(^122) Aptitude Test Problems in Physics
1.18. Let the initial winding radius be 4r. Then
the decrease in the winding area as a result of the
reduction of the radius by half (to 2r) will be
S = n (16r 2 — 4r 2 ) = 12nr 2 ,
which is equal to the product of the length I, of
the wound tape and its thickness d. The velocity v
of the tape is constant during the operation of the
tape-recorder; hence ll = vti, and we can write
12nr 2 = vtid. (1)
When the winding radius of the tape on the
cassette is reduced by half again (from 2r to r),
the winding area is reduced by n (4r 2 — r 2 ) =
3nr 2 , i.e.
3nr 2 = vt^2 d,^ (2)
where t 2 is the time during which the winding ra-
dius will be reduced in the latter case. Dividing
Eqs. (1) and (2) termwise, we obtain
t 2 = —ti 4 — = 5 min.
1.19. Let us go over to the reference frame fixed to
ring 0' In this system, the velocity of ring 0 is
vi/cos a and is directed upwards since the thread is
inextensible, and is pulled at a constant velocity
vi relative to ring 0' Therefore, the velocity of
ring 0 relative to the straight line AA' (which is
stationary with respect to the ground) is
2 sine (a/2)
V2 = VI =vl
cos CC cos a
and is directed upwards.
1.20. By the moment of time t from the begin-
ning of motion, the wedge covers a distance s --=
at 212 and acquires a velocity owed =-- at. During
this time, the load will move along the wedge over
the same distance s, and its velocity relative to the
wedge is urea = at and directed upwards along the
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