Answer:
[tex]l_{2} = 4106 \times 10^{-3} = 4.1 cm[/tex]
Explanation:
GIVEN
potential difference = 0.3 V
and given length = 62 -51 = 11 cm
question
how far apart on the wire are two points that differ in potential by 112 m V?
potential difference = 112 m V
by using formula [tex]V \alpha l[/tex]
[tex]\frac{V_{1}}{V_{2}} = \frac{l_{1}}{l_{2}}[/tex]
[tex]l_{2} =?[/tex]
[tex]l_{2} = \frac{V_2}{V_{1}} l_{1}[/tex]
put the values we get
[tex]l_{2} = 4106 \times 10^{-3} = 4.1 cm[/tex]
The wire are two points that differ in potential will be "4.1 cm" far apart.
According to the question,
Potential difference, V = 0.3 V
Length, l = 62 - 51
= 11 cm
We know the relation,
→ Potential difference (V) ∝ Length (l)
then,
→ [tex]\frac{V_1}{V_2} =\frac{l_1}{l_2}[/tex]
or,
l₂ = [tex]\frac{V_2}{V_1}[/tex] l₁
By substituting the above values,
= 4106 × 10⁻³
= 4.1 cm
Thus the above approach is appropriate.
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