Question

A student has a resistance wire of 1 ohm. If the length of this wire is 50 cm, to what length

he should stretch it uniformly so as to obtain a write off 4 Ohm's resistance? Justify your ans​

Answer 1

Given that, a student has a resistance wire of 1 ohm and the length of this wire is 50 cm.

We have to find that how much he should stretch it uniformly so as to obtain a wire of 4 Ohm's resistance.

We know that,

Resistance of a wire is directly proportional to it's length and inversely proportional to the area of cross-section. Where rho (p = resistivity) is a contact.

R = p l/A

Multiply and divide by length (l),

R = p l/A × l/l

R = p l²/Al [ Area × Length = Volume ]

R = p l²/V

As per given condition we are talking about two wires. So,

R1/R2 = (l1)²/(l2)²

Substitute the known values in the above formula,

1/4 = (50)²/(l2)²

(1)²/(2)² = (50)²/(l2)²

1/2 = 50/l2

l2 = 100 cm

Answer 2

$\bigstar$ $\sf{\purple{\underline{\underline{\blue{To\:Find:-}}}}}$

• To what length student should stretch the wire uniformly so as to obtained a “4 Ohm” resistance .

$\bigstar$ $\sf{\purple{\underline{\underline{\blue{SOLUTION:-}}}}}$

☞ we have know that,

$\tt{\red{\boxed{R\:=\:{\dfrac{{\rho}l}{A}}\:}}}$

$\tt{\implies\:(R)\:{\alpha}\:({\dfrac{l}{A}})\:}$

☞ If a wire is stretched or drawn out or folded, Area varies but volume remains constant .

$\tt{\implies\:(R)\:{\alpha}\:(l^2)\:}$

☞ GIVEN:-

• Resistance (R1) = 1 Ohm
• Length (l1) = 50 cm
• Resistance (R2) = 4 Ohm
• Length (l2) = ?

☞ Now we have know that

$\tt{\red{\boxed{{\:\dfrac{R_1}{R_2}}\:=\:{\dfrac{{l_1}^2}{{l_2}^2}}\:}}}$

$\tt{\implies\:{\dfrac{1}{4}}\:=\:{\dfrac{({50})^2}{{l_2}^2}}\:}$

$\tt{\implies\:l_2\:=\:{\sqrt{10000}}\:}$

$\tt{\implies\:l_2\:=\:100\:c.m\:}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:100\:c.m\:}}}}$