- ✓Reach the ground in $4$ second
- BBegin to move down after being released
- CHave a displacement of $50 m$
- DCover a distance of $40 m$ in reaching the ground
$s =\frac{1}{2} \times 1.25 \times 8 \times 8\; m / s =40 m$
now, $40=-10 t +\frac{1}{2} \times 10 \times t ^2$
$5 t^2-10 t-40=0$
$t^2-2 t-8=0$
hence, $t=4 s$
$s=40 \;m$.
displacement $=40\; m$
Just after being released the stone has an upward velocity, so it will move upwards first.
distance in an upward direction before stopping $d$.
$d =\frac{ v ^2- u ^2}{2 g }=\frac{10^2-0^2}{2 * 10}=5\; m$
and distance $= s +2 d =50 m$
So the distance covered is $50\; m$ and the displacement is $40\; m$.
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K. The heat capacities per mole of an ideal monatomic gas are $C_v=\frac{3}{2} \ R, C_p=\frac{5}{2} R$, and those for an ideal diatomic gas are $C _{ v }=\frac{5}{2} \ R , C _{ P }=\frac{7}{2} R$.
$1.$ Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be :
$(A)$ $550 \ K$ $(B)$ $525 \ K$ $(C)$ $513 \ K$ $(D)$ $490 \ K$
$2.$ Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be:
$(A)$ $250 \ R$ $(B)$ $200 \ R$ $(C)$ $100 \ R$ $(D)$ $-100 \ R$
Give the answer question $1$ and $2.$