575 - Physics Questions Answers
The acceleration (a) of a body moving along x-axis is given by a=4x3 ,where a is in m/s2 and x is in metre.if at x=0 the velocity of body is 2m/s,then find its velocity at x=4m?
a = 4x3
so dv/dt = 4x3
or dv/dx * dx/dt = 4x3
or dv/dx * v = 4x3
or vdv = 4x3dx
or ∫vdv = 4∫x3dx
or v2/2 = 4x4/4 + C
at x = 0, v = 2
so 4/2 = 0+C
so C = 2
on putting the value of C
we get v2/2 = x4+ C
now put x = 4 and get v
The position x of a particle moving along x-axis at time t is given by equation t=√x +4 , where x is in meter and t is in seconds. Find the position of particle when its velocity is zero.
given that x = (t-4)2
so dx/dt = 2(t-4)
or v = 2t-8
now for v=0, t=4
put t=4 and get x
A body projected upwards is at same height form ground at t=3s and t=7s . Find the maximum height attained by it. ( take g= 10 m/s^2 ).
A body projected upwards is at same height form ground at t=3s and t=7s. It will gain max. height after 5 second
so maximum height = gt2/2 = 10*5*5/2 = 125
another direct method for finding total distance = 5+15+25+35+45 = 125
these are the distances in 1st, 2nd, 3rd and etc sec.
What would be the Potential energy of two points charges q &-q which r separated by a dist d ( IF POTENTIAL AT MID PT IS TAKEN TO BE ZERO)
potential energy = -kqq/d
a particle , with an inital velocity v0 in a plane , is subjected to a constant acceleration in the same plane . then in general ,
the path of the particle would be
1) circle
2) ellipse
3) parabola
4) hyperbola
parabola
for understanding this make components of v parallel and perpendicular to a
a train is moving with a velocity of 30m/s. When brakes are applied , it is found that the velocity reduces to 10m/s in 240m.
When the velocity of the train becomes zero , the total distance travelled is??
270m
by v2 = u2-2as
100 = 900-2*a*240
or 800 = 480a
or a = 80/48 = 10/6 = 5/3
so total distance upto rest is u2/2a = 900*3/2*5 = 270m
A CARNOT ENGINE WORKING B/W 270 C & --1230C HAS EFFICIENCY η . IF TEMP OF SINK IS DECREASED BY 50K THE EFFICIENCY BECOMES??
ANS 4η/3
according to the given condition η = 1-(150/300) = 0.5
now after decreasing temperature of sink by 50K
we get the new efficiency = 1-(100/300) = 2/3 = 4*0.5/3 = 4η/3
in a thermodynamic process on 2 moles of a monoatomic gas work done on gas is 100J & change in temp is 30 C . The change in internal energy of the gas in this process is??
change in internal energy = nCvdT = 2*[R/(γ-1)]*30 = 60*8.31/[(5/3)-1]
while measuring the speed of sound by performing a resonance column expt , a student gets d first resonance conditn at a column length of 18 cm during winter . Repeating d same expt during summer , she measure d column length to be x cm for the second resonance. THEN
1)18>X
2)X>54
3) 54>X>36
4) 36>X>18
ANS (2)
AIEEE 2008
since l α λ α v/n n is frequency
on increasing temperature, v increases so λ increases
for first resonance l = λ/4
and for third resonance L = 3λ'/4
and λ' > λ so x will be more than 54
angle b/w vectors (i^+j^)and(j^+k^) is
1) 60o
2) 90o
3) 300
4) 450
use the formula A.B = |A||B|cosθ