Question

When radiation falls on the cathode plate, a current of $2\mu A$ is recorded in the ammeter. Assuming that the vacuum tube setup follows Ohm/s law, the equivalent resistance of vacuum tube operating in this case when jockey is at end P is

• A. $8\times {10}^8\Omega$
• B. $16\times {10}^6\Omega$
• C. $8\times {10}^6\Omega$
• D. $10\times {10}^6\Omega$

Answer 1

The correct option is C $8\times {10}^6\Omega$

Current through wire is $I_1$

$I_1\approx \frac{20}{8+2}=2A$

In $200I$

$IR+(I+I_1)2=20(R+2)I+4=20(R+2)I=16(R+2)I=16(R+2)=\frac{16}{I}=\frac{16}{2\mu A}=8\times {10}^6\Omega R\approx 8\times {10}^6\Omega$