n 1 : The measure mass and volume of a body are 22.42 g and 4.7 cm³ respectively with possible errors of 0.01 g and 0.1 cm³. The maximum error in density is :Answer 1 :Given :m = 22.42 gv = 4.7 cm³∆m = 0.01 g∆v = 0.1 cm³We know that, when two quantities are MULTIPLIED or divided, the relative error in the result is the sum of the relative errors in the multipliers.Hence, ∆R/R = ∆m/m + ∆v/vLet's find the relative error of mass :⇒ 0.01/22.42⇒ 0.00044 gNow, finding the relative error of velocity :⇒ 0.1/4.7⇒ 0.02 cm³Therefore, ∆R/R = ∆m/m + ∆v/v⇒ ∆R/R = 0.00044 + 0.02⇒ ∆R/R = 0.02044Finding the percentage error in density : 0.02044 × 100⇒ 2.044 % = 2%OPTION B = 2%Question 2 : If a physical quantity is represented by x = [M^aL^bT^-c] and if the percentage error in the measurement of M, L and T are α%, β% and γ% respectively, then the TOTAL percentage error in x is :Answer 2 :Here, the quantities M and L are in multiplication and the quantity T is divided.Hence, the final percentage error = (∆x/x)% = a(∆M/M)% + b(∆L/L)% + c(∆T/T)%We are already given that the percentage error of M, L and T are α%, β% and γ% respectively.Hence, (∆x/x)% = (aα + bβ + cγ)%Option A = (aα + bβ + cγ)%Question 3 : If the units of force, energy and velocity are 10 N, 100 J and 5 m/s; the units of length, mass and time will be :Answer 3 :The DIMENSIONAL Formula of :Force = [MLT^-2]Energy = [ML²T^-2]Velocity = [LT^-1]Given,F = [MLT^-2] = 10 NE = [ML²T^-2] = 100 JV = [LT^-1] = 5 m/sDividing Energy by Force, [ML²T^-2]/[MLT^-2] = [L]Therefore, 100/10 = 10 m.Now, V = [LT^-1] = 5 m/s = 10*T^-1 = 5 m/s⇒ 10/5 = T⇒ T = 2 sAt last, F = [MLT^-2] = 10 N⇒ M*10*2^-2 = 10⇒ M = (10 × 2²)/10⇒ M = 4 kgHence, Option (B) = Units of length, mass and time will be : 10 m, 2 s and 4 kg respectively.Question 4 : The least count of a stopwatch is 0.01 s. The time period of 100 oscillations of the pendulum is 100 s. The percentage error in the time period will be :Answer 4 :Number of oscillations = 100Time taken = 100 sThe error in 100 oscillations = 0.01 sSo, Time period for 100 oscillations = 100 s ± 0.01 sHence, Percentage error = ∆t/t × 100 = 0.01/100 × 100⇒ Percentage error = 0.01 %Hence, Option A = 0.01%.