Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

### The average translational kinetic energy of $N_2$ ....

The average translational kinetic energy of $N_2$ gas molecules at _ _ _ _ $^\circ C$ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 Volt. (Given $k_b = 1.38 \times 10^{-23} J/K$) (Fill the nearest integer)

Solution

K.E. of electron =  Translational K.E. of $N_2$ molecules

$\therefore eV = \frac {3}{2} kT$

$\therefore 2 \times 1.6 \times 10^{-19} \times 0.1 = 3 \times 1.38 \times 10^{-23} \times T$

$\therefore T = \frac {1.6 \times 10^3 }{3 \times 0.69 } \approx 773 K = 500 ^\circ C$

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, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$