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)

If ‘T’ is the surface tension of a fluid, the energy needed to break a liquid drop of radius ’R‘ into 64 equal drops would be?

Solution:

The required energy to break the bigger droplet into 64 smaller droplets=change in surface areaX surafce tension(T)

Change in surface area= 64X4πr^2-4πR^2------------(1)

                              64X4/3πr^3=4/3πR^3

                                         =>  r=R/4

                                         =>  r^2=R^2/16-----------------(2)

Substituting value of r^2 in (1) from(2),

                                         Change in surface area= 12πR^2

So, the energy required to break the bigger droplet into 64 smaller droplets= (12πR^2)T----Ans

The temperature of a gas contained in a closed vessel increases by 1 degree Celsius when pressure of the gas is increased by 1%. The initial temperature of the gas is ?

Solution: according to Boyle's law, PV=nRT
Let, initially the gas had pressure=P, volume=V, No. of moles=n and temperature=T degree Celsius
Therefore, the initial temperature in Kelvin scale is T + 273degree
After increasing 1 degree Celsius the pressure increased by 1%. Hence, the new temp. is=T+1+273
                                                                                                                                             =T+274K
and new pressure=(101/100)*P
In summery, PV=nR(T+273)So, PV/nR=T+273--------------1
                     (101/100)*PV=nR(T+274),So (101/100)*PV/nR=T+274-------------2
substituting PV/nR value from equation 1 in equation 2,
(101/100)*(T+273)=T+274
T=-173 degree=100K(-173+273)---answer

Two consecutive terms in the expansion of (3+2x)^74 whose coefficients are equal are ….?

cos-1 ([x2-1]/[x2+1]) + tan-1[(2x)/(x2-1)] =2π/3, then x=?

Q. cos-1 ([x²-1]/[x²+1]) + tan-1[(2x)/(x²-1)] =2π/3, then x=?

Solution: Let,

cos-1 ([x²-1]/[x²+1]) =θ-------------------1

→cos θ= (x²-1)/(x²+1)

→cos² θ=(x²-1)²/(x²+1)²

→1- cos²θ=1-(x²-1)²/(x²+1)²

              =[(x²+1)²-(x²-1)²]/ (x²+1)²

              = [2x²X 2]/(x²+1)²

→Sin²θ= 4x²/(x²+1)²

→sin θ=[2x]/(x²+1)

→sin θ/ cos θ=2x/(1-x²) =tan θ

→θ=tan^-12x/(1-x²) --------------------------2

From 1 and 2,

→cos-1 ([x²-1]/[x²+1]) + tan-1[(2x)/(x²-1)] =2π/3

→2 θ= 2π/3

→θ= π/3

so, cos θ= (x²-1)/(x²+1) =cos π/3

 → (x²-1)/(x²+1) =1/2

→X²+1=2-2x²

→3x²-1=0

→ (√3x+1) (√3x-1) =0

So, x=±1/√3(Ans)

sin-1 x+sin-1 y+sin-1 z=π then x2+y2+z2+2xyz=?

sin-1 x+sin-1 y+sin-1 z=π then x2+y2+z2+2xyz=?

Solution:

Let A = sin^-1(x), B= sin^-1( y) , C = sin^-1( z ) ,

=> X=Sin A, Y=Sin B, Z=Sin Z 

then given A+B+C = pi/2,

=>A+B= pi/2-C

Apply cos on either side  

=>Cos(A+B) = Cos (pi/2 - C) = sin C

=>Cos A. Cos B - sin A. sin B = sin C

=>Cos A. cos B = sin A. sin B + sin C

=>√ (1 - x²) √(1 - y²) = xy+z [squaring both  sides]

=>1 - x² - y² + x² y² = x² y² + z² + 2xyz

=> x^2 +y^2 +z^2 +2xyz = 1 (Answer)