Question

A force 1000 N stretches the length of a hanging wire by 1 mm. What force is required to stretch a wire of the same material and length but having four times the diameter of 1 mm?

Answer 1

Explanation:

so pressure = force upon area

so 1000n = 1

1

so answer is 1000 n

Answer 2

The required force to stretch a wire of the same material and length is $16\times10^3N$.

Explanation:

The Young’s modulus  is given as

$Y=\frac{Fl}{A\Delta l}$

Here, A is the area of the wire, l is the initial length, F is the applied force and $\Delta l$ is the change in the length.

As per question after stretches the wire, the length will remain same so,

$\frac{F_1}{F_2} =\frac{A_1}{A_2}$                          ( As wires of the same material)

$\frac{F_1}{F_2} =\frac{r_1^2}{r^2_2}$

Substitute the given values, we get

$\frac{1000N}{F_2} =(\frac{r}{4r})^2$

$F_2=16\times10^3N$

Thus,the required force to stretch a wire of the same material and length is $16\times10^3N$.

#Learn More: young modulus.

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