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Question 5. A particle is projecteil upwards with velocity u in a, resisting medium whose resistance varies as the velocity. Find relation, [Kanpur B.Sc. 2009, 13], between time and velocity., ....ile is poing un. the resistance (Kv) is, 1

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Question 11. If V be the terminal velocity of a particle of mass m, falling againsl a resistance proportional to the square of the velocity, prove, that ihe kinelic energy acquired in falling a distnace h from rest is, mva, -- e- 24h/v², [Kanpur B.Sc. 2012, 17], 2

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Question 13. A particle moving in a resisting medium is acted upon, by a central force µ/r". If the path be an equiangular spiral of angle a, whose pole is at the centre of force, show that the resistance is :, n - 3 u cos ., 2, 3

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Question 11. If V be the terminal velocity of a parlicle of mass m, talling against a resistance proporlional to the square of the velovity, prove, Eat the kinetic energy acquired in ſalling a distnace h from rest is, [Kanpur B.Sc. 2012, 17], Solution. Terminal velocity is the velocity in downward course at, the time when acceleration becomes zero. Therefore V being given the, terminal velocity we have from the equation, d'x, di? =8 - Ku?, ... (1), () --g -- KV²,, or, (2), K, So the equation of motion, dx, K., dv, =g|1, diº, or, U --, g 1, dx, du, U --, dx, 1) 24, Integrating it, we have, 2gx, 2v dv, + C,, 2gx, - log (V2 - u) + C., .. (3), Initially, x 0 and u 0. So C = log V 2., 2gx, log-, v2 - v, Hence equation (3) becomes, u2 - v? (1 - e 2gx/V°, )., If Vi is the velocity at a distance h from the starting point, then, Whence, (4), e have, vỉ = v(1 e 2gh/v², Therefore, kinetic energy in falling a distance h from rest is, - e, 4

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Question 13. A particle moving in a resisting medium is acted upó, by a central force µ/r". If the path be an equiangular spiral of angle, whose pole is at the centre of force, show that the resistance is :, n - 3 H cos a, 2, [Kanpur B.Sc. 2015, Solution. Suppose R is the resistance per unit mass. Then th, equations of motion along the tangent and normal are, dv, cos o – R, ds, v2, cos (Gr - 0), sin o., and, 0 cot a., For the equiangular spiral r=, = ae°, we know that, O = a and, p =r sin a,, i.e.,, r = p. cosec a., Therefore,, dr, =r cose a., p =r, dp, O', Substituting these values ofp and o in (2), we get, v2 = E sin a·r cosec a =, v? = H, -- ], -1, ie.,, Differentiating with respect to s, this gives, %3D, - 1), dv, 20, ds, -(n ~ 1) H dr, pr ds', dr, since, ds, (n – 1)µ, ds, dv, i.e., v, %3D, Cos O,, = CoS O = cos a., dv, in (1), we get, ds, Substituting this value of u, (n - 1)u, |, COs a =-, 2,", cos x - R,, i.c., (n- 1)H, 3, COS O+, H cos a, CoS Q =