23. A train of mass M is moving on a circular track of radius 'R' with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be [Q.59/RK_BM/COM] [Made 2004 ] ,dva su ftldk nzo;eku Mgs,d leku osxVls o dkkdkj iFk ftlch fakt;k'R'S eas xfr dj jgh gSA Vesuch EbbZV sdchifjf/kohvk/khOSAVsudkjs[kh; lasxgkXKA 2MV (1) 'kwU; O (4) MV (1) Mais 0 (2") 2 MV (3) MVR

A toy train of mass M is moving on a circular track of radius R with constant speed v. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

As a charged particle 'q' moving with a velocity vec(v) enters a uniform magnetic field vec(B) , it experience a force vec(F) = q(vec(v) xx vec(B)). For theta = 0^(@) or 180^(@), theta being the angle between vec(v) and vec(B) , force experienced is zero and the particle passes undeflected. For theta = 90^(@) , the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal force (mv^(2)//r) . For other values of theta (theta !=0^(@), 180^(@), 90^(@)) , the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle that carries a charge of magnitude q and has a mass 4 xx 10^(-15) kg is moving in a region containing a uniform magnetic field vec(B) = -0.4 hat(k) T . At some instant, velocity of the particle is vec(v) = (8 hat(i) - 6 hat(j) 4 hat(k)) xx 10^(6) m s^(-1) and force acting on it has a magnitude 1.6 N If the coordinates of the particle at t = 0 are (2 m, 1 m, 0), coordinates at a time t = 3 T, where T is the time period of circular component of motion. will be (take pi = 3.14 )

As a charged particle 'q' moving with a velocity vec(v) enters a uniform magnetic field vec(B) , it experience a force vec(F) = q(vec(v) xx vec(B)). For theta = 0^(@) or 180^(@), theta being the angle between vec(v) and vec(B) , force experienced is zero and the particle passes undeflected. For theta = 90^(@) , the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal force (mv^(2)//r) . For other values of theta (theta !=0^(@), 180^(@), 90^(@)) , the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle that carries a charge of magnitude q and has a mass 4 xx 10^(-15) kg is moving in a region containing a uniform magnetic field vec(B) = -0.4 hat(k) T . At some instant, velocity of the particle is vec(v) = (8 hat(i) - 6 hat(j) 4 hat(k)) xx 10^(6) m s^(-1) and force acting on it has a magnitude 1.6 N If the coordinates of the particle at t = 0 are (2 m, 1 m, 0), coordinates at a time t = 3 T, where T is the time period of circular component of motion. will be (take pi = 3.14 )

The path of a charged particle in a uniform magnetic field depends on the angle theta between velocity vector and magnetic field, When theta is 0^(@) or 180^(@), F_(m) = 0 hence path of a charged particle will be linear. When theta = 90^(@) , the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius r = (mv)/(qB) . The time period for circular path will be T = (2pim)/(qB) When theta is other than 0^(@), 180^(@) and 90^(@) , velocity can be resolved into two components, one along vec(B) and perpendicular to B. v_(|/|)=cos theta v_(^)= v sin theta The v_(_|_) component gives circular path and v_(|/|) givestraingt line path. The resultant path is a helical path. The radius of helical path r=(mv sin theta)/(qB) ich of helix is defined as P=v_(|/|)T P=(2 i mv cos theta) p=(2 pi mv cos theta)/(qB) Two ions having masses in the ratio 1:1 and charges 1:2 are projected from same point into a uniform magnetic field with speed in the ratio 2:3 perpendicular to field. The ratio of radii of circle along which the two particles move is :

The path of a charged particle in a uniform magnetic field depends on the angle theta between velocity vector and magnetic field, When theta is 0^(@) or 180^(@), F_(m) = 0 hence path of a charged particle will be linear. When theta = 90^(@) , the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius r = (mv)/(qB) . The time period for circular path will be T = (2pim)/(qB) When theta is other than 0^(@), 180^(@) and 90^(@) , velocity can be resolved into two components, one along vec(B) and perpendicular to B. v_(|/|)=cos theta v_(^)= v sin theta The v_(_|_) component gives circular path and v_(|/|) givestraingt line path. The resultant path is a helical path. The radius of helical path r=(mv sin theta)/(qB) ich of helix is defined as P=v_(|/|)T P=(2 i mv cos theta) p=(2 pi mv cos theta)/(qB) Two ions having masses in the ratio 1:1 and charges 1:2 are projected from same point into a uniform magnetic field with speed in the ratio 2:3 perpendicular to field. The ratio of radii of circle along which the two particles move is :

Read the following passage carefully and answer the questions The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's alpha- scattering experiment. If an alpha- perticle is projected from infinity with speed v, towards the nucleus having z protons then the apha- perticle which is reflected back or which is deflected by 180^(@) must have approach closest to the nucleus. It can be approximated that alpha- particles collides with the nucleus and gets back. Now if we apply the energy conservation at initial and collision point then: ("Total Energy")_("initial") = ("Total Enregy")_("final") (KE)_(i) +(PE)_(i)=(KE)_(f)+(PE)_(f) (PE)_(i) =0, "since"PE of two charge system separated by infinite distance is zero, finally the particle stops and then starts coming back. 1/2m_(alpha)v_(alpha)^(2) + 0=0 + (kq_(1)q_(2))/(R) rArr1/2m_(alpha)v_(alpha)^(2)=k(2exxze)/(R)rArrR=(4kze^(2))/(m_(alpha)v_(alpha)^(2) Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it . Experiments show that the average radius R fo a nucleus may be written as R=R_(0)(A)^(1//3) where R_(0) = 1.2xx10^(-15)m A= atomic mass number R=radius of nucleus Radius of a particular nucleus is calculated by the projection of alpha- particle from infinity at a particular speed. Let this radius be the true radius. If the radius calculation for the same nucleus is made by , alpha- particle with half of the earlier speed then the percentage error involved in the radius calculation is :