IIT JEE , AIEEE Forum - Articles Category , Community shelf - Few selected problems in physics with solutions Part 1

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1. Find the moment of inertia of an equilateral triangle of side

about its median.
Solution
Consider a horizontal rod like element of thickness dx at a distance x from the base of the triangle
Width of that element = 2(l-x) tan30 = (2/

3) (

3L/2-x)
Mass of that element = (4M/

3 L² ) (2/

3) (

3L/2-x) dx

MI of this elemental rod
dI = (1/12)[(4M/

3 L² ) (2/

3) ((

3L/2)-x) dx] [(2/

3) ((

3L/2)-x)]²

I = (8M/27 L²)₀

^L' ((

3L/2)-x)³dx
where L' is the length of the median
L' = (

3L/2)

I = (1/24)ML²

2. Two bodies A and B are thrown upwards simultaneously with the same speed. The mass of A is greater than the mass of B. If air exerts a constant and equal force of resistance on these two bodies, then which mass will go higher?
Solution
Acceleration of the bodies
a_A = g+R/m_Aa_B = g+R/m_BHeight attained by the bodies
H_A = u² /2a_AH_B = u² /2a_B

Let m_A >m_Bg+R/m_A < g+R/m_Bu² /2a_A > u² /2a_BH_A >H_BHence the heavier mass attains greater height.

3. A ball falls from the roof of a building. A man standing in front of a one meter high window in the building notes that the ball takes 0.1 seconds to fall from the top to the bottom of the window. The ball continues to fall and strike the floor. On striking the floor the ball rebounds with the same speed with which it hits the floor .If the ball reappears at the bottom of the window 2 seconds after passing the bottom on its way down ,find the height of the building.
Solution
Let the height of the building be h.
Let the topmost point of the building be A, The top of the window be B, the bottom of the window be C and the bottom most point of the building be D.

Height of the window, L = 1 m
Time taken by the ball to cross the window, T = 0.1 s
Time taken by the ball to travell from C to D and back to D again, T' = 2 s.
Acceleration, a = -g

Let the velocites of the ball at B & C be v_A & v_B respectively.
Then
v_B = v_A + aT

v_B = v_A - gT ----------------(1)
Also,
v_B² = v_A² + 2gL ---------------(2)

Solving equations (1) & (2) for v_B &v_A
v_B = -(gT + 2L/T)/2 = -10.5 m/s (Assuming g = 10m/s² )
v_A = (gT - 2L/T)/2 = -9.5 m/s

Let the distance from A to B be h₁, from C to D be h₂v_{A² =}2gh₁

h₁ = 4.5125 m

Now it can be shown that a ball dropped from a given height takes equal times to drop to the ground from that height and to come back to that height from the ground.
Hence time taken to travel from C to D
T' = 1 s

Hence,

-h₂= v_B T' - (1/2)gT' ²

h₂= 15.5 m

Hence the height of the building = h₁ + L + h₂=21.0125 m

4. Two projectiles have the same range R and heights H₁ & H₂ . Prove that R = 4

H₁H₂
Solution
As the ranges are same, they are projected at say,

and 90 -

4H₁ = Rtan

4H₂ = Rtan(90-

) => 4H₂ = Rcot

then,
16H₁H₂ = R²

R = 4

H₁H₂

5. A body thrown upwards from the tower take time t₁ to reach the ground. When thrown with the same speed downwards it takes time t₂ to reach the ground. If dropped from the tower, it takes time t to reach the ground. Prove t =

t₁t₂
Solution
or the ball thrown up
-h = ut₁ - (1/2)gt₁²
h = -ut₁ + (1/2)gt₁² --------(1)

For the ball thrown down
-h =- ut₂ - (1/2)gt₂²
h = ut₂ + (1/2)gt₂² --------------(2)

For the freely falling ball
-h = - (1/2)gt² -----------(3)

Multiply (1) by t₂ & (2) by t₁ and add
h(t₁ + t₂) = (1/2)gt₁t₂ ((t₁ + t₂)
Cancelling (t₁ + t₂) and substituting the value of h from (3)
(1/)gt² = (1/2)gt₁t₂
t =

t₁t₂6. A uniform chain of length L and weight W is hanging vertically from ends A & B which are close together. At a given instant the end B is released. Find the tension at A when B has fallen a distance x (x<L).
Solution
The left portion of the string is a variable mass system in which mass is entering.
By Merchersky's equation,
m(dv/dt) = F_ext + u_rel (dm/dt) = F_ext + u_rel (dm/dx)(dx/dt)
As the left portion of the chain is in equilibrium,
0 = {T-(m/L)[(L/2)+(x/2)]}j +(-

2gxj)(m/L)(

2gx)
T = (mg/L)[(L/2) + (5x/2)]
= (W/2)[1+(5x/L)]

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