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Question: A block has dimensions and. Ratio of maximum resistance to minimum resistance between any pair of op...

A block has dimensions and. Ratio of maximum resistance to minimum resistance between any pair of opposite faces of the block is:
(a) 9 : 1
(b) 1 : 9
(c) 18 : 1
(d) 1 : 6

Explanation

Solution

In this solution, we are going to use the formula of resistance that shows the relationship of Resistance with length and area. We must know that, Resistance of any conductor is directly proportional to its length and inversely proportional to its area.

Complete Step by Step Answer: Given:
Dimensions of block are:

a=1a = 1cm,
b=2b = 2cm, and
c=3c = 3cm
We get maximum resistance Rmax{R_{\max }} when the length of conductor(l)\left( l \right)is maximum and Area of cross-section(A)\left( A \right)is minimum.
Similarly, for minimum resistanceRmin{R_{\min }}the length of conductor(l)(l)should be minimum and area of cross-section(A)\left( A \right)is maximum.
From the given dimensions of the block:
Maximum length:
lmax=c =3  cm  {l_{\max }} = c \\\ = 3\;cm \\\
Minimum length:
lmin=a =1  cm  {l_{\min }} = a \\\ = 1\;cm \\\
Maximum area of cross-section:
Amax=b×c Amax=2×3 Amax=6  cm2  {A_{\max }} = b \times c \\\ {A_{\max }} = 2 \times 3 \\\ {A_{\max }} = 6\;c{m^2} \\\
Minimum area of cross-section:
Amin=a×b Amin=1×2 Amin=2  cm2  {A_{\min }} = a \times b \\\ {A_{\min }} = 1 \times 2 \\\ {A_{\min }} = 2\;c{m^2} \\\
The formula to calculate resistance RRis given by:
R=ρlA      ..........(1)R = \rho \dfrac{l}{A}\;\;\;..........(1)
Where ρ\rho is the resistivity of the block. The value of resistivity is constant for a material.
ll is length of conductor, andAAis area of cross-section.
Now, substituting values of lmin{l_{\min }}and Amax{A_{\max }} in Equation (1), to getRmin{R_{\min }}:
Rmin=ρlminAmax       Rmin=ρ16        .......(2)  {R_{\min }} = \rho \dfrac{{{l_{\min }}}}{{{A_{\max }}}}\;\;\; \\\ {R_{\min }} = \rho \dfrac{1}{6}\;\;\;\;.......(2) \\\
And, substituting the value of lmax{l_{\max }} and Amin{A_{\min }} in Equation (1), to getRmax{R_{\max }}:
Rmax=ρlmaxAmin       Rmax=ρ32        .......(3)  {R_{\max }} = \rho \dfrac{{{l_{\max }}}}{{{A_{\min }}}}\;\;\; \\\ {R_{\max }} = \rho \dfrac{3}{2}\;\;\;\;.......(3) \\\
Dividing equation (3) with equation (2), we get,

RmaxRmin=ρ32  ρ16   RmaxRmin=91  \dfrac{{{R_{\max }}}}{{{R_{\min }}}} = \dfrac{{\rho \dfrac{3}{2}\;}}{{\rho \dfrac{1}{6}\;}} \\\ \dfrac{{{R_{\max }}}}{{{R_{\min }}}} = \dfrac{9}{1} \\\

Therefore, the ratio of maximum resistance to minimum resistance between any pair of opposite faces of the block is 9 : 1.
So option (a) is the correct answer.

Note: While substituting the values of ll (length of conductor) and AA(area of cross-section) you don’t need to convert the values into its S.I units because as we are calculating the ratio, the corresponding units get cancelled out. Also, make sure that you calculate the ratio of maximum resistance to minimum resistance and not the minimum resistance to maximum resistance.