• Assume ideal behaviour of both gases.
• Let the mixture of gas contain x gram of Carbon monoxide and y gram of carbon dioxide.
• Let the total volume occupied by the mixture be V l
• Now, since the density of the mixture is 1.5g/l, the mass of mixture is 1.5V g
• i.e. x+y = 1.5V......(1)
• further since PV=nRT ,  n = PV/RT
• i.e. no. of moles of mixture = PV/RT
• i.e. x/28 + y/44 = n....(2)
• Solve for x and y in terms of V
• since P= 73/76 atm, R=0.0821, T=300K
• we have n = 0.04V (approx)
• x/28 + y/44 = 0.04V....(3)
• solve (1) and (3) simultaneously in for x,y in terms of V.
• 49.28V = 44x+28y...(4)
• solve for x and y in terms of V using 1 and 4
• calculate the composition hence

The diazonium salt test

Methylamine is a primary amine so cooling and treatment with NaNO₂/HCl will liberate nitrogen gas only.

In case of aniline, treatment with NaNO₂/HCl after cooling leads to dizotisation, which forms a diazonium salt. Addition of a alkaline beta-naphthol(2-naphthol in NaOH) produces a red/orange dye.

Methylamine, being a primary amine, would not give a dye.It'll evolve a continuous stream of nitrogen gas when you cool it and then add NaNO₂/HCl

Consider C

Resolve the Tension into horizontal and vertical components,

U have,

T2sin(theta2)=mg

T2cos(theta2)=mg

implies theta2 = 45 degree

Also using these we get T2=sqrt(2)mg

Now considering the block B

T1sin(theta1)=T2sin(theta2)

and, T1cos(theta1)=mg+T2cos(theta2)

Using these, we get T1=sqrt(5)mg

Hence option C

it's a no-brainer.

since the series also has an (x-x) =0

so value of product=0

Answer =0

Doppler effect is a result of relative motion between source and receiver along the line joining them.  So we need to consider the component of source and receiver velocity along the line joining them. If these components are zero or are such that their vector sum is zero, then there is no doppler effect.

So when the source and receiver are both in motion, doppler effect is still observed because there is still some relative motion along the joining them, even if the velocities are perpendicular.

However this wont be the case if the velocities are parallel. In that case there is no relative motion along the line joining source and observer.

Upward force exerted by Table= 10N

Therefore by Newton's third law of motion, the block exerts an equal downward force on the table

hence A

Further, since the net upward force is greater than the net downward force(weight), the body has an upward acceleration.

This is as if the system is placed in a lift accelerating upward. Hence the contact force is greater than the body's weight.

so A and D are correct.

Crap.

You guys would do a lot better to abandon such nonsense and study.

Let m and n be the roots of the first equation

Therefore, we can find a number, say k, such that km and kn are the roots of the second equation

 now m+n = -b/a

k(m+n) = -q/p

i.e. k=aq/bp

further  mn=c/a

and k²(mn) = r/p

so  use k= aq/bp and mn=c/a

so that a²q²/b²p²(c/a)  = r/p

i.e. caq² = rpb²

Yes there is. But this is how it is derived. It's easier to remember this than cram the formula.

In general in the cosine series were the angles are in AP, such as this, you multiply and divide by the sine of the common difference and then simplify using the defactorization formulae.

Multiply and divide by sin(2pi/7)

so we have {sin(2pi/7)(cospi/7) + sin(2pi/7)cos(3pi/7) + sin(2pi/7)cos(5pi/7)}(1/sin(2pi/7)

 = {1/2sin(2pi/7)} { sin(3pi/7) + sin(pi/7) + sin(5pi/7) - sin(pi/7) + sin(pi) - sin(3pi/7)}

since sin(pi-x) = sinx

the expression becomes

= {0.5/sin(2pi/7)} { sin(pi-5pi/7} = {0.5/sin(2pi/7)}{sin(2pi/7)} = 0.5

Consider the discriminant of the equation,

D= 16(a²b²c²d²) - 4(a⁴+b⁴)(c⁴+d⁴)

  =  16(a²b²c²d²) - 4(a⁴b⁴) -4(a⁴d⁴) - 4(b⁴c⁴) - 4(b⁴d⁴)

 = 4a²b²(c²d²-a²b²) + 4a²d²(b²c²-a²d²) + 4b²c²(a²d²-b²c²) + 4c²d²(a²b²-c²d²)

= -4(a²b²-c²d²)² - 4(b²c²-a²d²)²

Clearly D<=0

The roots can be real only for D=0 , that is when the two terms are simultaneously equal to zero.

So the roots have to be equal if they are real.

The proof is using the Maclaurin Series

f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ............... ad inf.

So cosx = 1 - x²/2! + x⁴/4!+........ad inf.

and sinx = x-x³/3! - x⁵/5! + ......ad inf.

and e^x = 1+x + x²/2! + x³/3! + x⁴/4! + .......................ad inf.

Now we have to prove that, cosx + isinx = e^ix

LHS = cosx+isinx

Use the above expansions, and multiply the sine expansion by "i" .

RHS = e^ix,

Expand using the above expression, and substitute ix instead of x.

You will see that  LHS = RHS.

Straightforward.

Apply the AM>=GM inequality.

For 10 numbers, a²,b²,c²,d²,ab,bc,ca,ad,bd,cd,

(a²+b²+c²+d²+ab+bc+ca+ad+bd+cd)/10 >= {a²b²c²d²(ab)(bc)(ca)(ad)(bd)(cd)}^1/10

                                                                       >= {(abcd)²(abcd)³}^1/10

                                                                     >=1  (since abcd=1)

Therefore, LHS>=10

As per the HT mint survey,

11th best government college in India, after 7 IITs,ITBHU,Anna and Jadavpur. Ranked higher than NITs and ISMU.

It's got one of the best chemical technology research programmes in the world. Its research output is much higher  than even IITs.

Its Chemical Engineering dept. has been ranked by UGC as the best in India.

But not many people know about it, surprisingly enough.

This year, as far as I remember, chem.engg cutoffs for open cat. were

  AIR 9000 odd (thru AIEEE)

and 187/200(MHT-CET)

There are about 75 seats in all for chemical engineering.(including reserved.)

there are 22 seats to be filled  through AIEEE in chemical engg.

The rest are filled through MHTCET. 

If you are really interested in chemical engineering, or are getting chemical engineering everywhere else, then ICT is the place for you. (Except possibly  IITs)

according to most people, in case of chem.engg., UDCT~IIT B> any other college in India.

Also UD has an excellent reputation with foreign universities and strong linkages with the chemical industry. Mukesh Ambani  studied here.

.

The electric field on the surface of a uniformly charged hollow sphere is discontinuous.

But its value ( magnitude) is the right hand derivative of the potential function at the surface.

i.e. Kq/R²

Straightforward.

Work done is = nRT where T is the rise in temperature, n  is the no. of moles of gas.

So Work=R(100)

=100R?

Ya, bbye :)

 Lol, I don't think this has any physical meaning at all. Volume is the space occupied by a body, When volume becomes negative, what are you trying to say? More space is going to be created? LOL

See, questions like negative mass, negative volume are meaningless in a physical sense. Don't get confused by looking at the mathematics. Physicists first see things and then use the math, not the other way round

Convert the equation into the standard form y²=4ax

After some manipulation, you see that

(y-1)²=2(x+2)

Shifting the origin to the point (1,-2)

Y=y-1 and X=x+2

Y²=2X

Now the position of the given point w.r.t. new coordinate system is,

X₁=x₁+2=2+2=4

and Y₁=y₁-1=6-1=5

So: (4,5)

Now insert this in the original equation, clearly S₁>0(=25-8=17)

Outside.