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.
| 51. |
If L,R, C denote inductance, resistance and capacitance, respectively .Then dimensions of `(1)/(R^(2)C)` areA. `[M^(0)L^(0)T]`B. `[MLT^(-1)]`C. `[A^(-1)M^(0)L^(0)T^(-1)]`D. `[M^(0)L^(0)T^(0)]` |
|
Answer» Correct Answer - d `(L)/(R ) =` time constant of `L-R` circuit `M^(0)L^(0)T]` `RC =` time constant of `R-C` circuit `= [M^(0)L^(0)T]` The dimensions of `(L^(2))/(R^(2)C) = ((L)/(R)) xx (1)/(RC) = ([M^(0)L^(0)T])/([M^(0)L^(0)T]) = [M^(0)L^(0)T^(0)]`. |
|
| 52. |
The radius of a sphere is `1.41`. its volume to an appropring number of significant figure isA. `11.73 cm^(3)`B. `11.736 cm^(3)`C. `11.7 cm^(3)`D. `117 cm^(3)` |
|
Answer» Correct Answer - c Radius of the sphere ,`r = 1.41 cm (3 "significant")` Volume of the sphere `V = (4)/(3) pi r^(3) = (4)/(3) xx 3.14 xx(1.41)^(3)` `= 11.736cm^(3)` Rounded off up to`3` significant) figure `= 11.7cm^(3)` |
|
| 53. |
The fundamental unit of quantity of metter isA. `kg`B. `gram`C. `mol`D. `"tonne"` |
| Answer» Correct Answer - c | |
| 54. |
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.A. `(3.5 +- 0.6 )ms^(-1)`B. `(3.5 +- 0.3)ms^(-1)`C. `(6.1 +- 0.6 )ms^(-1)`D. `(6.1 +- 0.3 )ms^(-1)` |
|
Answer» Correct Answer - b Here `s = (13.8 +- 0.2)m` `T = (4.0 +- 0.3)s` `:. "Velocity" v = (s)/(t) = (13.8)/(4.0) = 3.45 ms^(-1)` (Rounded off to first place of decimal `:. (Delta v)/(v)= (Delta s)/(s)+(Delta t)/(t)= (0.2)/(13.8) + (0.3)/(4.0)` `= (0.8+4.14)/(13.8 xx 4.0) = (4.94)/(13.8 xx4.0) = 0.0895` or `Delta v = v xx 0.0895 = 3.45 xx 0.0895 = 0.3087` `:. "velocity" = (3.5 +- 0.3)ms^(-1)` |
|
| 55. |
If there is a positive error of ` 50%` in the measurement of velocity of a body , find the error in the measurement of kinetic energy.A. `25%`B. `50%`C. `100%`D. `125%` |
|
Answer» Correct Answer - d Kinetic energy `E= (1)/(2) mv^(2)` `:. (Delta E)/(E) xx 100 = (v^(-2) - v^(2))/(v^(2)) xx 100` `:. (Delta E)/(E) xx 100 = 125 %` |
|
| 56. |
Write the dimensions of `a//b` in the relation `P = ( a - t^(2))/( bx)` , where `P` is the pressure , `x` is the distance , and `t` is the time .A. `[M^(2)LT^(-3)]`B. `[MT^(-2)]`C. `[LT^(-3)]`D. `[ML^(3)T^(-1)]` |
|
Answer» Correct Answer - b `(a - t^(2))/(b) = P x rArr [(a - t^(2))/(b)] = [Px]` `rArr [(a)/(b)] = [Px] = ml^(-1)T^(-2)L = Ml^(-2)` |
|
| 57. |
A physical quantity `P` is given by `P = (A^(3) B^(1//2))/( C^(-4) D^(3//2))`. Which quantity among `A, B , C, and D` brings in the maximum percentage error in `P`?A. AB. BC. CD. D |
|
Answer» Correct Answer - c Quantity C has maximum power So it bring maximum error in P. |
|
| 58. |
Find the dimensional formula of (a) coefficient of viscosity `eta` (b)charge `q` (c ) potention `V` (d) capacitance `C` and ( e) resistance `R` Some of the equations containing these quantities are `F= -etaA [[Delta v )/(Delta l]], q = It. U = VIt, q = CV and V = IR ` where A denotes Area, the v the velocity , I is the length ,I the electric current, t the time and U the energy .A. `[ML^(2)T^(-2)]`B. `[ML^(-1)T^(1)]`C. `[ML^(-2)T^(-2)]`D. `[M^(0)L^(0)T^(0)]` |
|
Answer» Correct Answer - b `F = - eta A(Delta v)/(Delta z) rArr [eta] = [ML^(-1)T^(-1)]` As `F = [MLT^(-2)] A = [L^(2)], (Delta v)/(Delta z)= [T^(-1)]`. |
|
| 59. |
Temperature can be expressed as a derived quantity in terms of any of the followingA. Length and massB. mass and timeC. Length , mass and timeD. None of these |
|
Answer» Correct Answer - d Because temperature is a fundamental quantity |
|
| 60. |
The sum of the numbers 436.32, 227.2 and 0.301 in appropriate singnificant figures isA. `663.821`B. `664`C. `663.8`D. `663.82` |
|
Answer» Correct Answer - b The sum of the number can be calculated as `663.82` arothmetically The number with least decimel place is `227.2` is correct to nly one decimel place The final result should therefore be rounded off one decimal place i.e. `664` |
|
| 61. |
Assertion: if two physical quantities have same dimension, then they can be certainly added or subtracted because Reason: if the dimension of both the quantities are same then both the physical quantities should be similar .A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - a The unit for the base quantities are callled the base unit The unit of all physics quantities can be expressed as combinations of the base units. |
|
| 62. |
The resistance of a metal is given by `R = V//I` , where `V` is potential difference and `I` is current . In a circuit , the potential difference across resistance is `V = ( 8 +- 0.5) V` and current in resistance , `I = ( 4 +- 0.2)A`. What is the value of resistance with its percentage error?A. `4 +- 16.25 %`B. `4 +- 6.25 %`C. `4 +- 10 %`D. `4 +- 8 %` |
|
Answer» Correct Answer - a We know that `R = V//l = 8//2 = 4 Omega` percentage error in `V` is `(0.5)/(8) xx 100 = 6.25` percentage error in `l` is `(0.2)/(2) xx 100 = 10` Now, add up the percentage error and get the answer. |
|
| 63. |
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. `0.1 s`B. `0.11 s`C. `0.01 s`D. `1.0 s` |
|
Answer» Correct Answer - b Average value `= (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/(5)` `= 2.62sec` Now `|Delta T_(1)| = 2.63 - 2.62 = 0.01` `|Delta T_(2)| = 2.62 - 2.56 = 0.06` `|Delta T_(3)| = 2.62 - 2.42 = 0.20` `|Delta T_(4)| = 2.71 - 2.62 = 0.09` `|Delta T_(5)| = 2.80 - 2.62 = 0.18` Mean absoltude `Delta T = (|Delta T_(1)| + |Delta T_(2)| + |Delta T_(3)| + |Delta T_(4)| + |Delta T_(5)|)/(5)` `= (0.54)/(5) = 0.108 = 0.11sec` |
|
| 64. |
If `X= A xx B and Delta X Delta A and Delta B` are maximum absolute error in X ,A and B respectively , then the maximum relative in X is given byA. `Delta X = Delta A + Delta B`B. `Delta X = Delta A - Delta B`C. `(Delta X)/(X) = (Delta A)/(A) - (Delta B)/(B)`D. `(Delta X)/(X) = (Delta A)/(A) + (Delta B)/(B)` |
|
Answer» Correct Answer - d When two quantities are multiplied, their maximum relative errors areadded up Hence `(Delta X)/(X)= (Delta A)/(A)+ (Delta B)/(B)` |
|
| 65. |
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.A. `5.0 +- 11%`B. `5.0 +- 1%`C. `5.0 +- 6%`D. `1.25 +- 5%` |
|
Answer» Correct Answer - a 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 relativedensity `= ("weight in air")/("weight loss in water")` i.e.`R.D= (5.00 +- 0.05)/(1.00 +- 0.1)` Now reletive density with max peemissible error `= (5.00)/(1.00) +- ((0.05)/(5.00) + (0.1)/(1.00)) xxx 100 = 5.0 +- (1 +10)%` `= 5.0 + 11%`. |
|
| 66. |
Which one of the following pairs does have the same dimension?A. Work and energyB. Angule and strainC. Relative density and refractive indexD. Plank constant and energy |
|
Answer» Correct Answer - d `["Plank constant"] = [ML^(2)T^(-1)] and [Energy] = [ML^(2)T^(2)]` |
|
| 67. |
An athlletic coach told his team that muscle times speed equals power. What dimesions does he view for muscle?A. `MLT^(-2)`B. `ML^(2)T^(-2)`C. `MLT^(2)`D. `L` |
|
Answer» Correct Answer - a According to problem "muscle" xx "speed" = "power" `:. "Muscle" = ("power")/("speed") = (ML^(2)T^(-1))/(LT^(-1)) = MLT^(-2)` |
|
| 68. |
Assertion: When percentage error in the meansurement of mass and velocity are `1%` and `2%` respectively the percentagwe error in K.E. is `5%`. Reason: `(Delta K)/(K) = (Delta m)/(m) = (2 Delta v )/(v)`.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - a Kinetic energy `K = (1)/(2) mv^(2)` Taking log on both sides In `K = (1)/(2) + "In" m + 2 "in" v` Differentiating both we get or `(Delta K)/(K) = (Delta m)/(m) + 2 (Delta v )/(v) ` `(Delta K)/(K)xx 100 = (Delta m)/(m) xx 100 + 2 xx (Delta v)/(v)xx 100 ` `= 1% = (2 xx 2%) = 3%` |
|
| 69. |
A physical quantity `X` is represented by `X = (M^(x) L^(-y) T^(-z)`. The maximum percantage errors in the measurement of `M , L , and T`, respectively , are `a% , b% and c%`. The maximum percentage error in the measurement of `X` will beA. `(alpha a + beta b - gamma c)`B. `(alpha a + beta b + gamma c)`C. `(alpha a - beta b + gamma c)`D. Zero |
|
Answer» Correct Answer - b `(Delta x)/(x) xx 100 = a(Delta M)/(M) xx 100 + b(Delta L)/(L)xx 100 + c (Delta T)/(T) xx 100` `= a alpha + b beta + c gamma` |
|
| 70. |
Assertion: the quantity `(1//sqrt(mu_(0) epsilon_(0)))` is dimensionally equal to velocity and numerical equal to velocity of light. Reason : `mu_(0)` is permeability of free space and `epsilon_(0)` is the permitivity of free space.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - b Both assertion and reason are true but reason is not the correct explanation of assertion `[epsilon_(0)] = [M^(-1)L^(-3)T^(4)I^(2)] , [mu_(0) = [MLT^(-2)I^(-2)]` ` rArr(1)/(sqrt(mu_(0)// 4 pi) xx 4 pi E_(0)) = sqrt((9 xx 10^(9))/(10^(-2))) = sqrt(9 xx 10^(16)` `= 3 xx 10^(8) m//s` therefore `(1)/(sqrt(mu_(0) epsilon_(0))` has dimension of velocity and numerical equal to velocity of light. |
|
| 71. |
Assertion : When we change the unit of meansurement of a quantities its numerical value change. Reason : smaller the unit of meansurement smaller is its numerical value.A. If both the asseration and reason are true and reason is a true explanation of the asseration.B. If both the asseration and reason are true but the reason is not the correct explanation of asseration.C. If the asseration is ture but reason is false.D. If both the asseration and reason are false. |
|
Answer» Correct Answer - c We know that `Q = n_(1) u_(1) = n_(2)u_(2)` are the two unit of meansurements of the quantity `Q` and `n_(1), n_(2)` are their respectively numerical value from relation `Q_(1) = n_(1) u_(1) = n_(2)u_(2) ,nu = "constant" rArr n prop 1//u` i.e., smaller the unit of meansurement , greater is the numerical value. |
|
| 72. |
Assertion: When we change the unit of measurerment of a quantity its numerical value changes. Reason: Smaller the unit of measurement smaller is its numerical value.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - c We know that `Q = n_(1) u_(1) = n_(2)u_(2)` are the two unit of meansurements of the quantity `Q` and `n_(1) ,n_(2)` are their respectively numerical value from relation `Q = n_(1) u_(1) = n_(2)u_(2) ,n u = "constant" rArr n prop 1//u` i.e., smaller the unit of meansurements greater is the numerical value. |
|
| 73. |
Which of the following represents a volt?A. `"Joule"//"second"`B. `"watt"//"ampere"`C. `"watt"//"columb"`D. `"coulomb"//"joule"` |
|
Answer» Correct Answer - b `("watt")/("ampere")= "volt"` |
|
| 74. |
Presure gradient has the ssame dimension as that ofA. Velocity gradientB. Potential gradientC. Energy gradientD. None of these |
|
Answer» Correct Answer - d Velocity gradient `= (v)/(x) = [LT^(-1)]/([L]) = [T^(-1)]` Potential gradient `= (V)/(x) = ([ML^(2)T^(-3)A^(-1))/([L]))` `= [MLT^(-3)A^(-1)]` Energy gradient `= (E)/(x) = ([ML^(2)T^(2)])/([L]) = [MLT^(-2)]` and pressure gradient `= (P)/(x) = ([ML^(-1)T^(-2)])/([L]) = [ML^(-2)T^(-2)]` |
|
| 75. |
A wire has a mass `0.3+-0.003g`, radius `0.5+-0.005mm` and length `6+-0.06cm`. The maximum percentage error in the measurement of its density isA. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - d Density , `rho= (M)/(V) = (M)/(pui r^(2)L)` `rArr (Delta rho)/(rho) =(Delta M)/(M) + 2 (Delta r)/(r ) + (Delta L)/(L)` ` = (0.003)/(0.3)+ 2 xx (0.005)/(0.5) + (0.06)/(6)` `= 0.01 + 0.02 + 0.01 = 0.04` `:.` Perecentage error `= (Delta rho)/(rho) xx 100 = 0.04 xx 100 = 4%` |
|
| 76. |
A wire has a mass `0.4 +- 0.004g` and length `8 +- 0.08(cm)` The maximum percentage error in the measurement of it density is `4%` The radius of the wire is `r +- Delta r` find `Delta r`A. `0.02r`B. `0.01 r`C. `0.03r`D. `0.1r` |
|
Answer» Correct Answer - b `rho= (m)/(pi r^(2)L)` `(Delta rho)/(rho) xx 100[((Delta m)/(m)+ 2 (Delta r)/(r ) + Delta L)/(L)]xx 100` `= (0.01 + 2(Delta r )/(r ) + 0.01) xx 100 = 4`(given) `or `100(Delta r)/(r ) = 1` `Delta r = 0.01r` |
|
| 77. |
Two numbers `a = 0.92` and `b = 0.08` are given.The number ofsignificant figure present in the result after the following operation `a+b, a-b,a xx b` and `a//b`respectively areA. `2,2,2,2`B. `3,3,2,2`C. `3,2,1,1`D. `3,2,2,2` |
|
Answer» Correct Answer - c `a +b = 1.00, a- b= 0.84` in case of multiplication division result result will contain least number of significant figure in `0.08` there is only one significant figure |
|
| 78. |
If `L=2.331cm`,`B=2.1cm`, then `L+B=A. `4.431 cm`B. `4.43 cm`C. `4.4 cm`D. `4 cm` |
|
Answer» Correct Answer - c Given ,`L = 2.331 cm = 2.3` (correct up to one decimel places) since minimum significant figure is `2` and `b = 2.1 cm` `:. L + B = 2.3+ 2.1 = 4.4 cm` |
|
| 79. |
The length and breadth of a metal sheet are `1.124m and 0.002m` respectively .The area of the sheet up in four currect significant figure isA. `9.3782m^(3)`B. `9.37m^(3)`C. `9.378248 m^(3)`D. `9.378m^(3)` |
|
Answer» Correct Answer - d According to rule of significant figures ●any zeros that are between non - zero also consider significant ● all none -zero digit significant. Hence , breadth of metel `b = 3.002m` has four significant figure `:.` Length of metal `= 3.124 m` has four significant figures. Since length and breadth have four significant therefore `"Area" = "length" xx "breadth"` `= 3.124 xx 3.002 = 9.378248` will also have four significant figures. Hence, `"area" = 9.378m^(2)` |
|
| 80. |
Dimensions of an unknown quantity ,`phi = (ma)/(alpha) log (1 +(alpha l)/(ma))` where `m =` mass, `a=` acceleration and `l = "length"` areA. `[MLT^(-2)]`B. `[MT^(-2)]`C. `[M^(0)LT^(0)]`D. `[ML^(-3)]` |
|
Answer» Correct Answer - c Legarithen has no dimensions `:. 1 +(alpha t)/(ma) = "dimensionless"` ` :. (alpha t)/(ma) = [M^(0)L^(0)T^(0)]` `rArr alpha = (ma)/(l) = ([MLT^(-2)])/([L])= [MT^(-2)]` `:.` Dimensions of `phi = "Demension of" (ma)/(alpha)` `= ([MLT^(-2)])/([MLT^(-2)]) = [L] = [M^0)LT^(0)` |
|
| 81. |
The currect voltage relation of a diode is given by `1 = (e^(van v//T) -1 )mA` where the applied volied `V` is in volts and the tempetature `T` is in degree kelvin if a student make an error meassurting `+- 01 V` while measuring the current of `5 mA at 300 K` what be the error in the value of current in `mA`A. `0.5 mA`B. `0.05 mA`C. `0.2 mA`D. `0.02 mA` |
|
Answer» Correct Answer - c Given `5 = e^(1000_(T)^(L)) - 1m` `rArr e^(1000) (V)/(T) = 6`…(i) Again, `1 = e^(1000_(T)^(L)) - 1` `(dl)/(dV) = e^((1000)/(T) V) (1000)/(T)` `dl = (1000)/(T) e ^((1000)/(T))Vdv` Using (i) `Delta I = (1000)/(T) xx 6 xx 0.01= (60)/(T) = (60)/(300) = 0.2 mA` |
|
| 82. |
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 and reason is the correct explation of assertion.B. If both assertion but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - b The number of significant figure in `1.202` is four and the number of significant figure inn `0.0024` is two. All non zero digits are significant. All zero occurring where the decimal point is. If the number is less then one, the zero on the ring of decimal point and to the left of first non - zero digit are not significant. |
|
| 83. |
The value of resistance is `10.845 Omega` and the current is `3.23 A` . On multiplying them , we get the potential difference in terms of significant figures?A. `35 V`B. `35.0 V`C. `3.029 5 V`D. `35.03 V` |
|
Answer» Correct Answer - b The final result be `3` significant figure |
|
| 84. |
The equation of stationary wave is `y = A sin kt cos omega` ,where y and x in second choose the correct optionA. the dimensions of A and k are sameB. the dimensions of `A,k and omega `are sameC. the dimensions of `k and omega` are sameD. the dimensions of `(kx) and (omega)` are same |
|
Answer» Correct Answer - d `kx and omega t` have dimensions of angle (i.e. `[M^(0)L^(0)T^(0)]`) Hence kt and omega t both are dimensionless |
|
| 85. |
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` |
|
Answer» Correct Answer - a Here `R_(1) = (100 +-3)Omega` `R_(2) = (200 +-4)Omega` The equaivalent resistent in parallel combination is `(1)/(R_(p))= (1)/(R_(1)) + (1)/(R_(2))` `(1)/(R_(p))=(1)/(100) +(1)/(100)= (1)/(100)` `R_(p) = (200)/(3) = 66.7 Omega` This error in equaivalent resistent is given by `(DeltaR_(p))/(R_(1)^(2)) + (DeltaR_(2))/(R_(2)^(2))` `DeltaR_(P) = DeltaR_(1) ((R_(p))/(R_(1)^(2)))^(2) + DeltaR_(2)((R_(p))/(R_(2)))^(2)` `=3((66.7)/(100))^(2) + 4 ((66.7)/(200))^(2) = 1.8 Omega` Hence, the equivalent resistance along with error parallel combination is `(66.7 +- 1.8) Omega`. |
|