Q 1 :    

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to                    [2024]

  • t

     

  • t2

     

  • t2/3

     

  • t3/2

     

(4)     

           dWdt=PdW=Pdt

            F·ds=Pdt

           mdvdsdt=Pdt

          0vvdv=Pm0tdtv22=Ptm

           Hence,  vtdsdt=Ct, C is a constant

           0sds=C0tt1/2dt

          s=C23t3/2st3/2

           displacement of proportional to (t)3/2

 



Q 2 :    

A body of mass 2 kg begins to move under the action of a time dependent force given by F=(6ti^+6t2j^)N. The power developed by the force at the time t is given by           [2024]

  • (6t4+9t5)W

     

  • (3t5+6t5)W

     

  • (9t5+6t3)W

     

  • (9t3+6t5)W

     

(4)     

          F=(6ti^+6t2j^)N

          F=ma=(6ti^+6t2j^)

          a=Fm=(3ti^+3t2j^)

          v=0tadt=3t22i^+t3j^

         P=F·v=(9t3+6t5)W

 



Q 3 :    

A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force F=(2i^+3j^)N, it moves to a new point Q having coordinates (6, 10) m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of :          [2025]

  • 13 : 6

     

  • 6 : 13

     

  • 1 : 2

     

  • 4 : 13

     

(2)

F=(2i^+3j^)N; displacement S=(3i^+6j^)m

Acceleration, a=Fm=(2i^+3j^4)m/s2

                              S=ut+12at2

 3i^+6j^=4u+12(2i^+3j^4)·16=4u+4i^+6j^

or u=14(3i^+6j^4i^6j^)  u=14i^

Average Power <P>=F·St

                                                =(2i^+3j^)·(3i^+6j^)4watt=6watt

                                  v at t=4sec = (12i^+34j^)×4=(2i^+3j^)

                                      Pins=F·v=(2i^+3j^)(2i^+3j^)=13 W

 <P>Pins=613

 



Q 4 :    

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., dmdtv. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?          [2025]

  • P2v3

     

  • Pv

     

  • Pv

     

  • P2v5

     

(4)

Power = F·v and F=dpdt=Rate of change of linear momentum

              F=v·dmdt=Kv3/2, K is constant

Power (P)=(Kv)3/2·(v)=Kv5/2

So, P2v5



Q 5 :    

An object of mass 1000 g experiences a time dependent force F=(2ti^+3t2j^)N. The power generated by the force at time t is:        [2025]

  • (2t3+3t3)W

     

  • (2t2+18t3)W

     

  • (3t3+5t5)W

     

  • (2t3+3t5)W

     

(4)

m = 1000 gram = 1 kg

F=(2ti^+3t2j^)

dvdt=Fm=(2ti^+3t2j^)

 0vdv=0t(2ti^+3t2j^)dt

 v=t2i^+t3j^

Power, P=F·v

P=(2ti^+3t2j^)·(t2i^+t3j^)

P=(2t3+3t5)W