Two ends of an inextensible string passing over two fixed pulleys with midpoint connected to a block are pulled down with constant speed ve making the block to move up. Then which of the following options is correct?
A) $v_m = v_e $
B) $v_m > v_e $
C) $v_m < v_e $
D) Data is insufficient to conclude
Continue reading Part B: Two Ends of an Inextensible String ……
Hydrazine formed from Urea is further oxidised to $N_2$ as shown below:
$N{H_2} – \mathop {\mathop C\limits^\parallel }\limits^O – N{H_2} \xrightarrow[\text{Hoffmann Bromamide Degradation}]{\text{NaOBr}} N{H_2} – N{H_2} \longrightarrow {N_2}$
Assuming 1 dL of blood sample contains 30 mg of Urea, calculate the volume of $N_2$ gas obtained at NTP from the sample. Continue reading $N{H_2} – CO – N{H_2}\xrightarrow{NaOBr} {N_2}$
[18]-crown-6 has:
A) 18 Carbon & 6 Oxygen atoms
B) 6 Carbon & 18 Oxygen atoms
C) 12 Carbon & 6 Oxygen atoms
D) 6 Carbon & 12 Oxygen atoms
Key
Crown ether named [T]-crown-O, means T is the total number of atoms in the ring and O is the number of oxygen atoms in the ring.
For, [18]-crown-6 there are 6 Oxygen atoms and a total of 18 atoms.
So, number of Carbon atoms = 18 – 6 = 12
Hence, option (C).
On rewriting we have the series as,
$\frac{1}{{1 \times 2 \times 3}} + \frac{1}{{2 \times 3 \times 4}} + …………….. + \frac{1}{{100 \times 101 \times 102}}$
$ = \sum\limits_{r = 1}^{100} {\frac{1}{{r(r + 1)(r + 2)}}} $
$ = \frac{1}{2}\sum\limits_{r = 1}^{100} {\left[ {\frac{1}{{r(r + 1)}} – \frac{1}{{(r + 1)(r + 2)}}} \right]} $ Continue reading $\scriptscriptstyle \frac{1}{{102 \times 101 \times 100}} + \frac{1}{{101 \times 100 \times 99}} + ……… + \frac{1}{{3 \times 2 \times 1}} = ?$
${t_r} = \frac{1}{{4{r^2} – 1}}$
$ = \frac{1}{{(2r – 1)(2r + 1)}} = \frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)$
${S_n} = \sum\limits_{r = 1}^n {\frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)} $
${S_n} = \frac{1}{2}\left[ {\left( {\frac{1}{1} – \frac{1}{3}} \right) + \left( {\frac{1}{3} – \frac{1}{5}} \right) + \left( {\frac{1}{5} – \frac{1}{7}} \right) + ……. + \left( {\frac{1}{{2n – 1}} – \frac{1}{{2n + 1}}} \right)} \right]$
${S_n} = \frac{1}{2}\left[ {1 – \frac{1}{{2n + 1}}} \right] = \frac{n}{{2n + 1}}$
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Consider a small bucket full of water tied to a string whirled around in vertical circle of radius r without water falling down. At the topmost position when the speed of the inverted bucket is v,
(A) $\frac{{{v^2}}}{r} = g$ necessarily
(B) $\frac{{{v^2}}}{r} < g$ necessarily
(C) $\frac{{{v^2}}}{r} > g$ necessarily
(D) None of the options given
Solution
Consider water as the system. Let, m be the mass of water. The weight mg acts downwards and the normal reaction N also acts downwards.
$mg + N = \frac{{m{v^2}}}{r}$
$ \Rightarrow \frac{{m{v^2}}}{r} \ge mg$
$ \Rightarrow \frac{{{v^2}}}{r} \ge g$
Hence, Option (D).
