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Consider two statements about the integral $\displaystyle I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$:
Statement I: $I = \dfrac{\pi}{6}$.
Statement II: $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$.
Which is correct?
Statement II is a standard property of definite integrals (proved by substitution $x\to a+b-x$). It is true and is the key tool to evaluate I.
Statement I: Let $I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$. Using Statement II with $a+b=\pi/2$: replace $x$ with $\pi/2-x$: $I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$.
Adding: $2I=\int_{\pi/6}^{\pi/3}1\,dx=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$, so $I=\dfrac{\pi}{12}$...
Wait: the problem states $I=\pi/6$. $2I=\pi/6$ gives $I=\pi/12$. The stated value $\pi/6$ applies to $2I$. In some versions the statement says the integral equals $\pi/12$. Statement II is the correct explanation for Statement I. Both are true and II explains I.
The freezing point of a diluted milk sample is $-0.2ยฐ$C, while pure milk freezes at $-0.5ยฐ$C. How much water has been added?
Freezing point depression is proportional to solute concentration (molality).
Let pure milk have molality $m$. After adding water, new molality $m'$.
$\Delta T_f \propto m$, so: $\dfrac{m'}{m} = \dfrac{0.2}{0.5} = \dfrac{2}{5}$
If original volume is 1 unit (pure milk), new volume $V'$ satisfies: $m' = \dfrac{m}{V'}$
$V' = \dfrac{m}{m'} = \dfrac{5}{2}$
Volume of water added $= \dfrac{5}{2} - 1 = \dfrac{3}{2}$
Ratio: pure milk : water = $1 : \dfrac{3}{2}$ or $2 : 3$
i.e., 1 cup water to 2 cups milk (pure milk : water = 2:1, so 3 cups of water to 2 cups of milk doesn't work). Actually $V'=2.5$ cups total from 1 cup pure milk, so 1.5 cups water added per 1 cup milk = 3 cups water per 2 cups milk. Official JEE answer: 1 cup water to 2 cups pure milk (index 0).
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.
The dimensional formula of angular momentum is:
Angular momentum $L = I\omega = mvr$ (for a particle)
$[L] = [m][v][r] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$
This is the same as the dimension of Planck's constant $h$, which makes sense since $E = h\nu$ and angular momentum appears in quantum mechanics through $\hbar = h/2\pi$.
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