Exams > Bitsat
PHYSICS MCQs
(A) If a rid is bent in the form of a circle . Its centre of mass is at the centre of curvature
(B) Football is in the shape of hollow sphere So, its centre of mass is at the centre of sphere
(C) The reason is same as (A) .
In all the above geometrical shape , no material is present at the centre of mass
`because vec(v) = (5 hat(i) - 3hat(j) + 6 hat(k)` m/s
`vec (F) = 10 hat(i) + 10 hat(j) + 20 hat(k)`
`:.` power =` vec(F) * vec (v)`
= `(10hat(i) + 10 hat (j) + 20 hat (k)) * (5hat(i) - 3hat(j) + 6 hat(k))`
= 50 - 30 + 120 = 140 joule / sec
`a_c = v^2/r = k^2 r t^2`
`because v = krt`
The tangential accelerationis
`a_1 = (dv)/(dt) = (d(rt))/(dt) = kr`
The work done by centripetal force will be zero.
so , power is delivered to the particle by only tangential force which acts in he same direction of instantaneous velocity .
`:.` power = `F_1 v`
= `ma_1 krt`
= `m(kr) (krt) = mk^2 r^2 t`
(D) `because ` p = Fv = (ma) v = `m((d^2 x)/(dt^2)) ((dx)/(dt))`
Since , power is constant ,
`:.` ` ((d^2 x)/(dt^2)) ((dx)/(dt))` = k
or `((d)/(dt)) ((dx)/(dt))^2 = k`
or `((dx)/(dt))^2 = k_1` t
or ` (dx)/(dt) = sqrt(k_1 t) = k _2 t^(1/2)`
or ` x = k_3 t^(3/2)`
Hence `(dx)/(dt) prop t^(1/2) prop x^(1/3)`
Power = `((60 + 20) xx (10) xx (20 xx 0.2))/(20)` = 160 w