Question

A steel wire of radius 0.4 x l0⁻³ m and length 1 m is tightly clamped between points A and B which are separated by 1 m and in the same horizontal plane. A mass is hung from the middle point of the wire such that the middle point sags by 2 cm from the original position. Compute the mass of body. (Y of steel= 20 x l0^10 N/m^2)

Answer 1

❏ Question:-

@ A steel wire of radius 0.4 x l0⁻³ m and length 1 m is tightly clamped between points A and B which are separated by 1 m and in the same horizontal plane. A mass is hung from the middle point of the wire such that the middle point sags by 2 cm from the original position. Compute the mass of body.

(Y of steel= 20 x l0^10 N/m^2).

❏ Solution:-

✦ Given:-

• radius(r) of steel wire = 0.4 × $10^{-3}$ m=0.0004 m

• length(l) of steel wire = 1 m

• distance(l')tell between point A & B = 1m

• displacement of middle point = 2 cm.

• Young modulus of steel = 20×$10^{10}$ N/m².

✦ To Find :-

• Mass of the body = ?.

✦Explanation :-

Let, AB be the the wire stretched between two points A and B and,

let, the mass of the body is W hung from displace O to C .

To Find the tension T we complete the parallelogram ACBDA. (see tha attachment).

Now,

➝ AB=1 metre=100 cm

➝ AO=OB=$\bf\frac{\cancel{100}}{\cancel2}\:$= 50 cm =0.5 m.

➝ OC=OD=2 cm =0.02 m.

➝ CD=OC+OD=(2+2)cm= 4 cm= 0.04 m.

Now,

➝ Area of cross-section

= π×(0.0004)² m²

= 5.03×$\bf10^{-7}$ m²

From the triangle of forces BCD giving the condition that C will be in equilibrium, we have.

$\sf\longrightarrow \frac{T}{W}=\frac{BC}{CD}$

$\sf\longrightarrow\boxed{ \large{T=W\times\frac{BC}{CD}}}$

• Now, From the Right angled triangle ∆BOC,

➝ BC²= OC²+OB²

➝ BC²= 2²+ 50²

➝ BC²= (4+2500)

➝ BC=$\bf\sqrt{2504}$

➝ BC=$\bf 50.04$ cm.

➝ BC= 0.5004 m

$\bf\therefore T=W\times\frac{0.5004}{0.04}$

$\sf\longrightarrow\boxed{\large{\red{ T=12.51 W \:\: Newtons.}}}$

Hence,

$\bf\therefore stress=\frac{T}{area}$

$\bf\longrightarrow stress=\frac{12.51W}{5.03\times10^{-7}}\:$

$\bf\longrightarrow stress=2.487\times10^{7}W$

&

$\bf\therefore strain=\frac{50.04-50}{50}$

$\bf\longrightarrow strain=0.0008$

Therefore,

$\bf\therefore Y=\frac{stress}{strain}$

$\bf\longrightarrow 20\times10^{10}=\frac{2.487\times10^{7}W}{0.0008}$

$\bf\longrightarrow \frac{20\times10^{10}\times0.0008}{2.487\times10^{7}}=W$

$\bf\longrightarrow mg= \frac{20\times10^{10}\times0.0008}{2.487\times10^{7}}$

$\bf\longrightarrow m= \frac{20\times10^{10}\times0.0008}{2.487\times10^{7}\times9.8}$

$\bf\longrightarrow \boxed{\large{\red{m=0.656\:kg}}}$ (approx).

∴ mass of the body = 0.656 kg= 656 gm.

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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