Math, asked by Anonymous, 3 months ago

 A mild steel wire of cross-sectional area 0.60 x 10^-2 cm2 and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.



last question ✌️​

Answers

Answered by Disha094
5

From the above figure,

Let x be the depression at the mid point i.e. CD=x.

In fig.,

AC=CB=l=0.5m ;

m=100g=0.100Kg

AD=BD=(l2+x2)21 

Increase in length, △l=AD+DB−AB=2AD−AB

=2(l2+x2)21−2l=2l(1+l2x2)−2l=2l[1+2l2x2]−2l=lx2

∴ strain =2l△l=2l2x2

If T is the tension in the wire, then

2Tcosθ=mg

or, T=2cosθmg

Here, cosθ=(l2+x2)21x=l(1+l2x2)21x=(1+21l2x2)x 

As, x<<l, so 1>>2l21x2 and 1+21l2x2≈1

Hence, T=2(1x)mg

As, x<<l, so 1>>2l21x2 and 1+21l2x2≈1

Hence, T=2(1x)mg=2xmgl

stress =AT=2Axmgl

Y=strainstress=2Axmgl×x22l2=Ax3mgl3

x=l[YAmg]31=0.5[20×10

Answered by REP0RTER
3

itna hard Question (༎ຶ ෴ ༎ຶ)

What is PHOTOSYNTHESIS puch le bhai

From the above figure,

Let x be the depression at the mid point i.e. CD=x.

In fig.,

AC=CB=l=0.5m ;

m=100g=0.100Kg

AD=BD=(l

2

+x

2

)

2

1

Increase in length, △l=AD+DB−AB=2AD−AB

=2(l

2

+x

2

)

2

1

−2l=2l(1+

l

2

x

2

)−2l=2l[1+

2l

2

x

2

]−2l=

l

x

2

∴ strain =

2l

△l

=

2l

2

x

2

If T is the tension in the wire, then

2Tcosθ=mg

or, T=

2cosθ

mg

Here, cosθ=

(l

2

+x

2

)

2

1

x

=

l(1+

l

2

x

2

)

2

1

x

=

(1+

2

1

l

2

x

2

)

x

As, x<<l, so 1>>

2l

2

1x

2

and 1+

2

1

l

2

x

2

≈1

Hence, T=

2(

1

x

)

mg

=

2x

mgl

stress =

A

T

=

2Ax

mgl

Y=

strain

stress

=

2Ax

mgl

×

x

2

2l

2

=

Ax

3

mgl

3

x=l[

YA

mg

]

3

1

=0.5[

20×10

11

×0.5×10

−6

0.1×10

]

3

1

=1.074×10

−2

m

=1.074 cm

Similar questions