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
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?
it depends on the speed with which it is thrown
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)]
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 ??
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
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
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
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
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)
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)
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
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
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
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
π/2 means it is pole so mg = 9 implies that m = 9/9.8
same will be the mass at centre of earth