You are provided with two wires marked P and C. a resistor \(R_s\) = 1\(\Omega\) and other necessary apparatus.
Connect \(R_s\) in the left-hand gap of the metre bridge, a length L= 100cm of wire P in the right-hand gap and the other apparatus as shown in the diagram: above
Determine the balance point B on the bridge wire AC
Measure and record AB =/s, and BC = /
Evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs
Repeat the procedure for four other values of L = 90, 80, 70 and 60cm. In each case obtain and record the value of |\(_{s}\) and |\(_{p}\) and evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs
Repeat the experiment with the second wire, Q. Obtain the value of |\(_{s}\) and |\(_{Q}\) for equal lengths of wire as used in wire P.
Evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs. In each case, tabulate your readings.
Plot a graph of \(R_2\) on the Vertical axis against \(R_1\) on the horizontal axis.
Determine the slope S, of the graph.
Evaluate the k = \(\sqrt s\).
State two precautions taken to ensure accurate results.
(b)i. Define the resistivity of the material of a wire.
ii. A galvanometer with a full-scale-deflection of 1.5 x10\(^{3}\). A has a resistance of 50\(\Omega\). Determine the resistance required to convert it into a voltmeter reading up to 1.5V.
Verified Correct Answer: .
Physical Principles & Numerical Solution
Table of values
L(cm)\(^{3}\)
Ls(cm)
Lp(cm)
\(R_1\)
100
5.3
94.7
17.90
90
6.5
93.5
14.40
80
7.0
93.0
13.30
70
8.5
91.5
10.80
60
9.5
90.5
9.50
L(cm)
Is\(^{1}\)
Lg
\(R_2\)
100
21.60
78.4
3.60
90
22.1
77.9
3.50
80
23.4
76.6
3.30
70
24.3
75.7
3.10
60
25.3
74.7
2.95
L(cm)
Ls(cm)
Lp(cm)
\(R_1\)
Ls\(^{1}\)cm\(^{3}\)
Lq
R\(^{2}\)
100
5.3
94.7
17.90
21.6
78.4
3.60
90
6.5
93.5
14.40
22.1
77.9
3.50
80
7.0
93.0
13.30
23.4
76.6
3.30
70
8.5
91.5
10.80
24.3
75.7
3.10
60
9.5
90.5
9.50
25.3
74.7
2.95
Slope (s) = of the graph = \(\frac{DR_{2}}{DR_{1}} = \frac{1.4}{8.0}\) = 0.175
K = \(\sqrt 5\) = 0.17555 = 0.42
Precautions:
Key was opened where readings were not being taken.
Tight connections were ensured
ParallaX errors were avoided in reading the metre rule.
Zero error was noted in reading the metre rule.
The jockey was not rolled on or allowed to scratch the wire.
(b) The resistivity of a material is defined as the resistance of a unit length of the material/wire of a uni cross-sectional area.
OR
P = \(\frac{AR}{L}\) where
A = cross-sectional area
R = eletrical resistance
L = length of the wire
V = Vg + \(V_R\) = IRg + IR = I (Rg + R) \(\therefore\) \(\frac{V}{I}\) = Rg+ R
= \(\frac{1.5}{1.5 \times 10^{-3}}\) = 50 + R = 10\(^{3}\) = 50 + R \(\therefore\) R = 950\(\Omega\)
Source: Myschool Past Questions Archive (Record #60114)