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 

Asked By: VISHAL PHOGAT
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Joshi sir comment

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??

Asked By: SARIKA SHARMA
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Joshi sir comment

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

Asked By: DANI
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Solution by Joshi sir

since body starts from height h at rest and ends in the same height so it will again be in rest 

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

Asked By: SARIKA SHARMA
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Joshi sir comment

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.

Asked By: VISHAL PHOGAT
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Joshi sir comment

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.

Asked By: VISHAL PHOGAT
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Joshi sir comment

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

Asked By: VISHAL PHOGAT
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Joshi sir comment

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.

Asked By: SARIKA SHARMA
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Joshi sir comment

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

 

Physics >> Mechanics part 1 >> Kinematics Medical Exam

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

Asked By: SARIKA SHARMA
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Joshi sir comment

 

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°

Asked By: AVANTIKA PURI
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Joshi sir comment

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

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