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)

A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water and its size is such that when it touches the sides of the conical vessel, it is just immersed. How much water will remain in the cone after the overflow?

A conical vessel of radius 6 cm and height 8 cm is filled with

water. A sphere is lowered into the water and its size is such

that when it touches the sides of the conical vessel, it is just

immersed. How much water will remain in the cone after the 

overflow? 

Solution :Let the radius of the spear=r and slant height of cone =h
Then r=A/S(The radius of a circle which is drawn inside a triangle) where A=area of trianle and S=semiperimeter of the triangle 
Here slant height=10cm
Area of trianle=48cm^2
Semiperimeter of the triangle=16cm
So radius=48/16=3cm
The left water content in the cone after submerging the spear=1/3x6^2x8-4/3x3^3=188.57cm^2(Answer)

Mohit went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was 1 and 1/2 times that from Chandigarh to Shimla. If the average speed for the entire journey was 49km/hr, what was the average speed from Chandigarh to Shimla?

Solution :
           

                    Delhi________________Chandigarh______________Shimla
Let the distance from Chandigarh to Shimla be X.km=d2
So, the distance from Delhi to Chandigarh be 3X/4.km=d1

Let the speed from Chandigarh to Shimla be y km/hr=S2
So,the speed from Delhi to Chandigarh =3/2y=S1

Let T1 be the time taken from Delhi to Chandigarh=(3x/4)/(3y/2)=1/2.x/y hrs
Let T2 be the time taken from Chandigarh to Shimla=x/y hrs
So total time of journey from Delhi to Simla=3/2xy

The total distance from Delhi to Simla=7x/4
The average speed from Delhi to Simla=(7x/4)/(3/2.x/y)
                                                               =>7y/6=49
                                                               =>y=42 km/hr(Answer)

There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. the second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. the approximate time required by each pipe to fill the tank simultaneously is

There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. The second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. The approximate time required by each pipe to fill the tank simultaneously is
Solution :
Let  the third pipe fills the tank in x hours
So, second pipe will fill the tank in x+3 hours
       first pipe will fill the tank in x+3+7=x+10 hours
The third pipe in 1hour will fill 1/x part of the tank
The second pipe in 1hour will fill 1/x+3 part of the tank
The first pipe in 1hour will fill 1/x+10 part of the tank

First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone ,so
(1/x+3) + (1/x+10)=1/x
=>x=(30)^1/2=5.4...=5 and 1/2 hours=Filling hour of third pipe
:. filling hour of first pipe=x+10=15 and 1/2 hrs
    filling hour of second pipe=x+3=8 and 1/2hrs(Answer)

___?____ times the sum of the squares of sides of a triangle is equal to four times the sum of the square of the medians of the same triangle?

____?____ times the sum of the squares of sides of a triangle is equal to four times the sum of the square of the medians of the same triangle?

Solution: Let, the be triangle be a equilateral triangle ABC as follows

AO, BM and CN are the medians of the triangle.
Here, they are all equal.
Three times of the  sum of the squares of sides of a triangle=3(3a^2)=9a^2=a^2-(a/2)^2........(1)
Four times the sum of the square of the medians of the same triangle=4[3{a^2-(a/2)^2}]=9a^2....(2)
Therefore, (1)=(2)......solved.