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The standard relation between these moduli is $\frac{9}{Y} = \frac{3}{\eta} + \frac{1}{K}$.
- Rearranging for $K$: $\frac{1}{K} = \frac{9}{Y} - \frac{3}{\eta} = \frac{9\eta - 3Y}{Y\eta}$
- $K = \frac{Y\eta}{9\eta - 3Y}$.
$LC = \frac{0.5}{100} = 0.005 \text{ mm}$.
Zero Error: Since the zero is 3 divisions below the mean line, it is a positive zero error. Zero Error $= +3 \times 0.005 = +0.015 \text{ mm}$.
Observed Reading: $5.5 + (48 \times 0.005) = 5.5 + 0.240 = 5.740 \text{ mm}$.
Actual Thickness: Observed Reading $-$ Zero Error $= 5.740 - 0.015 = 5.725 \text{ mm}$.
If $y = \sec(\tan^{-1}x)$, find $\dfrac{dy}{dx}$ at $x=1$.
Let $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $y = \sec\theta$.
$\frac{dy}{dx} = \sec\theta\tan\theta\cdot\frac{d\theta}{dx} = \sec\theta\tan\theta\cdot\frac{1}{1+x^2}$
Since $\tan\theta=x$ and $\sec\theta=\sqrt{1+x^2}$:
$\frac{dy}{dx} = \sqrt{1+x^2}\cdot x\cdot\frac{1}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$
At $x=1$: $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$
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