Question

13. A weight of 450 g produces an extension of 3.5 cm in a elastic string, find
(a) how much weight is required to produce an extension 1.75 cm.
5. Two numbers are in the ratio of 2:3. If 2 is added to both, the ratio becomes 3/4. Find the number.​

Answer 1

Question (1)

Given us the stress and strain as 450 g and 3.5 cm respectively.

So, Let's find the modulus of elasticity first,

⇒ Stress = k × Strain

⇒ k = 450 / 3.5

⇒ k = 128.57 g/cm

Now, In the second case, we have

• Strain = 1.75 cm
• k = 128.57 g/cm

We have to find, stress, Similarly

⇒ Stress = k × Strain

⇒ Stress = 128.57 × 1.75

⇒ Stress = 224.99

⇒ Stress ≈ 225 g

So, The required weight is 225 g

Question (2)

Given, Two numbers are in the ratio 2 : 3 but when 2 is added to both the numbers, the ratio becomes 3 : 4

Let us assume the numbers to be x and y respectively.

So, According to the question,

Case 1

• Ratio is 2 : 3

⇒ x / y = 2 / 3

⇒ 3x - 2y = 0 ...(1)

Case 2

• Ratio becomes 3 : 4 when 2 is added to both the numbers

⇒ (x + 2) / (y + 2) = 3 / 4

⇒ 4x + 8 = 3y + 6

⇒ 4x - 3y = -8 ...(2)

Multiply (1) by 3 and (2) by 2 and subtract (1) from (2),

⇒ 8x - 6y - (9x - 6y) = -8 - 0

⇒ -x = -8

⇒ x = 8

Substituting value of x in (1), we get

⇒ 3×8 - 2y = 0

⇒ 2y = 24

⇒ y = 12

Hence, The numbers are 8 and 12.

Answer 2

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Given us the stress and strain as 450 g and 3.5 cm respectively.

So, Let's find the modulus of elasticity first,

⇒ Stress = k × Strain

⇒ k = 450 / 3.5

⇒ k = 128.57 g/cm

Now, In the second case, we have

Strain = 1.75 cm

k = 128.57 g/cm

We have to find, stress, Similarly

⇒ Stress = k × Strain

⇒ Stress = 128.57 × 1.75

⇒ Stress = 224.99

⇒ Stress ≈ 225 g

So, The required weight is 225 g