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### The pressure acting on a submarine is $3 \times 10^5 Pa$ ....

The pressure acting on a submarine is $3 \times 10^5 Pa$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:

(Assume that atmospheric pressure is $1 \times 10^5 Pa$, density of water is $10^3 kgm^{-3}$, $g=10 ms^{-2}$)

(A) $\frac {3}{200}$%
(B) $\frac {5}{200}$%
(C) $\frac {200}{3}$%
(D) $\frac {200}{5}$%

Solution

We have, $P_1 = P_a + h_1 dg$ & $P_2 = P_a + h_2 dg$

% increase in pressure = $\frac {P_2 - P_1 }{P_1} \times 100 = \frac {(h_2 - h_1 ) dg }{P_1 } \times 100$

$\therefore$ % increase = $\frac {(2h - h) \times 10^3 \times 10 }{3 \times 10^5 } \times 100 = \frac {h}{30} \times 100$

Also, $P_1 = 3 \times 10^5 = 10^5 + h \times 10^3 \times 10$

$\therefore h = 20 m$

Now, % increase in pressure = $\frac {h}{30} \times 100 = \frac {20}{30} \times 100 = \frac {200}{3}%$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$