\nHow many seconds are there in a light fermi?
\n(a) 10-15<\/sup>
\n(b) 3.0 \u00d7 108<\/sup>
\n(c) 3.33 \u00d7 10-24<\/sup>
\n(d) 3.3 \u00d7 10-7<\/sup>
\nAnswer:
\n(c) 3.33 \u00d7 10-24<\/sup>
\nOne light fermi is time taken by light to travel a distance of 1 fermi ie. 10-15<\/sup>m
\n1 light fermi = \\(\\frac{10^{-15}}{3 \\times 10^{8}}\\) = 3.33 \u00d7 10-24<\/sup>s.<\/p>\n
Question 2.
\nWhich of the following pairs have same dimensional formula for both the quantities?<\/p>\n
- \n
- Kinetic energy and torque<\/li>\n
- Resistance and Inductance<\/li>\n
- Young\u2019s modulus and pressure<\/li>\n<\/ol>\n
(a) (1)only
\n(b) (2) only
\n(c) (1) and (3) only
\n(d) All of three
\nAnswer:
\n(c) (1) and (3) only<\/p>\nQuestion 3.
\nGive four dimensionless physical quantities.
\nAnswer:
\nAngle, Poisson\u2019s ratio, strain, specific gravity.<\/p>\nQuestion 4.
\nThe dimensions of plank constant are the same as those of______.
\nAnswer:
\nAngular momentum<\/p>\nQuestion 5.
\nA physical quantity P = \\(\\frac{\\sqrt{a b c^{2}}}{d^{3}}\\) measuring a, b, c and d separately with the percentage error of 2% , 3%, 2% and 1% respectively. Minimum amount of error is contributed by the measurement of
\n(a) b
\n(b) a
\n(c) d
\n(d) c
\nAnswer:
\n(b) a
\nP = \\(\\frac{\\sqrt{a b c^{2}}}{d^{3}}\\)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-1M-Q5.jpg\")
\nThe minimum amount of error is contributed by the measurement of a.<\/p>\nQuestion 6.
\nThe number of significant figures in 11.118 \u00d7 10-6<\/sup> is
\n(a) 3
\n(b) 6
\n(c) 5
\n(d) 4
\nAnswer:
\nAs per rules, number of significant figures in 11.118 \u00d7 10-6<\/sup> is 5.<\/p>\nQuestion 7.
\nWhat is the number of significant figures in 0.06070?
\nAnswer:
\n4.<\/p>\nQuestion 8.
\nIf f = x2<\/sup>, What is the relative error in f?
\nAnswer:
\n\\(\\frac{2 \\Delta x}{x}\\).<\/p>\nQuestion 9.
\nWhich of the following measurement is more accu\u00acrate?
\n(i) 7000m
\n(ii) 7 \u00d7 102<\/sup>m
\n(iii) 7 \u00d7 103<\/sup>m
\nAnswer:
\n(i) 7000 m<\/p>\nQuestion 10.
\nWhich of the following measurements is most, accurate?
\n(a) 5.0 cm
\n(b) 0.005 cm
\n(c) 5.00 cm
\nAnswer:
\n(c) Is most accurate because it has three significant figures. Greater is number of significant figures, more accurate is the measurement.
\n(a) has 2 significant figures
\n(b) has 1 significant figure.<\/p>\nQuestion 11.
\nName three physical quantities having same dimension.
\nAnswer:
\nWork, Energy, and Torque.<\/p>\nPlus One Physics Units and Measurement Tw0 Mark Questions and Answers<\/h3>\n
Question 1.
\nUsing dimensional analysis derive the relation F = ma. Where the symbols have the usual meaning.
\nAnswer:
\nForce on a body depends on mass(m), acceleration (a) an
\nF \u03b1 ma<\/sup>ab<\/sup>tc<\/sup>
\nM1<\/sup>L1<\/sup>T-2<\/sup> = Ma<\/sup>(LT-2<\/sup>)b<\/sup>Tc<\/sup>
\nM1<\/sup>L1<\/sup>T-2<\/sup> = Ma<\/sup>Lb<\/sup>T-2a+c<\/sup>
\nEquating the powers, we get a = 1 ,b = 1, -2b + c = -2, c = 0
\nF = m1<\/sup>a1<\/sup>t0<\/sup> = ma.<\/p>\nQuestion 2.
\nUse your definition to explain how simple harmonic motion can be represented by the equation y = a sin \u03c9t
\n(a) Show that the above equation is dimensionally correct
\nAnswer:
\nY = a sin \u03c9t
\nsin \u03c9t has no dimensions. Hence we get L = L
\nHence this equation is dimensionaly correct.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-2M-Q2.jpg\")
<\/p>\nQuestion 3.
\nFill in the blanks.<\/p>\n- \n
- The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m2<\/sup> (Write answer in 3 significant digits)<\/li>\n
- Im = ______ ly<\/li>\n<\/ol>\n
Answer:
\n1. Curved area = 2\u03c0l
\n= 2 \u00d7 3.14 (2 \u00d7 102<\/sup>) \u00d7 20 \u00d7 102<\/sup>
\n= 2.51 \u00d7 10-6<\/sup>m2<\/sup><\/p>\n2. l ly= 9.46 \u00d7 1015<\/sup> m
\nlm = \\(\\frac{l \\mathrm{ly}}{9.46 \\times 10^{15}}\\) \u2248 10-6<\/sup>ly.<\/p>\nQuestion 4.<\/p>\n
- \n
- Give a physical quantity with a unit and no dimension.<\/li>\n
- Arrange the following in the descending order.
\n1 light year, 1 parsec, 1 astronomical unit<\/li>\n<\/ol>\nAnswer:<\/p>\n
- \n
- Angle has no dimension. But it has unit.<\/li>\n
- 1 parsec, 1 light year, 1 astronomical unit.<\/li>\n<\/ol>\n
Question 5.
\nMagnitude of force F experienced by a certain object moving with speed V is given by F = KV2<\/sup>. Where K is a constant. Find the dimensions of K.
\nAnswer:
\nF = KV2<\/sup>
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-2M-Q5.jpg\")
<\/p>\nQuestion 6.
\nWhat is the maximum percentage error in the measurement of kinetic energy if percentage errors in mass and speed are 2% and 3% respectively?
\nAnswer:
\nE = \\(\\frac{1}{2}\\)v2<\/sup>
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-2M-Q6.jpg\")
\n% error in KE = % error in mass + 2 \u00d7 % error in speed
\n= 2% + 2 \u00d7 3% = 8%.<\/p>\nQuestion 7.
\nSolve the following with regard to significant figures.<\/p>\n- \n
- 5.8 + 0.125<\/li>\n
- 3.9 \u00d7 105<\/sup> – 2.5 \u00d7 104<\/sup><\/li>\n<\/ol>\n
Answer:
\n1. 5.8 + 0.125 = 5.925
\nRounding to first decimal point, we get 5.9<\/p>\n2. 3.9 \u00d7 105<\/sup> – 2.5 \u00d7 104<\/sup>
\n= 3.5 \u00d7 105<\/sup> – 0.25 \u00d7 104<\/sup>
\n= 3.65 \u00d7 105<\/sup>
\nRounding to first decimal place, we get 3.6 \u00d7 105<\/sup>.<\/p>\nQuestion 8.
\nWhat is maximum fractional error in
\ni) (a + b)
\nii) a – b
\niii) ab
\niv) \\(\\frac{a}{b}\\)
\nGiven \u2206 a and \u2206 b are absolute errors in measurements a and b.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-2M-Q8.jpg\")
<\/p>\nQuestion 9.<\/p>\n
- \n
- What is the fractional error in an<\/sup>? (Given absolute error in a is \u2206 a)<\/li>\n
- What is absolute error in the measurements according to least count?\n
- \n
- 3.0 kg<\/li>\n
- 25 s<\/li>\n
- 5.62 cm<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n
Answer:
\n1. n\\(\\frac{\\Delta a}{a}\\)<\/p>\n2. The measurements according to least count:<\/p>\n
- \n
- 0.1 kg<\/li>\n
- 1 s<\/li>\n
- 0.01 cm<\/li>\n<\/ul>\n
Plus One Physics Units and Measurement Three Mark Questions and Answers<\/h3>\n
Question 1.
\nA stone is thrown upwards from the ground with a velocity \u2018u\u2019.<\/p>\n- \n
- What is the maximum height attained by the stone?<\/li>\n
- Check the correctness of the equation obtained in (a) using the method of dimensional analysis.<\/li>\n<\/ol>\n
Answer:
\n1. H = \\(\\frac{u^{2}}{2 g}\\) ___(1)
\nu = u, v = o, a = -g, h = ?
\nWe can find maximum height using the equation
\nu2<\/sup> = u2<\/sup> + 2as
\n0 = u2<\/sup> + 2 \u00d7 -g \u00d7 H
\n2gh = u2<\/sup>
\nH = \\(\\frac{u^{2}}{2 g}\\)<\/p>\n2. Dimension of H = L
\nDimension of u = (LT-1<\/sup>)
\nDimensions of time (t) = T
\nDimension of g = (LT-2<\/sup>)
\nsubstituting these values in eq(1) we get
\nL = \\(\\frac{\\left(L T^{-1}\\right)^{2}}{\\left(L T^{-2}\\right)}\\)
\nL = L.<\/p>\nQuestion 2.
\nDerive an empirical relationship for the force experienced on the car in terms of mass of the car m, velocity v, and radius of the track r using dimensional analysis.
\nAnswer:
\nCentripetal force may depends on mass (m),radius(r) and velocity(v)
\nF \u03b1 ma<\/sup>rb<\/sup>vc<\/sup>
\nM1<\/sup>L1<\/sup>T-2<\/sup> = Ma<\/sup>Lb<\/sup>(LT-1<\/sup>)c<\/sup>
\nM1<\/sup>L1<\/sup>T-2<\/sup> = Ma<\/sup>Lb<\/sup>Lc<\/sup>T-c<\/sup>
\nM1<\/sup>L1<\/sup>T-2<\/sup> = Ma<\/sup>Lb+c<\/sup>T-c<\/sup>
\nEquating we get a = 1, b + c = 1, c = 2, b = -1
\nSubstituting these values in eq(1),we get
\nF = \\(\\frac{M V^{2}}{r}\\).<\/p>\nQuestion 3.
\nDimensional formula of a physical quantity indicate how many times fundamental quantity is involved in the measurement of the quantity.<\/p>\n- \n
- What is the dimensional formula of coefficient of viscosity?<\/li>\n
- Write any two drawbacks of dimensional analysis.<\/li>\n<\/ol>\n
Answer:
\n1. F = \u03b7A\\(\\frac{d V}{d x}\\)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-3M-Q3.jpg\")
<\/p>\n2. The method of dimensional analysis has the following drawbacks:<\/p>\n
- \n
- It gives no information about the dimensionless constant involved in the equation.<\/li>\n
- The method is not applicable to equations involv\u00acing trigonometric and exponential functions.<\/li>\n
- This method cannot be employed to derive the \u2022 exact form of the relationship if it contains sum
\nof two, or more terms.<\/li>\n - If the given physical quantity depends on more than three unknown quantities, the method fails.<\/li>\n<\/ul>\n
Question 4.
\nPrinciple of homogeneity is based on the fact that two quantities of same nature can be added.<\/p>\n- \n
- What do you mean by principle of homogeneity?<\/li>\n
- Velocity V depends on the time t as V = at2<\/sup> + bt + c. Find dimension of constants a, b, and c.<\/li>\n<\/ol>\n
Answer:
\n1. For the correctness of an equation, the dimensions on either side must be the same. This is known as the principle of homogeneity of dimensions.<\/p>\n2. V = at2<\/sup> + bt + c
\nM0<\/sup>L1<\/sup>T-1<\/sup> = aT2<\/sup> + bT + c
\nAccording to principle of homogenity, we get
\naT2<\/sup> = M0<\/sup>L1<\/sup>T-1<\/sup>
\na = \\(\\frac{\\mathrm{M}^{0} \\mathrm{L}^{1} \\mathrm{T}^{-1}}{\\mathrm{T}^{2}}\\)
\n= M0<\/sup>L1<\/sup>T-3<\/sup>.<\/p>\nQuestion 5.
\nIf x = a + bt + ct2<\/sup> where x is in meter and t in second.<\/p>\n- \n
- Find the dimensional formula of \u2018b\u2019.<\/li>\n
- If error in the measurement of time is 2%. What will be the error in x?<\/li>\n<\/ol>\n
Answer:
\n1. According to principle of homogeneity, the dimensions of both sides must be same.
\nie. L = a + bT + cT2<\/sup>
\nie : L = bT, b = L\/T<\/p>\n2.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-3M-Q5.jpg\")
\n% error in x = 3 \u00d7 %. error in ‘t’ = 3 \u00d7 2% = 6%.<\/p>\nQuestion 6.
\nA physical quantity P is related to four observables a, b, c as P = \\(\\frac{a^{3} b^{2}}{\\sqrt{c d}}\\). The % error in the measurement of a, b, c, and d are 1%, 3%, 4%, 2% are respectively.<\/p>\n- \n
- What do you mean by error in a measurement?<\/li>\n
- What is the % error in the measurement of P?<\/li>\n<\/ol>\n
Answer:
\n1. The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty is called error.<\/p>\n2.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-3M-Q6.jpg\")
\n% error in
\nP = 3 \u00d7 1 + 2 \u00d7 3 + 1\/2 \u00d7 4 + 1\/2 \u00d7 2
\n= 3 + 6 + 2+ 1
\nP = 12%<\/p>\nQuestion 7.
\nRahul measured the height of Ramesh in different trials as 1.67m, 1.65m 1.64m, and 1.63m.<\/p>\n- \n
- Find the mean absolute error?<\/li>\n
- Find the percentage error?<\/li>\n<\/ol>\n
Answer:
\n1. Arithametic mean,
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-3M-Q7.jpg\")
\namean<\/sub> = 1.645m = 1.65
\nabsolute error,
\n\u2206a1<\/sub> = amean<\/sub> – a1<\/sub>
\n\u2206a1<\/sub> = 1.65 – 1.67 = -0.02
\n\u2206a2<\/sub> = 1.65 – 1.65 = 0
\n\u2206a3<\/sub> = 1.65 – 1.64 = 0.01
\n\u2206a4<\/sub> = 1.65 – 1.63 = 0.02
\nMean absolute error
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/08\/Plus-One-Physics-Chapter-Wise-Questions-and-Answers-Chapter-2-Units-and-Measurement-3M-Q7.1.jpg\")
\n= 0.012<\/p>\n2. percentage error = \\(\\frac{\\Delta \\mathrm{a}_{\\text {mean }}}{\\mathrm{a}_{\\text {mean }}}\\) \u00d7 100
\n= \\(\\frac{0.012}{1.65}\\) \u00d7 100
\n= 0.75%.<\/p>\nQuestion 8.
\nIn a particular experiment Ramu used the relation F = AB + (P + Q) Y to calculate force.<\/p>\n- \n
- Which principle is used to check the correctness of the equation (1)<\/li>\n
- If the dimensional formula of Y is M0<\/sup>L1<\/sup>T-1<\/sup>, then find the dimensional formula of P<\/li>\n<\/ol>\n
Answer:
\n1. Principle of homogenity<\/p>\n2. F = AB + (P+Q)Y
\nF = AB + PY + QY
\nMLT-2<\/sup> = AB + PY+ QY
\nAccording to principle of homogeneity
\nMLT-2<\/sup> = PY
\nM1<\/sup>L1<\/sup>T-2<\/sup> = P M0<\/sup>L1<\/sup>T-1<\/sup>
\nie. P = \\(\\frac{M^{\\prime} L^{1} T^{-2}}{M^{0} L^{1} T^{-1}}\\) = M1<\/sup>T-3<\/sup><\/p>\nQuestion 9.<\/p>\n
- \n
- Which of the following is precise\n
- \n
- A vernier calliperse with 40 divisions on sliding scale<\/li>\n
- An optical instrument that can measure length of the order of wavelength of light.<\/li>\n<\/ul>\n<\/li>\n
- Is it possible to increase the accuracy of screw gauge by increasing the number of divisions on the head scale?<\/li>\n<\/ol>\n
Answer:
\n1. (i) L.C of vernier caliperse = \\(\\frac{1}{40}\\) = 0.025mm
\n= 0.025 \u00d7 10-3<\/sup>m
\n= 2.5 \u00d7 10-5<\/sup>m.<\/p>\n(ii) L.C of optical instrument = 6000A\u00b0
\n= 6000 \u00d7 10-10<\/sup>m
\n(Taking \u03bb of visible light = 6000\u00b0A)= 6 \u00d7 10-7<\/sup>m<\/p>\n2. Yes. Because L.C proportional to number of division on the headscale. So with the increase in number of divisions, the least count will increase. This leads to increase the accuracy of above screw guage.<\/p>\n
Plus One Physics Units and Measurement Four Mark Questions and Answers<\/h3>\n
Question 1.
\nIn an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate<\/p>\n- \n
- The mean value of the body<\/li>\n
- Mean absolute error<\/li>\n
- Percentage error<\/li>\n<\/ol>\n
Answer:
\n1. Mean value, Mmean<\/sub>
\n= \\(\\frac{2.52+2.53+2.51+2.49+2.54}{5}\\)
\n= 2.5g<\/p>\n2. Absolute error,
\nAbsolute error \u2206m1<\/sub> = |2.52 – 2.52| = 0
\n\u2206m2<\/sub> = |2.52 – 2.53| = 0.01
\n\u2206m3<\/sub> = |2.52 – 2.51| = 0.01
\n\u2206m4<\/sub> = |2.52 – 2.49| = 0.03
\n\u2206m5<\/sub> = |2.52 – 2.54| = 0.02
\n\u2234 Mean absolute error
\n\\(\\frac{0+0.01+0.01+0.03+0.02}{5}\\)
\n\u2206mmean<\/sub> = 0.014g<\/p>\n3. Percentage error = \\(\\frac{\\Delta \\mathrm{m}_{\\text {mean }}}{\\mathrm{m}_{\\text {mean }}}\\) \u00d7 100
\n= \\(\\frac{0.014}{2.52}\\) \u00d7 100 = 0.556.<\/p>\nQuestion 2.
\nWhile discussing the period of a pendulam, one of the student argued that period depends on the mass of the bob.<\/p>\n- \n
- What is your opinion?<\/li>\n
- How will you prove your argument dimensionally?<\/li>\n<\/ol>\n
Answer:<\/p>\n
- \n
- Period is independent of mass of the bob<\/li>\n
- The principle of homogeneity of dimensions also helps to derive a relationship between the different physical quantities involved; This method is known as dimensional analysis.<\/li>\n<\/ol>\n
The period of the simple pendulum may possibly depend upon:<\/p>\n
- \n
- The mass of the bob, m<\/li>\n
- The length of the pendulum, I<\/li>\n
- Acceleration due to gravity, g<\/li>\n
- The angle of swing, q<\/li>\n<\/ul>\n
Let us write the equation for the time period as t = k ma<\/sup> lb<\/sup> gc<\/sup> qd<\/sup>
\nwhere, k is a constant having no dimensions; a, b, care to be found out.
\nThe dimensions of, t = T1<\/sup>
\nDimensions of. m = M1<\/sup>
\nDimensions of, l = L1<\/sup>
\nDimensions of, g = L1<\/sup>T-2<\/sup>
\nAngle q has no dimensions (since, q = arc\/radius = L\/L)
\nEquating the dimensions of both sides of the equation, we get,
\nT1<\/sup> = Ma<\/sup>Lb<\/sup> (L1<\/sup>T-2<\/sup>)c<\/sup>
\nie. T1<\/sup> = Ma<\/sup>Lb+c<\/sup>+ T-2c<\/sup>.
\nThe dimensions of the terms on both sides must be the same. Equating the powers of M, L and T.<\/p>\na = 0; b + c = 0; -2c = 1
\n\u2234 c = –<\/sup>\\(\\frac{1}{2}\\), b = –<\/sup> c = \\(\\frac{1}{2}\\)
\nHence, the equation becomes,
\nt = kl1\/2<\/sup>g-1\/2<\/sup>
\nie, t = k\\(\\sqrt{1 \/ g}\\)
\nExperimentally, the value of k is found to be 2p.<\/p>\nPlus One Physics Units and Measurement Five Mark Questions and Answers<\/h3>\n
Question 1.
\nIn an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate<\/p>\n- \n
- The mean value of the mass of the ring<\/li>\n
- The absolute error in each measurement<\/li>\n
- Mean absolute error<\/li>\n
- Relative error<\/li>\n
- Percentage error<\/li>\n<\/ol>\n
Answer:
\n1. The mean value of the mass of the ring.
\nMmean<\/sub> = \\(\\frac{2.52+2.53+2.51+2.49+2.54}{5}\\) = 2.52g.<\/p>\n2. The absolute error in each measurement.
\n\u2206m1<\/sub> = Mmean<\/sub> – m1<\/sub> = 2.52 – 2.52 = 0.00
\n\u2206m2<\/sub> = Mmean<\/sub> – m2<\/sub> = 2.52 – 2.53 = -0.01
\n\u2206m5<\/sub> = Mmean<\/sub> – m5<\/sub> = 2.52 – 2.54 = -0.02<\/p>\n3. mean absolute error = |\u2206m1<\/sub>| + |\u2206m2<\/sub>|………..+|\u2206m5<\/sub>|<\/span>
\n= 0.014<\/p>\n4. Relative error = \u03b4m = \\(\\frac{\\Delta \\mathrm{m}_{\\text {mean }}}{\\mathrm{m}_{\\text {mean }}}=\\frac{.014}{2.52}\\) = 0.00555<\/p>\n
5. Percentage error \u03b4m \u00d7 100 = 0.555%.<\/p>\n
Plus One Physics Units and Measurement NCERT Questions and Answers<\/h3>\n
Question 1.
\nFill in the blanks:<\/p>\n- \n
- The volume of a cube of side 1 cm is equal to______m3<\/sup>.<\/li>\n
- The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is equal to____(mm)2<\/sup>.<\/li>\n
- A vehicle moving with a speed of 18km h-1<\/sup> covers_____m in 1s.<\/li>\n
- The relative density of lead is 11.3.Its density is_____g cm-3<\/sup>or_____kgm-3<\/sup>.<\/li>\n<\/ol>\n
Answer:
\n1. V = (1 cm)3<\/sup>
\n= (10-2<\/sup>m)3<\/sup>
\n= 10-6<\/sup>m3<\/sup> - The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is equal to____(mm)2<\/sup>.<\/li>\n
- The volume of a cube of side 1 cm is equal to______m3<\/sup>.<\/li>\n
- Which of the following is precise\n
- What is absolute error in the measurements according to least count?\n
- What is the fractional error in an<\/sup>? (Given absolute error in a is \u2206 a)<\/li>\n
- Im = ______ ly<\/li>\n<\/ol>\n
- The curved surface area of a solid cylinder of radius 2 cm and height 20 cm is_____m2<\/sup> (Write answer in 3 significant digits)<\/li>\n