302 - Mechanics part 1 Questions Answers
two cars of masses m &M are moving in circles of radii r & R their speeds are such that they make complete circle in same time t. ratio of centripetal acceleration
1) m:M
2)R:r
3)1:1
4) MR:mr
aieee 2012 is the ans is (4)
since both make complete circle in same time t so angular velocities of the two will be same
now use F = mrω2
a cyclist is moving in a circle of radius 20m with velocity 5ms-1mass =60kg of both boy & cycle find static friction on tyre
v2/rg = µ
so µ = 5*5/20*10 = 1/8
Fs = µ * N = 1/8 * 60*10 = 600/8 = 75 N
µ is the coefficent of static friction, Fs is static friction and N is normal reaction
This answer is given by SHUBS and is correct answer
CENTRIPETAL & CENTRIFUGAL FORCES ALWAYS ACT IN PAIR. IS D STATEMENT IS CORRECT IF YES THEN Y &IF NOT THEN ALSO GIVE REASON
if you are observer in the ground then centripetal will work but if observer is in the rotating frame then centrifugal force will work. These two force does not form a pair
IN THAT QUES OF CENTRE OF MASS BOTH THE OPTION ARE GIVEN THEN IN EXAM HOW WE KNOW THAT WHICH ONE IS CORRECT. AS WE USE Vm /M+m =V OF PLANK & m= mass of man
I could not understand, what the question is,
THE MAXm TENn IN STRING OF A SIMPLE PENDULUM IS 1.2TIMES THE MINm TENSION. IF a IS ANGULAR AMPLITUDE THEN a WILL BE
A)cos-1(4/5)
B)COS-1(3/4)
C)cos-1(15/16)
d) cos-1(7/8)
max tension at bottom point = mg+mv2/r
min tension at extreme point = mgcosθ
according to the given condition mg+mv2/r = 1.2mgcosθ (1)
and by energy conservation 1/2mv2 = mg(r-rcosθ) (2)
solve now answer will be C
IN THE PREVIOUS QUESn THAT WAS BALL IS DROPPED FROM HEIGHT 5M ACC. DUE TO GRAVITY IS NOT KNOWN & WE HAVE TO FIND = IT LOSES ITS VEL. BY WHICH FACTOR WHEN IT RISES TO 1.8M
IN THE SOLn SUBMITTED BY SOMEONE HE HAS TAKEN ACC. WHILE FINDING VEL EQN= =V2-U2=2AS AS CONST. SO I WANT TO KNOW WHY IN THIS QUES ACC. IS TAKEN CONST. WHILE THERE IS NO MENTn IN QUES
answer by Nikhil is correct
consider once agian
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.
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