At $ x = \frac{\pi^2}{4} $, $ \frac{d}{dx} \left( \operatorname{Tan}^{-1}(\cos \sqrt{x}) + \operatorname{Sec}^{-1}(e^x) \right) = $At $ x = \frac{\pi^2}{4} $, $ \frac{d}{dx} \left( \operatorname{Tan}^{-1}(\cos \sqrt{x}) + \operatorname{Sec}^{-1}(e^x) \right) = $
Two mercury drops of radii $ r $ and $ 2r $ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is } $S$ and take $9^{2/3} = 4.326$)
An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength of the electron? (Assume $ h = 6.626 \times 10^{-34}$ J·s , $ m_e = 9.1 \times 10^{-31}$ $\text{kg} $, $ e = 1.6 \times 10^{-19}$ $\,$ $\text{C} $.)
The preferred reagent for the following conversion is $CH_3CH_2COOH \rightarrow CH_3CH_2COCl$
The IUPAC name of the following molecule is
The isotherms of an ideal gas at $ T_1, T_2, T_3 $ along with their slopes (m) (in the brackets) are shown here. If $ T_1 > T_2 > T_3 $, then the correct order of slopes of these isotherms is:
