Messages posted by yahiyafirdous

Let deceleration when going up is a1 and acceleration when coming downward a2

Let the block is thrown upward with velocity v, and distance travelled upward is s

.......................(1) when going upward.

.........................(2)

.................................................(3)

t2=2t1

Implies that

Put value of a2 and a1 we get

= 0.16

Let us apply priciple of conservation of energy.

...................(1) where m1 is mass of cylinder when radius is r/2 and Ic is moment of Inertia about an axis passing through point of contact.

m1= m/4

Now put m1 and Ic in equation (1) and find w

Then V=wr/2 you will get:

Ans

V=LdI/dt +IR

Solve above equation for I

where E is magnetic energy at any intant t

where Em is max energy and Im is max current

Hence E= Em/4

Engineering Mechanics by Bear n Jonson

Ha/Hb=a2/a1 where a1 decceleration of A and a2 is that of B

Since a2<a1

Hence Hb>Ha

Ans is C

Mass B will go higher than A

Force inversely proportion to the square root of time

where k is a constant and m is mass

The resultant velocity of end B is along a direction perpendicular to the rod. Since block is moving in horizontal direct with contstant velocity, hence

Miss Shivani

Its my wise advice to you to kindly first develop fundamentals. You must be able to solve such simple problems on your own.

At Earth surface:

At height h:

From above equation we find h=3R

Linear velocity V=wR

E1 =

Total KE= E2 = 0.5( ) where R is radius of ball(not radius of gayration)

E1/E2 / () =

Total Energy=

Since all forces are internal , hence apply conservation of linear momentum on center of mass.

Let mass of the whole system is m

m(0)= mv

v=0 where v is velocity of COM

.................................(1)

.....................................................................................(2)

W= is work done against friction force; W=JH where H heat generated; H=St" where S=sp heat, t" = temperature

.......................(3)

Now put the value of w, alpha, T in above equation and integrate, u will get the relationship between temperature

and time t

Frictional force due to elementary part shown in figure :

Torque due to this mass is given by:

V1=t1..................(1)

S1= (1/2) ....................(2)

S3= () /8...........................(3)

0= V1 -4t3

t3=(t1)/4 ...................................(4)

S2=V1(t-t1-t3)=t1(t-t1-t1/4)

...............(5)

Now S=S1+S2+S3

t=5t1/8 + 200/t1

Differentiate above equation wrt t1 and equate it to zero

You will find S2 = 0

Hence Required time is

t=t1+t3

Ans

Let side of cube is a. Since F is less than force of friction, the cube will not have translation motion, but it can be rotated about the edge shown in figure.

Let required distance is d; Taking moment of all forces about the edge shown:

Fa=mg d

d= a/3 Ans

At the heighest point man will feel weightless if weight of the man provide centripetal force to keep it in circular motion

mV^2/R=mg

V=Square Root of(Rg)=14.14 m/sec

Hence 15>V>14

Case(1)

Let velocity of man relative to earth is V1,

V1=V+u where V is velocity of buggy after both of them jump off simultanieously.

Applying conservation of momentum:

0=2m(V+u) + MV

V= - 2mu/(M+2m) Ans

Case(2)

When first man jumps off th buggy, velocity of buggy is V1

0=m(V1+u) + (M+m)V1 [ applying same logic as in case 1)

V1= - mu/(M+2m)

When second man jumps off, velocity of buggy becomes say V2

(m+M)V1=m(u+V2)+MV2

Solve for V2 and substitute the value of V1 you will get:

V2= - m(2M+3m)u/[(m+M)(2m+M) Ans

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