Question

There is a metal block of dimensions 20 x 10 x 15 cm. What is the ratio of the maximum and minimum resistance of the block?

Answer 1

Given:

There is a metal block of dimensions 20 x 10 x 15 cm.

To find:

Ratio of maximum and minimum resistance of the block.

Calculation:

Maximum resistance be r1;

$\rm{r1 = \rho \times ( \dfrac{l}{area} )}$

$= > \rm{r1 = \rho \times ( \dfrac{20}{10 \times 15} )}$

$= > \rm{r1 = \rho \times ( \dfrac{20}{150} )}$

$= > \rm{r1 = \rho \times ( \dfrac{2}{15} )}$

$= > \rm{r1 = 0.133\rho }$

Minimum resistance be r2;

$\rm{r2= \rho \times ( \dfrac{l}{area} )}$

$= > \rm{r2 = \rho \times ( \dfrac{10}{20 \times 15} )}$

$= > \rm{r2 = \rho \times ( \dfrac{10}{300} )}$

$= > \rm{r2 = \rho \times ( \dfrac{1}{30} )}$

$= > \rm{r2 = 0.03 \rho }$

Required ratio ;

$\sf{ \therefore \: \dfrac{r1}{r2} = \dfrac{0.133}{0.03} }$

$\sf{ = > \: \dfrac{r1}{r2} = 4.44 }$

$\sf{ = > \: r1 : r2 = 4.44 :1 }$

So, final answer:

$\boxed{ \blue{ \sf{ \: r1 : r2 = 4.44 :1 }}}$

Answer 2

Answer:

Hey mate here is ur answer

Explanation:

resistance of a wire is given as R = ρL/A

where ρ is specific resistance it depends on material.

so, resistance of wire is directly proportional to its length and inversely proportional to its cross sectional area.

i.e., R ∝ L/A

for minimum resistance,

taking L = 2cm, A = 10cm × 5cm

so, R(min) = ρ(2cm)/(10cm × 5cm).....(1)

for maximum resistance,

taking L = 10cm, A = 2cm × 5cm

so, R(max) = ρ(10cm)/(2cm × 5cm).....(2)

from equations (1) and (2),

R(max)/R(min) = (2cm)/(10cm) × (2cm × 5cm)/(10cm × 5cm)

= 1/5 × 1/(2 × 1)

= 1/10

hence, ratio of maximum and minimum resistance is 1 : 10