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Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
To do :
We have to find the final position of the hour hand of a clock in each case.
Solution :
We know that,
Number of hours in $360^o$ revolution $=12$ hours
Therefore,
(a) The hour hand of a clock starts from 6 and turns through 1 right angle.
This implies,
It turns by $90^o$.
Number of hours in $90^o$ revolution $=12\times\frac{90^o}{360^o}$ hours
$=3$ hours
The final position of the hand of the clock $=6+3=9$
So, the hand will move 3 hours and stop at 9.
(b) The hour hand of a clock starts from 8 and turns through 2 right angles.
This implies,
It turns by $180^o$.
Number of hours in $180^o$ revolution $=12\times\frac{180^o}{360^o}$ hours
$=6$ hours
The final position of the hand of the clock $=8+6=14$
$14=12+2$
The clock hand will cross the 12 mark and reach the 2 mark
So, the hand will move 6 hours and stop at 2.
(c) The hour hand of a clock starts from 10 and turns through 3 right angles.
This implies,
It turns by $270^o$.
Number of hours in $270^o$ revolution $=12\times\frac{270^o}{360^o}$ hours
$=9$ hours
The final position of the hand of the clock $=10+9=19$
$19=12+7$
The clock hand will cross the 12 mark and reach the 7 mark
So, the hand will move 9 hours and stop at 7.
(d) The hour hand of a clock starts from 7 and turns through 2 straight angles.
This implies,
It turns by $360^o$.
Number of hours in $360^o$ revolution $=12\times\frac{360^o}{360^o}$ hours
$=12$ hours
The final position of the hand of the clock $=7+12=19$
$19=12+7$
The clock hand will cross the 12 mark and will reach the 7 mark.
So, the hand will move 12 hours and stop at 7.