572 - Physics Questions Answers
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
in the above ques, last time u found out potential at center but according to ques we hav eto find out on the surface and ans. is 3
if the answer is 3 V then 2r should be the distance of q from surface not from centre.
first confirm
In the ques previously submitted by someone which was as follow
the electric field vector is given by E= a(x)1/2 i^ . find the φ thru a cube bounded by surface x=l, x=2l, y=0, y=l , z=l, z=0.
Sir , my query is that in the solutn submitted the leaving flux at x=2l is given as( 2l)1/2 X l2 . but as the electric field is itself along x then why 2l will be used as along x axis it will be zero because cosθ used is perpendicular to surface with electric field axis thus cos90 =0??
θ should be measured between E and perpendicular drawn on the surface
a small body starts sliding from a height h down an inclined groove passing into a half circle of radius h/2. find the speed of the bodywhen it reaches the highest point of its trajectory
since body starts from height h at rest and ends in the same height so it will again be in rest
A BLOCK OF WOOD IS KEPT ON THE FLOOR OF A STATIONARY ELEVATOR. D ELEVATOR BEGINS TO DESCEND WITH AN ACCLN 12m/s2 . IF g =10 , THEN DISPLACEMENT OF BLOCK W.R.T. GROUND DURING FIRST 0.2 SEC. AFTER D START WILL BE ??
ANS 0.2m
s = ut + 1/2 at2
so s = 0 + 1/2 10(0.2)2
so s = 5(0.04) = 0.2
since elevator is moving with an acc. greater than g so we should take it a free fall
4 particles each of mass m and charge q are held at the corners of a square of side a . they are released at t=0 and move under mutual repulsice force. find the speed of any particle when its distance from the center is double.
potential energy in initial condition = 4*k*q*q/a + 2*k*q*q/a√2
potential energy in final condition = 4*k*q*q/2a + 2*k*q*q/2a√2
and kinetic energy in the final condition = 4*mv2/2
now use energy conservation
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
potential at the centre due to q = kq/2r = 9 V
potential at centre due to surface charge = 0 V
total potential = 9 V
now at surface potential due to q = kq/r = 18 V
but it should be 9 V totally so due to surface induction, potential will be -9V
the electric field vector is given by E=a(x)½ iˆ.find the flux through a cube bounded by surfaces x=l ,x=2l, y=0 ,y=l, z=0, z=l
entering flux at x=l will be a(l)1/2 * l2
leaving flux at x=2l will be a(2l)1/2 * l2
so net flux will be the difference of these 2 flux
A body of mass 32kg is suspended by a spring balance from the roof of a vertically operating lift going downward from rest . At that instant the lift covered 20m and 50m , the spring balance showed 30 kg & 36 kg respt. the vel. is increasing at 20m & decreasing at 50m. Plzz sir explain this.
for downward journey
first case
32g-30g = 2g = 32a so a = g/16
second case
32g-36g = -4g = 32a so a = -g/8
so it is acceleration upto 20 m and retardation after that
d benches of a gallery in a cricket stadium r 1m wide & 1m high . a boy strikes d ball at a level of 1m above d ground & hits a sixer. d ball starts at 35m/s at an < of 530 with d horizontal . d benches r perpendicular to plane of motion & d first bench is 110m from d boy. ON WHICH BENCH WILL THE BALL HIT??
ANS 6TH
we know that y = xtanθ - gx2/2u2cos2θ
according to the given conditions
x = 110+n
y = n here n is the number of bench in which ball strikes
θ, u and g are given
A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P and R is about
1. 143° 2. 127° 3. 120°
since initially bomb is in rest so initially momentum of the bomb = 0i+0j+0k
after explosion total momentum = 30mi+40mj+mv here v is a vector
now by momentum conservation 0 = 30mi+40mj+mv so v = -30i-40j
so vector along P is i and along R is -3i-4j
so by A.B = |A||B|cosθ
we get i.[-3i-4j] = 1*5*cosθ
so -3/5 = cosθ so θ = 127