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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. |
If units of length, mass and force are chosen as fundamental units, the dimensions of time would be :A. `M^(1//2) L^(-1//2) F^(1//2)`B. `M^(1//2) L^(1//2) F^(1//2)`C. `M^(1//2) L^(1//2) F^(-1//2)`D. `M^1 L^(-1//2) F^(-1//2)` |
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Answer» Correct Answer - (c ) `F = M^1 L^1 T^(-2)` `:. T^2 = (M^1 L^1)/(F) or T = M^(1//2) L^(1//2) F^(-1//2)` |
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| 52. |
Which of the following units denots the dimensions `ML^2 // Q^2` where Q denots the electric charge ?A. Henry (H)B. `H//m^2`C. Weber (Wb)D. `Wb//m^2` |
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Answer» Correct Answer - (a) Henry is unit of self- indcutance (L). From `e = (L dI)/(dt)` `L = (edt)/(dI) = ((W//q)dt)/(dI)` `=[ML^2 T^(-2)]T/(QA) = ([ML^2 T^(-2)]T)/(Q Q//T)` `i.e., = Henry (H) = ML^2 // Q^2` |
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| 53. |
The dimensions of `[ML^(-1) T^(-2)]` may correspond toA. Work done by forceB. Linear momentumC. PressureD. Energy per unit volume |
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Answer» Correct Answer - (c,d) Work `= ML^2 T^(-2),` Linear momentum `=MLT^(-1)` pressure `= ML^(-1) T^(-2),` energy per unit volume `=("Energy")/("Volume") = (ML^2T^(-2))/(L^3) = ML^(-1)T^(-2)` Hence option (c ), (d) are correct. |
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| 54. |
Experiments reveal that the velocity `v` of water waves may depend on their wavelength `lambda`, density of water `rho`, and acceleration due to gravity `g`. Establish a possible relation between `v and lambda , g, rho`.A. `upsilon^2 = k lambda^(-1) g^(-1) d^(-1)`B. `upsilon^2 = k lambda g`C. `upsilon^2 = k lambda dg`D. `upsilon^2 = k lambda^3 g^(-1) d^(-1)` |
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Answer» Correct Answer - (b) `k lambda g= L (LT^(-2)) = L^2 T^(-2) = (LT^(-1))^2 = L^2 T^(-2) = (LT^(-1))^2 = upsilon^2` |
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| 55. |
The frequency (V) of an oscillatingj drop may depends upon radius (r ) of the drop density `(rho)` of liquid and the surface tension (S) of the liquid. Deduce of formula dimensionally. |
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Answer» Correct Answer - `v = k sqrt((S)/(rho r^3))` Let `v = k r^a rho^b s^c….(i)` `[M^0L^0T^(-1)] = L^a(ML^(-3))^b (MT^(-2))^c`, `[M^0L^0T^(-1)] = M^(b+c) L^(a -3b) T^(-2a)` `:. b + c = 0 …(iii)` a-3b = 0 ….(iii) `-2c = -1, :. C =(1)/(2) ` From (ii),` b = -c =-(1)/(2)` from (iii), `a =3b =-(3)/(2)` Put in (i) to get the required formula. |
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| 56. |
How may metric tone are there in a teragrm? |
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Answer» Correct Answer - `10^6` 1 Teragram `= 10^(12) g`: 1 metric ton = `10^3 kg = 10^6g` `(1"Tergram")/(1 "metric ton") = (10^(12)g)/(10^6g) = 10^6` :. 1 millisecond `=10^3` microsecond. |
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| 57. |
For a glass prism of refracting, angle `60^@,` the minimum angle of deviation, `D_m` is fonud to be `36^@` with a maximum error of `1.05^@,` when a beam of parallel light is insident on the prism. Find the range of experimental value of refactive index `mu.` it is known that refractive index `mu` of material of prism is given by `mu =(sin(A+D_m)/(2))/(sinA//2)` |
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Answer» `Here, A = 60^@, D_m = (36^@+-1.05^@),mu = ?` Taking `(D_m) = 36^@ 1.05^@ = 37.05^@`, `mu_1 = sin(A+D_1)/((2)/(sin A//2)) = (sin(60+37.05)^@//2)/(sin 60^@//2) = (sin69.52^@)/(sin30^@) = (0.7492)/(1//2)` `=1.4984 = 1.50("rouding off to two decimal places")` Taking`(D_m)_2 = 36^@ -1.05^@ = 34.95^@`, `mu_2 = (sin(A+D_m)_2)/(2)/(sinA//2) = (sin(60+34.95)^@//20)/(sin60^@//2) = (sin47.48^@)/(sin30^@) = (0.7370)/(1//2) = 1.4740` `=1.47("rounding off to two decimal places")` Hence, `1.47_mu1.50`with a mean value of 1.49 |
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| 58. |
Physics involves the study ofA. humansB. birds and animalsC. plantsD. nature and natural phenomena |
| Answer» Correct Answer - (d) | |
| 59. |
What are the five main branches of Physics? |
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Answer» These are : Mechanics, Heat & Thermodyna- mics , Electromagneitsm , Theory of Relativity, Quantum Mechanics. |
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| 60. |
Science, like any knowledge, can be put to good or bad use, depending on the user. Given below are some of the applications of science, Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized : (a) Mass veccination against small pox to curbe and finally eradicate this disease from the population. (this has already been successfuly done in india). (b) Televison for eradication of illiteracy and for mass communication of news and ideas. (c ) Prental sex determination. (d) Computers for increase in work efficiency. (e ) Putting artificial stellites into orbits around the Earth. (f) Development of nuclear weapons.(g) Development of new and powerful techniques of chemical and biological warface.(h) Purificiation of water for drinking. (i) Plastic surgery. (j) Cloning. |
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Answer» (a) Mass vaccination is good (b) Television for eradication of illiteracy and for mass communciation of news and ides is really good. (c ) Prenaltal sex determination is not bad. But people are misuing it. They must be educated to avoid its misuse in creating imbalance between the male and female population. (d) Computers for increases in work efficiency are good, (e ) Putting artificial satellites in work orbits around the earth is a good development. (f) Development fo nuclear weapons is bad as they are the weapons of mass destruction. (g) Development of new and powerful techniques of chemical and biological warface is real bad us these weapons are for destructio of makind. (h) Purification of water for drinking is good, (i) Plastic surgery is good, (j) Cloning is also good. |
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| 61. |
In a vernier callipers, 10 vernier scale divisions coincide with 9MSD each of value 1mm. While measuring length of a cylinder using this calliper, main scale reading is 5.1 cm and 7th vernier division coincides with any main scale division. Whenm diameter of cylinder is measured, main scale reading is 1.7 cm and 3rd vernier division coincides with any main scale division. Calculate curved surface area of the cylinder with correct number of significant figures. |
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Answer» Correct Answer - `28.1 cm^2` V.C = 1 MSD -1 VSD `= (1-(9)/(10))MSD` `=(1)/(10)MSD = (1)/(10)mm =0.1mm` `=0.01cm` `I = (5.1xx7xx0.01)cm =5.17cm` `D = (1.7+ 3xx0.01)cm =1.73cm` curved surface area `=2pi rxx I =pi DI` `=3.14(1.73)5.17cm^2` `A = 28.084cm^2` Rounding off to three significant digits, `A = 28.1cm^2` |
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| 62. |
The velocity `upsilon` of a particle depends upon time t, according to the equation `upsilon = a+ bt +(c )/(d +t)` Write the dimensions of a,b,c, and d. |
| Answer» Correct Answer - `[LT^(-1)]`, [LT^(-2)], [L], [T]` | |
| 63. |
The velocity of sound `(upsilon)` in a gas depends upon coefficint of volume elesticity E of the gas and density d of the gas. Use method of dimensions to derive the formula for `upsilon.` |
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Answer» Let `upsilon = KE^a d^b …(i)` where K is dimensionless constant of proportionality. `:. [ M^0 L^1 T^(-1)] = (ML^(-1) T^(-2))^a (ML^(-3))^b` ` = M^(a +b) L^(-a -3b) T^(-2a)` Applying the principle of homogeneity of dimensions, we get `a +b = 0 , - a-3b b =1 , -2a = -1` `:. a = (1)/(2), b = -a =-(1)/(2)` From (i), `upsilon = KE^(1//2) d^(-1//2) =Ksqrt(E//d)` |
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| 64. |
Assuming that the mass m of the largest stone that can be moved by a flowing river depends upon the velocity `upsilon,` of water, its density `rho` and acceleration due to gravity g, then m is directly proportional toA. `upsilon^4`B. `upsilon^6`C. `upsilon^5`D. `upsilon` |
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Answer» Correct Answer - (b) Let `m = k upsilon^a rho^b g^c `, where k is dimnsionless constant. `[M^1 L^0 T^0] = (LT^(-1))^a (ML^(-3))^b (LT^(-2))^c` `M^b L^(a -3b+c) T^(-a-2c)` Applying the principle of homogeneity of dimensions, we get `b= 1 , - a-2c = 0 or c = (-a)/(2)` ` a- 3b + c = 0` `a - 3- a//2 =0` or `a =6` Hence, `m prop upsilon^6` |
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| 65. |
The energy of a system as a function of time `t` is given as `E(t) = A^(2)exp(-alphat)`, `alpha = 0.2 s^(-1)`. The measurement of `A` has an error of `1.25%`. If the error In the measurement of time is `1.50%`, the percentage error in the value of `E(t)` at t = 5 s` isA. 0.01B. 0.02C. 0.03D. 0.04 |
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Answer» Correct Answer - (d) Given `E (t) =A^2 exp (- alpha t), alpha = 0.2 s^(-1),` `(dA)/(A) = 1.25%, (dt)/(t) = 1.50%, t =5 sec., (dE)/(E ) = ?` Taking log of both sides, we get `log E = log A^2 e^(-alpha t)` `log E = log A^2 + log e^(-alphat)` or `log E =2 log A - alpha t` Differentiating both sides, we get `(dE)/(E ) = +- (2dA)/(A) +- alpha dt ........ (i)` `:gt (dt)/(t) =1.50%, t= 5 sec.` `:. (dt)/(5) = 1.50%, dt = 7.50%` From (i), `(dE)/(E ) = +- 2(1.25) + -0.2xx7.5` `= +- 2.5 +- 1.5 = +- 4%` |
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| 66. |
The percentage errors of measurement in a, b, c and d are 1%. 3%, 4% and respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off 2% the result? e resuuiit |
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Answer» Here, `P =(a^3 b^2)/(sqrtc d)` Maximum fractional error in P is given by `(Delta P)/(P) = +-[3(Delta a)/(a) +(Deltab)/(b) +(1)/(2) (Deltac)/(c ) +(Deltad)/(d)] = +-[ 2((1)/(100))+2((3)/(100))+(1)/(2)((4)/(100))+(2)/(100)] = +- (13)/(100) =+-0.13` `"Percentage error in" P = (DeltaP)/(P)xx100 = +- 0.13xx100 = +- 13%` As the result (13%error) has two significant figures, therefore, if P turns out to be 3.763, the result would be rounded off to 3.8. |
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| 67. |
A physical quantity `x` is calculated from ` x = ab^(2)//sqrt(c )`. Calculate the percentage error in measuring `x` when the percentage errors in measuring a , b , and c are 4 , 2 , and 3%, respectively . |
| Answer» Correct Answer - `+- 9.5%` | |
| 68. |
The time dependence of a physical quantity P is given by `P=P_(0) exp (-alpha t^(2))`, where `alpha` is a constant and t is time. The constant `alpha`A. is dimensionlessB. has dimensions `[T^(-2)]`C. has dimensions `[T^(2)]`D. has dimensions of `p` |
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Answer» Correct Answer - B `p = p_(0) exp (-alphat^(2))` As powers of exponential quantity is dimesnionless, so `alphat^(2)` is dimensions. or `alphat^(2) =` dimensionless `= [M^(0)L^(0)T^(0)]` `:. alpha = (1)/(t^(2)) = (1)/([T^(2)]) = [T^(-2)]` |
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| 69. |
If the dimension of a physical quantity are given by `M^a L^b T^c,` then the physical quantity will beA. pressure if `a = b, b =- 1, c =- 2`B. velocity is `a = 1,b = 0, c =- 1`C. acceleration if `a = 1, b =1, c =- 2`D. force if `a = 0, b =- 1, c =-2` |
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Answer» Correct Answer - A (i) Dimensions of velocity `= [M^(0)L^(1)T^(-1)]` Here, `a = 0, b = 1, c =- 1` (ii) Dimensions of acceleration `[M^(0)L^(1)T^(-2)]` Here, `a = 0, b = 1, c =- 2` (iii) Dimensions of force `= [M^(1)L^(1)T^(-2)]` Here, `a = 1,b = 1,T =- 2` (iv) Dimensions of pressure `=[ML^(1)L^(-2)T^(-2)]` `:.` Here, `a = 1, b =- 1,c =- 2` `:.` The physical quantity is pressure. |
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| 70. |
The mean length of an object is 5 cm. Which of the following measurements is most accurate?A. 4.9 cmB. 4.805 cmC. 5.25 cmD. 5.4 cm |
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Answer» Correct Answer - (a) Here, x =5 cm `Deltax_1 =5- 4.9 = 0.1 cm, Delta x_2 =5- 4.805 = 0.195cm` `Delta x_3 = 5-5.25 = - 0.25 cm, Delta x_4 = 5-5.4 = - 0.4 cm` Choice (a) is most accurate. |
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| 71. |
Which of the following lengths measured is most accurate and why ? (a) 500.0cm (b) 0.0005 cm (c ) 6.00 cm |
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Answer» The measurement (b) 0.0005 cm is most accurate as it is correct upto fourth place of decimal. |
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| 72. |
The mass of a body is measured by two persons is 10.2 kg and 10.23 kg. Which is more accurate and why ? |
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Answer» The value m = 10.23 kg is more accurate, being correct upto 2nd place of decimal. |
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| 73. |
Which of the following length measurment is most accurate and why ? (i) 4.00 cm (ii) 0.004 mm (iii) 40.00 cm. |
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Answer» `(i) (Deltax)/(x) = (0.01)/(4.00) = 0.0025` (ii) `(Deltax)/(x) = (0.001)/(0.004) = 0.25` (iii) `(Deltax)/(x) =(0.01)/(40.00) = 0.00025` The last observation has the last fractional error and hence it is most accurate. |
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| 74. |
Why do we have different units for same physical quantity? |
| Answer» The value of a given physical quantity may vary over a wide range. To express the quantity in proper format, we may need different units. Further, a given quantity may be expressed in terms of different quantites, will have same dimensions. This would lead to different equivalent units of the same quantity. | |
| 75. |
In dimension of circal velocity `v_(0)` liquid following through a take are expressed as `(eta^(x) rho^(y) r^(z))` where `eta, rhoand r `are the coefficient of viscosity of liquid density of liquid and radius of the tube respectively then the value of `x,y` and `z` are given by |
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Answer» Correct Answer - `upsilon_c prop =(eta)/(rho r)` Let `upsilon_c = k r^a rho^b eta^c` `[M_0 L^1 T^(-1)] = L^a [ML^(-3)]^b [ML^(-1) T^(-1)]^c` `=M^(b+c) L^(a-3b-c)T^(-c)` `:. B+c = 0, a-3b -c =1, -c =-1`. Simplify to get the required relation. |
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| 76. |
In dimension of circal velocity `v_(0)` liquid following through a take are expressed as `(eta^(x) rho^(y) r^(z))` where `eta, rhoand r `are the coefficient of viscosity of liquid density of liquid and radius of the tube respectively then the value of `x,y` and `z` are given byA. 1, 1,1B. 1,-1,-1C. `-1, -1,1`D. `-1, -1 ,-1` |
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Answer» Correct Answer - (b) Critical velocity `upsilon_c = k eta^x rho^y r^x….(i)` Expression for critical velocity, `upsilon_c = (k eta)/(rho(2r))= (k eta)/(2 rho r) ….(ii)` From (i) and (ii) x = 1, y =-1 ,z = -1 |
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| 77. |
A radar signal is beamed towards a planet and its echo is recived 7 minutes later. If the distance between the planet and earth is `6.3xx10^(10)m,` calculate the speed of the signal. |
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Answer» Correct Answer - `3xx10^8 m//s` `t = 7min = 7xx60s, x =6.3xx10^(10)m, upsilon = ?` `As upsilonxxt = x+x.` `upsilon = (2x)/(t) = (2xx6.3xx10^(10))/(7xx60) = 3xx10^8m//s` |
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| 78. |
Draw a schematic arrangement of the Geiger Marsden experiment. How did the scattering of `alpha` particles by a thin foil of gold provide an important way to determine an upper limit on the size of nucleus? Explain briefly. |
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Answer» Both the nucleus and an alpha particle are positively charged. When an alpha particle approcaches a nucleus, it is repelled, the velocity and hence KE of `alpha` particle goes on decreasing. in turn, potential energy of the system goes on increasing. At a cartain distance, the entire KE of `alpha` particlel is converted into P.E. the alpha particle can no longer go closer to the nucleus. This distance of closet approach of `alpha` particle gives us the approxiate size of the nucleus. |
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| 79. |
How many degress are there in one radian ? |
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Answer» `pi radian = 180^@` ` 1 radian = (180^@)/(pi) = (180xx7)/(22) = (630)/(11) = 57.3^@` |
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| 80. |
The density of a cube is measured by measuring its mass and length of its side. If the maximum errors in the measurements of mass and length are 3% and 2% respectively. Then the maximum error in the measurement of density is :A. 0.07B. 0.05C. 0.09D. 0.03 |
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Answer» Correct Answer - (c ) Here, `(Delta m)/(m) = 3%, (DeltaL)/(L) = 2%` `As density, rho = (mas(m))/(volume(V)) = (m)/(L^3)` `(Delta rho)/(rho) = (Deltam)/(m) + (3 DeltaL)/(L) = 3% + 3(2%)` `=3% + 6% = 9%` |
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| 81. |
The volume of a cube of side 10 cm. is … `m^3.` (ii) A vehicle moving with a speed of `36kg h^(-1)` covers …. M in 1 sec. (iii) The density of water at `4^@C is ……g//cc or ……kg//m^3.` |
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Answer» (i) As `v = L^3` and `L = 10 cm = 10^(-1)m` `:. V = (10^(-1)m)^3 m^3` (ii) `upsilon = (36km)/(hr) = (36xx1000m)/(60xx60s) = 10m//s.` (iii)density = `(mass)/(volume) = (1g)/((1 cm)^3)` `= 1g//cc = (10^(-3)kg)/((10-2m)^3 ) = 10^3kg//m^(3)` |
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| 82. |
Significant figures in the meaured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of measurement and vice-verse, In addition or subtraction, the number of decimal places in the result should equal the smallest number of decimal places in any term in the operation. In multiplication and division, the number of significant figures in the product or in the quatient is the same as the smallest number of significant figures in any of the factors. With the help of the compreshension given above, choose the most appropriate alternative for each of the following questions : The circumference of the circle of diameter 1.06 m with correct number of significant figures isA. 3.33 mB. 3.33142 mC. 3.3 mD. 3m |
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Answer» Correct Answer - (a) Here, d = 1.06 m, Circumference ` =2pi r = pi d` `=(22)/(7)xx1.06 = 3.3314` =3.33m (with 3 significant figures) |
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| 83. |
Significant figures in the meaured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of measurement and vice-verse, In addition or subtraction, the number of decimal places in the result should equal the smallest number of decimal places in any term in the operation. In multiplication and division, the number of significant figures in the product or in the quatient is the same as the smallest number of significant figures in any of the factors. With the help of the compreshension given above, choose the most appropriate alternative for each of the following questions : The area enclosed by a circle of diameter 1.06 m with correct number of significant figures isA. `0.88 m^2`B. `0.883 m^2`C. `1.88m^2`D. `0.882026 m^2` |
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Answer» Correct Answer - (b) Here, `d= 2 r= 1.06m, r= (1.06)/(2) = 0.530 m` Area enclosed `=pi r^2=(22)/(7) (0.530)^2` `=0.88282 m^2 = 0.883 m^2` (with 3 significant figures)` |
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| 84. |
In the relation `C = v lambda` true dimensionally ? |
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Answer» Yes, `C = LT^(-1)` and `v lambda =((1)/(T)) (L) = LT^(-1)` |
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| 85. |
Show dimensionally that the relation `t = 2pi((I)/(g))` is incorrect, where I is length and t is time period of a simple pendulum , g is acc. Due to gravity. Find the correct form of the relation, dimensionally |
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Answer» `RHS = 2pi((I)/(g)) = (L)/(LT^(-2))` `=T^2 != t(LHS)` :. Formula is incorrect Let `t =k l^a g^b …..(i)` `[M^0L^0T^1] = L^a (LT^(-2))^b = L^(a +b) T^(-2b)` Using principle of homogeneity of dimensions, ` a +b =0, -2b =1, b = -(1)/(2)` `:. a =-b =- (-(1)/(2)) = (1)/(2)` From (i), `t = kl^(1//2) g^(-1//2) = k sqrt((I)/(g))` |
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| 86. |
Given `F = (a//t) + bt^2` where F denotes force and t time. The diamensions of a and b are respectively :A. `[MLT^(-1)] and [MLT^(-4)]`B. `[LT^(-1)] and [T^(-2)]`C. `[T] and [T^(-2)]`D. `[LT^(-2)] and [T^(-2)]` |
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Answer» Correct Answer - (a) Here a//t and `bt^2` both should represent the force, `:. F = (a)/(t)` `:. a = Fxx t = MLT^(-2)xxT = MLT^O(-4)` `Also, F =bt^2,` `:. b = (F)/(t^2), i.e., b= (MLT^(-2))/(T^2) = MLT^(-4)` `b =MLT^(-4)` |
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| 87. |
A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?A. Error `Delta T` in measuring T, the time period, is 0.05 secondsB. Error `DeltaT` in measuring T, the time period, is 1 secondC. Percentage error in the determination of g is 5%D. Percentage error in the detremination of g is 2.5% |
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Answer» Correct Answer - (a,c) Here `Delta t = 1 sec, t = 40 s, I = 1m,` `T = 2 sec., Delta T =?` `(Delta T)/(T) = (Delta t)/(t) = (1s)/(40s) = (1)/(40)` `DeltaT = ((1)/(40)xx T) = (1)/(40)xx2 = 0.05 sec` Also, `t = 2pi sqrt((I)/(g))` (time period of simple pendulum) `:. g= (4pi^2L)/(t^2)` `:. (Deltag)/(g) =(Delta L)/(L) + (2 Delta t)/(t)` As L is exactly known, `DeltaL = 0` `:. (Delta g)/(g) = (2 Delta t)/(t) = 2xx (1)/(40) = (2)/(40)` `% error (Delta g)/(g)xx100 = (2)/(40)xx100 = 5%` |
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| 88. |
An object 1 mm square is projected and its image on the screen is 1cm square. Calcullate the linear magnificaiton. |
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Answer» Correct Answer - 10 Areal magnification ` = ((1cm)^2)/((1mm)^2) = ((10mm)^2)/(1mm^2) = 100` :. Linear magnification `= sqrt100 =10` |
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| 89. |
Rad corpuscles of human blood stream are known to be flattened discs. Blood count shows `RBC_s` of sdhte order of `5xx10^6` in each cubic millimeter of blood. If the adult body contains 5 litres of blood, what is the order of total number of red corpuscles it contains? |
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Answer» Correct Answer - `10^(13)` Number of `RBC_s in 1mm^3 = 5xx10^6` Total volume fo blood = 5 litres `5xx10^3xx10^3 mm^3` Total number of `RBC_s = 5xx10^6xx5xx10^6` `2.5xx10^(13) ~~10^(13)` |
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| 90. |
Find the value of 100 J on a system which has 20 cm, 250g and half minute as fundamental units of length, mass and time. |
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Answer» Correct Answer - `9xx10^6 new units` Joule is SI unit of work ` = [ML^2T^(-2)]` `:. a = 1, b =2, c= -2` To conver SI system to new system, we write `M_1 =1kg, L_1 =1m, T_1 = 1s, n_1 =100J` `M_2 = 250g , L_2 =20cm ,` `T_2 =(1)/(2) min = 30s,n_2 = ?` `n_2 = n_1 ((M_1)/(M_2))^a ((L_1)/(L_2))^b ((T_1)/(T_2))^c` `=100((1kg)/(250g))^1((1m)/(20cm))^2 ((1sec)/(30sec))^(-2)` `=100xx((1000g)/(250g))((100cm)/(20cm))^2xx30^2` `= 100xx4xx5^2xx30^2` `=9xx10^6 "new units"]` |
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| 91. |
If the unit of force is 1 kN, unit of length 1 km and unit of time is 100s, what will be the unit of mass? |
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Answer» Correct Answer - `10^4kg` Here, `F = MLT^(-2) =1kN = 1000N` L = 1 km = 1000m, T = 100s, M =? From `F = MLT^(-2) = 1000,` `M = (1000)/(LT^(-2)) = (1000T^2)/(L) = 1000(100)^2/(1000)` ` =10^4kg` |
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| 92. |
Choose the pairs of quantities which have same dimensions : Impulse, force, work, momentum, moment of force, tension. |
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Answer» (i) Impulse and momentum have the same dimensions : `[M^1 L^1 T^(-1)]` (ii) Force and tension have the same dimensions :`[M^1 L^1 T^(-2)].` (iii) Work and moment of force have the same dimensions : `[M^1 L^2 T^(-2)]` |
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| 93. |
If size of an atom `(~~10^(-10)m)` is scaled up to 1m, what would be the size of nucleus `(~~ 10^(-14) m)?` |
| Answer» Correct Answer - `10^(-4)m` | |
| 94. |
If the size of a nucleus `(~~10^(-15)m)` is scaled up to the tip of a sharp pin `(~~ 10^(-5)m),` what roughly is the size of an atom? |
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Answer» `"Magnification" = ("size of tip of pin")/("size of nucleus")` `=10^(-5)/(10^(-15)) = 10^(10)` `"As actual size of atom" `~~ 10^(-10)m` `"and it is magnified" 10^(10)` times, :. `"Apparent size of atom" = 10^(10) xx10^(10)` `=10^0 = 1m` |
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| 95. |
If an atom of size `10^(-10)`m were enlarged to the size of the earth `(~= 10^7m),` how large would its nucleus be ? Take size of nucleus ` =10^(-14)`m. |
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Answer» Correct Answer - `10^3m` `("Size of nucleus")/("Size of atom") = (x)/(10^7) = (10^(-14))/(10^(-10)) = 10^(-4)` `x = 10^(-4)xx10^7 = 10^3m` |
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| 96. |
IF the size of an atom ( = 1Å) were enlarged to the tip of a sharp pin `(~= 10^(-5)m)`, how large would the height of mout everest `(~= 10^4m)` be? |
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Answer» Correct Answer - `10^9m` Magnification =` (10^(-5)m)/(1 Å) = (x)/(10^4m)` or `(10^(-5)m)/(10^(-10)m) = (x)/(10^4m)` `x = 10^5xx10^4 m =10^9m` |
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| 97. |
By the use of dimensions, show that energy per unit volume is equal to pressure. |
| Answer» `("Energy")/("Volume") = ("work")/("area" xx "length") = ("force" xx "distance") /("area" xx "length")` =` (M LT^(-2)xxL)/(L^2 xxL) = [ M L^(-1)T^(-2)] = pressure.` | |
| 98. |
The mass of an electron is `9.11xx10^(-31)kg.` How many electrons would make 1kg? |
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Answer» Here, mass of an electron `=9.11xx10^(-31)kg,` total mass = 1kg `:. "Number of electrons = ""total mass"/("mass of each electron")` `=1/(9.11xx10^(-31)) = 1.1xx10^(30)` |
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| 99. |
Ten drops of olive of radius 0.20 mm spread into a circular film of radius 14.6 cm. on the surface of water. Estimate the size of an oil molecule. |
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Answer» Here n =10, `r =0.20 mm = 2xx10^(-4)m,` `R = 14.6cm = 14.6xx10^(-2)m.` Let d be the diameter of oil molecule = thickness of layer volume of olive oil = area xx thickness of layer `n xx(4)/(3) pi r^3 = piR^2 xxd` `d = (4)/(3)n (r^3)/(R^2) = (4)/(3) xx(10(2xx10^(-4))^(3))/((14.6xx10^(-2))^2)` `=5xx10^(-9)m` |
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| 100. |
In an experiment, two capacities measured are `(1.3 +- 0.1)muF` and `(2.4+- 0.2)muF.` Calculate the total capacity in parallel with percentage error. |
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Answer» Here, `C_1 = (1.3 +- 0.1)muF and C_2 = (2.4 +-0.2)muF.` In parallel, `C_p = C+1 +C_2 = 1.3 +2.4 = 3.7muF, Delta C_p = +-(DeltaC_1 + DeltaC_2) = +-(0.1 +0.2_ = +-0.3` `% "age error" =l +- (0.3)/(3.7)xx100 = +- 8.1%` Hence, `C_p = (3.7 +- 0.3)muF = 3.7muF +- 8.1%` |
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