Lab Phy 3rd 4th
Lab Phy 3rd 4th
Lab Phy 3rd 4th
AIM OF THE EXPERIMENT: To measure the cross-sectional area of a wire by using Screw
gauge
APPARATUS REQUIRRED
1. Screw Gauge
2. A piece of wire
3. Geometry box
THEORY:
D2
Area = R 2
4
Diameter of the wire is calculated by the screw gauge using the formula
C.S.R = (Difference between ICSR and FCSR) X Least Count = (ICSR ~ FCSR) X Least Count
Pitch: It is the distance covered by the circular scale on the linear scale by one complete rotation
Pitch = 0.5 mm
Pitch 0.5
Least Count = 0.01 mm
Total no. of divisions on the circular scale 50
Zero Error: If the zero of the Circular scale does not coincide with the reference line of the linear
scale, then zero error arises.
Where X is the no. of circular scale divisions which have either crossed or fall short of main scale
Positive Zero Error: if the zero of the circular scale remains below the reference line or the zero of
the linear scale, then it is known as positive zero error. It is to be subtracted from the observed
value.
Negative Zero Error: if the zero of the circular scale remains above the reference line or the zero
of the linear scale, then it is known as negative zero error. It is to be added to the observed value.
PROCEDURE:
1. The pitch and the least count of the screw gauge were determined.
2. The linear scale and ordinary scale was standardised.
3. When the wire was placed between the two gaps, initial circular scale reading (ICSR) was
noted.
4. The wire was removed and the no. of complete rotations was counted while closing the
gaps.
5. When the gap is closed, the Final Circular Scale Reading (FCSR) was noted.
6. To get the difference, the following condition was followed.
i. If I > F, then difference = I – F
ii. If I < F, then difference = n + I – F
7. The pitch scale reading and circular scale reading were found out and their sum gives
diameter of the wire.
8. Then the above procedure is repeated at least 10 times to find the length of the cylinder.
9. Mean diameter of the wire was determined.
10. Then the cross-sectional area of the wire was obtained using the formula.
OBSERVATION:
No. of Pitc LC ICSR No. Of FCS Differen PSR CSR Diameter Mean
Observatio h In complete R ce (P * (I~F * = PSR+ In
n P in m rotations I~F N) LC) CSR mm
mm m N in in mm In
mm mm
CALCULATION:
D2
Area = R 2
4
CONCLUSION:
After performing the above experiment and taking necessary readings, the value of the cross-
sectional area of the wire is found to be ____________ mm2.
EXPERIMENT - 4
AIM OF THE EXPERIMENT: To measure the volume of an irregular lamina by using Screw
gauge
APPARATUS REQUIRRED
1. Screw Gauge
2. Irregular plane lamina
3. Graph paper
THEORY:
V = S x t cm3
Thickness of the irregular lamina is calculated by the screw gauge using the formula
C.S.R = (Difference between ICSR and FCSR) X Least Count = (ICSR ~ FCSR) X Least Count
Pitch: It is the distance covered by the circular scale on the linear scale by one complete rotation
Pitch = 0.5 mm
Pitch 0.5
Least Count = 0.01 mm
Total no. of divisions on the circular scale 50
Zero Error: If the zero of the Circular scale does not coincide with the reference line of the linear
scale, then zero error arises.
Where X is the no. of circular scale divisions which have either crossed or fall short of main scale
Positive Zero Error: if the zero of the circular scale remains below the reference line or the zero of
the linear scale, then it is known as positive zero error. It is to be subtracted from the observed
value.
Negative Zero Error: if the zero of the circular scale remains above the reference line or the zero
of the linear scale, then it is known as negative zero error. It is to be added to the observed value.
PROCEDURE:
1. The pitch and the least count of the screw gauge were determined.
2. The linear scale and ordinary scale was standardised.
3. When the irregular lamina was placed between the two gaps, initial circular scale reading
(ICSR) was noted.
4. The lamina was removed and the no. of complete rotations was counted while closing the
gaps.
5. When the gap is closed, the Final Circular Scale Reading (FCSR) was noted.
6. To get the difference, the following condition was followed.
i. If I > F, then difference = I – F
ii. If I < F, then difference = n + I – F
7. The pitch scale reading and circular scale reading were found out and their sum gives
diameter of the wire.
8. Then the above procedure is repeated at least 10 times to find the thickness of the irregular
lamina.
9. Mean thickness of the lamina was determined.
10. Then the given lamina was placed at three different places on the graph paper and its
outline is drawn at each location with the help of a pencil.
11. The outlines were numbered as figure 1, figure 2 and figure 3.
12. The total number of big square divisions (area 1cm2) and small square divisions (area 0.01
cm2) enclosed inside the outline boundary were counted.
13. Then the total area of the lamina and the mean area were calculated as per the tabulation.
14. The volume of the irregular lamina was calculated using the formula.
OBSERVATION:
CALCULATION:
CONCLUSION:
After performing the above experiment and taking necessary readings, the value of the volume of
the irregular lamina is found to be ____________ cm3