Soham Shanbhag

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A loop of light inextensible string passes over smooth small pulleys A and B. Two masses m and M are attached to the string at the respective positions. Then the condition between m, M and n so that m and M will cross each other will be?

The masses are point masses and the pulley has negligible radius and are at the same elevation

This is how I did it

Method I]

The final position is dictated by the total length of the string (L(1 + n)), the distance between pulley and m(or M) will be (L(1 + n) / 2).

We find the distance traveled by m by Pythagoras theorem as L√[(n + 3)(n - 1)] / 2 = x

Total distance between the blocks in the initial position is L√[(n - 1)(n + 1)]

We also find the distance travelled by M in the same time as L√[(n - 1)(n + 1)] - x

velocity at that time of the blocks will be non zero

Applying Work energy theorem,

mgx - Mg(L√[(n - 1)(n + 1)] - x) > 0

=> mgx > Mg(L√[(n - 1)(n + 1)] - x)

=> m / M > (L√[(n - 1)(n + 1)] - x) / x

=> m / M > (L√[(n - 1)(n + 1)] - L√[(n + 3)(n - 1)] / 2) / L√[(n + 3)(n - 1)] / 2

=> m / M > (2√[(n + 1) / (n + 3)]) - 1

Which is the correct answer

Method II]

We note that the masses meet at the centre of mass of the system

mx - M(L√[(n - 1)(n + 1)] - x) = 0 [If we define origin as the final location in method I]

=> m/M = (L√[(n - 1)(n + 1)] - x) / x

=> m / M = (2√[(n + 1) / (n + 3)]) - 1 [For the bodies to meet]

=> m / M > (2√[(n + 1) / (n + 3)]) - 1 [For the bodies to continue their motion]

Note that this is the same answer as in method I

Now I see there is a flaw in the second method that the centre of mass keeps changing its position which is governed by external forces gravity and reaction force on the pulleys.

In solving the problem I have assumed that the centre of mass of the bodies does not change.[That is why m1x1 + m2x2 = 0 (final position of the bodies is zero)]

If the position of Centre of mass is not same as the initial position then I should have got an different answer. So this means that the centre of mass doesn't change its position inspite of external forces.

Can someone explain why?? Or maybe the centre of mass goes down and comes up.

Also new methods are welcome

The near point of an hypermetropic eye is 80 cm.

What is the power of the lens required to correct the vision.

Given:

Assuming eye is perfectly spherical

Diameter of eye = 2.3 cm

Near point of normal eye = 25 cm

If a + bi is a root of a quadratic equation with real coefficients then a - bi is also a root.

This gives us a equation:

(x - (a + bi))(x - (a - bi)) = 0

=> x² - 2ax + (a² + b²) = 0.................(1)

Now given equation (1) we can easily find the roots of all equations.

METHOD A]

Example:

x² - 8x + 17 = 0

gives a = (-8) / (-2) = 4

and a² + b² = 17

=> 16 + b² = 17

=> b² = 1

=> b = ±1

Hence 4 + i and 4 - i are two roots of the equation

We all use formulas like x = (-b ± √D)/2a for finding roots (Applicable to ax² + bx + c = 0)

I want to verify whether the above method (mentioned in (1) and method (A)) has any limitations.

I want to make sure as I may use it for solving problems

Thanks for any answers and explanations

1) If out of n things p are alike, q are alike, r are alike and s are alike and (n - p - q - r - s = t) are different, then prove that the total no of combinations will be (p + 1)(q + 1)(r + 1)(s + 1)(2^t) - 1

2) If we have to distribute 5 chocolates in 4 people, we know that there are

i) 4^5 ways if the chocolates are distinguishable

ii) C[8,3] ways if they are same.

What if the chocolates are partly identical.

i.e. say out of 5 you have 3 identical. I would like a general formula here not the answer to the question.

A cylindrical beaker of height = 30 cm and diameter = 20 cm contains two immiscible liquids mercury and water of R.D. 13.6 and 1 respectively. A solid sphere of diameter = 14 cm floats in such a way that half the volume is dipped in mercury and the other half in water. The top of the sphere touches the uppersurface of water.

a) Find the density of the solid.

b) What happens if we add some more water to the beaker without disturbing the equilibrium. (will the body rise\dip more\no change)

c) If we add some other liquid of density less than that of mercury and more than that of water what will happen.(As compared to (a))

Given 4 numbers a, b, c, d such that d - c = c - b = b - a and a² + b² + c² = d

Find the value of a + b + c + d

1. Two Boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least two of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box have the same colour and have the same size (radius).

2. For all positive real numbers a, b, c prove that

a/(b + c) + b/(a + c) + c/(b + a) ≥ 3/2

3.Find the remainder when 2^1990 is divided by 1990

4. prove that the radius of the excircle opposite angle A of a triangle is (area)/(semiperimeter - BC)

Please try to answer most number of questions and show steps.

We have been recently taught how to find the magnetic lines of forces in a straight current carrying conductor using maxwell's right hand grip rule or corkscrew rule. But where does the north pole lie in such a case when the lines are circular?

Find k so that x² + 2x + k is a factor of 2x^4 + x³ - 14x² + 5x + 6.

I have a line in my book stated as

"In congruent modulo theorem, we cannot divide by a number, always.

Example. We have 14 ≡ 2 (mod 4), but we cannot say 7 ≡ 1(mod 4).

In fact 7 ≡ 3 (mod 4) or 7 ≡ -1 (mod 4)

If a ≡ b (mod m) and d is the common divisor of a and b s.t. g.c.d. (d,m) = 1, then we can have a/d ≡ b/d (mod m). Only in this condition, we can divide by a number."

Prove that (d,m) = 1 or division cannot be done.

First prove the theorem and then give examples if you wish to.

Number 73 is in base 10

When you write it in base 5 you get

2 * 25 + 4 * 5 + 3 * 1

= 243

Now in the similar way how would you write 23 base 10 in base 12.

How do you compute it.

I have a triangle ABC of Area 200 cm²

There is a incircle incribed in it with radius = 8 cm

Find the perimeter of triangle.

Prove that n^(16) - 1 is divisible by n for (n,17) = 1

i.e. n and 17 are co-prime

ABDC is a square.

Angle OCD = Angle ODC = 15°

O is equidistant from C and D

AOD and BOC are not collinear

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Prove that triangle ABO is an equilateral triangle

ABC is a triangle with D,E and F as the midpoints of AB, BC and AC respectively. The medians meet at a point G forming 6 triangles inside the triangle ABC. Prove that the circumcentres of the 6 triangle formed lie on a circle or centres are con-cyclic

Preference to pure geometric proof for BA. Guidance welcomed if complete proof is too long. Any other method (like coordinate geometry, trigonometry) or reference to sites also OK.

I am in IX standard and it would help if explanations are in simple mathematical language.

I am learning geometry and hence would like reference to informative sites.

Thank you

In a certain 4 digit no, if a pair of 1st 2 digit numbers and a pair of last 2 digit numbers is exchanged, the new number obtained is 5 more than twice the original number. Find the original number.

note: This question was asked in an scholarship exam for 8th standard and so please answer this question using only simple methods. DO NOT USE CALCULUS OR GRAPHS AS FAR AS POSSIBLE.