575 - Physics Questions Answers

Mam/Sir,  Please Help Me to Understand The Symmetry method And Star-Delta Method To find Eqivalent Resistance?

 

 

Asked By: GAUTAM SAGAR
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Joshi sir comment

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) 

Two Particles move in a uniform gravitational field  with an acceleration g. At The Initial  moment the particles were located at one point and moved with velocities v1= 3.0ms-1 and v2 = 4.0ms-1 horizontally in opp. Direction.Find The distance between the particles at the moment when their velocity vector become mutually Perpendicular?

Asked By: GAUTAM SAGAR
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Joshi sir comment

see you tube video of iitbrain by the name I E Irodov problem 1.11

 

A Train Accelerates from rest at a constant rate α for sometime and then it retards to rest at a constant rate β. If te total distance covered by the particle is x, then what is the maximum velocity of the train ?

Asked By: GAUTAM SAGAR
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Joshi sir comment

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

 

Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.

(1) Ma/ 12

(2) 2/3 x Ma2

Asked By: SWATI KAPOOR
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Joshi sir comment

let the moment of inertia about a diagonal = I

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

now calculate I

Two sphere are rotating about their own axis if axis of rotation of these axis are perpendicular to each other than angular acceleration of one with respect to other
Asked By: MANISH KUMAR SAHU
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Joshi sir comment

let the angular velocities are ω1 ω2

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

A taxi leaves the station X for Station Y every 10 min. Simulataneously, a taxi also leaves the station Y for station X Every 10min. The Taxis move at the same constant speed and go from X to Y or Viceversa in 2Hours. How Many taxis coming from the other side meet each taxi enroute from Y To X?

Asked By: GAUTAM SAGAR
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Joshi sir comment

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 

A train accelerates from rest at a constant rate α for distance x1 and time t1.After that it retards to rest at constant rate β for distamce x2 and time t2.Which of the following relation is correct?

A. x1/x2=α/β=t2/t1 B.   x1/x2=β/α=t2/t1
C. x1/x2=α/β=t1/t2 D.  x1/x2=β/α=t1/t2

 

Asked By: GAUTAM SAGAR
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Joshi sir comment

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

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

on comparing v we get  αt= βt2

so β/α = t1/t2   (1)

and x1/x2αt12/βt2= β/α   (2)        by using (1)

A ball of mass 1 kg is projected with a velocity of 20√2 m/s from the origin of an xy coordinates axis system at an angle 45° with x-axis (horizontal). The angular momentum of the ball about the point of projection after 2s of projection is [take g=10 m/s2] (y-axis is taken as vertical)

Asked By: SWATI KAPOOR
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Joshi sir comment

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.

= mvxy - mvyx = 1*20*20 - 1*0*40 = 400

A disc of mass 3kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

Asked By: SWATI KAPOOR
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Joshi sir comment

mgh = 1/2 mv2 + 1/2 Iω2

I for the disc about centre = 1/2 MR

and for pure rolling v = Rω

calculate v then 1/2 mv2

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion

Asked By: SWATI KAPOOR
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Joshi sir comment

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 

now put t in (1) for getting v

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