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Sagot :
Solution :
We know, resistance is given by :
[tex]R = \dfrac{\rho l}{A}[/tex]
[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]
Now, we know mass of wire is given by :
[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]
Hence, this is the required solution.
The mass of the wire will be "15.00 g".
Given:
- Length of wire, l = 3.5 m
- Resistance, R = 0.125 Ω
The resistance will be:
→ [tex]R = \frac{\rho l}{A}[/tex]
or,
→ [tex]A = \frac{\rho l}{R}[/tex]
By substituting the values, we get
[tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]
[tex]= 4.816\times 10^{-7} \ m^2[/tex]
hence,
The mass will be:
→ [tex]Mass = Density\times Volume[/tex]
[tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]
[tex]= 0.01500 \ kg[/tex]
[tex]= 15.00 \ g[/tex]
Thus the above answer is right.
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