
To determine the minimum length PQ required to just glow the LED, we must ensure the voltage across the LED is 1.8 V. The circuit involves a potential divider comprised of a resistive wire PR, with a total voltage of 20 V from the DC source. The zener diode, with a breakdown voltage of 3.2 V, ensures a stable voltage across PR. The LED needs 1.8 V to glow.
First, calculate the voltage drop across the zener and LED once the LED starts glowing:
Total Voltage across PQ + QR = 20 V
1.8 V is across LED and the zener does not conduct below 3.2 V.
To make LED glow, 1.8 V (LED) + 3.2 V (Zener) = 5 V across PQ is needed.
Now that we have determined a 5 V drop is needed to just start the LED glowing, PQ must drop 5 V of the total 20 V.
Given, PR = 20 cm, we determine the position of Q such that voltage drop across PQ is 5 V.
Voltage across PQ/Total Voltage = Length of PQ/Total Length of PR
\( \frac{5 \text{ V}}{20 \text{ V}} = \frac{\text{Length of PQ}}{20 \text{ cm}} \)
Cross-multiplying gives:
\(\text{Length of PQ} = \frac{5}{20} \times 20 \text{ cm} = 5 \text{ cm}\)
The minimum length PQ is 5 cm. This solution falls within the specified range. Therefore, the correct answer is 5 cm.
The total voltage across $PR$ is:
\[V_{PR} = 20 \, \text{V}.\]
The resistive wire $PR$ has a total length of:
\[\ell_{PR} = 20 \, \text{cm}.\]
The voltage across $QR$ is determined by the zener diode:
\[V_{QR} = 3.2 \, \text{V}.\]
The voltage across $PQ$ is:
\[V_{PQ} = V_{PR} - V_{QR} = 20 - 3.2 = 16.8 \, \text{V}.\]
The fraction of voltage across $PQ$ relative to $PR$ is:
\[\frac{V_{PQ}}{V_{PR}} = \frac{16.8}{20} = \frac{1}{4}.\]
Using the proportionality of voltage and length:
\[\ell_{PQ} = \frac{1}{4} \times \ell_{PR}.\]
Substitute $\ell_{PR} = 20 \, \text{cm}$:
\[\ell_{PQ} = \frac{1}{4} \times 20 = 5 \, \text{cm}.\]
Thus, the minimum length of $PQ$ to just glow the LED is:
\[\ell_{PQ} = 5 \, \text{cm}.\]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,




What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)