plz solvE Q- 6 OF pg-no.193 andQ-14, and 40 of pgno 196 h.c verma-1

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Discussion Response Post to: plz solvE Q- 6 OF pg-no.193 andQ-14, and 40 of pgno 196 h.c verma-1

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saharsha kumar keshkar (3397)

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6. radius = 10 cm = 0.1 m Angular velocity = 20 rad/s

=>Linear velocity on the rim = (omega)r = 20 × 0.1 = 2 m/s

=>Linear velocity at the middle of radius = (omega)r/2 = 20 × (0.1)/2 = 1 m/s.?

14. The moment of inertia about the center and perpendicular. to the plane of the disc of radius r and mass m is = mr^2.

According to the question the radius of gyration of the disc about a point = radius of the disc.

Therefore mk^2= ½ mr^2 + md^2(K = radius of gyration about acceleration point, d = distance of that point from the centre)

=> K^2 = r^2/2 + d^2 => r^2 = r^2/2 + d^2 (? K = r) => r2/2 = d^2 => d = r/ root 2 .

40. Acc. to the question, m1 = 200 g, I = 1 m, m2 = 20 g

Therefore, (T1 × r1) ? (T2 × r2) ? (m1f × r3g) = 0 => T1 × 0.7 ? T2 × 0.3 ? 2 × 0.2 × g = 0 => 7T1 ? 3T2 = 3.92 ?(1) T1 + T2 = 0.2 × 9.8 + 0.02 × 9.8 = 2.156 ?(2) (1) and (2) gives 10 T1 = 10.3 => T1 = 1.038 N = 1.04 N Therefore T2 = 2.156 ? 1.038 = 1.118 = 1.12 N.

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