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)