To find the percentage error in the physical quantity \( Q \) given by the relation \(Q = \frac{a^4 b^3}{c^2}\), we use the formula for percentage error in a derived quantity:
Let's analyze each of the terms individually:
Therefore, the total percentage error in \( Q \) is:
| Percentage error in \( Q \) | = \( 12\% + 12\% + 10\% \) |
| = \( 12\% + 12\% + 10\% = 34\% \) |
Hence, the percentage error in \( Q \) is \(34\%.\)
Therefore, the correct answer is 34%.
Given: \(Q = \frac{a^4 b^3}{c^2}\)
The percentage error in \( Q \) can be calculated using the rule for propagation of errors in multiplication and division.
Step 1. Calculate the fractional error in \( Q \):
\(\frac{\Delta Q}{Q} = 4 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + 2 \frac{\Delta c}{c}\)
Step 2. Substitute the given percentage errors:
\(\frac{\Delta Q}{Q} \times 100 = 4 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + 2 \left( \frac{\Delta c}{c} \times 100 \right)\)
\(= 4 \times 3\% + 3 \times 4\% + 2 \times 5\%\)
\(= 12\% + 12\% + 10\%\)
Step 3. Calculate the total percentage error in \( Q \):
\(\text{Percentage error in } Q = 12\% + 12\% + 10\% = 34\%\)
Thus, the percentage error in \( Q \) is 34%.
The Correct Answer is: 34%
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 : \(A_2 \rightleftharpoons 2A\)