but how did you get this: "T1=t2e^u(pi)."

Can you please explain?

actually, now I feel what vikrant said is wrong.

Hi,

 

Consider a fixed pulley of mass M. A light inextensible string pass over it. The ends of a string are connected to blocks of mass m1 and m2 respectively.  Let T1 and T2 be the tensions in string on the left and right side respectively. What is the tension of the string passing over the rim of the pulley??

Hi,

 

I am a XI student preparing for JEE.. I dont want to go down in boards but it isnt my proiority. I havr enrolled in fittjee class room proram.  

 

Can you please advice on me what needs to be done? I am following H.C. Verma and fiitjee material for physics. for maths and chemistry I have only done my fiitjee material.. Is that enough?Which books do I need to follow for maths and chemistry???

 

 I feel nervous that what i have done is not enough. Also, I get frustrated when I am not able to do a particular problem and often end up wasting a lot of time.. I keep getting distracted easily. please advice me..

Hi,

 

Find the number of positive integral solutions to the equation

 

thanks for your time.

a,b and c are correct.

I think question meant : Find the work done in slowly pulling down the system by 8cm. Only then, I am getting the solution you had provided.

 

Let a₁ and a₂ be the extensions in the springs at equilibirum.

 

Start by drawing the FBDS:

 

m₂g = k₂a₂ --- (1)

k₁a₁ = m₁g + k₂a₂ --- (2)

 

a₂ = 0.02m and a₁ = 0.02m.

 

Let x₁ and x₂ be additional elongations causes by pulling m₂ by L = 8cm. The additional forces on  m1 are equal and are in opposite directions.

 

Thus you can infer that k₁x₁ = k₂x₂  --- (3)

Also, use the constraint x₁ + x₂ = L  --- (4)

 

Solve (3) and (4) to obtain the values of x₁ and x₂ as 0.06m and 0.02m respectively.

 

Apply work energy theorem now. Initial and final kinetic energy is 0.

 

W_g + W_p + W_s = 0 --- (5) where W_p is the work done by pulling force.

 

W_s = U₁ - U₂ where U₂ = 1/2 k₁(a₁ + x₁)² + 1/2 k₂(a₂ + x₂)²

 

Put in the values, and you get W_p = 1.2 J.

By the way, if the above is not clear, you might want to refer "Concepts of Physics - H.C Verma). I remember seeing a quite similar problem in it long back.

 

cheers.

 

 

Hi,

 

A uniform rod of mass 'm' and length 'L' is initially held in vertical position with one end touching the smooth horizontal surface. Now, it is released. Then, find the velocity of the centre of mass when it makes an angle  = 37 with the vertical.  [Solution : (27gL/260)^1/2]

 

Can someone please help me out with this one? Thanks for your time.

 

x = 5t³ + 9t² + 2t +4 m

 

1) Differentiating displacement w.r.t to time, you can obtain the velocity of the car as a function of time.

 

dx/dt = 15t² + 18t + 2 = v

Plugging in t=2.7, v = 159.9

 

2) Same procedure again. Differentiate velocity with respect to time to get the acceleration of the car.

 

dv/dt = 30t + 18

a = 99m/s² at t=2.7s.

 

3) Average acc = change in vel / time = 18.8/15 = 1.25m/s²

 

4) Use equation of motion :  v = u + at

    a = (v - u)/t = 21.1/10 = 2.1m/s²

 

5) Use 2nd equation of motion : s = s₀ + ut + (at²⁾/2

    s = 12(9) + 1.16(81) = 201.96

 

6) 18.8 m/s?? (given)

 

7) Once again use 2nd equation of motion: s = s₀ + ut + (at²⁾/2

    But do take note that, u = +19 since the helicopter is moving up and a = -g since gravity acts downward.

 Plug in the values, you get 234m.

 

8) As I stated in the previous question, the package initially has a velocity 19m/s in the upward direction. Due to the influence of gravity, this velocity decreases, reaches 0 and then becomes negative. The maximum height of the package above the ground is at the instant when its velocity becomes 0.

 

Use v² = u² + 2ah

   0 = 19² + 2(-g)h

 

9) The policeman catches the speeder when the following condition is satisfied:

 

S_s - S_p = 0

where S_s is the displacement of the speeder and S_p is the displacement of the police man.

Use s = ut + 1/2 at²  to calculate their displacements and a time 't'

 

49t - (3.27t²)/2 = 0

Solving,t = 30s.

 

10) The question isnt very clear. Does the rock fall from a height of 65m under the influence of gravity? If so,

 

65 = (1/2)(gt²)

t = 3.6s.

 

The runner can run a distance = 10(3.6) = 35m.

 

I hope I have been helpful. Do contact me if any of the above details are not clear. By the way, I have plugged in the value of g as 10m/s² in all the problems. If you wish to be more accurate, take g as 9.8m/s².

 

Cheers.