\(t_1=\frac{\sqrt2⋅\frac{H}{2}}{g}=\frac{\sqrt{H}}{g}\)
\(t_2=\frac{\sqrt{2H}}{g}−t_1\)
\(⇒t_2=\frac{\sqrt{2H}}{g}−\frac{\sqrt{H}}{g}\)
\(⇒t2=√Hg{√2−1}\)
\(⇒t2=(√2−1)t1\)
So, the correct option is (D): t2 =(√2−1)t1
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)