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The history of print media in the subcontinent begins with colonial journalism. The textbook states that the Hickey's Bengal Gazette, also known as the Calcutta General Advertiser, started by James Augustus Hickey in 1780, “is regarded as the first regular publication from the Indian soil.” It was a two-sheet newspaper notorious for writing about the private lives of Company officials and for mounting bold attacks on Governor-General Warren Hastings and the Chief Justice โ attacks that landed Hickey in prison twice (first a 4-month term with a Rs.500 fine, then a one-year term with a Rs.5,000 fine). Hickey's willingness to challenge colonial authority despite persecution makes him a pioneer of subcontinental journalism. Note: William Bolts had attempted to start a newspaper in 1776 but was stopped by the East India Company's Court of Directors before he could publish.
For $x\in\left(0,\dfrac{1}{4}\right)$, the derivative of $\tan^{-1}\!\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)$ equals $\sqrt{x}\cdot g(x)$. Find $g(x)$.
Write $\frac{6x\sqrt{x}}{1-9x^3} = \frac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}=\frac{2t}{1-t^2}$ where $t=3x^{3/2}$.
So $\tan^{-1}\!\left(\frac{2t}{1-t^2}\right)=2\tan^{-1}(t)=2\tan^{-1}(3x^{3/2})$
Differentiate: $\frac{d}{dx}[2\tan^{-1}(3x^{3/2})] = \frac{2}{1+9x^3}\cdot 3\cdot\frac{3}{2}x^{1/2} = \frac{9\sqrt{x}}{1+9x^3}$
Since this equals $\sqrt{x}\cdot g(x)$: $g(x)=\dfrac{9}{1+9x^3}$
The plasma membrane is selectively permeable mainly due to the presence of:
Integral membrane proteins (e.g., channels, carriers, pumps) control selective transport. Lipids form the barrier but without proteins, permeability is limited to small nonpolar molecules.
Choose the correct sentence:
With the auxiliary verb 'did' (past tense of 'do'), the main verb must be in its base form: 'know', not 'knew'. 'She didn't know' is correct.
First, calculate the volume of nitrogen at STP ($P_1V_1/T_1 = P_2V_2/T_2$):
- $P_1 = 715 - 15 = 700 \text{ mm Hg}$ [cite: 3, 53]
- $V_1 = 50 \text{ mL}$, $T_1 = 300 \text{ K}$ [cite: 18]
- $V_{STP} = \frac{700 \times 50 \times 273}{300 \times 760} \approx 41.9 \text{ mL}$ [cite: 19]
- Mass of $N_2 = \frac{28 \times 41.9}{22400} \approx 0.0524 \text{ g}$ [cite: 54]
- $\% \text{ Nitrogen} = \frac{0.0524}{0.3} \times 100 \approx 17.46\%$ [cite: 56]
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