A physical quantity C is related to four other quantities p, q, r and s as follows $ C = \frac{pq^2}{r^3 \sqrt{s}} $ The percentage errors in the measurement of p, q, r and s are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of C will be _______ %.
To find the percentage error in the measurement of C, we first need to understand how errors propagate through the given formula. The expression for C is:
C = \(\frac{pq^2}{r^3 \sqrt{s}}\)
The percentage error formula for a product or quotient involving powers, given a function like \(Z = \frac{A^m B^n}{C^p D^q}\), is:
\(\frac{\Delta Z}{Z} \times 100 \approx m \frac{\Delta A}{A} \times 100 + n \frac{\Delta B}{B} \times 100 + p \frac{\Delta C}{C} \times 100 + q \frac{\Delta D}{D} \times 100\)
Applying this to our expression:
The percentage error in C is:
\(\frac{\Delta C}{C} \times 100 \approx 1 \cdot 1\% + 2 \cdot 2\% + 3 \cdot 3\% + 0.5 \cdot 2\%\)
Calculating this gives:
Add these errors together:
1% + 4% + 9% + 1% = 15%
Thus, the percentage error in the measurement of C is 15%, which is within the expected range of 15,15.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
