JEE physics

Sponsor Area

Question
CBSEENPH11020269

From the tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is

  • 2gH = n2u2

  • gH = (n-2)2u2

  • 2gH = nu2(n-2)2u2

  • gH = (n-2)2u2

Solution

C.

2gH = nu2(n-2)2u2

Time is taken to reach the maximum height t1 = u/g
If t2 is the time taken to hit the ground,
i.e, negative straight H space equals space ut subscript 2 minus 1 half gt subscript 2 superscript 2
straight t subscript 2 space equals space nt subscript 1
minus straight H space equals space straight u nu over straight g minus 1 half straight g fraction numerator straight n squared straight u squared over denominator straight g squared end fraction
minus straight H space equals space nu squared over straight g minus 1 half fraction numerator straight n squared straight u squared over denominator straight g end fraction
2 gH space equals nu squared left parenthesis straight n minus 2 right parenthesis

Sponsor Area

Question
CBSEENPH11020270

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

  • 2g/3

  • g/2

  • 5g/6

  • g

Solution

B.

g/2

For the mass m,
mg-T = ma

As we know, a = Rα    ... (i)
So, mg-T = mRα
Torque about centre of pully
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get 
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

Question
CBSEENPH12039449

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is 

  • 1 over 6 straight m
  • 2 over 3 straight m
  • 1 third straight m
  • 1 half straight m

Solution

A.

1 over 6 straight m

A block of mass m is placed on a surface with vertical cross section, then

tan space straight theta space equals space dy over dx space fraction numerator straight d open parentheses begin display style straight x cubed over 6 end style close parentheses over denominator dx end fraction space equals space straight x squared over 2
At limiting equilibrium, we get
μ = tan θ
0.5 = x2/2
⇒ x2 =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1
straight y space equals space fraction numerator left parenthesis 1 right parenthesis cubed over denominator 6 end fraction space equals 1 over 6
When x =-1
straight y space equals space fraction numerator left parenthesis negative 1 right parenthesis cubed over denominator 6 end fraction space equals fraction numerator negative 1 over denominator 6 end fraction
So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.

Question
CBSEENPH11020271

When a rubber band is strecthed by a distance x, it exerts a restoring force of magnitude F = ax +bx2, where a and b are constants. The work done in stretching are unstretched rubber-band by L is

  • aL2 +bL2

  • 1 half left parenthesis aL squared plus bL cubed right parenthesis
  • aL squared over 2 space plus bL cubed over 3
  • 1 half space open parentheses aL squared over 2 plus bL cubed over 3 close parentheses

Solution

C.

aL squared over 2 space plus bL cubed over 3 straight U subscript straight f minus straight U subscript straight i space equals space minus straight W space equals space minus space integral subscript straight i superscript straight f straight F. dr
Given, F = ax +bx2
We know that work done in stretching the rubber band by L is 
|dW|= |Fdx|
vertical line straight W vertical line space equals space integral subscript 0 superscript straight L left parenthesis ax space plus bx squared right parenthesis dx
space equals space open square brackets ax squared over 2 close square brackets subscript straight O superscript straight L space plus space open square brackets bx cubed over 3 close square brackets subscript straight O superscript straight L
equals open square brackets aL squared over 2 minus fraction numerator ax space left parenthesis 0 right parenthesis squared over denominator 2 end fraction close square brackets space plus space open square brackets fraction numerator straight b space straight x space straight L cubed over denominator 3 end fraction minus fraction numerator straight b space straight x space left parenthesis 0 right parenthesis cubed over denominator 3 end fraction close square brackets
vertical line straight W vertical line space equals space aL squared over 2 plus bL cubed over 3

Question
CBSEENPH11020272

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension

  • angular momentum is conserved

  • angular momentum changes in magnitude but not in the direction

  • angular momentum changes in direction but not in magnitude

  • angular momentum changes both in direction and magnitude

Solution

C.

angular momentum changes in direction but not in magnitude