Question

The resistance between the oppośsite faces of 1 meter-cube is found to be 1 2. If its length is increased to 2 metres, its volume remaining unchanged, then resistance between the opposite faces of its length is

Answer 1

Surface area of cube = 6∗(s2)=A6∗(s2)=A.
- Volume of cube = s3=Vs3=V.
- Since A=2VA=2V, so 6s2=2∗(s3)6s2=2∗(s3) -> simplify, we get s=3s=3.
- Volume of cube with s=3s=3 -> s3=27s3=27.

Answer 2

Answer:

The resistance between the opposite faces of its length is 48Ω.

Explanation:

The resistance is formulated as follows:-

$\mathrm{R\propto \frac{l}{A} }$               (1)

Where,

R=resistance of the cube faces

l=length of the faces

A=area of the cube

The volume of the cube is unchanged. So,

$\mathrm{V_1=V_2}$              (2)

V₁=initial volume

V₂=final volume

We can also write equation (2) as,

$\mathrm{A_1l_1=A_2l_2}$                     (as volume is area × length)

$\mathrm{\frac{A_1}{A_2}= \frac{l_2}{l_1}}$               (3)

We can also write equation (1) as,

$\mathrm{\frac{R_1}{R_2} =\frac{l_1A_2}{l_2A_1}}$      (4)

Using (3) in equation (4) we get;

$\mathrm{\frac{R_1}{R_2} =\frac{l_1^2}{l_2^2}}$               (5)

Inserting all the values which are given in question into equation (5) we get;

$\mathrm{\frac{12}{R_2} =\frac{1^2}{2^2}}$

$\mathrm{\frac{12}{R_2} =\frac{1}{4}}$

$\mathrm{R_2=12\times 4}$

$\mathrm{R_2=48\Omega}$

The resistance between the opposite faces of its length is 48Ω.

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