575 - Physics Questions Answers
two trains each 100m long ,oving in opposite directions ,cross each other in 8sec .if one is moving twice as fast than other ,then the speed of the faster train is?
Crossing means to cover the length of both the trains
so according to the given conditions 2v+v = (100+100)/8
or 3v = 25 or v = 25/3
so speed of faster train = 50/3 m/s
there is a small block of mass m kept at the left end of a larger block of mass M and length l.The system can slide on a horizontal road.The sysytem is started towards right with an initial velocity v. The friction coefficeient between the road and the bigger block is μ and that between the block is μ/2. Find the time elapsed before the smaller block separates from the bigger block?
Since there is no force for accelerating the larger block so larger block will be retarded due to this smaller block will move in forward direction
for the process equations are
0 + (µ/2)mg - µ(M+m)g = Ma (1)
(µ/2)mg - ma = ma1 (2)
take the value of a from eq (1) to eq (2) and get a1
then l = 0t + (1/2)a1 t2
a wedge of mass M' is connected with a hanging block of mass M with a light string and a massless pulley. and a small block of mass m is placed on the top of te wedge. find the mass M of the hanging block which will prevent the smaller block from slipping over the wedge . All the surfaces are smooth .
According to the given diagram Mg - T = Ma, T = (M'+m)a and macosθ = mgsinθ
solve these equations.
please solve q. no. 33 on page no. 82 of hc verma (part -1 newtons laws of motion)......
Dear Amit
this book is not available here so please send the question.
There is some amount of water in the beaker, and a vertical rod is passing through the centre of the circular base of the cylindrical beaker which is rotated with angular velocity ω , then water takes a particular shape(miniscus is formed). Find the equation of the miniscus of water when the mid point of the miniscus just touches the bottom surface of the beaker?
after making the diagram, complete the miniscus as a sphere, then following conditions will be obtained
1) h = R(1-cosθ) here h, R and θ are height of miniscus, radius of the sphere and contact angle.
2) R = r/cosθ, r is the radius of beaker
3) tanθ = g/rw2
4) πr2hρ - πR3ρcos2θ = 2πrTcosθ
There are 4 equations, remove R, θ and T and get a relation between h and r in terms of g, w, ρ and π
what is meant by rate of change of transverse magnification
transverse magnification m = v/u so rate of change of trasverse magnification is dm/dt. calculate by differentiation
what is the main use of capacitors
Capacitor :- A capacitor is a device to store energy
Focsl length of lens depends on colour of light?
Yes, f depends on µ and µ depends on colour of light
the ratio of intensities between two coherent sound sources is 4:1. the difference of loudness in decibel between maximum and minimum intensities, when they interfere in space is
(1) 10 log2
(2) 20 log3
(3) 10 log3
(4) 20 log2
I1/I2 = 4/1 so a1/a2 = 2/1 so amax/ amin = 2+1/2-1 = 3/1 so Imax/ Imin= 9/1
now loudness L = 10 log(I/I0)
so difference of loudness = L1-L2 = 10 log (9I/I) = 10 log9 = 20 log3
two coherent light sources A and B with seperation 2λ are placed the x-axis symmetrically about the origin. they emit light of wavelength λ. obtain the position of maximas on a circle of large radius R lying in the xy plane and with the center at origin
for maxima path diffrence should be integral multiple of λ.
It is given that two coherent light sources A and B with seperation 2λ are placed on the x-axis symmetrically about the origin, and we have to get the points of maxima in the circle so at the point where circle cuts the X axis (path diffrence = 2λ) and where circle cuts the Y axis (path difference = 0) will be the point of maxima. Besides it one point will also be present in the circle in first quadrant where path difference will be λ so it will be a maxima also. it means total 8 points are there