Q 1 :    

A body of m kg slides from rest along the curve of a vertical circle from point A to B in a frictionless path. The velocity of the body at B is ______.

(given, R=14 m,  g=10 m/s2 and 2=1.4 )       [2024]

  • 21.9 m/s

     

  • 10.6 m/s

     

  • 19.8 m/s

     

  • 16.7 m/s

     

(1)

By conservation of mechanical energy, decrease in P.E. = increase in K.E.

mg(R+R2)=12mvB2-02gR(1+12)=vB2

vB2=2×10×14×(1+11.4)

vB=20×24=43021.9 m/s



Q 2 :    

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (θ) of thread deflection in the extreme position will be ______.                   [2024]

  • tan-1(2)

     

  • 2tan-1(12)

     

  • tan-1(12)

     

  • 2tan-1(15)

     

(2)

Loss in kinetic energy = Gain in potential energy

12mv2=mg(1-cosθ)

v2=2g(1-cosθ)

Acceleration at lowest point =v2

Acceleration at extreme point =gsinθ

 v2=gsinθ

 sinθ=2(1-cosθ)

tanθ2=12θ=2tan-1(12)



Q 3 :    

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m, making 10 rpm. The tension in the string when the stone is at the lowest point is (if π2 = 9.8 and g = 9.8 m/s2)               [2024]

  • 17.8 N

     

  • 8.82 N

     

  • 97 N

     

  • 9.8 N

     

(4)

Given that, m=900 gm=9001000 kg=910 kg

r=1 m

ω=2πN60=2π(10)60=π3 rad/sec

T-mg=mrω2    T=mg+mrω2

T=910×9.8+910×1(π3)2

           =8.82+910×π29=9.80 N



Q 4 :    

A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes a half-circle, reaching the topmost position B. The ratio of kinetic energies (K.E)A(K.E)B is ______.            [2024]

  • 3 : 2

     

  • 5 : 1

     

  • 2 : 5 

     

  • 1 : 5

     

(2)

Apply energy conservation

12mVL2=12mVH2+mg(2L)

 VL=5gL

So, VH=gL

(K.E)A(K.E)B=12m(5gL)212m(gL)2=51