Q: Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is

aligned & 10 MSD =1 ~cm ; 1 MSD =0.1 ~cm \\ & 7 MSD =10 VSD \\ & 1 VSD =0.07 ~cm \\ & Reading =2 MSD - VSD \\ & =0.2 ~cm -0.07 ~cm =0.13 ~cm aligned Ans. Option (C)

Q: In order to measure the internal resistance r 1 of a cell of emf ε , a meter bridge of wire resistance R 0 = 50 Ω , a resistance R 0 2 , another ce

Resistance of potential wire is R 0 = 50 Ω Resistance of 100 cm wire = 50 Ω So, Resistance of 72 cm wire = 50 100 × 72 = 36 Ω Current, I = ε 2 14 + 25 = ε r 1 + 75 ⇒ r 1 = 3 Ω

Question 1

Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given. To obtain the value of d, the diameters of the two wires are measured with a screw gauge. The screw gauge has a

Accepted Answer

The least count (LC) of the screw gauge is given by: LC = Pitch Number of divisions on CS = 0.5 100 = 0.005 mm For Wire-1, the measured reading is: Reading _1 = LS + CS LC Reading _1 = 0.5 + 42 0.005 = 0.5 + 0.210 = 0.710 mm The actual diameter of Wire-1 is 0.650 mm . The zero error of the screw gauge is: Zero Error = Measured Reading - Actual Value = 0.710 - 0.650 = +0.060 mm For Wire-2, the measured reading is: Reading _2 = LS + CS LC Reading _2 = 1.5 + 95 0.005 = 1.5 + 0.475 = 1.975 mm The ac

Question 2

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is

Accepted Answer

aligned & 10 MSD =1 ~cm ; 1 MSD =0.1 ~cm \ & 7 MSD =10 VSD \ & 1 VSD =0.07 ~cm \ & Reading =2 MSD - VSD \ & =0.2 ~cm -0.07 ~cm =0.13 ~cm aligned Ans. Option (C)

Question 3

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0 . 5 &#160; mm . The circular scale has  100 divisions and for one full rotation of the circul

Accepted Answer

Note: This question is given bonus by JEE council. In one rotation 2 divisions of the main scale are crossed. Therefore, the least count of the screw gauge is L C = 2 &#215; 0 . 5 100 = 0 . 01 &#160; mm And the zero error is  4 &#215; L C = 0 . 04 &#160; mm Reading-1 R 1 = MSR + LC &#215; CSR - Zero &#160; error = 2 + 0 . 20 - 0 . 04 mm = 2 . 16 mm Reading-2 R 2 = 2 + 0 . 16 - 0 . 04 mm = 2 . 12 mm Therefore, average reading  R m = R 1 + R 2 2 = 2 . 14 &#160; mm Average mean error&nbsp

Question 4

In order to measure the internal resistance  r 1  of a cell of emf  &#949; ,  a meter bridge of wire resistance  R 0 = 50 &#160; &#937; , a resistance  R 0 2 , another ce

Accepted Answer

Resistance of potential wire is  R 0 = 50 &#160; &#937; Resistance of  100 &#160; cm  wire  = 50 &#160; &#937; So, Resistance of  72 &#160; cm  wire  = 50 100 &#215; 72 = 36 &#160; &#937; Current, I = &#949; 2 14 + 25 = &#949; r 1 + 75 &#8658; r 1 = 3 &#160; &#937;

Question 5

The smallest division on the main scale of a Vernier calipers is  0 . 1 &#160; cm . Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left

Accepted Answer

Given,  10 V S D = 9 M S D Least count of the vernier caliper,  L C = M S D - V S D &#8658; L C = M S D - 9 M S D 10 = M S D 10 = 0 . 1 &#160; cm 10 = 0 . 1 &#160; mm &#8658; L C = 0 . 01 &#160; cm Case I: When sphere is not present between jaws then, main value  =  Zero error  = M . S . R + L . C . &#215; V . S . R Zero error  = - 1 + 0.1 &#215; 10 - 6 &#160; mm = &#8722; 0.04 &#160; cm Case II: When sphere is present between jaws then, main value  = 3.1 + 0.0

Experimental Physics — Previous Year Questions & Solutions