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Good solution Magiclko.
Cauchy, you have two solutions: One with an arbitrary origin and one with origin at one of the given points.
Understand this, a vector result is independent of the origin unless the origin itself moves.
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On applying the given condition of R>>r, we get the flux to be zero, which is plausible, as the cone shrinks to make the solid angle enclosed zero.
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As per the question,  is a function of r. So we can't take it out of the integral and integrate.
Hence the only way is to differentiate it and get the result.
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I can't draw. However I will try to make the explanation clear. Taking the Y axis to be +ve up, and origin at the point of projection. Let the height of the tower br H.
v0 = u
v = -3u
a = -g
 y = (-H -0) = -H
v2 = v02 + 2a(y)
9u2 = u2 + 2(-g)(-H)
H = 4u2/g
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The block is attached to a spring fixed to the top of the inclined plane, right??
So it starts moving down. It stops and then comes back up. So the distance travelled by the block is greater than the final extension of the spring.
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Understand this first:
The plate is non conducting. Hence the filed lines won't be perpendicular to it.
With the point charge as the center construct a sphere of radius  R2+r2 so that the boundary of the plate lies on the sphere.
Total flux through this sphere is  = q/
Consider the curved potion of the sphere below the plate.
Its area is 2 (R2+r2)(1-cos )
where is the semi vertical angle of the cone formed by the plate at the point charge. cos = R/ R2+r2
Then the solid angle subtended by the plate at the point charge is
2(R2+r2)(1-cos)/R2+r2) = 2(1-cos)
Now, a solid angle of 4 contains a flux q/
Hence a solid angle of 2(1-cos) contains flux [(q/)/4]2(1-cos)
=(q/2)(1-cos)
=(q/2 )(1-R/ R2+r2)
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In that region of space consider a imaginary sphere of radius r. Let the volume charge density be .
Then,
= Surface Integral E.ends = Surface Integral Ar2er.ends
where en is the outward normal unit vector and er is the radial unit vector. On the surface of a sphere they are the same.
= ASurface Integralr2ds = ASurface Integralr2(2rsin rd) = Ar4 0pi sind
= 4Ar4
ds is taken as a ring on the surface of the sphere which subtends a cone of semi vetical angle and is of thickness rd.
By Gauss' Theorem
= (Volume integral  dV)/ = (0r 4r2dr)/
Hence
4Ar4 = (0r 4r2dr)/
Differentiating both sides
16Ar3 = 4r2/
= 4Ar
 r
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For solving the question I am using the result for electric filed due to a cylinder.
E =  R/r
You can find this by using Gauss' Theorem
The figure isn't very clear. I am assuming that there is a slit parallel to the axis of the cylinder of width dl.
In the slit, assume that there is filled +ve and -ve chrges of charge density  and - respectively.
Due to the +ve charge density the cylinder is now a whole one. Hence the system is that of a complete cylinder of charge density and a thin wire of charge density - & width dl.
I am assuming a general point, whose position vector from the cylinder axis isn't passing through the slit.
E = E+ + E- = (R/r)er - ( /2 r2-R2)(r/ r2-R2er+ R/ r2-R2 et)
where - is the linear charge density of the wire and er is the radial unit vector pointing from the axis of the cylinder outwards and et is the tangential unit vector.
Now, if H be the height of the cylinder, then,
total charge on the wire = -H = -(dl)H
= (dl)
Hence,
E = (( R/r) - ((dl)r/2 (r2-R2)))er+  R/2(r2-R2)et
=(2R(r2-R2) - (dlr2)/2r(r2-R2)er+ (dl)R/2 (r2-R2) et
Now, if the point is such that the position vector from the cylinder axis passes through the slit,
In the slit, assume that there is filled +ve and -ve chrges of charge density and - respectively.
Due to the +ve charge density the cylinder is now a whole one. Hence the system is that of a complete cylinder of charge density and a thin wire of charge density - & width dl.
E = E+ + E- = (R/r)er - /2rer
where - is the linear charge density of the wire and er is the radial unit vector pointing from the axis of the cylinder outwards.
Hence,
E = ((R/r) - (dl)/2r)er = (/2r)(2R - dl)e
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Assume 7 other similar cubes joined together with the given one so that the charge is at the centre of a large cube of side 2a. Now, each face of the new cube is equidistant from the charge and hence the flux through each of the faces of the large cube is same.
6 = q/
= q/6
Now the three outer faces of the original cube are 1/4 in size than the new cube's face. Hence flux through them
' = q/24
Along the other three faces, no electric lines of force cuts through them. Hence flux through the other three faces which meet at the point where the charge is placed is zero
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Here the object is virtual.
Let the mirror be facing towards left.
Then,
u = -10 cm
f = -20 cm
1/v + 1/u = 1/f
v = 20 cm
The image is in front of the mirror & real.
For the second case,
u = -30 cm
1/v + 1/u = 1/f
v = -60 cm
The image is behind the mirror and hence virtual
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If the packets are thrown at the moment when the trains are side by side, then both train move with the same speeds as before. Neither accelerates.
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Take the normal NSEW directions (N up, E right, etc). Assume XY axes such that+ve X is along east and +Y along north. Let the origin of the coordinate system be at a point on the south shore from where the man starts.
Speed of man wrt water, vMW = 10 m/min
Velocity of water wrt to earth, vWE = 5 i m/min
Let velocity of man wrt eath be vME.
Let the width of the river be d.
(a) Let the man try to swim in a direction making with the X axis.
vMW = 10(cos i + sin j)
vME = vMW + vWE = (10cos + 5) i + 10sin j
Time taken by the man to cross, T = d/sin
T is minimum if sin is maximum, ie, = /2
Hence the man should swim perpendicular to the current if he is to reach in shortest possible time.
(b) The shortets possible path is pependicular to the river. For the man to follow this path, vME must not have any velocity component along X axis.
10cos + 5 = 0
= 1200
Hence the man should swim in a direction making 1200 the +ve X direction
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Draw the diagram. Take any arbitrary point as the origin. Let the position vectors of A, B, C, D be a, b, c, d respectively.
Position vector of E, e = (a + c)/2
Position vector of F, f = (b + d)/2
AB + AD + CB + CD = (b - a) + (d - a) + (b - c) + (d - c) = 2(b+d-a-c)
= 4[(b+d)/2 - (a+c)/2] = 4(f - e) = 4EF
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Lets try it without LH rule
First transform as follows
(1-sinx)/sin2x = (1- sinx)/2sinxcosx =2(cosecx - 1)secx
x pi/2 2(cosecx - 1)secx = 2x(1 - 1)x0 = 0
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Pearson'd guide is a good book. Solve it thoroughly and you will definitely benefit.
But I would suggest MCQ by Bharati Bhawan. Its more oriented towards JEE. But if you can't get it, absolutely nothing wrong with Pearson.
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