0 cm)` ` = 325 m s^(-1)` So `V = (325 +- 1.3) m s^(-1)`. |
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| 86. |
The mass of the liquid flowing per second per unit area of cross section of the tube is proportional to `P^(x) and v^(y)` , where `P` is the pressure difference and `v` is the velocity , then the relation between x and Y isA. ` x = y `B. ` x = -y`C. ` y^(2) = x`D. ` y = -x^(2)` |
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Answer» Correct Answer - B `(M)/(At) prop P^(x) v^(y)` rArr `ML^(-2) T^(-1) = [ML^(-1)T^(-2)]^(x) [L^(1) T^(-1)]^(y) = M^(x) L^(-x + y )T^(-2 x - y)` ` x = 1 , -x + y = -2 , and -2x -y = -1` From here, we get `y = -1` . Thus , `x = -y`. |
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| 87. |
The length , breadth , and thickness of a block are measured as `125.5 cm , 5.0 cm , and 0.32 cm `, respectively .Which one of the measurement is most accurate? |
| Answer» Relative error in measurement of length is minimum . So this measurement is most accurate. | |
| 88. |
If `P` represents radiation pressure , `C` represents the speed of light , and `Q` represents radiation energy striking a unit area per second , then non - zero integers `x, y, z` such that `P^(x) Q^(y) C^(z)` is dimensionless , find the values of `x, y , and z`. |
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Answer» `[ P^(x) Q^(y) C^(z)] = M^(0) L^(0) T^(0)` Substituting the dimension of each quantity in the given expression , `[ML^(-1) T^(-2)]^(x) [MT^(-3)]^(y) [LT^(-1)]^(z) = [ M^(x+y) L^(-x + z) T^(-2x - 3y -z)] = M^(0) L^(0)T^(0)` By equating the power of `M , L , and T `on both sides , we get ` x + y = 0 , -x + z = 0 , and -2x -3y -z = 0` By solving , we get ` x = 1, y = -1 , and z =1`. |
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| 89. |
Dimensionsal methods provide three major advantages in verification , deviation , and changing the system of units . Any empirical formula that is derived based on this method has to be verified and propportionality constants found by experimental means . The presence or absence of certain factors - non dimensional constants or variables - cannot be identified by this method . So every dimensionally correct relation cannot be taken as perfectly correct. If `alpha` kilogram , `beta` meter , and `gamma` second are the fundamental units , `1 cal` can be expressed in new units as `[ 1 cal = 4.2 J]`A. `alpha^(-1) beta ^(2) gamma`B. `alpha^(-1) beta^(-2) gamma`C. ` 4.2 alpha^(-1) beta`D. ` 4.2 alpha^(-1) beta^(-2) gamma^(2)` |
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Answer» Correct Answer - D ` 1 cal = 4.2 J = 4.2 kg m^(2) s^(-2) = n_(2) (alpha kg) (beta m)^(2) (gamma s)^(-2)` `rArr n_(2) = 4.2 alpha ^(-1) beta^(-2) gamma^(2)` So , `1 cal = (4.2 alpha^(-1) beta^(-2) gamma^(2))` new units |
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| 90. |
If velocity (V) , force (F), and energy (E) are taken as fundamental units , then find the dimensional formula for mass. |
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Answer» Let `M = V^(a) F^(b) E^( c ) ` Putting dimensioons of each quantities on both sides `[M] = [LT^(-1)] ^(a) [MLT^(-2)]^(b) [ML^(2)T^(2)] c` Equating powers of dimensions , we have ` b + c =1 , a+b + 2 c = 0 , and -a -2b- 2c = 0` Solving these equations , ` a = -2 , b = 0 , and c= 1`. So ` M = [ V^(-2) F^(0) E]` |
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| 91. |
If the persent units of length , time and mass `( m , s, kg)` are changed to `100 m , 100 s, and (1)/(10) kg `, then how will the new unit of force change ? |
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Answer» Unit of velocity `= m s^(-1)`, In new system `= ( 100 m)/( 100 s) = ( m)/( s)` (same) Unit of force `= ( kg xx m)/(s^(2))` In new system `= ( 1)/( 10) kg xx ( 100 m xx 100 m)/( 100 s xx 100 s) = (1)/( 10) ( kg xx m^(2))/( s^(2))` Unit of pressure `= ( kg) /( m xx s^(2))` In new system `= (1)/( 10) kg xx (1)/(100) m xx (1)/( 100 s xx 100 s) = 10^(-7) ( kg)/( m xx s^(2))` |
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| 92. |
Dimensionsal methods provide three major advantages in verification , deviation , and changing the system of units . Any empirical formula that is derived based on this method has to be verified and propportionality constants found by experimental means . The presence or absence of certain factors - non dimensional constants or variables - cannot be identified by this method . So every dimensionally correct relation cannot be taken as perfectly correct. The time period of oscillation of a drop depends on surface tension `sigma` , density of the liquid `rho` , and radius ` r ` . The relation isA. ` sqrt (( rho r^(2))/( sigma))`B. ` sqrt (( r^(2))/( rho sigma))`C. ` sqrt (( r^(3) rho)/( sigma))`D. ` sqrt (( rho sigma)/( r^(3)))` |
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Answer» Correct Answer - C Let `T prop sigma ^(a) rho^(b) r^(c )` `M^(0) L^(0) T = (MT^(-2))^(a) (ML^(-3))^(b) L^(c )` Equating the powers of `M: 0 = a + b rArr b = -a` `L : 0 = -3b + c rArr c = 3b` `T : 1 = -2a rArr a = -(1)/(2), b = (1)/(2) , c = (3)/(2) rArr T = k sqrt((r^(3)rho)/(sigma))` |
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| 93. |
The equation of a wave is given by `Y = A sin omega ((x)/(v) - k)`, where `omega` is the angular velocity and `v` is the linear velocity. Find the dimension of `k`. |
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Answer» According to principle of dimensional homogenity , `[k] = [(x)/(v)] = [(L)/(LT^(-1))] = [T]` |
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| 94. |
The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`.A. ` M^(1) L^(7//2) T^(-2)`B. ` M^(1) L^(11//2) T^(-2)`C. ` M^(1) L^(5//2) T^(-2)`D. ` M^(1) L^(9//2) T^(-2)` |
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Answer» Correct Answer - B Here `x^(2)` has the dimensiions of `L^(2) , B = [L^(2)]` Also `M L^(2) T^(-2) = (AL^(1//2))/(L^(2))` or `A = ML^(7//2 T^(-2))` `:. A xx B = M L^(1//2 T^(-2))` |
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| 95. |
The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`. |
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Answer» From the principle of dimensional homogenity , `[x^(2)] = [ B]` :. `[B] = [L^(2)]` As well as `[U] = ([A] [x^(1//2)])/( [x^(2)] + [B]) rArr [ ML^(2)T^(-2)] = ([A] [L^(1//2)])/([L^(2)])` :. `[A] = [ ML^(7//2)T^(-2)]` Now `[AB] = [ ML^(7//2)T^(-2)] xx [L^(2)] = [ML^(11//2)T^(-2)]` |
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| 96. |
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is `0.5mm` and there are `50` divisions on the circular scale. The reading on the main scale is `2.5mm` and that on circular scale is `20` divisions. if the measured mass of the ball has a relative error of `2%`, the relative percentage error in the density isA. `0.9%`B. `2.4%`C. `3.1%`D. `4.2%` |
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Answer» Correct Answer - C Least count `= ( 0.5)/( 50) = 0.01 mm` Diameter of ball D `= 2.5 mm + ( 20) (0.01) = 2.7 mm` `rho = (M)/( Vol) = (M)/((4)/( 3) pi (( D)/(2))^(3))` `(( Delta rho)/( rho)) = ( Delta m)/( m) + 3 ( Delta D)/( D) = 2% + 3((0.01)/( 2.7)) xx 100% = 3.1%` |
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| 97. |
The relative density of a material is found by weighing the body first in air and then in water . If the weight in air is `( 10.0 +- 0.1) gf` and the weight in water is `( 5.0 +- 0.1) gf`, then the maximum permissible percentage error in relative density isA. `1`B. `2`C. `3`D. `5` |
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Answer» Correct Answer - D Relative density ` = (W_(a))/(W_(a) - W_(w)) , rho = (W_(a))/( w)` where ` rho` is relative density , `W_(a)` is weight in air , and `w` is loss in weight. `(Delta rho)/( rho) = ( Delta W_(a))/( W_(a)) - ( Delta w)/(w)` For maximum error , `( Delta rho)/( rho) = (Delta W_(a)) /( W_(a)) + ( Delta w)/( w)` For maximum percentage error , `( Delta rho)/( rho) xx 100 = ( Delta W_(a))/( W_(a)) xx 100 + ( Delta w)/( w) xx 100` Given ` Delta W_(a) = 0.1 gf and W_(a) = 10.0 gf` ` w = 10.0 - 5.0 = 5.0 gf` `Delta w = Delta W_(a) + Delta W_(w) = 0.1 + 0.1 = 0.2 gf` `( Delta rho)/( rho) xx 100 = (( 0.1)/( 10.0)) xx 100 + (( 0.2)/( 5.0)) xx 100 = 1 + 4 = 5 ` |
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| 98. |
The relative density of material of a body is found by weighting it first in air and then in water . If the weight in air is `( 5.00 +- 0.05) N` and the weight in water is `(4.00 +- 0.05) N`. Find the relative density along with the maximum permissible percentage error. |
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Answer» Weight in air ` = ( 5.00 +- 0.05) N` weight in water `= ( 4.00 +- 0.05 ) N` Loss of weight in water `= ( 1.00 +- 0.1 ) N` Now relative density `= ( "weight in air")/( "weight loss in water")` i.e., `RD = ( 5.00 +- 0.05) /( 1.00 +- 0.1)` Now relative density with maximum permissible error `= (5.00) /( 1.00) +- (( 0.05)/( 5.00) + (0.1)/(1.00)) xx 100` `= 5.0 +- (1+10)% = 5.0 +- 11%` |
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| 99. |
A diamond weighs `3.71 g`. It is put into a box weighing `1.4 kg`. Find the total weight of the box and diamond to the correct number of significant figures. |
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Answer» Correct Answer - `~ 1.4 kg` `3.17 g = 0.00371 = 1.40371 kg ~ 1.4 kg` [ correct up to one place of decimal] |
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| 100. |
The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of he body in air is `W_(1) = 8.00 +- N` and the weight in water is `W_(2) = 6.00 +- 0.05 N`, then the relative density `rho_(r ) = W_(1)//( W_(1) - W_(2))` with the maximum permissible eror isA. `4.00 +- 0.62%`B. `4.00 +- 0.82%`C. `4.00 +- 3.2%`D. `4.00+- 5.62%` |
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Answer» Correct Answer - D Relative density `rho_(r ) = (W_(1))/( W_(1) - W_(2)) = ( 8.00) /( 8.00 - 6.00) = 4.00` `( Delta rho_(r))/( rho_(r )) xx 100 = ( Delta W_(1))/( W_(1)) xx 100 + ( Delta (W_(1) - W_(2)))/( W_(1) - W_(2)) xx 100` ` = ( 0.05)/( 8.00) xx 100 + ( 0.05 + 0.05)/( 2) xx 100 = 5.62%` :. `rho_(r ) = 4.00 +- 5.62%` |
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