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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.

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