575 - Physics Questions Answers

fig 1 and fig 2 are explaining symmetry method. 

divide the circuit in two parts by a line symmetry, then calculate eq. resistance between A and O by series parellel rules then for getting eq. resisitance between A and B double the answer between A and O

Delta star is the method to convert the circuit into an equivalent circuit in which series parallel rules can be applied 

By this method a set of three resistances R1, R2 and R3 can be converted to 3 new resistances 1, 2 and 3

where 1 = R1R2/(R1+R2+R3), 2 = R1R3/(R1+R2+R3) and 3 = R2R3/(R1+R2+R3) 

for first part of motion  

v² = 2αx₁   (1)

similarly for second part of motion 

v²= 2βx₂     (2)

now by given conditions 

x = x₁ + x₂

so x = v²/2α + v²/2β

so x = v²/2[1/α + 1/β]

now solve

let the moment of inertia about a diagonal = I

then by perpendicular axis theorem 2I = 1/6 Ma²

now calculate I

let the angular velocities are ω₁,  ω₂

then angular acceleration of one w. r. t. other =  ω₁​ ω₂​

the answer depends on the condition that where taxi meets with the first taxi coming from opposite side.

If we consider that at the starting point taxi meets with the first taxi coming from other side, then after every 5 min it will meet with a taxi from opposite side. This will be due to the double relative speed of two taxi moving opposite. Total time length of the path = 2 hrs = 120 min

so no. of taxi = 120/5 + 1 = 25

if first taxi meet at the point a little ahead the starting point then no of taxi = 120/5 = 24

similarly other cases 

for first part x₁ = αt₁²/2 and v = αt₁

similarly for second part x₂ = βt₂²/2 and v = βt₂

on comparing v we get  αt₁ = βt₂

so β/α = t₁/t₂   (1)

and x₁/x₂ = αt₁²/βt₂² = β/α   (2)        by using (1)

initial vertical and horizontal velocities are 20 m/s and 20 m/s

after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s

and y and x displacements are 20 m and 40 m

so about point of projection angular momentum after 2 sec.

= mv_xy - mv_yx = 1*20*20 - 1*0*40 = 400

mgh = 1/2 mv² + 1/2 Iω²

I for the disc about centre = 1/2 MR² 

and for pure rolling v = Rω

calculate v then 1/2 mv²

for translational motion 

0-μMg = Ma so a = -μg

so after time t velocity v = u-μgt  (1)

for rotational motion 

about centre, μMgr = Iα = 2Mr²α/5

so α = 5μg/2r

so ω = 0 + αt

so ω = 5μgt/2r  (2)

for pure rolling

v = rω implies that u-μgt = r5μgt/2r

implies that u = 7μgt/2

so t = 2u/7μg