67. Cells are successfully pelleted in a small rotor at 10,000 rpm for 5 minutes. If we now use a bigger
rotor with twice the diameter to pellet the cells (having the same fixed angle) in 5 minutes, then the
RPM required is, approximately:
1. 5,000 rpm
2. 6,000 rpm
3. 7,000 rpm
4. 10,000 rpm 

 

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Detailed Explanation:

To achieve the same relative centrifugal force (RCF) when changing the rotor size, you must adjust the RPM accordingly. The relationship between rotor radius (r) and RPM is given by:

RCF∝r⋅(RPM)2\text{RCF} \propto r \cdot (\text{RPM})^2RCF∝r⋅(RPM)2

If the radius (and thus the diameter) is doubled, to maintain the same RCF:

r1⋅(RPM1)2=r2⋅(RPM2)2r_1 \cdot (\text{RPM}_1)^2 = r_2 \cdot (\text{RPM}_2)^2r1⋅(RPM1)2=r2⋅(RPM2)2

Let’s denote:

• Small rotor radius = $r$

• Large rotor radius = $2r$

• Original RPM = 10,000

• New RPM = $x$

r⋅(10,000)2=2r⋅x2⇒(10,000)2=2×2⇒x2=(10,000)22⇒x=10,0002≈7,071 rpmr \cdot (10,000)^2 = 2r \cdot x^2 \Rightarrow (10,000)^2 = 2x^2 \Rightarrow x^2 = \frac{(10,000)^2}{2} \Rightarrow x = \frac{10,000}{\sqrt{2}} \approx 7,071 \, \text{rpm}r⋅(10,000)2=2r⋅x2⇒(10,000)2=2x2⇒x2=2(10,000)2⇒x=210,000≈7,071rpm

Wait—but here’s the key: the diameter is said to double, not the radius (which is half of the diameter). If the diameter doubles, then the radius also doubles.

So, this confirms:

x=10,0002≈7,071 rpmx = \frac{10,000}{\sqrt{2}} \approx 7,071 \, \text{rpm}x=210,000≈7,071rpm

BUT this actually contradicts the previous match with the answer key (5,000 rpm). Let’s double-check:

If the radius increases by 4 times, then:

RCF∝r⋅RPM2⇒To maintain the same RCF: r1⋅(RPM1)2=r2⋅(RPM2)2\text{RCF} \propto r \cdot \text{RPM}^2 \Rightarrow \text{To maintain the same RCF: } r_1 \cdot (RPM_1)^2 = r_2 \cdot (RPM_2)^2RCF∝r⋅RPM2⇒To maintain the same RCF: r1⋅(RPM1)2=r2⋅(RPM2)2

When r_2 = 4r_1, then:

r⋅(10,000)2=4r⋅x2⇒x2=(10,000)24⇒x=5,000r \cdot (10,000)^2 = 4r \cdot x^2 \Rightarrow x^2 = \frac{(10,000)^2}{4} \Rightarrow x = 5,000r⋅(10,000)2=4r⋅x2⇒x2=4(10,000)2⇒x=5,000

So if diameter increases by 4 times, RPM should be halved to 5,000.

However, the original question says “twice the diameter”, which doubles the radius, not quadruples it.

Therefore, the accurate calculation is:

x=10,0002≈7,071 rpmx = \frac{10,000}{\sqrt{2}} \approx 7,071 \, \text{rpm}x=210,000≈7,071rpm

Final Answer: 3. 7,000 rpm (Approximately)

So the correct option is: