secre tat err de 2 parallax was avoided when taking ™Y reading from the meter rule. APPARATUS: Vernier calipers, metal cylinder. : ip other to estimate the volume of a given cylinder, the length and Gameter of the cylinder is measured as followed. | Least Count, = I Main Scale Division - I Venire Scale Division Venire | Constant or Accuracy : ISD = MS.Dx Noofiisionsonthemcinscale : Noofdivisiosonthe Vernierscole ee ee ee es Where VSD = Venire Scale Division xe: homarsaier = Main Scale Divisi Some times in solving calculations izvolving cali given (this is eee ee ving calipers, an error may be {Add the value of such er to your fl anower i Par exxor but subtract if it is a positive error. if it is 2 negative } Preblems day's lectures carelessly picked up 2 ee his hostel after the curiosity, he picked up a metre rule and venire caliper to ‘dew, Our of | dimensions of the metal piece on getting to the laborat If the following observations were made by him. ory the following day. Length of the metal piece = 3m Qne main scale division = Imm One Vernier scale division = = 0.98mm Main scale reading = 3.8mm Vernier coincidence =. is What should be the diameter and the volume of the metal pj | 87|Pagef 4 Least Count = IMS.D-Lv.s.D Imm - 0.98mm 0.020mm MAS.R + (Least Count x Vernier 3.8 + (0.02 x 18) selneldeane) 3.8m + 0.36mm 416mm = 0.416cem 7 . Volume ofaCylinder = nr d = 314x (2418) 500m . 2 = 408m? NB: Take units into consideration, know when you are using mm or em. A traveling microscope has a main scale graduated in 0.5mm. 50 vernier divisions coincide with 49 main scale divisions, What is the vemier constant and the reading if the main scale reads 7.45cm at a venire coincidence of” 29cm division. Solution : Noofdivisionsonthemainscale ee MS Dey ofdivisiosontheVernlerscale 1ySD= O.Smmese i 0.49mm -. Vernier constant = IMSD-LVS.D =0.5mm — 0.49mm =0.0lm =0.001lcm Diameter Reading on the vemier caliper = MSR + (Least Count x Venre Coincidence) = —-'7.45cm + (0.001cm x 29) = —-7,54cm x 0.029em = = -7.479em 88|Pageshould be put into consideration. Pitch: This can bé said to be the distance moved by the head after one complete rotation. 7 A 7 Pitch = distance moved by the head /no of rotation Least Count or Accuracy = Pitch/no of divisions : Thickness or diameter = Pitch scale reading + —‘Fractignal reading (No of rotatighs x pitch) (LC |} head scale coincidence) PROBLEMS ole | ‘An engineering student who picked up 8 cylindrical metal piece, chopped off part of the diameter of the metal piece at one end with a lathe machine. Finally, he made the following observations while trying to measure the thickness of the “zxrchopped portion with a micrometer screw gauge. Number of divisions on the head 50 Distance moved by the head after one complete rotation 0.05mm. After 28 rotations with the head, the head coincidence with the pitch is 10. What is accuracy of the gauge and the thickness of the unchopped portion. Solution Pitch = —-0.05mm Least Count = ewe Noofdivisions = — 0.05mm 50 = 0.00imm = — 0.0001cm Thickness = (No of rotation x Pitch) + (L.C x Head scale coincidence) (28 x 0.05) + (0.001 x 10) 7 14mm + 0.0mm O.l4em + —0.001em - 0.141em The pitch of a micrometer screw ii many divisions Pitch i gauge is Imm. How required inthe ead sale to measure up tothe ascureey of 0200S c 92|Pagepiation \ccuracy as Pitch Noofdivisions " No of divisions = __ Oem 0.0005cm = 200 divisions The pitch of a screw gauge is Imm and there are 260 divisions | circular scale. What is the accuracy of the instrument? Solution Accuracy = Pitch . No of divisions a 0.1lem 250 - = 0.0004cmAfter that, then calculate the slope of the graph and also measure the mass of the metre rule and the length of the thread L, Hence the moment of inertia can then calculated according to the equation. T=2 Mg Square both sides 4IL T? = 4n? xo ed? Tq? _ \on7IL Mg But for a graph of T against 1/d, slope=Td (Stopey = 1S pe Mz 6 i = Mg x(Slopey Source of Error 1on'L . i. Faulty apparatus ii. Zero error iii, Error due to human in capacitance iv. The meter rule can measure only up to an accuracy of 0.lem Precautions i. It was ensured that the meter rule was vertically suspended ii. — The windows were closed to avoid air resistance iii. The meter rule was displaced only by a small. angle QUESTION Is it possible to conduct this experiment by changing one length one length (1) of thread? If yes state the variables for plotting the graph. If no why? Solution | it is possible to conduct the experiment by changing one length of the thread. In this case a graph of T against | is plotted. PROBLEMS An i with bifilar suspension was Performed to determine Moment of inertia. From the theory: 95; PageEXP MC3 BIFILAR SUSPENSION Aim: Determination of moment of inertia using the bifilar Suspension, Apparatus: Two heavy stands and clamps, two threaded co; stop watch. mS mete Tie Nature of the Graph The_experiment is carried out and the value of d-and the time for) complete oscillations are recorded. Then a graph h of T aoe | “a is plotted, this gives rise to a stright line graph. “ud M4]Pace‘After that, then calculate the slope of the graph and also measure the mass of the metre rule and the length of the thread L. Hence the moment of inertia can then calculated according to the equation. \ Square both sides Tn det tll Mgd’ Pd = 16n77L But for a graph of T against 1/d, slope = Td 16n7IL (Stope)’ = “Fe i Mg x(Slope)_ Source of Error Vent “i, Faulty apparatus ii. Zero error iii. | Error due to human in capacitance iv. The meter rule can measure only up to an accuracy of 0.lem Precautions i: It was ensured that the meter rule was vertically suspended ii. | The windows were closed to avoid air resistance iii. | The meter rule was displaced only by a small.angle QUESTION Is it possible to conduct this experiment by changing one len: length (1) of thread? If yes state the variables for plotting the graph. If why? Solution Yes, it is possible to conduct the experiment by changing one \e of the thread. In this case a graph of T” against | is plotted. PROBLEMS An experiment with bifilar suspension was performed to determine my of inertia, From the theory: . %8|Page qae Term [ale Mgd’ . If the rod is Placed through small Wh, inclination of the thread is 9, which of the foto Solution . Ifa slope of 4x Was obtained, th, ce » the Moment of Inertia Will be 1 = Mgx(slope) 16x77 (4x)? 1 = Max 67 1a Meslxn? 16x77 =k e T€“L” is varied in the experiment; the table of the graph would be Solution From the equation, the table of the Braph could be: T against JE r against L EXP.MC4 it with Cantilever ; oe cote of the Young’s Modulus from the period of vibration of a loaded cantilever, a3 Gea Apparatus: metre rule with weight rigidly attached to one end, G-clamp, screw-gauge, stop watch. ch by the Nature of the Lew the loaded bum fully oh eg of te ad Geelaney with nie length (L) projecting om i -C! 96|Pageyp vibrate and the periodic time T for 20 complete oscillations with various {Lis recorded, Se against L is plotted thus given a straight line graph oe to the equation r=2e|Me® IE Squaring both sides we have m= ite UE . rae . @ Ls But 12 mat = slope of the graph. Hence we have, - 4n2M x12! 0 NP KE E= 162°M bd? x slope The last equation is used to compute the Young Modulus Source of Error Refer to page Precautions i. It was ensured that the loaded beam was firmly clamped ii The beam was displaced lightly with a small angle 0 , iii, Error due to parallax was avoided when taking from the metre rule. 7)PaneQUESTIONS Given T; = 4n?M13/318 discuss what nature of the graph expected, Does your graph conform to the relation (explain), if not why. Solution For a graph of T; against L?, a strai; i 2 ag . ight line graph is obtained. This ; from one equation’ of the graph the to parameters have a Vineae rejee® as seen = (40M a (54) L+0 y=mx+C PROBLEMS eIn an experiment with cantilever, the vari . rectangular beam of length L. if 6 and Cae eagle lend affixed to 4 * thickness of the rectangular beam. What is the Young Modules sia and of the graph is 42°. Sif the slope Solution : i: Young Modulus E is given by 5-6? . bd? xslope _ 16M ~ bdx4x? Au bd 1647xL? bd?x4n? Ae elf the sae in question above is kept constant while lengin Lis varied, what will be the Young’s Modulus, if a slope of 2n? is obtained. E= 162°M? bd°xslope 16x7M bd?x4x* Re 7EXP,MC5 Experiment with Spiral Spring A. Elastic constant of spiral spring ‘Aim: To determine the elastic constant of the given spiral spring and a value for the acceleration due to gravity g, ' Apparatus: Spiral spring, weight hanger, meter rule, two clamps and slotted weights, heights pointer, stop watch. ea Nature of the Graph The different masses which gives rise to corresponding extensions are recorded as the experiment is being carried out. NB: The student is advised to be fast and be time conscious while performing this experiment as he/she is expected to plot two graphs within a limited period of time. 2 A graph of F against the extension is plotted. Then the slope of the graph is used to obtain the elastic constant of the spring. F = Ke (Hookes Law) F (sop) =. K Foy ecm B. Determination of Acceleration due to Gravity Using the same apparatus, the student performs another experiment by ulowing the spring to make simple harmonic motions. The one t for 20 Complete oscillators and the tension in each case as recorded.oN traph of T) against M Is plotted hence ravity) can be calculated according to the equation. | rane [Me saa a. ® * Where T = pertod M . tension in the spring . : . soodoeton dae ed from Hooke: By using both sides we have ie “Lew, pte died & Hence for a graph of T’ against M the slope is safzid q 4nd ge Hen¢e q can becalculated Source of Error , Refer to page 3 Precautions Error due to parallax was avoided when measuring the extension The exp. was performed twice with inc sing nal decreasin ensure accuracy. tl in a QUESTION The slope of the graph represents the elastic constant of the spring PROBLEMS Acceleration due to gravity g can be determined with experiment involveg the use of spiral spring. From the theory T =2z. pend & The spring constant d can be obtained from Hookes law as either. > The inverse of the slope of graph of extension against tension Or > The slope of graph of tension against extension. 100|PageIf, going by the theory in the Question above, the mass is varied, what will be the acceleration due to Btavity if the slope of 2n is obtained? Solution reas [td 8 regi? & Fora graph of T? ag Pied nm g sate & ButS=2r- x4 g “.g= 2nd ‘ When a mass m, is hung from lower end of a spring of negligible mass, the spring extends by x. what is the period (T) if the mass is set into vertical oscillations.(2010) Solution ainst m slope equals _ of Wns m Ts2r)™ tt But F =kx (Hookes Law) Ks x > T2209 | mq Ts aft q 101| Paceape main scale of a spectrometer is graduated in 0.5°, The length of thirty venire divisions is observed to be equal to twenty nine divisions on the mail scale. The least count of the instrument is jution oo 1° is equivalent to 60! <. 0,5" is equivalent to x ~_ 0.5*x60 > x =30' VSD =MS.Dx-Noefaivisionsonthemainscale s ‘NoofdivisiosontheVernierscale 2 IVS.D =30'x— 3 = 29" . ‘Least Count = 1MS.D-LV.S.D 30' -29! v While using the spectrometer above, the following measurements were — made: E Main Scale = 104° Venire coincidence = 12 The reading on the instrument is therefore given by: Solution 7 : Reading on the an = Main Scale Reading + (Least Count x Venire Coincidence) 10.49 + (1x 12) 1049412) ie Bui 7 a - Where x= 2a 0°24 +12 = fore + 12 6" : we ‘ — value of one main scale division is O.tem*and a value- of ‘crn for each venire scale division. Ifthe vernier caliper has a positive 0.09756 ‘05cm, what is the accuracy of the instrument and the thickness of error of OF ie main scale reading is 2.9Sem and the vernier coincidence is w ou1.M.8.D ~ L.V.8.D 0.1m - 0.0975em - 0,0025em Thickness of the wire = Main scale reading + (L.C x Venire = 2.950m + (0.0025em x 12) “ineldence) = 2.950m + 0.03cm = 2.98 This is not yet the final answer because a positive error of 0.05cm is given in the question. Although most students will hurry to shade, just to take your time. Positive error is 0.05cm Since it is a positive error we subtract it from the main reading i .. Thickness = — 2.98cm - 0.05cm = 2.93cm This is now your final answer The value of one main scale division is 0.1cm and a venire of 0.0975em for each verlire scale division. If the vernier caliper has a negative zero error of 0.05cm, what is the accuracy of the instrument and the accurate thickness of a wire when the main scale reading is 2.95cm and the vernier coincidence is 12. Solution = Least Count = LM.S.D-1V.S.D a 0.1 - 0.975 = 0.0025 Thickness of the wire = = M.S.R+(L.C x Venire coincidence) = 2,95ém + (0.025cm x 12) = 2.98cm This case a negative error of 0.05cm is given hence the accurate thickness of the wire equals the original thickness + the negative error. (We “add in this case since it is negative error) Accurate thickness © = —-2.98 + 0.05 - = . 3.03cm Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale division is 0.49mm. What is the least count of the instrument in millimeter and in centimeter units. Which of the following instruments has a reading accuracy of 0.0S5mm? 90|PageSolution LeastCount= = 1.M.§.p— LV.8.D 7 naan = 0.0975cm TH .0025em ickness of the wire: = Main scale reading + (L.C x Venire Coincidence 2.9Sem + (0,0025cm x 12) ? = 2.95cm + 0,03cm This is not yet the final ancora given ° ¢ final answer because a positive error of 0, i ieee : .05cm is given i a ion. Although most students will hurry to shade, just tothe ew si Positive error = 0.05em ; ince it is a positive error we subtract it from the main readi ~. Thickness = — 2.98cm - 0.0Sem wie ; : = * 2.93cm This is now your final answer The value of one main scale division is 0.1cm and a venit ‘ nain . ire of 0. each verlire scale division. If the vernier caliper has a negative pi eee 0.05cm, what is the accuracy of the instrument and the accurate. thickness of a wire when the main scale reading is 2.95cm and the vernier coincidence is 12. Solution = Least Count= LM.S.D-LV.S.D . = 01-0975 = 0.0025 M.S.R + (L.C x Venire coincidence) 2.95ém + (0.025em x 12) eon ay 2.98cm ' : This case a negative error of 0.05cm is given hence the accurate thickness of the wire equals the original thickness + the negative error. (We “add in this case since it is negative error) Accurate thickness = 2.98 + 0,05 - = 3.03cm is 0.5mm and the Suppose the value of one division of vernier scale value of one main scale division is 0.49mm. What is the least count of the instrument in millimeter and in centimeter units. Which of the following instruments has a reading accuracy of 0.0Smm?_—. : 90|Page Thickness of the wirecoum IMS.D = LV.8.D. gaat count = 0.8mm = 049mm =0.01mm = 0.001em eis the vernier caliper that has an accuracy of0.05mm. Te these on your own before looking at the solution The length of 9 divisions of the main scale of a venire caliper is observed to be equal to 10 divisions on the venire scale. The value of 1 main scale division is Imm while measuring the length of a rod; the main scale reading was 2.70cm with the 5” division on the vernier scale coinciding with a division on the main scale. If the instrument has a positive zero error of 0.05cm, the Least Count and the length of the rod are Solution LV.s.D = MSD .x No of divisions on the main scale No of divisions on the venire scale IvSD= mm ; = 09mm : LeastCount= LMS.D-LV.S.D How tl So 8 3 Main scale reading + (L.C x venire coincidence) 2.70cm + (0,01em x 5) 2.70cm + 0,05cm 2.75em 0.05cm 2.75cm — 0.05cm 2.7om B, THE USE OF MICROMETER SCREW GUAGE ‘Aim: To measure the diameter of a given Wire by using a screw gauge and calculate the volume of the wire, o1[Page Length of the rod Positive error /. Final reading ranunanadns