2006-09-09 16:41:45 · 12 answers · asked by Arun 1 in Science & Mathematics ➔ Physics
If by cross sectional area, you mean the cross section of the metallic wire (here I am assuming that the wire has a cylindrical shape with a circular cross section and length L), then the answer is NO, because all you've done is increase the length of the wire from L to 2L. The cross section remains the same.
PS - since you didn't mention anything about keeping the volume of the wire the same, so I assumed that it's okay if it changes.
2006-09-09 17:12:28 · answer #1 · answered by PhysicsDude 7 · 2⤊ 0⤋
Yes it will affecvt the crossectional area.
Since wire is same only redrawn therefore
volume in both cases will be same.
Let original area = A
New Area = A'
original length= L
New Length = 2L
Original volume = New Volume
(original area)*(Original length)=(new area)*(new length)
A*L = A' * (2L)
A=2A'
A' = A/2
Thur area is reduced to half of original area
2006-09-10 00:13:43 · answer #2 · answered by sunny 2 · 0⤊ 1⤋
ofcourse,the cross sectional area will be reduced .on account of drawing.
the deforming or drawing will keep the volume constant.
initial volume is 3.14/4*d*d*L
volume after drawing is 3.14/4*d1*d1*2L,
both the volumes will be equal.
d*d*L =d1*d1*2L
d*d=d1*d1*2
0.7072*d =d1, where d is dia of wire when it had a length of L,
and d1 is dia of wire after drawing it to length 2L.
we observe that diameter of wire rod has reduced and we know that the cross sectional area of length 2L will be
3.14/4*(0.7072*0.7072)d, it will be half of area when the length was 2L.
2006-09-13 04:04:18 · answer #3 · answered by kailash s 2 · 0⤊ 0⤋
Absolutely! The volume would have to stay the same, so if the length is doubled, the cross sectional area must be halved.
2006-09-09 23:45:21 · answer #4 · answered by kris 6 · 1⤊ 1⤋
mass of the wire remains constant when we stretch it. density depends only on temperature. mass=density*volume. that means volume is also constant. so area*lenght is constant. let A be area then
A*L=A(new)*2L
=> A(new)=A/2.
2006-09-10 04:17:04 · answer #5 · answered by imfamouspersonality 1 · 0⤊ 0⤋
it wont,ifu dont consider the volume.now, if u walt to keep the volume constant,the area of cross section will decrease accordingly
2006-09-10 02:14:10 · answer #6 · answered by nithesh s 1 · 1⤊ 0⤋
Yes it will. The cross-sectional area will be half.
2006-09-09 23:51:13 · answer #7 · answered by j_son_06 5 · 1⤊ 1⤋
yes of course,the same wire is lenthened.so its cross section area will definitely change
2006-09-10 01:40:09 · answer #8 · answered by Aravinth t 1 · 0⤊ 1⤋
of course! the volume remains the same..
initial volume=final volume
L*A =2L* A'
therefore new area(A')=A/2
2006-09-10 00:24:20 · answer #9 · answered by hp-here4u 2 · 0⤊ 1⤋
If A wire of length L is stretched to length 2L, the the volume remains constant.
V(initial)=V(final),
pi*r (initial)*L=pi*r (final)*2L
i.e. r(final)= r(initial)/sqrt of 2
Now crossectional area is given by,
pi*R squred(initial)=pi*R squared(final)
Therefore , AREA(FINAL)=pi*R squared(final)
=pi*R squared(initial)/2
=AREA(INITIAL)/2
2006-09-10 01:19:22 · answer #10 · answered by Alpha 1 · 0⤊ 0⤋