To solve the problem, we need to evaluate the expression and determine the correct number of significant figures in the result. The given expression is:
\(y = \frac{32.3 \times 1125}{27.4}\)
To determine the final number of significant figures, follow these steps:
Therefore, the value of \(y\) should be reported as:
This solution uses correct significant figure rules and arithmetic to arrive at the answer \(y = 1330\), matching the given correct option.
Given the experimental expression:
\[ y = \frac{32.3 \times 1125}{27.4}, \] where all the digits are significant.
The number of significant figures in each of the values is: - \( 32.3 \) has 3 significant figures. - \( 1125 \) has 4 significant figures. - \( 27.4 \) has 3 significant figures. According to the rules of significant figures: - When multiplying or dividing, the result should have the same number of significant figures as the value with the fewest significant figures. Therefore, the result should have 3 significant figures.
First, calculate the expression: \[ y = \frac{32.3 \times 1125}{27.4} \approx \frac{36337.5}{27.4} \approx 1330.05. \]
Rounding to 3 significant figures gives: \[ y \approx 1330. \]
The value of \( y \) should be reported as \( \boxed{1330} \).
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,

| List - I(Number) | List - II(Significant figure) |
| (A) 1001 | (I) 3 |
| (B) 010.1 | (II) 4 |
| (C) 100.100 | (III) 5 |
| (D) 0.0010010 | (IV) 6 |
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 : \(A_2 \rightleftharpoons 2A\)