Physics MCQs for NEET — Practice Questions with Answers

$ I_{ring} = MR _1^ 2$ As Volume and Mass remain same $ I_{solid } = { 2 \over 5} MR_2^2 $ $ R_2 \lt \lt R_1 $

Volume of disc = $ V_1 = \pi R_1 ^2 . t = { \pi R_1^3 \over 6 } $ $ \therefore { R_1^3 \over 6} = { 4 \over 3} R_2 ^3 $ Volume of Sphere = $ = { 4\over 3} \pi R_2^3 $ $ R_1^3 = 8 R_2^3 $ $ \therefore R_1 = 2R_2 $ $ I_1 = M.I of disc = I = { 1 \over 2} MR_1^2 $ $ I_2 = M.I of sphere = { 2 \over 5} MR_2^2 = { mR_1^2 \over 5 \times 2 } = { I \over 5} $

$M.I of disc = {1 \over 2} MR^2 = { 1 \over 2 } M \left( { M \over \pi \rho } \right) = { 1 \over 2} { M^2 \over \pi t \rho } $ As their mass & thickness are some $ I \alpha { 1 \over \rho } $

Let $I_Z $ be the M.I of square plote about the axis passing through the centre and perpendicular to the plane of square, hence according to Perfendicular axis theorm. $ I_Z = I _{AB } + I_{AB} Also I_Z = I_{CD} + I_{C'D'} $ As axis are symmetric $ I_{AB} = I_{A'B'} = { I_z \over 2 } $ And $I_{CD} = I_{C'D'} = { I_Z \over 2 } $ So we can say that $I_{AB} = I_{A'B'} = I_{CD} = I_{C'D'} =I $

Let the radius of complete disc is a & that of small disc is b After small disc is cut from complete disc let the C.M. shift to $O_2$ at distance $x_2$ flem original centre O. The Position of new C.M. is givenly let 6 is mass percunitarea. $ X_ {cm} = { -6 \pi b^2 \times 1 \over 6 \pi a^2 - 6 \pi b^2 } $

Let the mass of an element of length dx of the rod located at a distance X away from left and is $ { M \over L } dx $ , the x cordinate of the C.M. is given by. $ Total mass of rod = \int_0^1 Ax .dx = {AL^2 \over 2 } $ $ x_{cm} = { 1 \over M } \int xdm = { 1 \over \left( { AL^2 \over 2 } \right)} \int _0^1 x ( Ax dx ) $

By Conservation of Energy P.E. of rod = Rotational K.E. $ M.g {1 \over 2} sin d = { 1 \over 2} I \omega^2 = { 1 \over 2} { mL^2 \over 3} \omega^2 $ $ \therefore \omega = \sqrt { 3 g sin \alpha \over L } $ As here l = 2 L $ \omega = \sqrt { 3g sin \alpha \over 2L } $

Initial angular momentum of ring $ = I \omega = MR^2 \omega $ Final angular momentum of ring and particles $ = ( MR^2 + 2 mR^2 ) \omega' $ As No external forque so According to Law of conservation of angular momentum. $ MR^2 \omega = ( MR^2 + 2 mR^2 ) \omega' $ $ \therefore \omega' = { wM \over (M + 2m ) } $

As it is head-on elastic collision between two idential balls there fore they will exchange their linear vecocity is A comes to rest and B starts moving with linear velocity V. As there is no friction any where, forque on both the spheres about their centre of mass is zero and their angular velocities remains unchanged $ \therefore \omega_A = \omega and \omega = 0_B $

$ I = 0.3 x^2 + 0.7 ( 1.4 - x)^2 $ For minimum work moment of inertia of the system should be minimum is $ { dI \over dx} = 0 = 03 \times 2x -0.7 \times 2 ( 1.4 - x) = 0 $