When 700ml of water is added in a buffer solution containing 0.01M of CH3COOH and 0.1M CH3COONa then the pH of final solution is(Ka of CH3COOH=1.8X10^-5,log 1.8=0.25)

When 700ml of water is added in a buffer solution containing 0.01M of CH3COOH and 0.1M CH3COONa then the pH of final solution is(Ka of CH3COOH=1.8X10^-5,log 1.8=0.25)

Solution :
Step 1: Addition of 700 ml of water does not affect pH because
Step 1:The Henderson-Hasselbalch equation for the pHpH of a buffer solution of the monoprotic acid HAHA is given by
pH=pKa+log[A−][HA]
pH=pKa+log⁡[AX−][HA]
Since concentration appears in both the numerator and denominator of the fraction [A−][HA][AX−][HA] and pKapKa is constant (at a fixed temperature), it appears that dilution of the solution with pure H2OHX2O would not change the pHpH. However, since
pH=−log[H+]
pH=−log⁡[HX+]
the moles of H+HX+ must increase in order for pHpH to stay constant upon dilution.
Where is this additional H+HX+ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid

Step 2:pH=pKa+log10([A−][HA])pH=pKa+log10⁡([A−][HA])

where  pKa=−log10KapKa=−log10⁡Ka
1.[CH3COOH]=0.01M,[CH3COONa]=0.1M and Pka=4.75 and log 10[ CHCOO-]/[CH3COOH]=1
So,Ph=5.75(Answer )

A ball thrown from ground with speed u at an angle ϴ with the horizontal, at the same instant from the same position a boy runs along the ground with speed [3^ (1/2)/2]u to catch the ball. If the boy can catch the ball, the angle of projection ϴ is

A ball thrown from ground with speed u at an angle ϴ with the horizontal, at the same instant from the same position a boy runs along the ground with speed u[3^ (1/2)/2] to catch the ball. If the boy can catch the ball, the angle of projection ϴ is

solution. Horizontal range of the projectile = The distance travelled by the boy, to catch the ball

               Horizontal range of projectile= u^2sin2ϴ/g----------1

               Distance travelled by the boy= [3^ (1/2)/2]u.2usinϴ/g-------------2

Equating 1 and 2,

u^2sin2ϴ/g=[3^ (1/2)/2]u.2usinϴ/g

          cosϴ=3^ (1/2)/2

               ϴ=30 degree(Ans)

A cylinder containing water stands on a table of height H A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of water in the cylinder is

A cylinder containing water stands on a table of height H A small hole is punched in the

side of cylinder at its base. The stream of water strikes the ground at a horizontal distance

R from the table. Then the depth of water in the cylinder is 

Solution :

Let t is time taken by the jet to travel the vertical height H from bottom of the cylinder and v is the velocity of stream of water,

H=1/2gt^2------------1

t=(2H/g)^1/2-------------2

v=(2gh)^1/2-----------3

The horizontal distance R travelled by stream=vt=[(2gh)^1/2][(2H/g)^1/2]----------------4

solving equation 4, we get h(depth of water in the cylinder)=R^2/4H(ans)

A train M leaves Meerut at 5AM and reaches Delhi at 9AM.Another train leaves Delhi at 7AM and reaches Meerut at 10.30AM.What time do the two trains cross each others?

Q.A train M leaves Meerut at 5A.M. and reaches Delhi at 9A.M..Another train leaves Delhi at 7A.M. and reaches Meerut at 10.30A.M..What time do the two trains cross each others?

Solution :
Let the distance between the stations =D km
Then  the speed of train from Meerut side=D/4 km/hr and  the speed of train from Delhi side =2D/7 km/hr
Let, after t hours the Delhi to Merrut train starts ,two trains meet each other then
 After x hours, the distance covered by Delhi to Meerut train =2Dx/7 km ------1
As the train from Meerut side has started 2 hours before ,the distance covered by Meerut to Delhi train is=(x+2)D/4----------2
 From1 and 2,
 (x+2)D/4+2Dx/7 km =D
Then solving the equation ,x=14/15hrs=56 miniutes which corresponds to 7:56A.M.(Ans)