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Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is (in $\Omega$)

$A$ current of $2 \, A$ flows in an electric circuit as shown in the figure. The potential difference $(V_R - V_S)$, in volts ($V_R$ and $V_S$ are potentials at $R$ and $S$ respectively) is:

$e^{\log (\cosh^{-1} 2)}$ is equal to :

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

$2 \pi - (\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65})$ is equal to

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