Question

The hour hand of a clock is $7\text{ cm}$ long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is :

• A. $\left(\frac{35}{4}\right)^\circ$
• B. $\left(\frac{35}{2}\right)^\circ$
• C. $35^\circ$
• D. $70^\circ$

Hint:

A useful thumb rule for clocks:
The hour hand moves at a rate of $30^\circ$ per hour, or $0.5^\circ$ per minute.
The minute hand moves at a rate of $360^\circ$ per hour, or $6^\circ$ per minute.
For 1 hour and 10 minutes, the hour hand moves:
\[ (1 \times 30^\circ) + (10 \times 0.5^\circ) = 30^\circ + 5^\circ = 35^\circ \]
This breakdown makes mental calculation extremely quick.

Answer 1

The Correct Option is C

Step 1: Track the hour hand's actual position on the dial at both times, instead of multiplying by the elapsed minutes.
Rather than finding the total minutes elapsed and multiplying by the speed, we find where the hour hand points at 7:00 a.m. and where it points at 8:10 a.m., measured from the 12 mark, and subtract.

Step 2: Recall the hour hand's speed.
The hour hand covers $360^\circ$ in 12 hours, which is $30^\circ$ per hour, and since each hour has 60 minutes, that is $0.5^\circ$ per minute.

Step 3: Find the position at 7:00 a.m.
At 7:00 a.m., exactly 7 hours have passed since 12:00, with no extra minutes. So the hour hand's angle from the 12 mark is:
\[ \text{Position at 7:00} = 7 \times 30^\circ = 210^\circ \]

Step 4: Find the position at 8:10 a.m.
At 8:10 a.m., 8 full hours and an extra 10 minutes have passed. The 8 hours contribute $8 \times 30^\circ = 240^\circ$, and the extra 10 minutes contribute $10 \times 0.5^\circ = 5^\circ$. So:
\[ \text{Position at 8:10} = 240^\circ + 5^\circ = 245^\circ \]

Step 5: Subtract the two positions to get the angle swept.
\[ \text{Angle swept} = 245^\circ - 210^\circ = 35^\circ \]

Step 6: Final answer.
The angle swept by the hour hand is $35^\circ$, which is option (C).
\[ \boxed{35^\circ} \]