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When an elastic material with Young’s modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is- a)YS/2
- b)S2Y/2 [1989]
- c)S2/2Y
- d)S/2Y
Correct answer is option 'C'. Can you explain this answer?
When an elastic material with Young’s modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is
a)
YS/2
b)
S2Y/2 [1989]
c)
S2/2Y
d)
S/2Y
|
Aniket Chawla answered |
Energy stored per unit volume
stress (stress / Young' s modulus) (stress)2 /( Young ' s modulus)
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]- a)Length = 100 cm, diameter = 1 mm
- b)Length = 200 cm, diameter = 2 mm
- c)Length = 300 cm, diameter = 3 mm
- d)Length = 50 cm, diameter = 0.5 mm
Correct answer is option 'D'. Can you explain this answer?
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]
a)
Length = 100 cm, diameter = 1 mm
b)
Length = 200 cm, diameter = 2 mm
c)
Length = 300 cm, diameter = 3 mm
d)
Length = 50 cm, diameter = 0.5 mm
|
Akshat Chavan answered |
So, extension, 
[∵ F and Y are constant]
The ratio of 
is maximum for case (d).
Hence, option (d) is correct.
is maximum for case (d).Hence, option (d) is correct.
Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If they are pulled by the same force, then increase in length will be in the ratio- a)2 : 1
- b)1 : 4 [1988]
- c)1 : 8
- d)8 : 1
Correct answer is option 'C'. Can you explain this answer?
Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If they are pulled by the same force, then increase in length will be in the ratio
a)
2 : 1
b)
1 : 4 [1988]
c)
1 : 8
d)
8 : 1
|
Nilotpal Gupta answered |
We know that Young's modulus
Since Y, F are same for both the wires, we have,
So, ℓ1 : ℓ2 = 1 : 8
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Dl on applying a force F, how much force is needed to stretch the second wire by the same amount?- a)9F
- b)6F
- c)4F
- d)F
Correct answer is option 'A'. Can you explain this answer?
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Dl on applying a force F, how much force is needed to stretch the second wire by the same amount?
a)
9F
b)
6F
c)
4F
d)
F
|
Ambition Institute answered |
Young’s modulus, 
Since initial volume of wires are same and their areas of cross sections are A and 3A so lengths are 3l and l respectively. For wire 1,
Since initial volume of wires are same and their areas of cross sections are A and 3A so lengths are 3l and l respectively. For wire 1,
For wire 1,
For wire 2, let F ′ force is applied
From eqns (i) and (ii),
⇒ F' = 9F
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