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)

If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?

Q.If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?
Solution:

 As shown in the graph the given two lines, their slopes, the perpewndicular distance between two lines and the coordinates of various points which would immideatly help solving this question given,
              The perpendicular distance between two lines(AB)=√5

               The value of tanθ= 1/2 = slope of two lines
In the triangle ABC,      tanθ=1/2=[√5]/BC, Therefore, BC=2√5
                                       AC^2    =AB^2+BC^2
                                       (k-3)^2 =(√5)^2+(2√5)^2
                Therefore,                k  = 8(ans)


                                                                                                                      -Mimansa

       

Square ABCD ha side length 13, and points E and F are exterior to the square such that BE=DF=5 and AE=CF=12. Find EF2

Q. Square ABCD ha side length 13, and points E and F are exterior to the square such that BE=DF=5 and AE=CF=12. Find EF².

Solution:

Complete the sqaure EGFH,then find out (EF)²=(17√2)² =578(Ans)