Hello candidate,

The work done by a force is calculated as the force applied multiplied by the distance travelled by the body by that applied force in the same direction or at a specific angle.

The work done by a variable force depends as a dependence of force on the displacement travelled which is basically solved using the method of integration from the initial point to the final point of displacement.

Hope you found it helpful. If you have any further queries feel free to post it here!!

Hello Aspirant,

Hope you are doing well!!

It is quite difficult to make integration sign, In place of integration sign I'll put |, and in place of power I'll put **  when you are solve in your notebook you can place | sign to integration sign and ** to power sign.

Let X+Y = v

1 + dY/dX = dv/dX

dY/dX= dv/dX-1

dY/dX = sin (X + Y) + cos (X + Y)

dv/dX - 1 = sin v + cos v

dv / (1+cos v + sin v) = dX

Integrate both side

| dv / (1 + cos v + sin v ) = | dX | dv / ( 1 + ((1 - tan**2 (v/2)) / (1 + tan** 2 (v/2)) + ((2 tan (v/2) / (1 + tan **2 (v/2))

= |dX | sec**2 (v/2) dv / (2(1+ tan(v/2))

=|dx log (1 + tan (v/2)

= X+ c log ( 1+ tan (X + Y) / 2) = X + c

I hope this will help you.

Feel free to ask any query.

Dear Student,

• The work done by a constant force magnitude F on a point that moves a displacement d in the direction of the force is the product of , W = Fd.
• Integration method can be utilized to calculate work done by a variable force and work done by a constant force

Dear aspirant,

Work is the energy transferred from the object via the application of force along the displacement.

Power is basically the amount of energy transferred per unit time.

Work done by a variable force is the integral of F.dx where F is the force and dx is the displacement.

Hi.

This law exhibits the short-run production functions in which one factor varies while the others are fixed.

Also, when you obtain extra output on applying an extra unit of the input, then this output is either equal to or less than the output that you obtain from the previous unit.

The Law of Variable Proportions concerns itself with the way the output changes when you increase the number of units of a variable factor. Hence, it refers to the effect of the changing factor-ratio on the output.

Hello student,

The linear equation in one variable is an equation which can be expressed in the form such as 'ax + b = 0' (a, b and c are real numbers) where a and b are the two integers , and x has just one solution also which is a variable .

Taking example of one such equation :

4x+9=-11

When you solve this equation, it will have only one solution.

These type of equations are also known as first degree equations because the exponent on the variable is 1.

I hope the answer is clear to you. Thank you.

Hello aspirant,

Here is the answer to your question.

Marginal costing shows that the meaning of profit gets changed, but here profit means contribution.

Hence, Selling Price= 10 per unit-Variable Cost=6 per unit contribution which is equal to 4 per unit total contribution= 4 per unit * 100000 units= 4 lakhs

So, final profit= 4 lakhs - 2 lakhs (Fixed Cost= 2 lakhs.

Hope, it helps you.

Hello aspirant,

The answer will be-

Assume slope of line be m

(2,3) are fixed points

Equation of line:-

y-3=m*(x-2)

y-3=mx-2m

mx-y+3-2m=0

Foot of perpendicular from origin (0,0) is-

(2-0)/m=(y-0)/(-1)= -(0+3-2m)/{0.5^(1+m^2)}

m= -x/y

Now, mx-y+3-2m=0

(-x/y)x-y+3-2(-x/y)=0

x^2+y^2-2^x-3^y=0

Centre=(1, 3/2)

Radius= 0.5^(1^2+(3/2)^2-0) = √13/2

Diameter = 2*Radius = 2* (√13/2) = √13

Hope, it helps you.

Hello!

31-(a) Transition metals show variable oxidation state because of their valence electron in two orbitals. Difference between the two orbitals is too less, hence both the energy levels can form bonds.

(b) Zinc (Zn) have higher tendency to loose electron when compared to that of Cupper(Cu)

(c) Mn shows +7 oxidation state with oxygen is because they are capable of forming double bonds, multiple bonds. When it comes to the case of fluorine they will become unstable.

32) Tollens is a test that is used to distinguish between aldehydes and ketones. Tollens reagent is used for silver mirror test. Aldehydes gets oxidized in Tollens and ketone wont. Tollens have ammoniacal content.                                                    EQUATION- 2AgnO3 + 2NaOH --- Ag2O(s) + 2NaNO3 + H2O.

ALL THE BEST. Hope the answer was helpful.

Thank You