Question

A tensile stress of 50 N/m produces a change of 0.1 m in a wire of length 1 m. The young
modulus of elasticity is?

A) 50
B) 500
C)o
D) 5000


​

Answer 1

$\large\underline{\red{\sf \pink{\bigstar} Given:-}}$

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• Tensile stress = 50 N/m

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• Change in confrigation =∆l = 0.1 m

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• Orginal confrigation = ?

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\purple{\sf \orange{\bigstar} To Find:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

• Young modulus of elasticity

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\green{\sf \orange{\bigstar} Theory:-}}$

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Young modulus of elasticity (Y):

⠀⠀⠀⠀⠀⠀⠀⠀

• It is defined as the ratio of normal stress to the longitudinal strain within the elastic limit.

⠀⠀⠀⠀⠀⠀⠀⠀

• Y = stress/strain

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• Stress = F/A

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• Strain = change in configuration/ original configuration= ∆l/l

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$\large\underline{\blue{\sf \red{\bigstar} Solution:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

• We have to find the young modulus

⠀⠀⠀⠀⠀⠀⠀⠀

• We know  that,

⠀⠀⠀⠀⠀⠀⠀⠀

• Young modulus = stress/strain

⠀⠀⠀⠀⠀⠀⠀⠀

• Young modulus = (F/A)/(∆l/l)

⠀⠀⠀⠀⠀⠀⠀⠀

• F/A = 50 and l = 1 and ∆l = 1

⠀⠀⠀⠀⠀⠀⠀⠀

• Then put  the given values

⠀⠀⠀⠀⠀⠀⠀⠀

• Young modulus = [ 50/(0.1) ]/1

⠀⠀⠀⠀⠀⠀⠀⠀

• = 50 × 10

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• $\large\underline{\orange{\sf \pink{\bigstar} 500}}$

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$\large\underline{\purple{\sf \orange{\bigstar} Hence :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

The correct option is B)500

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\orange{\sf \pink{\bigstar}More \:\:about\:\: the \:\:topic :-}}$

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⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\pink{\sf \orange{\bigstar} Tensile \:\:stress :-}}$

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⠀⠀⠀⠀⠀⠀⠀⠀

• the length of the rod is increased on application deforming force over and then stress produced in the rod is called tensile stress.

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\green{\sf \orange{\bigstar} Strain :-}}$

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• the ratio of change in configuration to the original configuration is called this strain

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$\large\underline{\blue{\sf \orange{\bigstar} Note :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\large\underline{\red{\sf \orange{\bigstar} kindly \:\: go \:\: through \:\: the \:\:extra \:\: information .}}$

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Answer 2

Answer:

$\large\underline{\red{\sf \pink{\bigstar} Given:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

Tensile stress = 50 N/m

⠀⠀⠀⠀⠀⠀⠀⠀

Change in confrigation =∆l = 0.1 m

⠀⠀⠀⠀⠀⠀⠀⠀

Orginal confrigation = ?

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\purple{\sf \orange{\bigstar} To Find:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

Young modulus of elasticity

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\green{\sf \orange{\bigstar} Theory:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

Young modulus of elasticity (Y):

⠀⠀⠀⠀⠀⠀⠀⠀

It is defined as the ratio of normal stress to the longitudinal strain within the elastic limit.

⠀⠀⠀⠀⠀⠀⠀⠀

Y = stress/strain

⠀⠀⠀⠀⠀⠀⠀⠀

Stress = F/A

⠀⠀⠀⠀⠀⠀⠀⠀

Strain = change in configuration/ original configuration= ∆l/l

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\blue{\sf \red{\bigstar} Solution:-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

We have to find the young modulus

⠀⠀⠀⠀⠀⠀⠀⠀

We know  that,

⠀⠀⠀⠀⠀⠀⠀⠀

Young modulus = stress/strain

⠀⠀⠀⠀⠀⠀⠀⠀

Young modulus = (F/A)/(∆l/l)

⠀⠀⠀⠀⠀⠀⠀⠀

F/A = 50 and l = 1 and ∆l = 1

⠀⠀⠀⠀⠀⠀⠀⠀

Then put  the given values

⠀⠀⠀⠀⠀⠀⠀⠀

Young modulus = [ 50/(0.1) ]/1

⠀⠀⠀⠀⠀⠀⠀⠀

= 50 × 10

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\orange{\sf \pink{\bigstar} 500}}$

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\purple{\sf \orange{\bigstar} Hence :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

The correct option is B)500

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\orange{\sf \pink{\bigstar}More \:\:about\:\: the \:\:topic :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\pink{\sf \orange{\bigstar} Tensile \:\:stress :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

the length of the rod is increased on application deforming force over and then stress produced in the rod is called tensile stress.

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\green{\sf \orange{\bigstar} Strain :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀

the ratio of change in configuration to the original configuration is called this strain

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\blue{\sf \orange{\bigstar} Note :-}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\red{\sf \orange{\bigstar} kindly \:\: go \:\: through \:\: the \:\:extra \:\: information .}}$

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