Title 3.EXPERIMENTAL PHYSICS ( 66 - 87 ).pdf ✅

Jr Chemistry E/M GENERAL PHYSICS 6 6 NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED - VOL-IIX NISHITH Multimedia India (Pvt.) Ltd., SYNOPSIS 1. Least count of a screw guage = pitch of screw number of divisions on circular scale 2. Least count of Vernier calliper = 1 main scale division - 1 vernier scale division 3. In both screw gauge and vernier calliper actual reading = observed reading - zero error = observed reading + zero correction 4. Determining the value of ‘g’ using a simple pendulum Theoretically 2 2 L T=2 4 g L g T     where L = length of the thread + radius of the bob 5. Determining Young’s Modulus of a given wire by “Searle’s Method”: To determine Young’s Modulus, we can perform an ordinary experiment. Lets hang a weight ‘m’ from a wire from Hook’s law: 0 2 0 mg x l Y x mg A l  r Y                If we change the weight, the elongation of wire will increase proportionally. If we plot elongation v/s mg, we will get a straight line By measuring its slope and equating it to 0 2 l  r Y       , we can estimate Y . 6. Determining specific heat capacity of an unknow liquid using calorimeter: Figure shows the Regnault’s appratus to determine the specific heat capacity of a unknown liquid. A solid sphere of known specific heat capacity 1 S having mass m1 and initial temperature 1 , is mixed with the unknown liquid filled in a calorimeter. Let masses of liquid and calorimeter are m2 and m3 respectively, specific heat capacities are 2 S and 3 S and initially they were at room temperature 2 . When the hot sphere is dropped in it, the sphere looses heat and the liquid calorimeter system takes heat. This process continues till the temperature of all the elements becomes same (say  ). Heat lost by hot sphere = m s1 1 1    Heat taken by liquid and calorimeter = m s m s 2 2 2 3 3 2           If there were no external heat loss Heat given by sphere = Heat taken by liquid - Calorimeter system m s m s m s 1 1 1 2 2 2 3 3 2                 Get     1 1 1 3 3 2 2 2 2 m s m s s m m         By measuring the final (steady state) temperature of the mixture, we can estimate 2 s : specific heat capacity of the unknow liqid. To give initial temperature 1  to the sphere, we keep it in steam chamber "O" , hanged by thread. Within some time (say 15 min) it achieves a constant temperature 1 . GENERAL PHYSICS
NISHITH Multimedia India (Pvt.) Ltd., 6 7 JEE - ADVANCED - VOL-IIX CURRENT ELECTRICITY NISHITH Multimedia India (Pvt.) Ltd., GENERAL PHYSICS Now the calorimeter, filled with water (part C) is taken below the steam chamber, the wooden removable disc D is removed, and the thread is cut. The sphere drops in the water calorimeter system and the mixing starts. If sp.heat capacity of liquid s2  were known and that of the solid ball s1  is unknow then we can find      2 2 3 3 2 1 1 1 m s m s s m         7. Determining speed of sound using resonance tube: Resonance tube is a kind of closed organ pipe. So its natural frequencies are 3 5 , , ...... 4 4 4 eq eq eq V V V l l l If the air column resonates with a tuning fork of frequency 0 f then   0   0 2 1 2 1 4 4 eq eq V V n f l n l f      eq l l e   where l is length of the air column in tube and e is end correction. 1 4 0 V l e f   , 2 0 3 4 V l e f   soV f l l   2 0 2 1   Max Permissible Error in speed of sound due to error in 0 1 2 f l l , , : for Resonance tube experiment V f l l   2 0 2 1   ln ln 2 ln ln V f l l     0 2 1   max. permissible error in speed of sound =   0 2 1 max 0 2 1 dV l l f V f l l              8. Verification of Ohm’s law using voltmeter and ammeter Ohm’s law states that the electric current I flowing through a conductor is directly proportional to the P.D. (V) across its ends provided that the physical conditions of the conductor (such as temperature, dimensions, etc.) are kept constant. Mathematically. V I or V IR   Here R is a constant known as resistance of the conductor and depends on the nature and dimensions of the conductor. 9. METER BRIDGE: Meter bridge is a simple case of Wheatstone’s- Bridge and is used to find the unknown Resistance. The unknown resistance is placed in place of R, and in place of S , a known resistance is used, using R.B. (Resistance Box). There is a 1m long resistance wire between A and C. The jockey is moved along the wire. When R l S l 100     then the Bridge will be balanced, and the galvanometer will gives zero deflection. " "l can be measured by the meter scale. The unknown resistance is 100 l R S l   ............(1) If length of unknown wire is L and diameter of the wire is d, then specific resistance of the wire 2 4 d R L          from eq.(1)   2 4 100 d l S L l           End Corrections In meter Bridge circuit, some extra length comes (is found under metallic strips) at end points A and C. So some additional length   and  should be included at ends for accurate results. Hence in place of l we use l  and in placed of 100 l , we use 100  l  (where  and  are called end corraction). To estimate  and  , we use known resistance R1 and R2 at the place of R and S in meter Bridge. Suppose we get null point at 1 l distance then 1 1 2 1 100 R l R l       ...................(i) Now we interchange the position of R1 and R2 and get null point at 2 l distance then 2 2 1 2 100 R l R l       .................(ii)
Jr Chemistry E/M GENERAL PHYSICS 6 8 NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED - VOL-IIX NISHITH Multimedia India (Pvt.) Ltd., Solving equation (i) and (ii) get 2 1 1 2 1 1 2 2 1 2 1 2 100 R l R l R l R l and R R R R          These end corrections   and  are used to modify the observations INDEX ERROR: In u - v method, we require the distance between object or image from the pole (vertex) of the mirror (actual distance). But practically we measure the distance between the indices A and B. (Observed distance), which need not exactly coinsides with object and pole, there can be a slight mismatch called index error, which will be constant for every observation. Index error = Observed distance - Actual distance (Just like zero error in screw gauge, it is the excess reading). To determine index error, mirror and object needle and placed at arbitery possition. For measuring acual distance, a knitting needle is just fitted between the pole of mirror and object needle “O”. The length of knitting needle will give the actual object distance while the seperation between indices A and B at that instant is the observed distance. So index error is - e = Observed distance - Actual distance =Seperation between indices A and B - Length of knitting needle once we get e, in every observation, we get Actual distance = Observed distance (seperation between the indices) - Excess reading (e) *There is an another term, Index corraction which is invert of index error. Index correction = - index error LEVEL -V SINGLE ANSWER QUESTIONS 1. The length of a rectengular plate is measured by a meter scale and is found to be 10.0 cm. Its width is measured by vermier callipers as 1.00 cm. The least count of the meter scale and varmier callipers are 0.1cm and 0.01cm respectively (Obeviously). Maximum permissible error in area measurement is (A) 2 0.2 cm (B) 2 0.1cm (C) 2 0.3 cm (D)Zero 2. In the previous question, minmum possible error in area measurement can be (A) 2 0.02 cm (B) 2 0.01cm (C) 2 0.03 cm (D)Zero 3. For a cubical block, error in measurement possible error in area measurement os sides is 1% and error in measurement of mass is 2%, then maximum possible error in density is (A)1% (B)5% (C)3% (D)7% 4. TO estimate ‘g’ 2 2 4 L from g T        , error in measurement of L is 2% and error in measurement of T is 3%. The error in es- timated ‘g’ will be (A) 8% (B) 6% (C) 3% (D) 5% 5. The least count of astop watch is 0.02 sec- ond. The time of 20 oscillations of a pendu- lum is measured to be 25 seconds. The per- centage error in the time period is (A)16% (B)0.8% (C)1.8% (D)8% 6. The dimensions of a rectangular block mea- sured with a vernier callipers having least count of 0.1 mm is 5 10 5 mm mm mm   . The maximum percentage error in measure- ment of volume of the block is (A)5 % (B)10 % (C)15 % (D) 20 % 7. An experiment measures quantities x y z , , and then t is calculated from the data as 2 3 XY t Z  . If percentage error in x y and z , are respectively 1%, 3%, 2% , then percent- age error in t is: (A)10 % (B) 4 % (C) 7 % (D)13 %
NISHITH Multimedia India (Pvt.) Ltd., 6 9 JEE - ADVANCED - VOL-IIX CURRENT ELECTRICITY NISHITH Multimedia India (Pvt.) Ltd., GENERAL PHYSICS 8. The external and internal diameter of a hollow cylinder are measured to be. The thickness of the wall of the cylinder is (A)0.34 0.02   cm (B)0.17 0.02   cm (C)0.17 0.01   cm (D)0.34 0.01   cm 9. The mass of a ball is 1.76 kg. The mass of 25 such balls is (A) 3 0.44 10  kg (B) 44.0 kg (C) 44 kg (D) 44.00 kg 10. Two resistors R1 24 0.5   and R2 8 0.3   are joined in series. The equivalent resistance is (A)32 0.33   (B)32 0.8   (C) 32 0.2   (D)32 0.5   11. In ohm’s law experiment, potential drop across a resistance was measured as v  5 0. volt and current was measured as i amp  2 00 . . Find the maximum permissible error in resistance. (A) 1.5% (B) 2.5% (C) 1% (D) 5% 12. Read the normal screwgauge main scale has only mm marks. Circular scale has 100 division. In complete rotation, the screw advances by 1 mm. (A) 11 mm (B) 11.65 mm (C) 11.650 mm (D) 11.6 mm 13. In a complete rotation, splindle of a screw gauge advances by 1 2 mm. There are 50 divisions on circular scale. The main scale has 1 2 mm marks 1 is graduated to mm 2        If a wire is put between the jaws, 3 main scale divisions are clearly visible, and 20th division of circular scale coincides with the reference line. Find diameter of wire in correct S.F. (A) 1.7 mm (B) 1.70 mm (C) 3.40 mm (D) 3.20 mm 14. Read the vernier (A) 15.6 mm (B) 15.60 mm (C) 14.6 mm (D) 14.4 mm 15. Read the special type of vernier (A) 13.6 mm (B) 13.60 mm (C) 12.6 mm (D) 12.60 mm