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### An empty LPG cylinder weighs 14.8 kg ....

An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside the cylinder is _ _ _ _ atm. (Nearest integer)
(Assume LPG to be an ideal gas)

Solution

Weight of gas before use = 29.0 - 14.8 = 14.2 kg

Weight of gas after use = 23.0 - 14.8 = 8.2 kg

Now, $\frac {P_1}{n_1} = \frac {P_2}{n_2}$

$\therefore \frac {3.47}{14.2} = \frac {P_2}{8.2}$

$\Rightarrow P_2 = \frac {3.47 \times 8.2}{14.2} \approx 2 atm$

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$