576 - Physics Questions Answers
a small block of mass m is kept on a rough inclined surface of inclination o * fixed in an elevator the elevator goes up with a uniform velocity v &block doesn't slide on the wedge the w.d by the force of fricition on block in t time is?
ANS) mgtsin2o*
work done by the force of fricition on block in t time = fvtcos(90-θ) = mgsinθvtsinθ 
TWO EQUAL MASSES ARE ATTACHED TO TWO ENDS OF A SPRING CONST. k THE MASSES ARE PULLED OUT SYMMETRICALLY TO STRECH SPRING BY A LENTH x OVER ITS NATURAL LENTH. WORK DONE BY THE SPRING ON EACH MASS IS -1/4kx2. HOW
total work done on the spring = 1/2kx2
so work done by the spring = -1/2kx2
this work is done on 2 masses so work done by the spring on 1 mass = -1/4kx2
A SYSTEM OF m1 &m2 THE PARTICLE m1 IS PUSHED TOWARDS THE CM OF PARTICLES THRU DIST. d BY WHAT DIST. WOULD m2 MOVE SO AS TO KEEP MASS CENTRE OF PARTICLES AT THE ORIGINAL POSITION
SIR, I WANT TO KNOW Y ans is m1d/m2 WHY NOT m1d/ m1+m2?
with centre of mass as reference point m1d = m2x so x can be calculated.
IF MORE AIR IS PUSHED IN A SOAP BUBBLE THE PRESSURE IN IT WILL INCREASE OR DECREASE . PLS GIVE REASON ALSO
since inside soap bubble pressure is given by the formula P0 + 4T/R and on pushing air R will increase so pressure will decrease.
Q Q1=1.5C , R1=2M, Q2 =3C&R2=1M WHEN THESE SPHERICAL CAPACITER TOUCHED EACH OTHER THEN CHARGE WILL MOVE OR NOT . IF MOVE THEN HOW MUCH
(MANIPAL 2012)
on touching, the new ratio of charges = x/y = 2/1 this is by the formula q1/q2 = C1/C2 = R1/R2
and x+y = 1.5+3 = 4.5
or 2y+y = 4.5
or y = 1.5
so 1.5 C will move from second to first.
Q IF A BODY IS PROJECTED UPWARD WITH SPEED 1/2 OF ESCAPE SPEED OF EARTH THEN MAXn HEIGHT ATTAINED BY IT IS=?
ANS=R/3 WHERE R=RADIUS OF EARTH
escape velocity = √(2GM/R) so 1/2 of escape = 1/2√(2GM/R)
so according to given condition on applying energy conservation we will get
energy at start = energy at the heighest point
or -GMm/R + mv2/2 = -GMm/R1 + 0
or - GMm/R + 2GMm/8R = -GMm/R1 + 0
solve now for finding R1 then find R1-R
IN QUESTn < B/W 2 NON ZERO VECTOR A &B 120* RESULTANT GIVEN IS GREATER WHILE ACC. TO IT SHD BE LESS SO I REQUEST YOU SIR TO AGAIN CHECK . QUES FROM HC VERMA
i am confirmed, option B will be correct answer
Q A UNIFORM SPHERICAL SHELL GRADUALLY SHRINKS MAINTAING ITS SHAPE. THE GRAVITATn POTENTIAL AT THE CENTRE WILL DECREASE. Y DOESN'T IT SHD REMAIN CONST. =?
Formula for gravitational potential at centre is -3/2 GM/R so if R will decrease V will increse numerically but - is present so decreases
Q IF IN THE SAME QUESTn OF BAROMETER THE ELEVATOR IS ACCELERATING UPWARD READS 76cm. THEN AIR PRESSURE IN ELEVATOR IS >76cm PLS EXPLAIN THE CONCEPT BEHIND SUCH CASES.
when elevator will move in upward direction the extra Hg level will feel a down force and given that in movable condition it is 76 cm. Hg so really it will be more than 76 cm.
Q A CURVE IS REPRESENTED BY y=sinx IFx IS CHANGED FROM PIE/3 TO PIE/3+PIE/100, FIND APPROX. CHANGE IN x?
Q A CAR IS GOING DUE NORTH AT A SPEED =50km/h IT MAKES A 90* LEFT TURN WITHOUT CHANGING SPEED THE CHANGE IN VEL. OF CAR IS=? (ANS 70km/h towards south west ) i want to know that here we are just changing the directn so only directn. shd. b change but here magnitude is also changing HOW?
FLUIDS
Q A BAROMETER KEPT IN N ELEVATOR READS 76cm WHEN IT IS AT REST . IF ELEVATOR GOES UP WITH INCREASING SPEED READING WILL BE <76 cm . IS REASON IS THAT PRESSURE IS DIRECTLY PROPTn TO HEIGHT ?
I think we have to caculate change in y
since y = sin x
so dy = cosx dx = cosπ/3* π/100 = π/200
the initial velocity of car = 50j and after turn velocity = -50i here i and j are the unit vectors along east and north direction
so change in velocity = -50i - 50j
its mod = 50 √2 = 70
Because in barometer the extra Hg level will experience a downward force so Hg level will decrease.