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A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two springs is nearly
- a)0.3 N/mm
- b)15 N/mm
- c)5 N/mm
- d)3.3 N/mm
Correct answer is option 'D'. Can you explain this answer?
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**Explanation:**
To find the equivalent spring stiffness of the two springs in series, we need to consider the combined effect of the two springs.
Let's denote the stiffness of the first spring as k1 = 10 N/mm and the stiffness of the second spring as k2 = 5 N/mm.
**Step 1: Finding the total stiffness of the two springs in series**
When springs are connected in series, the total stiffness is given by the inverse of the sum of the inverses of the individual stiffness values.
1/k_total = 1/k1 + 1/k2
Substituting the given values:
1/k_total = 1/10 + 1/5
1/k_total = (1/10 + 2/10)
1/k_total = 3/10
k_total = 10/3 N/mm
**Step 2: Rounding to the nearest value**
Since the answer choices are given in N/mm, we need to round the calculated value to the nearest N/mm.
The calculated value of k_total is 10/3 N/mm, which is approximately 3.33 N/mm.
Rounding this value to the nearest N/mm, we get the equivalent spring stiffness as 3.3 N/mm.
Therefore, the correct answer is option 'D' (3.3 N/mm).
To find the equivalent spring stiffness of the two springs in series, we need to consider the combined effect of the two springs.
Let's denote the stiffness of the first spring as k1 = 10 N/mm and the stiffness of the second spring as k2 = 5 N/mm.
**Step 1: Finding the total stiffness of the two springs in series**
When springs are connected in series, the total stiffness is given by the inverse of the sum of the inverses of the individual stiffness values.
1/k_total = 1/k1 + 1/k2
Substituting the given values:
1/k_total = 1/10 + 1/5
1/k_total = (1/10 + 2/10)
1/k_total = 3/10
k_total = 10/3 N/mm
**Step 2: Rounding to the nearest value**
Since the answer choices are given in N/mm, we need to round the calculated value to the nearest N/mm.
The calculated value of k_total is 10/3 N/mm, which is approximately 3.33 N/mm.
Rounding this value to the nearest N/mm, we get the equivalent spring stiffness as 3.3 N/mm.
Therefore, the correct answer is option 'D' (3.3 N/mm).
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A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two springs is nearlya)0.3 N/mmb)15 N/mmc)5 N/mmd)3.3 N/mmCorrect answer is option 'D'. Can you explain this answer?
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