575 - Physics Questions Answers

chock coil is used to

1) increase current

2) increase voltage

3) decrease current

4) decrease voltage

here we know power= IV  & chock coil is used to reduce the power loss , so here what vyl be the answer

Asked By: SARIKA SHARMA
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Joshi sir comment

chock coil in ac is the device equivalent to rheostat in dc, so it is used to increase current.

If a particle (strong enough to survive in space) is released in space from earth. Where it will go? What about its motion? Will it revolve around the sun or remain stationary?

Asked By: PRINCE KUMAR
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Joshi sir comment

it depends on the speed with which it is thrown

smiley a large open tank has two holes in the wall. one is a square hole of side L at a depth Y from the top and the other is a circular hole of radius R At a depth 4Y  from the top . when the tank is completely filled with H2O , the quantities of H2O flowing / sec from both holes are same . then R IS equal to??

ans L / [√(2π)]

Asked By: SARIKA SHARMA
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Joshi sir comment

speed of water from any hole = √2gh

so volume of water per sec. = A√2gh

for square hole volume/sec = L2√2gY

for circular hole volume/sec = πR2√2g4Y

compare and get answer

when a tuning fork vibrates with 1.0 m  or  1.05m long wire of a sonometer, 5 beats per second are produced in each case. what will be the frequency of the  tuning fork ??

Asked By: SARIKA SHARMA
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Joshi sir comment

for sonometer wire

l = λ/2

or 2l = v/n                                    here n represents frequency

or n = v/2l

for first wire of 1 m length n= v/2

for second wire of 1.05 m length n=v/2.1

let frequency of tuning fork = n1

so according to the given condition v/2 = n1+ 5

and v/2.1 = n1 - 5

solve now

 

Physics >> Optics >> Refraction Medical Exam

if ω1 &ω2 are dispersive powers of lenses of focal length f1 & f2 respectively . then the condition of chromatic aberration for two lenses in contact is

1) ω1/ω2 = f2/f1

2) ω1/ω2 = f1/f2

3) ω1/ω2= - f1/f2

4) ω2/ω1= f1/f2

Asked By: HIMANSHU
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Joshi sir comment

3 is the true condition and can be proved

the half life of radium is 1620 year & its atomic weight is 226gm/mole. the number of atoms that will decay from its 1g sample per second is

1) 3.16 x 1010

2)3.6 x 1012

3) 3.1 x 1015

4) 31.1 x 1015

Asked By: HIMANSHU
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Joshi sir comment

for radioactivity we know that it is a first order process

so dN/dt = -λN

so dN/1 = -[0.6932/(1620*3.1*107)]*(1/226)*6.023*1023

solve now

smiley An open knife edge of mass m is dropped from a height h on a wooden floor. if the blade penetrates a dist s into the wood , the avg. resistance offered by wood to blade.  ANS mg(1+h/s)

Asked By: SARIKA
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Joshi sir comment

velocity at the time of strike of the knife at the wood = √(2gh)

now equation for penetration on the wood is R-mg = ma            here R is reaction by the wood

so R = m(g+a) 

here it is given that s is the distance penetrated by the knife on the wood so 02 = (√2gh) - 2as so a = 2gh/2s = gh/s

on taking this value of a 

we get R = m(g+gh/s) = mg(1+h/s)

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A at road of 200km & B at 100km road. A with vel 20m/s & B with 7.5m/s . Then wat will b d time when B overtake A.

1)  24s

2) 40s

3) 80s

4)nothing can be predicted

Asked By: SARIKA SHARMA
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Joshi sir comment

data are irrelevent

A LIFT IS MOVING  UPWARD WITH ACC 2m/s2 .WHEN IT GAIN THE VELOCITY 4m/s THEN AT THAT MOMENT A BALL IS PROJECTED BY 300 (wrt floor to lift) WITH VEL. 4m/s RELATIVE TO LIFT . FIND HOW MUCH TIME WILL IT TAKE TO 1) TO REACH GROUND &   2) TO THE FLOOR OF LIFT AGAIN

Asked By: SARIKA SHARMA
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Joshi sir comment

 

since velocity of the ball is 4 m/s relative to lift so 

time taken by the ball to reach the floor of lift T = 2usinθ/g = (2*4*1/2)/10 = 0.4 sec

height of the lift at the time of dropping the ball = (v2-u2)/2g = 16/20 = 8/10 = 0.8 m = h 

and vertical component of the velocity of ball with respect to the ground = 6 m/s = u

so use -h = ut - gt2/2

 

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

A STONE WEIGHS 9N AT A PLACE ON THE SURFACE OF EARTH HAVING LATITUDE π/2 . ITS MASS AT CENTRE OF EARTH WILL BE (kg)

 

1) 10/ 37√2

2 ) 10√2

3) 5

4) 15

Asked By: SARIKA SHARMA
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Joshi sir comment

π/2 means it is pole so mg = 9 implies that m = 9/9.8

same will be the mass at centre of earth

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