A body comes and sits suddenly on a circular rotating table the quantity which conserved is
A. angular velocity
B. angular momentum
C. linear momentum
D. angular acceleration
Correct Answer - A Torque, `tau =(dL)/(dt)` where, L = angular momentum If `tau =0`, then `(d L)/(dt)=0` i.e. L = constant
2 Answers 5 viewsCorrect Answer - C `r_(2) onega_(1)^(2)` `r_(2)=(r_(1) omega_(1)^(2))/(omega_(2)^(2))=(4r_(1) omega_(1)^(2))/(4 omega_(1)^(2)` `r_(2)=r`
2 Answers 1 viewsCorrect Answer - D `r_(2) omega_(2)^(2)=r_(1) omega _(1)^(2)` `r2=r_(1)omega_(1)^(2)/(omega_(2)^(2))=(9xxomega_(1)^(2))/(9 omega_(1)^(2))=1 cm` `r2=1cm`.
2 Answers 1 viewsCorrect Answer - A `(1)/(2)mv^(2)=(1)/(2)I omega^(2)`
2 Answers 1 viewsCorrect Answer - C `I_(1)omega_(1)=I_(2)omega_(2)` `I_(1)omega_(1)=2I_(1)omega_(2)` `therefore omega_(2)=(500)/(2)=250` rpm
2 Answers 4 viewsCorrect Answer - B `KE_("roll")=KE_("rot")` `(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))=(1)/(2)m omega^(2)` `mv^(2)(1+(k^(2))/(R_(2)))=(mR^(2))/(4)omega^(2)` `omega =(2v)/(R )`
2 Answers 1 viewsCorrect Answer - A `L = I omega = mK^(2)omega`
2 Answers 4 viewsCorrect Answer - D Orbital speed `v_(c )=sqrt((GM)/(r ))`, Gravitational force `prop (1)/(r^(2))`, c.p. acceleration `=(v^(2))/(r )` All the three decrease as the orbital radius is increased. But the gravitational P.E....
2 Answers 1 viewsCorrect Answer - C (c ) : In a photon-particle collision such as photon-electron collision, the total energy and total momentum are conserved. However of photons may not be conserved in...
2 Answers 2 viewsCorrect Answer - D Angular acceleration, `alpha=(2pi(n_2-n_1))/t=(2pi(3300/60))/10` `=11pi=(11pixx180)/pi=1980 degs^(-2)`
2 Answers 1 views