576 - Physics Questions Answers
a uniformrope of lenth 10 m and mass 12 kg hangs verticaly from a rigid support . a block of mass 4 kg is attached to the free end of the rope . atrans verse wave of wavelenth 00.03 m is prodused at the lower end . wavelenth of pulse when it reaches upper end will be?
at top point tension in the string = 16g
and at bottom point T = 4g
and v = √(T/m) here m is mass per unit length which is same throughout the rope.
first calculate v at both the points
frequency will be same throughout
so use the formula v = nλ twice
once at bottom and once at top point
then by diving these 2 formulae we will get the result
try it.
why is electric field of earth is in order of 100v/m any reason by formulae
emf generated between the extremes of equator line = vBl = RωBl
so E = RωBl/l = RωB = 2πRB/T
now calculate
please, write tibs about tidal effect and it's fourmula ? Is it depend inversly on the cube of the distance? and is tidal effect of the moon's pull grater than the tidal effect of Sun?
The tidal force is a secondary effect of the force of gravity and is responsible for the tides. It arises because the gravitational force per unit mass exerted on one body by a second body is not constant across its diameter, the side nearest to the second being more attracted by it than the side farther away. Stated differently, the tidal force is a differential force. Consider three things being pulled by the moon: the oceans nearest the moon, the solid earth, and the oceans farthest from the moon. The moon pulls on the solid earth, but it pulls harder on the near oceans, so they approach the moon more causing a high tide; and the moon pulls least of all on the far oceans (on the other side of the planet), so they stay behind more, causing another high tide at the same time. If we imagine looking at the Earth from space, we see that the whole Earth was pulled, but the near oceans more and the far oceans less; the far oceans stayed behind since they are pulled less (since they are farther away).
Tidal effects become particularly pronounced near small bodies of high mass, such as neutron stars or black holes, where they are responsible for the "spaghettification" of infalling matter. Tidal forces create the oceanic tide of Earth's oceans, where the attracting bodies are the Moon and, to a lesser extent, the Sun.
Formula
(axial) 
why the direction of electric dipole moment is from -ve to +ve
It is the direction of electric field due to a dipole at axial point.
a hollow sphere of mass M (in K.G.) radius R (in m.) is rotating with angular frequency omega(in radsec) . it suddenly stops rotating and 75% of the K.E. converted into heat energy . if S joulekg-kelvin is the spesific heat of material of the sphere, the rise of temperature of the sphere is ?
Rotational K. E. = 1/2 I ω2
so 0.75 [1/2 I ω2] = M S dT
and I = 2/3 M R2
now solve
2 comparable masses m1 and m2 r orbiting about their bodycenter o at distance r1 and r2 from their body center respectively with a common period T. the squar of time period ?
According to the diagram
Gm1m2/(r1+r2)2 = m1r1ω2 = m2r2ω2
compare first part with either second or third part of the given equation for getting ω2 then find T2
how can we calculate ratio of intensities of two sound waves having unequal amplitudes (one having double the amplitude of the other - both are sinusodial waves) and the graph is displacement vs time .
Intensity of a wave depends on two power of amplitude and two power of frequency. Check the graph carefully for getting idea about frequency and details of amplitude are given in the question.
The activity of a freshly prepared radioactive sample is 10^10 disintegrations per second, whose mean life is 10^9 sec.
The mass of an atom of this radioisotope is 10 ^ -25 kg. The mass (in mg) of the radioactive sample is
dN/dt = -λN
1010 = -[1/109] N λ = 1/mean life
so N = 1019 disintegration
so mass of radioisotope = 1019 *10-25 = 10-6 Kg = 1 mg
A transparent sheet of paper of refractive index 'n' and thickness 't' is pasted on one of the slits in YDSE which uses monochromatic light of wavelength 'x' . How many fringes will cross through the centre if the paper is removed ?
total path difference created by the paper = (n-1)t
geometrical path difference = yd/D
so position of central maxima will be obtained by eq. (n-1)t = yd/D
here y is position of central bright, d is slit appeture and D is distance between slits and screen.
so y = (n-1)tD/d
width of maxima = xD/d , x is wavelength of light
so no. of fringes shifted = [(n-1)tD/d]/[xD/d] = (n-1)t/x
we taken a equal length of wire like copper and maganim which one is more thicker ?
length are equal and metals are different
but none of the two is a deciding factor for the thickness of wire, so question is irrelevent