A piece of ice with a stone frozen into it, is floating in a glass vessel filled with water. If the ice melts completely, then level of water in the vessel will

Solution :
Volume of melted ice=water  is less than ice ,then stone volume would remain unchanged ,therefore ,the level of water would fall

In ΔABC, tan(∠CAB) = , and the altitude from A divides CB into segments of length 3 and 17, then the area of ΔABC is

In ΔABC, tan(∠CAB) = , and the altitude from A divides CB into segments of length 3 and 17, then the area of ΔABC is
Solution :

Mnemonic for macro nutrients

See see head master not observing possible passing student(CC HM Not Observing Possible Passing Students)

C-CARBON
C -CALCIUM
H-HYDROGEN
M -MAGNESIUM
N-NITROGEN
O-OXYGEN
P-PHOSPHORUS
P-POTASSIUM
S -SULPHUR 

ABC is a triangle, the incircle touches the sides BC, CA and AB at D, E and F respectively. BD, CE and AF are consecutive natural numbers. I is the incentre of the triangles. The radius of the incircle is 4 units.

ABC is a triangle, the incircle touches the sides BC, CA and AB at D, E and F respectively. BD, CE and AF are consecutive natural numbers. I is the incentre of the triangles. The radius of the incircle is 4 units. The sides of the triangle are  13,14 and 15 unit ,prove

Solution :
Step1:Follow this figure

Step2:You can easily find that BC=2X+1,AC=2X+3 and AB=2X+2.....1
Step3:Find the area of ABC triangle by Heron's formula =[(3x+3)(x)(x+1)(x+2)]^(1/2).....2
Step4:Add area of three triangles namely BDC,CDA and ABD by using their base and perpendicular =12(x+1) square unit......3
Step5:(3x+3)(x)(x+1)(x+2)=144(X+1)^2........4
Step6:Solving 4 ,would give x=6,the sides of triangles are proved to be 13,14 and 15(Answ)

The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

Solution :Steps 

Step 1:Follow this figure 

Step 2:

Step 3:

The eqaution of line which is perpendicular to a given line ax+by+c=0 is bx-ay+k=0 .This deduction would clarify the value of (x,y)=(αβ,-αβ) is the point where perpendicular from C to AB intersect line II.

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)

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)